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[Dover Books on Mathematics] L. I. Volkovyskii, G. L. Lunts, I. G. Aramanovich - A Collection of Problems on Complex Analysis (1991, Dover Publications)

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A COL LEC TIO N
OF PROBLEMS ON
IS
LYS
COM PLE X ANA
L. I. Volko vyskii , G. L. Lunts ,
.and I.G. Aram anovi ch
A Collection of Problems
on
Complex Analysis
A Collection of Problems
on
COMPLEX ANALYSIS
L.I. VOLKOVYSKII, G.L. LUNTS,
I. G. ARAMANOVICH
Translated by
J. BERRY
Translation edited by
T. KOVAR!, PH.D.
DOVER PUBLICATIONS, INC.
New York
Copyright © 1965 by Pergamon Press Ltd.
All rights reserved under Pan American and International Copyright
Conventions.
Published in Canada by General Publishing Company, Ltd., 30 Lesmill
Road, Don Mills, '.Ibronto, Ontario.
This Dover edition, first published in 1991, is an unabridged and unaltered republication of the work first published by Pergamon Press,
Oxford, in 1965 as Volume 68 in the International Series of Monographs
in Pure and Applied Mathematics. This book is an edited translation of
the original Russian Sbornik zadach po teoriyi funktsii kompleksnogo
peremennogo, published in 1960 by Fizmatgiz, Moscow. The Dover
edition is published by special arrangement with Pergamon Press,
Headington Hill Hall, Oxford OX3 OBW England.
Manufactured in the United States of America
Dover Publications, Inc., 31 East 2nd Street, Mineola, N. Y. 11501
Library of Congress Cataloging-in-Publication Data
Volkovyskii, L. I. (Lev Izrailevich)
[Sbornik zadach po teorii funkfsii kompleksnogo peremennogo.
English]
A collection of problems on complex analysis IL.I. Volkovyskii, G.L.
Lunts, LG. Aramanovich; translated by J. Berry; translation edited by
T. Kovari.
p.
cm.
_
Translation of: Sbornik zadach po teorii funktsiI kompleksnogo
peremennogo.
Originally published: Oxford : Pergamon Press, 1965. (International
series of monographs in pure and applied mathematics ; v. 68).
Includes bibliographical references.
ISBN 0-486-66913-0
1. Functions of complex variables-Problems, exercises, etc.
2. Mathematical analysis-Problems, exercises, etc. I. Lunts, G. L.
(Grigorii L'vovich) II. Aramanovich, I. G. (Isaak Genrikhovich) III.
Berry, J. IV. Kovari, T. V. Title.
QA331. 7. V6513 1991
515' .9'076-dc20
91-40596
CIP
CONTENTS
l!ortJWO'l'd,
ix
CHAPTER I
Complex numbers and f1ll1Ctions of a complex variable
§ 1 Complex numbers (complex numbers; geometrical interpretation; stereographic projection; quaternions)
§ 2 Elementary transcendental functions
§ 3 Functions of a complex variable (complex functions of a real
variable; functions of a complex variable; limits and continuity)
§ 4 Analytic and harmonic functions (the Cauchy-Riemann
equations; harmonic functions; the geometrical meaning of
the modulus and argument of a derivative)
I
7
11
14
CHAPTER II
Conformal mappings connected with elementary functions
§ 1 Linear functions (linear furiptions; bilinear functions)
§ 2 Supplementary questions of the theory of linear transformations (canonical forms of linear transformations; some ap·
proximate formulae for linear transformations; mappings of
simply connected domains; group properties of bilinear transformations; linear transforma~ions and non-Euclidean geometry)
§ 3 Rational and algebraic functions (some rational functions;
mappings of circular lunes and domains with cuts; the func-
tion
!(z+ ~);
21
26
application of the principle of symmetry;
the simplest non-sohlicht mappings)
§ 4 Elementary transcendental functions (the fundamental tran·
scendental functions; mappings leading to mappings of the
strip and half-strip; the application of the symmetry principle; the simplest many-sheeted mappings)
§ 5 Boundaries of univalency, convexity and starlikeness
33
42
CHAPTER III
Supplementary geometrical questions. Generalised analytic functions
§ 1 Some properties of domains and their boundaries. Mappings
of domains
§ 2 Quasi-conformal mappings. Generalised analytic functions
51
55
CHAPTER IV
Integrals and power series
§ 1 The integration of fw1ctions of a complex variable
v
64
vi
CONTENTS
§
§
§
§
2
3
4
5
Cauchy's integral. theorem
Cauchy's integral formula
Numerical series
Power series (determination of the radius of convergence;
behaviour on the boundary; Abel's theorem)
§ 6 The Taylor series (the expansion of functions in Taylor series;
generating functions of systems of polynomials; the solution
of differential equations)
§ 7 Some applications of Cauchy's integral formula and power
series (Cauchy's inequalities; area theorems for univalent
functions; the maximum principle; zeros of analytic functions;
the uniqueness theorem; the expression of an analytic function
in terms of its real or imaginary part)
68
70
72
74
78
83
CHAPTER V
Laurent series, singularities of single-valued functions. Integral
functions
§ 1 Laurent series (the expansion of functions in Laurent series;
some properties of univalent functions)
§ 2 Singular points of single-valued analytic functions (singular
points; Picard's theorem; power series with singularities on
the boundary of the circle of convergence) \j
§ 3 Integral functions (order; type; indicator function)
89
92
CHAPTER VI
Various series of functions. Parametric integrals. Infinite products
§ 1 Series of functions
§ 2 Dirichlet series
§ 3 Parametric integrals (convergence of integrals; Laplace's
integral)
§ 4 Infinite products
102
106
108
111
CHAPTER VII
Residues and their applications
§ 1 The calculus of residues
§ 2 The evaluation of integrals (the direct application of the theorem
ofresidues; definite integrals; integrals connected with the inversion of Laplace's integral; the asymptotic behaviour of integrals)
§ 3 The distribution of zeros. The inversion of series (Rouch6's
theorem; the argument principle; the inversion of series)
§ 4 Partial fraction and infinite product expansions. The summation of series
116
118
141
147
vii
CONTENTS
CHAPTER VIII
Integrals of Cauchy type. The integral formulae of Poisson and Schwarz.
Singular integrals
§ 1 Integrals of Cauchy type
§ 2 Some integral relations and double integrals
§ 3 Dirichlet's integral, harmonic functions, the logarithmic
potential and Green's function
§ 4 Poisson's integral, Sohwarz's formula, harmonic measure
§ 5 Some singular integrals
152
159
163
167
174
CHAPTER IX
Analytic continuation. Singularities of many-valued character.
Riemann surfaces
§ 1 Analytic continuation
§ 2 Singularities of many-valued character. Riemann surfaces
192
199
§ 3 Some classes of analytic functions with non-isolated singularities
207
CHAPTER X
Conformal mappings (continuation)
§ 1 The Schwarz--Ohristoffel fotmula
§ 2 Conformal mappings involving the use of elliptic functions
211
229
CHAPTER XI
Applications to mechanics and physics
§ 1 Applications to hydrodynamics
§ 2 Applications to electrostatics
§ 3 Applications to the plane problem of heat conduction
243
258
271
ANSWERS AND SOLUTIONS
Chapter
Chapter
Chapter
Chapter
Chapter
Chapter
Chapter
Chapter
Chapter
Chapter
Chapter
I
n
m
IV
V
VI
VD
vm
IX
X
XI
274
285
307
310
317
324
328
339
349
374
395
FOREWORD
A Collection of Problema on Oomple:x: Analysis is intended mainly
for students of the physico-mathematical and mechanico-mathematical faculties of universities and also for students of the physicomathema.tical faculties of teachers training colleges.
The authors believe also that the collection will be useful for
those specialising in the mechanics of continuous media (hydrodynamics, the theory of elasticity) and electrical engineering, since
it contains a large number of problems either on the direct application of the theory of functions of a complex variable to the given
subject, or on questions forming their mathematical basis (conformal
mappings, harmonic functions, potentials, integrals of Cauchy type,
etc.).
Problems connected with the compulsory university course on
the theory of functions of a complex variable are arranged ma.inly
in Chapters I, II (§ 1, 3 and 4), IV, V, VII, X (§ 1).
The book also contains groups of problems which go beyond the
range of the syllabuses. Some of these may be used as the basis of
course work and as material for study in seminars on the theory of
functions of a complex variable.
It seems to us that the book deals with the main sections of the
theory of functions of a complex variable approximately in accordance with the curriculum.
For the convenience of those using the book, in the contents
besides the titles of the chapters and sections, the main groups of
problems contained in them have sometimes been enumerated (this
refers ma.inly to the basic course material).
It is assumed that the users of the collection are acquainted with
the corresponding sections of the course in the theory of functions
of a complex variable contained in the book by A. I. Markushevich
A Short O<YUrse on tke Theory of Analytic Functions, and therefore
as a rule references are not given to this textbook. If additional
material is used the necessary information of the source and also references to the literature are given. The following notation is used
for the books most :frequently cited:
[l] A. I. Markushevich, Theory of Analytic Functions (Teoriya
analitickeskikk funktsii), Gostekhizdat, Moscow, 1950.
ix
x
FOREWORD
[2] A. I. Markushevich, Short Oourae of the Theory of Analytic
Functions (Kratkii kura teorii analitickeakikh funktaii), Gostekhizdat,
Moscow, 1957.
[3] M.A. Lavrent'yev and B. V. Sha.bat, Methoda of the Theory
of Functions of a Oomplez Variable (Metody teorii funktaii kompleksnogo peremennogo), 2nd ed. Fizmatgiz, Moscow, 1958.
[4] I. I. Privalov, Introduction to the Theory of Functions of a
Oomplez Variable ( V vedenie v teoriyu funktaii kompleksnogo peremennogo), 10th ed. Fizmatgiz, Moscow, 1960.
All the hints for the solution of the problems are given in the
main text. The most difficult problems, the numbers of which are
marked by asterisks are provided with solutions located in the answers.
In compiling the collection use was made of textbooks and monographs by both Russian and foreign authors.
The manuscript of the book was reviewed in the Faculty of the
Theory of l!'unctions at Moscow University and the authors express
their sincere gratitude to the members of the faculty and especially
to B. V. Sha.bat for much valuable advice.
Much help in the checking and arrangement of some of the problems was given by the pupils of Professors L. I. Volkovyskii. B. A.
Vertgeim, S. Ya. Gusman, V. V. Dumkin, V. G. Mikhal'chuk, Yu. L.
Rodin, E. V. Starkova and I. I. Filimonova. I. I. Pesin took part
in the choice of the problems of§ 3, Chapter IX. A number of pr'oblems
were proposed by L. E. El'sgol'ts. Extremely important work was
done on the collection by the editor of the book N. A. Ugarova.
The authors express their deep obligation to all those mentioned.
THE
AUTHORS
CHAPTER I
COMPLEX NUMBERS AND FUNCTIONS
OF A COMPLEX VARIABLE
IN THIS chapter and generally everywhere in the book where nothing is
said to the contrary the following notations are used: z = m+iy =rail/>,
w = u+i11 = eelll (:ii, y, u, 11, r, (!, qi and 8 are real numbers, r;;;;,: O, e ~ 0);
Re z =:ii, Im z = y, arg z =qi, lzl = r, z= z-iy. In the absence of any
additional indications the principal value of the argument arg z is defined
by the inequalities -:ii< arg z ~:ii; the complex plane whose points are
represented by the complex number z, will be called the z-plane; usually the
terms "complex number z" and "point z" a.re used as synonyms.
§ I. Complex numbers
I. Perform the operations indicated:
2
(1) l/i;
(3) l-3i;
1-i
(2) l+i ;
(4) (l+iy3)s.
2. Find the modulus and argument of each of the following complex numbers (a and b are real numbers):
(1) 3i;
(6) 2-5i;
(2) -2;
(7) -2+5i;
(3) l+i;
(8) -2-5i;
(4) -1-i;
(9) bi (b =/: O);
(5) 2+5i;
(10) a+bi (a=/: 0).
3. Find all the values of the following roots and plot them:
Jf (1-i);
(1) yl;
(6)
(2> jli;
(3> V(-1);
(4) (-8);
c1> V(3+4i);
(8) j/(-2+2i);
(9) (-4+3i),
v
fl
(5) j/1;
4. Prove that both values of (z2- l) lie on the straight line passing
through the coordinate origin and parallel to the bisector of the
r
I
2
PROBLEMS ON COMPLEX ANALYSIS
internal angle of the triangle, with vertices at the points -1, I and
z, which passes through the vertex z.
5. Let m and n be integers. Prove that (ii'z)m has n/(n, m) different
values, where (n, m) is the greatest common divisor of the numbers
m and n. Verify that the sets of values of (yz )m and j/ (zDI) are identical if and only if (n, m) =I, that is, if n and mare coprimes.
6. Prove the following inequalities from geometrical considerations:
(I) lzt +z2J ~ Jz1l + Jz2i;
(2) Jz1-z2I ;?:: ] Jz1J-lz2i I;
(3) iz-IJ
< J izi-Il+lzlJarg zl.
In each case explain when the sign of equality applies.
7. Prove the identity
iz1+z2l 2 +iz1-z.l2 = 2(1Zil 2 +lzal2)
and explain its geometrical meaning.
8. Prove that if iz11 = iz2 J = IZsl, then
Z:J-Z2
I
z2
arg---=-arg-.
Zs-Z1
2
Z1
9. Prove that if z1+z2 +z3 = 0 and iz11 = iz21= IZsl = I, then
the points z1, z2, Zs are the vertices of an equilateral triangle inscribed
in the unit circle.
10. Prove that if Zi +z 2 +z3 +z4 = 0 and Jz11 = lz21= !Zs! = jz4 J,
then the points z1, zs, Zs· z4 are either the vertices of a rectangle or
pairs of them are identical.
11. Find the vertices of the regular n-gon if its centre is at the
point z = 0, and one of its vertices z1 is known.
12. The points z1 and Zs are adjacent vertices of a regular n-gon.
Find the vertex Zs• adjacent to z2 (Zs #: z1 ).
13. Given three vertices of a parallelogram z1, z2, Zs· find the
fourth vertex z4 , opposite to the vertex z2•
14. Under what conditions do three points z1, z2, z3 , no two of
which are coincident, lie on a straight line 1
15. Under what conditions do four points zt, z2, Zs• z4 , no two of
which are coincident, lie on a circle or straight line 1
COMPLEX NUMBERS
3
16*. The points z1, z8, • • • , z. lie on one side of a certain
straight line passing through the coordinate origin. Prove that
the points l/zt, 1/z2, ••• , 1/z. possess a similar property (with respect
to what straight linen and that Zi +z.+ ... +z. :#: 0; l/z1 + l/z2+ ...
+1/z. ¥: O.
17. Prove that if zt + z. + •.• + z. = 0, then any straight line
passing through the coordinate origin divides the points zt, z., ... , z.,
provided only that these points do not lie on this straight line.
18. Prove that any straight line passing through the centre of
gravity of the material points zt, z., ... , z. with masses fni, m.i•••• , m.,
divides these points provided only that they do not lie on this
straight line.
Explain the geometrical meaning of the relations given in problems
19-29.
19. lz-z0 1 < R; lz-z01 > R;
lz-z0 1 = R.
20. lz-21+1z+21=5.
21. lz-21-lz+21>3.
22. lz-z1I = lz-zal·
23. (1) Rez ~ O; (2) lmz < 0.
!4. 0 <Re (iz) < 1.
25. IX < arg z < f3;
IX < arg (z-z0) < f3
(-n < IX < f3 :.,;;; n).
26. lzl =Re z+l.
27. Re z+Im z < 1.
28. Im z-zi
= 0;
= 0.
Re z-zi
z-z2
lzl < arg z,
lzl < arg z,
z-z.
29. (1)
if 0 :.,;;; arg z < 211:;
(2)
if 0 < arg z :..;;; 211:.
In problems 30-33 it is required to determine the families of curves
in the z.plane defined by the corresponding equations.
1
1
30. (1) Re-= O;
(2) Im-= 0
(-oo < 0 < oo).
z
z
(2) Imz11 =0
31. (1) RezB=O;
32. , z-zt
z-z.
I= A
z-z1
33. arg-- =IX
z-z9
(A
(-oo<O<oo).
> 0).
(-n
<IX:.,;;; n).
34. (1) A family of curves in the z-pla.ne is defined by the equation
lz2 -ll =A
(A> 0).
4
PROBLEMS ON' COMPLEX ANALYSIS
For what values of A. do the curves of the family consist of a simple
curve and for what values do they split into two branches 1
(2) Answer the same questions for the family
lz2+az+bl = A.
(A.
> 0) .
35. The function a.rg z is defined uniquely at all points z ::/: 0, if
we put lzl-2:n; < a.rg z ~ lzl. What is the geometrical locus of the
points at which the continuity of the function a.rg 21 defined in this
way breaks down 1
36. What is the geometrical locus of the points at which the continuity of the function a.rg z, uniquely defined for any 21 ::/: 0 by the
inequalities
log jzl-2:n; < a.rg z
~log
lzl
breaks down 1
37. For z = 2 the initial value ofa.rgf(z) is ta.ken to be equal to 0.
The point z makes a complete rotation anticlockwise a.long a circle
with centre at the coordinate origin and returns to the point 21 = 2.
Considering that a.rgf(z) varies continuously during the motion
of the point z, find the value of a.rgf(2) after the given rotation, if:
(1) f(z) = y(z-1);
(2) f(z) =
(z-1);
V
(3) f(z) = y(z2 -l);
(4) f(z) = y(z2+2z-3);
(5) f(z) =
y(:+D.
Stereographic projection
38. Derive the formulae of stereographic projection expressing
the coordinates (~, rJ, C) of a point P of the sphere of diameter 1,
touching the z-pla.ne at the coordinate origin in terms of the coordinates (x, y) of the corresponding point z. Also express x and y in
terms of ~, rJ, C (the axes of ~ and 1/ a.re assumed to coincide with
the axes of x and y respectively).
REMARK. In problem 38 a correspondence is established between the points
of the complex plane and the points of a sphere of radius 1/2 touching this
plane. Another correspondence exists in which the sphere is of radius I and
the z-plane passes through its centre. See, for example, [I, Chapter I, § 5].
39. Wha.ta.retheima.geson thesphereofthepoints 1,-1, i, 1V2i 1
5
OOMl'LEX NUMBERS
40. What is the image on the plane of the parallel with latitude
p (- ; < p < ; ) 1 What corresponds to the north and south poles!
41. Find on the sphere the images:
(1) Of the rays arg z =ct;
(2) Of the circles !zl = r.
42. What is the position on the sphere of pairs of points symmetrical:
(1) With respect to the point z = O;
(2) With respect to the real axis;
(3) With respect to the unit circle 1
43. Subject to what conditions are the points z1 , and z9 the stereographic projections of two diametrically opposite points of the
sphere1
44. Which points of the sphere correspond to the points z and ..!..1
z
45. Find on the sphere the images of the domains defined by the
inequalities:
(l)lmz>O;
(4) Rez < 0;
(2) Imz < O;
(5) iz! < l;
(3) Rez > O;
(6) !z! >I.
46. What corresponds on the sphere to a family of parallel straight
lines on the plane 1
47. Prove that in stereographic projection, circles on the sphere
project into circles or straight lines on the plane. Which circles on
the sphere correspond to straight lines 1
48. Let K be a circle on the plane corresponding to the circle K'
on the sphere, let N be the north pole of the sphere, and let S be the
vertex of the cone touching the sphere along K' (it is assumed that
K' is not a great circle). Prove that the centre of the circle K lies on
the ray NS. Consider the case when K' is a great circle.
49. Prove that in stereographic projection, angles between curves
on the sphere are equal to the angles between their images on the
plane. Prove also that the sense of angles is preserved if on the
sphere the positive direction of rotation is defined with respect to
the inward normal.
liO. Find the length h(z, a) of the chord connecting points of the
sphere which correspond to the points z and a. Consider also the
case when a is the point at infinity.
6
PROBLEMS ON OOMPLEX ANALYSIS
51. Given two points z1 and z9 (one of them may be at infinity),
:find the geometrical locus of the points of the z-plane which correspond to a circle on the sphere equidistant from the images of the
given points.
Quaternions
Quaternionst are numbers which are linear combinations with real ooefB.
cients of four units-one being real and three imaginary t, J, k;
the multiplication of units being defined by the equations
,..... ,.. = k 1 =
I.i==i.I==i,
-1;
I.J=J.I=J,
l.k=k.I=k.
Quaternions with identical components (coefficients of the units) are con·
sidered to be equal. In the addition of quaternions the corresponding units
are added. Multiplication is subject to the associative and distributive laws.
Thus, for example,
(2i+43) (3J-2k) = (2i) (3j)+ (4J) (3J)+(2i)(-2k)+(4J) (-2k)
= 2.3.i.j+4.3.3.. +2(-2).i.k+4.(-2).J.k
= 6k-12+4J-8i.
The numbers q = a+bi+ci+dk and iJ = a-bi-cj-dk are said to be conjugate to one another. The number N(q) = a•+b•+cl+dl is called the norm
of the quaternion q.
52. Prove that in the quaternion system the operation of subtrac.
tion is defined uniquely and that for divisors different from zero
the operation of division is defined uniquely.
53. Prove that q1q9= ~ · q9.
54. Prove the equality
N (q1 q2) = N (q1) N (q2)
and using this equality show that the product of two integers each
of which can be represented as the sum of the squares of four integers
is also a number which can be represented as the sum of four squares.
Obtain this representation.
t For hyperoomplex numbers and qua temions, see for example, A Uniwr
sity Algebra, D. E. Littlewood. Heinemann's, London (1950).
7
OOMPLEX NUMBERS
55. Using quaternions construct the matrix
B1
B2
B3
B4
A1
( A2
A3
A4
01
02
03
04
D 1)
D2
D3
D4
with integer elements which has the same sum of squares of elements
along the rows, columns and principal diagonals and satisfies the
relations
AkA1+BkB1+0k01+DkD1
=0
(k :F l;
k, l
= 1, 2, 3, 4).
§ 2. Elementary transcendental functions
By definition
expz =es= e" (cosy+isiny);
COSZ=
els+e-ls
;
2
es+e-s
cosh z = - 2- ;
es-e-s
sinhz=---;
2
sinz
00s7;
sinhz
tanhz=--;
coshz
cosz
cotz=-.-;
coshz
cothz=--·
sinhz
tanz =
SlnZ
56. Using the definition of es, prove that:
(1) es1 .es1 = es1 +s1 ;
(2) e=+ 2" 1 =es.
(3) If e=+(I) =el' for every z, then
ro = 2nlci
(k = 0,
±
1, ± 2, ... ) .
The relation exp iq, =ell/>= cos tJ>+i sin q, (Euler's formula) makes it
possible to use the exponential form of notation z = relt/> for complex numbers
instead of the trigonometric form z = r (cos tJ>+i sin I/>). In what follows by
41 is usually understood the principal value of the argument, that is,
-n<tJ>~n.
57. Represent the following numbers in the exponential form:
1, -1, i, -i, l+i, 1-i, -l+i, -1-i.
58. Find e
:1:.!..1
2
et"' (k = 0, ±1, ±2, ... ) .
8
PROBLEMS ON COMPLEX ANALYSIS
59. Find the moduli and principal values of the arguments of
the following complex numbers:
eS+I;
eB-81;
e8+'1;
e-8-41;
-ae'.P (a>
o, It/>! : : : ; 3t);
e-1.p (It/>! ~ 3t).
80. Find the sums:
(1) l+cosx+cos2x+ ... +cosna:;
(2) sinx+sin2x+ ... + sinna:;
(3) cou+cosSz+ ... +cos (2n-l)x;
(4) sinx+sin3x+ ... +sin (2n-l)x;
(5) sinx-sin2x+ ... +(-1)•-lsinna;.
61. Find the sums:
(1) cou+cos (oc+ P>+ ... +cos (oc+np);
(2) sinoc+sin (oc+P>+ ... +sin(oc+np).
62. Starting from the definitions of the respective functions, prove:
(1) sinllz+cosBz = 1;
(2) sinz =cos(; -z);
(3) sin (Zi.+z2) = sinz1cosz2+cosz1sinz2;
(4) cos (z1 +zs) = COSZ1 COU2-sin2l1 sinz2;
2tanz
(5) tan2z= 1 ta 2 ;
- n z
(6) cosh(21i+z2) = coshz1cosh212+sinhZi,sinhz2.
63. Prove that if cos (z+ro) = cosz for every z, then ro = 23tlc
(le= 0, ±1, ±2, ... ).
64. Prove that:
(1) siniz = isinhz;
(3) taniz = itanhz;
(2) cosiz = coshz;
(4) cotiz = -icothz.
65. Express in terms of trigonometric and hyperbolic functions
with real arguments the real and imaginary parts and also the moduli
of the following functions:
(1) sinz;
(4) sinhz;
(2) cosz;
(5) coshz;
(3) tanz;
(6) tanhz.
66. Find the real and imaginary parts of the following expressions:
(1) cos(2+i);
(4) cot(:-ilog2);
(2) sin2i;
(5) coth (2+i);
(3) tan (2-i);
(6) tanh(log3+
~i).
COMPLEX NUMBERS
9
67. For each of the functions e", cosz, sinz, tanz, coshz, cothz,
find the set of points z, where it assumes:
(1) Real values;
(2) Purely imaginary values.
By definition,
I.ogz
= logr+it1>+2nik (k = 0, ±1, ±2, .•. ),
logz = logr-1-it/>{-n<tf>:o:;;.;n)
(logz is called the principal value of the quantity Logz).
68. Evaluate:
(1) Log4,
Log (-1), log (-1);
(2) Logi, logi;
l±i
(3) Log V2 ;
(4) Log (2-3i), Log (-2+3i).
69*. Find the error in the reasoning leading to I. Bernoulli's
paradox: (-z)2 = z2, hence 2 Log (-z) = 2 Log z, consequently,
Log(-z) = Logz(!).
70. The initial value of Im f(z) at z = 2 is assumed to be zero.
The point z makes a complete rotation anticlockwise along a circle
with centre at the point z = 0 and returns to the point z = 2.
Assuming that f(z) varies continuously during the motion of the
point z, find the value of Im f(z) after the given rotation, if:
(1) f(z) = 2Logz;
(2) f(z) =Log 1/z;
(3) f(z) = Logz-Log (z+l);
(4) f(z) = Logz+Log (z+l).
By definition, for any complex numbers a =F 0 and a
a"= exp {a Log a}
(1)
or, if as usualt by e" we understand exp z, then
a"= e"Loga
71. Find all the values of the following powers:
(1) 1>'2 ;
(5) i 1;
1-i )l+f
(2) (-2)>' 2 ;
(6) (
y2
21;
(3)
(4) 1-1;
(7) (3-4i)l+i
(8) (-3+4i) 1+i.
t By eqn.(l) a== exp {zLoge} =exp {:z:(l+2nik)}. However, if nothing
is said to the contrary, we shall consider that k = 0, that is, as usual e" = exp z.
10
PROBLEMS ON COMPLEX ANALYSIS
72. Show that in the case of a rational index (a.= m/n) the
general definition of the power z• is identical with the usual definition:
z"n"' =
(Y'z)"'
(see also problem 5).
73. Are the sets of values of the following identical:
a••,
(a•)11, (a2)•!
By definition, the equation w = cos-1 z is equivalent to the equation
z = cosw. The functions sin-1 z, tan- 1 z, cot-1 z and the inverse hyperbolic
functions cosh-1 z, sinh-1 z, tanh-1 z, coth- 1 z are defined similarly.
74. Prove the following equalities (all the values of the roots
a.re ta.ken):
(1) cos-1z = - i Log(z+y(zll-1));
(2) sin-1z = -iLogi(z+y(zll-1));
(3) ta.n-lz = ; Log((i+z)/(i-z)) =
(4) cot-lz =
:i
Log !+::;
~Log ((z-i)/(z+i));
(5) cosh-1z =Log (z+y(z2 -1));
(6) sinh-1z = Log(z+y(zll+l));
1
l+z
(7) ta.nh-1z = - Log--;
2
1-z
1
z+l
(8) coth-1z = - Log - - ·
2
z-1
75. Prove that for any value of cos-1z it is possible to choose
a value of sin-1z, such that the sum of these values is equal to n/2.
Prove a similar assertion for ta.n-lz and cot-1z.
REMA.BE. The equalities sin-lz+cos-lz = n/2 and tan- 1 z+cot- 1 z = n/2
are always understood in the sense indicated in the present problem.
76. Show that all the values of cos-1 z a.re contained in the
formula.
cos-lz = ±iLog((z+y(z11-1)),
where by y(zll - 1) is understood any one of its values.
77. (1) For what values of z a.re all the values of the functions
cos- 1 z, sin-1 z, and ta.n-1 z real!
11
FUNCTIONS OF A OOMI'LEX VARIABLE
(2) For what values of z does the function sinh-1 z assume
purely imaginary values 1
78. Find all the values of the following functions:
(1) sin-11/2;
(5) tan-1(1+2i);
(2) cos-11/2;
(6) cosh-12i;
(3) cos-12;
(7) tanh-1(1-i) .
(4) sin-1i;
§ 3. Functions of a complex variable
In problems 79-85 it is required to determine the curves defined
by the given equations.
79. z =I-it; 0 ~ t ~ 2.
80. z = t+it2; -oo < t < oo.
81. z = t 2 +it';
- 00 < t < oo.
3n
.
. )
n
82. z =a (cost+ismt; 2 ~ t ~2;
a> 0 ·
83. z = t+(i/t);
-oo < t < 0.
84. (1) z = t+iy'(I-t2);
- I ~ t ~ 1;
(2) z = -t+i y'(l-t2); -1 ~ t ~ 0 (the arithmetic value of
the root is taken).
85. (1) z =a (t+i-ie-");
-oo < t < oo,
a> O;
(2) z = ia+at-ibe-1 ;
0 ~ t ~·2n,
a> 0,
b > 0.
86. For the mapping w = z2 it is required:
(1) To find the images of the curves x = 0, y = 0, x = y,
lzl = R, argz =ex and explain which of them are transformed
one-one;
(2) To find the originals (in the z-plane) of the curves u = 0,
ti= 0 (w = u+it1).
87. For the mapping w =.!__find:
z
(1) The images of the curves x = 0, y = 0,
argz =ex, jz-11 = I;
(2) The originals of the curves u = 0, t1 = 0.
88. For the mappings w =
z+_.!_
z
and w =
z-.!_
z
lzl
= R,
find the
images of the circles lzl = R.
89. For the transformation w = z +(I/z) find on the Z·plane
the original of the rectangular net (u = 0, t1 = 0) of the w-plane.
12
PROBLEMS ON COMPLEX .ANALYSIS
90. What is the circle lzl = 1 transformed into in the mapping
= z/(1-z2) 1
91. For the mapping w = es find:
(1) The images of the curves x = 0, y = 0, x = y;
(2) The originals of the curves fl=() (0 ~ () < oo).
92. Find the transformations of the rectangular net (x = 0,
y = 0) of the z-plane by means of the functions:
w
(1) w
= z2 +z;
(2) w
= cothz;
(3) w
=e
21'.
93. Into what are the segments of the straight lines x = 0 and
the straight lines y = 0, which lie in the strip 0 ~ y ~ :n; transformed by means of the function w = e:r: +z 1
94. What corresponds in the z-plane to the polar net lwl = R,
arg w = ix in the transformations:
1
(1) w=e-=;
(2) w=e:r:•1
95. Find the limit points of the sets:
(1) z = 1+(-l)•n/(n+l)
(n = 1, 2, ... );
(2) z = 1/m+i/n (m, n are arbitrary integers);
(3) z = p/m+ i q/n (m, n, p, q are arbitrary integers);
(4) izl < 1.
96. Prove that from an infinite bounded sequence of points {zn}
it is possible to select a convergent subsequence.
97. Prove the following propositions:
(1) The convergence of the sequence {zn = Xn + iyn} is
equivalent to the simultaneous convergence of the sequences {xn}
and {Yn}·
(2) For the limit lim Zn '# 0, to exist, it is necessary and
n-+OO
sufficient that the limits lim lznl '# 0 and (for a suitable definition
n-+OO
of argzn) lim arg
11-+00
Zn
should exist. If rm
zn
is not a negative numbel'
n-+OO
it is possible, for example, to take -:n: < arg Zn ~ n.
In which cases is the convergence of the sequence {zn} equivalent
only to the convergence of the sequence {lznl} 1
98. On the basis of the statements of problem 97 prove:
(1) lim (I+z/n)n = ex(cosy+isiny);
n-+0a
(2) lim [n('j/(z)-1)] = logr+it/J+2nik
n-+oo
(k = 0, 1, 2, .•• ).
Jl'UNCTIONS OF A COMl'LEX VARIABLE
13
99*. The function f(z) defined in the neighbourhood of the point
z0 is said to be Heine continuous at the point z0 , if for any sequence
{z11}, which converges to z0 , the condition lim f(z 11 ) =f(z0 ) is satisfied;
11-+00
the same function is said to be Cauchy continuous, if for any e > O
there exists a c5(e) > 0 such that from the inequality Jz-z0 J < c5
it follows that Jf(z)-f(z0 )J < e. Prove that these definitions are
equivalent (see, for example, [l, Chapter I, § 3.6]).
.
Rez
z Re(~) zRez
100. The functions -z-' /ZI' lzj2' -Jz_J_ are defined for
z ..P 0. Which of them can be defined at the point z = 0 in such
a way that they are continuous at this point 1
101. Are the functions
1
1
(l) 1-z'
(2) l+zs
continuous inside the unit disk (JzJ
continuous 1
<
1)? Are they uniformly
1
102. (1) Prove that the function e -lif is uniformly continuous
in the disk JzJ
~
R with the point z = 0 deleted.
1
(2) Is the function e:O uniformly continuous in the same
domain!
1
0
<
(3) Is the function e:O uniformly continuous in the sector
Jzl ~ R, JargzJ ~ n/6?
1
103. The function w = e- z is defined everywhere except at
the point z == 0. Prove that:
(1) in the semicircle 0 < Jzl ~ 1, Jarg zJ ~ n/2 this function
is bounded but not continuous;
(2) inside the same semicircle the function is continuous
but not uniformly continuous;
(3) in the sector 0 < JzJ ~ 1, Jarg zJ ~ct < n/2 the function is uniformly continuous.
104. The function f(z) is uniformly continuous in the disk JzJ < 1.
Prove that for any point Con the circle JzJ = 1 and any sequence
z11 -+ C, Jz11 J < 1 there exists the limit lim f(z 11 ). Prove also that this
11-+00
limit does not depend on the choice of the sequence and that if f (z)
is defined on the boundary circle by means of passing to the limit,
it is continuous throughout the whole of the closed disk Jzl ~ 1.
14
PROBLEMS ON OOMPLEX ANALYSIS
§ 4. Analytic and harmonic functions
Tke Oaucky-Riemann equations
105. Verify that the Cauchy-Riemann equations are satisfied.
for the functions z•, ez, cos z, Logz and prove that
(z•)'
= nz•- 1 ,
(ez)' = ez,
(1 ogz) t
(cosz)' = -sinz,
1
=-.
z
106. Write down the Cauchy-Riemann equations in polar coor.
dinates.
107. Prove that if f(z) = u+iv is an analytic function and
s and n are perpendicular vectors, the rotation from the vector s
to the vector n through a right angle being anticlockwise, then
au
Ts=
av
an
au
av
Tri= -Ts
and
(a fas and a fan are the derivatives of the function of two real variables
with respect to the corresponding directions).
108. Prove that the functionf(z) = is not differentiable anywhere.
109. Prove that the function w = z Re z is differentiable only
at the point z = 0; find w' (0).
110. Prove that the Cauchy-Riemann equations are satisfied.
for the function f(z) = y(JxyJ) at the point z = 0 but the derivative
does not exist.
111. Prove the following assertions:
(1) If the limit
z
Jim [Re
.dz-+0
exists for the function w =
and v1 exist and are equal.
f (z)
(2) If the limit lim [rm
.dz....0
u1 and
~w]
LJZ
then the partial derivatives
Uic
~w] exists then the partial derivatives
LJZ
exist and u 1 = -Vic.
(3) If it is assumed in advance that the functions u and v
are differentiable then the existence of either of the limits mentioned
in parts (1) and (2) ensures the existence of the other and consequently the differentiability of the function f(z).
Vic
15
FUNOTIONS OF A OOMPLEX VARIABLE
112. At the point z0 the function w
= f(z)
possesses the following
properties: (1) u, v a.re differentiable; (2) the limit lim
.dz-+0
I~wl exists.
LJZ
Prove that either f(z) or f(z) is differentiable at the point z0 •
113. At the point z0 the function w = f(z) possesses the following
properties: (1) the functions u, v a.re differentiable; (2) the limit
lim arg
4z....0
~w
LJZ
exists. Prove that f(z) is differentiable at the point z0 •
114. f(z) = u+iv = efi- 9 is an analytic function. Prove that if one
of the functions u, v, (!., (J is identically equal to a constant, then the
function f(z) is also a constant.
Harmonie function8
A function u(m, y) which in some domain possesses continuous partial
derivatives up to the second order inclusive and satisfies Laplace's equation
aau aau
Liu== a:i;1+ay•=0,
is said to be a harmonic function. Two harmonic functions u(m, y) and v(m, y)
connected by the Cauchy-Riemann equations
au av
-=z-,
am
8y
are said to be conjugate.
115. Prove the following propositions:
n
(1) A linear combination of harmonic functions}; c1u1 (x, y)
i=l
is a harmonic function.
(2) H the inversion transformation
is applied to the argument of the harmonic function u(x, y) the
transformed function is harmonic.
(3) H the transformation
x = "' (~. 'YJ),
y = 'ljJ(~. 'YJ),
where <P and 'P a.re conjugate harmonic functions is applied to the
argument of the harmonic function u(x, y), the transformed function
is harmonic. (Hence, in particular, the preceding proposition follows.)
16
PROBLEMS ON OOMPLEX ANALYSIS
(4) Let u(x, y) and v(x, y) be conjugate harmonic functions
and let the Jacobian
~~=:=~-be different
from zero in a certain
domain. Then the inverse functions x(u, v) and y(u, v) are also harmonic and conjugate.
116. (1) Prove that every function u(x, y) harmonic in the simply
connected domain G has a family of conjugate harmonic functions,
differing from one another by an additive constant
(x, y)
v(x, y)
J-
=
au
au
ay dx+ ax dy+O.
(x,, y,)
Or,
c:Jn
Cf
0
FIG. 1
(2) Prove that if the domain G is multiply connected and
bounded by an external contour I'0 and by internal contours I'1 ,
I'2 , ••• , I'n (Fig. 1) (each of which can be shrunk to a point), then
the function v(x,y) may turn out to be many-valued and the genera]
formula for its values will be of the form
(x, y)
v(x, y) =
J
n
au
au
~
- ay ax+ ax dy+ L,.; ma::n:1,+0.
k-1
(x,, y 0)
The integral is ta.ken along a path which lies in the domain G,
the ma: are integers and
:n;k
=
J
au
au
--dx+-dy,
ay
ax
YA:
where the /'A: are simple closed contours each of which contains
within it a single connected part of the boundary (I'a:) (the numbers :n;"
are known as the periods of tke integral or the cyclical conBtants).
FUNCTIONS OF A OOMPLEX VARIABLE
17
For the function v(x, y) to be single-valued it is necessary and
sufficient that all the numbers nk should be zero.
REMARK. The contour I'0 may not exist provided that the function u(a:, y)
is harmonic at the point at infinity. By definition this means that the function
U(~, fJ), obtained from the function u(a:, y) by the inversion transformation
(see problem 115, (2)), will be harmonic at the coordinate origin. It can be
n
proved that in this case
2 :n:k =
0.
k=l
117. Assuming it to be Jinown that an analytic function is infi.
nitely differentiable, prove the following theorems:
(1) The real and imaginary parts of the analytic function
/(z) = u+iv are conjugate harmonic functions.
(2) The derivatives (of any order) of a harmonic function
are also harmonic functions.
118. (1) Is u 2 a harmonic function if u is a harmonic function 1
(2) Let u be a harmonic function. For what functions J is
the function f(u) also harmonic 1
119. Are lf(z)I, argf(z), log IJ(z)I, harmonic functions if f(z) is
an analytic function 1
a2u
a2u.
120. Transform Laplace's operator Liu = aaf' + ay2 mto the polar
system of coordinates (r, c/J) and find the solution of Laplace's
equation Liu= 0, which depends only on r.
121. Calculate for n = 1, 2, 3, 4 the harmonic polynomials Pn(x, y)
and qn(X, y), defined by the equation zn = Pn+iqn. Write down the
genera.I form of Pn and qn in polar coordinates.
Using the formulae of problem 116 in problems 122--126 find the
functions conjugate to the given harmonic functions in the domains
indicated.
122. u(x, y) = x2-y2 +x, 0 ~ lzi < oo.
x
123. u(x, y) = x2+y2 , 0 < lzl ~ CX>.
124. u(x, y) =
1
2 1og(x2+y2);
(a) in the domain obtained from the plane by the deletion
of the semi-axis: y = 0, -oo < x ~ O;
(b) in the plane with the coordinate origin deleted (O<lzl <oo).
1
125. u(x, y) = 2 1og(x2+1f)-log[(x-1)2 +y2]:
(a) in the plane with the points z = 0 and z = I deleted;
18
PROBLEMS ON COMPLEX ANALYSIS
(b) in the plane with the segment y = 0, 0 ~ z ~ 1 of the
real axis deleted;
(c) in the plane with the ray y = 0, 1 ~ z < oo, deleted.
n
126. u(z, y) =
!2
ixklog[(z-zk)2+(Y-Ykf1];
k-1
(a) in the plane with the deletion of the points Zi_, z2 , ••• , Zn
z; =F z1);
(b) in the plane with the deletion of a simp]e curve (that is,
one free from self-intersections) formed of rectilinear segments, connecting the given points.
127. Do analytic funotionsf(z) = u+iv exist for which:
z2-y2
(1) u = (z2+y2)2;
(2) v = log (z2+y2)-z2+y2;
(zk = zk+iyk;
11
(3) u = e"i°1
In problems 128-131 find the analytic functions f(z) = u+iv from
the given real or imaginary part.
y
128. u = z2-ya+5z+y- z2+y2 ·
129. u = ex(zcosy-ysiny)+2sinzsinhy+x3-3zy2 +y.
y
130. ·V = 3+z2-y2- 2 (z2+y2)
131. v =log (z2+y2)+z-2y.
REMARx. See also problems 569-572.
In problems 132-139 explain whether harmonic functions of the
given form (different from a constant) exist and if they exist find
them.
132. u = c/J(z).
133. u = cf>(ax+by) (a and b are real numbers).
134. u =
<P(!).
135. u = c/J(zy).
136. u = c/J(z2+y2).
z2+y2)
137. u=cf> (z-.
FUNOTIONS OF A OOMl'LEX VARIABLE
19
138. u = </>[x+y(x2+1f)].
139. u = ef>(x2+?f).
In problems 140-143 prove the existence of and find the analytic
functions f(z) from the given modulus or argument.
140. = (xt+ys)eX.
141. = e""cos8'[>.
142. 0 = ey.
143. 0 = ef>+rsinef>.
144. Prove that for the family of curves cf>(x, y) = 0, where cf> is
a twice continuously differentiable function, to be the family of
level lines of some harmonic function it is necessary and sufficient
that the ratio LJcf>/(gra.dcf>) 2 should depend only on ef>.
e
e
HINT. As a preliminary establish that the required harmonic function is of
the form u =/[•(a:, y)].
In problems 145-149 find analytic functions of which the real
part, the imaginary part, the modulus or the argument preserves
a constant value along any line of the corresponding family.
145. x = 0.
146. y = o.
147. y =Ox.
148. x11+y11 = 0.
149. xl+y2 =Ox.
Phe geometrical meaning of the modulUB and argument of a derivative
150. Mappings are made using the functions w = z2 and w = z8•
For each function find the angle of rotation (0) of the directions
issuing from the point z0 , and the magnification (k) at the following
points:
(1) 210 = 1; (2) z0 = -1/4; (3) z0 = l+i; (4) z0 = -3+4i.
151. Which part of the plane is compressed and which is stretched
if the mapping is effected by the functions:
(1) w = z2;
(4) w = e:r:;
(2) w = z8 +2z;
(5) w =log (z-1) 1
(3) w = l/z;
152. The domain G is mapped by means of the function f(z)
conformally and one-one onto the domain G'. Find formulae for
the calculation of the area S of the domain G' and the length L of
the arc onto which some arc l belonging to the domain G is mapped.
153. Find the length L of the spiral onto which the function e:I:
maps the segment y = x, 0 .:::;; x .:::;; 23t (see problem 91).
20
PROBLEMS ON OOMPLEX ANALYSIS
154. Find the area of the domain onto which the function e"
maps the rectangle 1 < z < 2, 0 ,.s;; y < 4.
155. Find the domain D, onto which the function e" maps the
rectangle 1 < z < 2, 0 < y < 8. Calculate the area of the domain
D by means of the formula obtained in the solution of problem 152
and explain why this formula gives an incorrect result.
OHAPTER II
CONFORMAL MAPPINGS CONNECTED
WITH ELEMENTARY FUNCTIONS
§ 1. Linear functions
Linear function!J
156. Find the linear function which maps the triangle with vertices at the points 0, 1, i onto a similar triangle with vertices at
0, 2, l+i.
157. Find the linear transformation with fixed point 1+2i whioh
transforms the point i into the point -i.
158. For the transformations given find the finite fixed point z0
(if it exists), the angle of rotation round it, (), and the magnification le.
Reduce these transformations to the canonical form w-z0 = A.(z-z0).
(1) w = 2z+l-3i;
(4) w-w1 = a(z-z1 ) (a =F O);
(2) w = iz+4;
(5) w = az+b (a =f:: 0).
(3) w = z+l-2i;
159. Find the general form of the linear transformation which
transforms :
(1) The upper ha.If-plane into itself;
(2) The upper half-plane into the lower ha.If-plane;
(3) The upper half-plane into the right ha.If-plane;
(4) The right ha.If-plane into itself.
Show that in each case the transformation is uniquely determined
by specifying a single pair of corresponding interior points or two
pairs of boundary points.
160. Find the general form of the linear transformation which
transforms :
(1) The strip 0 < z < 1 into itself;
(2) The strip -2 < y < 1 into itself;
(3) The strip bounded by the straight lines y = z and
y = z-1 into itself.
Explain which pairs of points can correspond to one another in
these mappings and in which case this correspondence determines
the mapping uniquely.
21
22
PROBLEMS ON OOMPLEX ANALYSIS
161. Find the linear function w(z), which maps the strip, contained between given straight lines onto the strip 0 < u < 1 with
the given normalisation:
(1) x =a,
x = a+k; w(a) = O;
(2) x=a,
x=a+k;
Imw(a+; +i)
(3) y ='/ex,
y
(4) y = kx+bi,
<
w(a+
;)= !
+i;
1.
= kx+b; w(O) = O;
y = kx+b 2 ; w(b1 ) =
0.
162. Find the linear function which maps the circle !z! < 1 onto
the circle !w-Wo! < R and is such that the centres of the circles
correspond to one another and the horizontal diameter passes into
a diameter ma.king an angle a with the direction of the real a.xis,
Bilinear functiona
163. For the function w = 1/z find the images of the following
curves:
(1) The family of circles x2+y2 = ax;
(2) The family of circles zs+ys =by;
(3) The pencil of parallel straight lines y = x+b;
(4) The pencil of straight lines y = kx;
(5) The pencil of straight lines passing through a given
point z0 =F 0;
(6) The para.bola. y = z2.
164. Explain into what the function w
forms:
= - 1-+k trans-
(1) The rectangular net x = 0, y = O;
(2) The polar net !z-z0 ! = R, a.rg (z-z0 ) =
z-z0
Cl.
Z-Z1
165. Given the function w = - - :
Z-Z2
=A (O <
A< oo) is a family of circles (the circles of Apollonius). For
a given A find the radius and the position of the centre of the
<
(1) Prove that the original of the family !w!
corresponding circle in the z-pla.ne.
(2) Find the original of the ray arg w = ().
(3) Construct the net in the z-plane which corresponds
to the polar net in the w- plane.
23
CONFOBMAL MAPPINGS
(4) Find the domain of the z-plane which corresponds
to the semicircle lwl < 1, Im w > 0.
In problems 166-170 explain what the domains indicated are
transformed into with the given mapping functions.
166. The quadrant :x:
> 0,
y
> O;
< 1,
Imz
z-i
z+i
w = -..
w = 22 z-:i.
+iz
168. The angle 0 < </> < :rc/4; w = z/(z-1).
169. The strip 0 < :x: < 1; (1) w = (z-1)/z; (2) w = (z-1)/(z-2).
170. The ring 1 < lzl < 2; w = z/(z-1).
171. Map onto the vertical strip 0 < Rew < 1:
(1) The half-plane Re z > 0 with the disk lz-d/21 d/2
deleted;
(2) The lune contained between the circles
167. The semicircle lzl
> O;
<
lz-d1 /21 = d1 /2, lz-d2 /21 = d2 /2
(d1
<
d2 );
(3) The exterior of the circles lz+d1 /21 = di/2, lz-d2/21
= d2 /2 so that w(d2 ) = 0.
172. Find the bilinear function which transforms the points -1,
i, 1+i respectively into the points:
(1) 0, 2i, 1-i;
(2) i, oo, 1.
173. Find the bilinear function which transforms the points -1,
oo, i respectively into the points :
(1) i, 1, l+i;
(2) oo, i, 1;
(3) 0, oo, 1.
174. Find bilinear functions from the following conditions:
(1) The points 1 and i are fixed and the point 0 passes into
the point -1;
(2) The points 1/2 and 2 are fixed and the point 5/4+3i/4
passes into oo;
(3} The point i is a double fixed point and the point 1 passes
into oo.
175. Find the bilinear function which transforms the points -1,
0, 1 respectively into the points 1, i, -1, and explain what the upper
half-plane becomes in this mapping.
176. Find the general form of the bilinear transformation whioh
transforms:
(1) The upper half-plane into itself;
(2) The upper half-plane into the lower half-plane;
(3) The upper half-plane into the right half-plane.
24
PROBLEMS ON COMPLEX ANALYSIS
177. Find the mapping of the upper half-plane into itself \\ith
the given normalisation:
(1) w(O) = 1, w(l) = 2, w(2) = oo;
(2) w(O) = 1, w(i) = 2i.
RBMABK. For mappings of the upper half·plane into itself for other normalisations see problem 188.
178. Find the function w(z) which maps the circle lzl < R onto
the right half-plane Re w > 0 in such a way that w(R) = 0,
w (-R) = oo, w (0) =I. What is the image of the upper semicircle in this mapping 1
Two points P 1, P 1 are said to be symmetrical with respect to the circle
K with centre 0 and radius R, if they lie on the same ray issuing
from 0 and
OP1 .0P1 = R 8 •
179. Find the points symmetrical to the point 2+i with respeot
to the cii-cles:
(1) lzl = 1;
(2) lz-il = 3.
180. Find the symmetrical image with respect to the unit circle
of the following curves:
(1) lzl = 1/2;
(4) lz-zol = l:<iol
(zo = xo+iyo):
(2) lai-11 = 1;
(5) lz-z0 1= y(lz0 12 -l) (jz0 1> 1);
(6) the hyperbola. x2-y2 = 1;
(3) y = 2;
(7) The perimeter of the rectilinear triangle with vertices
Zt• Z,, Z:1(Z1 '¥: 0).
181. Prove that for the symmetry of the points P 1 and P 2 with
respect to K it is necessary and sufficient that one of the following
two conditions should be satisfied:
(1) Every circle K 1 passing through the points Pv P 2 , is orthogonal
to K;
(2) For all points M of the circle K
MP1
--=Const
MP2
(that is, K is a circle of Apollonius with respeot to the points P 1
and P 2).
182. The function
w =
z-{3
e'•---=
z-{3
({3 = a+ib, b
> 0)
CONFORMAL MAPPINGS
25
maps the upper half-plane onto the unit disk.
(1) Find arg w(z) = O(z);
(2) Find w' ({3);
(3) Explain what part of the upper half-plane is compressed
in this mapping and what part is stretched.
183. Map the upper half-plane Im z > 0 onto the unit disk lwl < 1
in such a way that:
(1) w(i) = 0, arg w' (i) = -7'/2;
(2) w(2i) = 0, arg w' (2i) = 0;
(3) w(a+bi) = 0, arg w'(a+bi) = 0 (b > 0).
184. Map the upper half-plane Im z > 0 onto the disk lw-w0 1 < R
so that the point i passes into the centre of the disk and the derivative at this point is positive.
185. Map the disk izl< 2 onto the half-plane Rew>O in such
a way that w(O) = 1, arg w'(O) = 7'/2.
186. Map the disk iz-4il< 2 onto the half-plane v > u so that
the centre of the disk passes into the point -4, and the point 2i
on the circumference becomes the coordinate origin.
187. Find the general form of the bilinear function w(z) which
maps the disk izl <R onto the right half-plane Re w>O in such a way
that w(z1 ) = 0, w(z2 ) = oo, where Zi· z2 , are specified points on the
circumference izl = R. Construct the family of curves in the circle
lzl < R corresponding to the polar net in the half-plane Rew> 0.
HINT. It is possible to use either the general form of the transformation
of the half-plane into a disk or the general form of the bilinear transformation
for three pairs of corresponding points and the result of problem 8.
188. Find the function which maps the upper half-plane onto
itself BO that
w(a) = b,
argw'(a) =ix
(Ima>O, Imb>O).
HINT. As a preliminary map both specimens of the half-plane onto the
unit disk with the corresponding normalisation.
189. Map the upper half-plane onto the lower one so that w(a) =ii
and arg w'(a) = -7'/2 (Ima>O).
190. For the function
z-a
w = e1• - - (lal<l)
l-az
which maps the unit disk onto itself:
(1) Find argw(e1"') = O(cf>);
(2) Find w' (0) and w' (a);
26
PROBLEMS ON COMPLEX ANALYSIS
(3) Explain which pa.rt of the unit disk is compressed in
this mapping and which is stretched;
(4) Find max ldw/dz! and min ldw/dz! for !z! ~I.
191. Map the disk !z!<l onto the disk !w!<l so that:
(1) w (1/2) = 0, a.rg w' (1/2) = O;
(2) w(i/2) = 0, a.rgw'(i/2) = n/2;
(3) w(O) = 0, a.rg w' (0) = -n/2;
(4) w(a) =a, a.rg w'(a) = o:.
192. Map the disk !z! <R1 onto the disk !wJ <R2 so that w(a) = b,
arg w'(a) = o: (!a!<R1 , !bJ<R2 ).
193. Map the disk !z! < 1 onto the disk !w-1J<1 so that w(O)
= 1/2 and w(l) = 0.
194. Map the disk !z-2!<1 onto the disk !w-2i!<2 so that
w(2) = i and arg w' (2) = 0.
195. Find the general form of the bilinear function w(z), which
maps the disk Jzj <R onto itself subject to the following conditions:
(1) w(a) = 0 (!a!<R);
(2) w(a) = b (laJ<R, !b!<R);
(3) w(±R) = ±R.
196. Map the disk !zJ<l onto itself so that the specified points
z1 , z2, within the disk pass into the points ±a (Jal<l); find Jal.
19'7. Map the disk Jz! < 1 onto itself so that the real a.xis segment:
y == 0, O~x~a (a<l) passes into a real a.xis segment sym..
metrical about the coordinate origin. Find the length of the transformed segment.
198. Prove that in the mapping of a disk onto a disk the linear
transformation is uniquely determined by the specification of one
interior and one boundary point.
199. The unit disk is mapped onto itself in such a way that the
point z0 ~0 passes into the centre of the disk. Prove that then
a unit semicircle is mapped onto a unit semicircle if and only if its
ends lie on a diameter passing through the point z0 •
200. Construct the mapping of the unit disk onto itself in which
the original of the centre is on the real a.xis and the a.re O~</>~n/2
of the unit circle is mapped onto the following arcs:
(1) O~()~n/2;
(2) O~()~n/3;
(3) n/2~()~7n/6.
§ 2. Supplementary questions of the theory of linear transformations
Canonical formB of linear traneformationa
A bilinear transformation with one fixed point is said to be parabolic. A
OONFOBMAL MAPPINGS
27
parabolic transformation can be written in the canonical form
I
I
·--=--+h,
,,,_Zo
Z-Zo
if z.;i:oo, or
if z.=oo.
A bilinear transformation with two distinct fixed points z1 and :i:s has the
canonical form
w-z1
z-z1
z-z.
--=k--,
w-z.
w-z 1
=
k (z-z1 )
if z1 = oo; the transformation with two distinct fixed points is hyperbolic if
k> O, elliptic if k = elar: and ct;f: 0 and lOlllO!lromic if k = aelar:, where a;f: I
and ct#:O (ct and a real numbers, a>O).
201. Prove the following assertions:
(1) The general bilinear transformation
az+b
W==--
cz+d
can be reduced to the form
where
I~ ~I= 1.
(2) If a+<5 is a real number, the transformation is elliptic
when !a+'51 < 2, hyperbolic when la+<51 > 2, and parabolic when
1a+<5I = 2.
(3) If Im(0!+'5) '# 0, then the transformation is loxodromic.
202. Prove that if a linear transformation has two fixed points
the product of the derivatives at these points is equal to unity.
203. Find the circles which in the parabolic transformation
1
1
--=--+h
W-Zo
Z-Zo
pass into themselves.
204. Find the general form of the parabolic transformation of
the disk !zl < R into itself if the point R is fixed.
205. Prove the following properties of the hyperbolic transformation:
28
PROBLEMS ON OOMPLEX ANALYSIS
(1) Any circle passing through the two fixed points is transformed into itself, the direction of traversal being preserved.
(2) Any circle orthogonal to the circles passing through the
fixed points is transformed into a circle having the same property.
(This property follows directly from property (1).)
HINT.
As a preliminary consider the case when the fixed points are 0
and oo.
206. Prove the following properties of the elliptic transformation:
(1) Any circle orthogonal to the circles passing through two
fixed points is transformed into itself with preservation of the direction of traversal.
(2) A circular arc connecting the fixed points passes into a
circular arc connecting the fixed points and making an angle IX with
the first arc (IX= arg k).
207. (1) Prove that in the loxodromic transformation the properties (2) of the hyperbolic (see problem 205) and of the elliptic
transformation (see problem 206) are preserved.
(2) Prove that, provided that ix =f:: n (ix= argk), fixed circles
do not exist in the loxodromic transformation. Prove that if
oc = n, then the circles passing through the fixed points transform
into themselves with change of the direction of traversal.
208. Prove that in the loxodromic transformation w = ae1«z the
10110"
logarithmic spirals r = Ae...- {A > 0) pass into themselves.
209. Prove that the linear transformation
w=
z-a
e"·---1-az
(a=
lale1«, lal <
1),
which transforms the unit circle into itself can only be elliptic, hyperbolic or parabolic. Explain for what values of a each of the given
cases holds. Find the fixed points of the transformation and reduce
it to the canonical form.
Some approximate formul,ae for linear transformations
210. The upper half-plane is mapped onto the unit disk so that
the point z = ki (k > 0) passes into the centre of the circle. Find
the length I' of the image of the segment [O, a] of the real axis (a> 0)
and obtain linear approximate formulae for I' for small a/k and
for small k/a.
211. The unit disk is mapped onto itself in such a way that the
original of the centre of the disk, the point x 0 , is on the real axis.
Find the length I' of the image of the arc 0 ~ cf> ::( y of the unit
CONFORMAL MAPPINGS
29
circle (y::,;: n). How does the quantity I'/r vary depending on the
sign of x0 1
212. With the conditions of problem 211 obtain the formulae:
l+xo
(1) I'= - 1 - r+o (y8) for small 'Y;
-Xo
(2)
I'= n-ecot..l- ~cot2 1..+o(e8 )
2
2
2
for small e, where
l-x0 •
213. The unit disk is mapped onto itselfso that the point z0 = r0e1 •
passes into the centre. The points z1 = e1• 1 and z2 = e1•• lie on opposite sides of the diameter passing through z0 (c/> 0 < c/>1 < </> 2 ::,;: c/> 0 +n).
Assuming that the point z0 is situated close to the unit circle, prove
that the length I' of the image of the arc c/> 1 ::,;: cf> ::,;: </> 2 of the unit
circle satisfies the formula
8 =
where e = l-r0 •
Mappings of simply connected domains
214. Prove that if the linear mapping of the disk izl < 1 onto
itself does not reduce to a rotation, then no concentric ring with
centre at the coordinate origin passes into a concentric ring.
This proposition is a particular case of the following theorem:
For the conformal mapping of the ring r 1 < lzl < r 8 onto the ring R 1 < lwl
< R 1 it is necessary and sufficient that the condition R 1 /R1 = r 1 /r1 should
be satisfied. In this case the mapping function can only have two forms:
w = az or w = a/z. The mapping is uniquely determined by the specification
of a single pair of boundary points (see, for example, [3, Chapter II, § 3]).
REMARK.
2Ui. (1) Map the ring 2 < jzj < 5 onto the ring 4 < lwl < 10 so
that w(5) = -4.
(2) Map the ring 1 < jz-2il < 2 onto the ring 2 < jw-3+2il
< 4 so that w(O) = -l-2i.
The following theorem holdst:
Every doubly connected domain, the boundaries of which do not shrink
t See, for example, the paper: M. V. KELDYsH, (1939) Conformal representations of multiply connected domains (Konformnyye otobrazheniya mnogosvyaznykh oblastei), Uape'/chi, matem. naulc, vol. VI, 90-119.
30
PROBLEMS ON COMPLEX ANALYSIS
to points, can be conformally mapped onto a concentric ring with well-defined
ratio µ of the radii of the outer and inner circles (µ is the modulus of the doubly
connected domain).
216. Map the semicircle Rez > 0 with the disk lz-hl < R (h > R)
deleted onto the ring (! < lwl < 1 so that the imaginary axis passes
into the circle jwj = 1. Find !!·
HINT. Construct the circle with centre at the coordinat.e origin and ortho·
gonal to the circle lz-hl = R; then find the linear transformation, which trans•
forms the real axis and the constructed circle into two int.ersecting (orthogo·
nally) straight lines and verify that the given domain is then mapped into
the concentric ring. Prove that the centre of this ring is coincident with the
coordinate origin if the points of intersection of the constructed circle and
the real axis pass into 0 and oo.
217. Map the half-plane Rez > 0 with the disk jz-hj < 1, h > 1
deleted, onto the ring 1 < jwl < 2. Find h.
218. Map the eccentric ring bounded by the circles lz-31 = 9,
jz-81 = 16 onto the ring (! < jwj < 1. Find p.
219. Map the doubly connected domain bounded by the circles
lz-z1 j = r1 , jz-z 2 1= r 2 (lz2 -z1 j > r1 +r2 or lz2 -ztl < jr2 -r1 1) onto
a concentric oircular ring with oentre at the coordinate origin. Find
the modulus (µ) of the domain.
HINT. Find a pair of points symmetrical with respect to both circles and
map one of them onto 0 and the other onto oo.
REMARK. It is easily verified that the methods of solution recommended
in the hints to problems 216 and 219 a.re the same.
220. "Q'sing the solution of the preoeding problem find the moduli
of the doubly connected domains bounded by the given circles:
(1) jz-11=2, lz+ll = 5;
(2) lz-3il = 1, jz-4J = 2.
Group properties of bilinear transformations
The transformation T(z) = T 8 [T1 (z)] is said to be the product of T 1 and
= T 1 T 1 (the order is important since, generally
speaking, T 1T 1 of: T 1 T 1 ). The set G of transformations T forms a group if it
contains the product of every two transformations belonging to it and together
with the transformation T contains the transformation T-1 inverse to it.
A group consisting of the powers T1I and p-n of a single transformation T is
said to be cyclical. If the group G is formed from the transformations TI> T 8, ••• ,
Tn by the construction of all the inverse transformations and of all possible
products of the given transformations and their inverses, these transformations
are said to be generatora of the group G. The points obtained from the fixed
point z by means of all the transformations of the group G are said to be
equivalent or congruent with respect to the group G.
The fundamental domain of the group is an open set (connected or discon·
nected) which does not contain a single pair of points equivalent to one another
T 8 and is written in the form T
31
CONFORMAL MAPPINGS
with respect to the given group, and in the neighbourhood of every boundary
point of which there are points equivalent to the points of the set.
221. Let T 1 be the linear transformations:
la,
LI,=
db11· ¥: 0 (i = 1, 2, ... ).
c1z+d1
c, 1
Prove the following assertions:
(1) T = T 1T 2 is a linear transformation with determinant
LI = Ll 1 L1 2•
(2) The product of the transformations is associative, that is,
T,(z) = a1z+b1'
(T3 T 2)T1
=
T 3 (T2T 1 ).
(3) Every transformation T 1 has an inverse Tj 1, that is,
T 1Tl 1 = T'j' 1T 1 = J,
where J(z) = z is the identical transformation.
(4) The product of transformations, generally speaking, is not
commutative (give examples).
222. Prove that the transformations
1
1
T 3 = 1-z, T , = - - ,
z
1-z,
z-1
z
-T,,=--, T
a--z-1
z
form a group (the group of anharmonic ratioB).
223. Prove that the set of linear transformations which consists
of the rotation of a plane round the coordinate origin by angles which
a.re multiples of cc, forms a cyclical group. In which case does this
group consists of a finite number of transformations 1
224. (1) Prove that the set of transformations of the form
w = (az+b)/(cz+d) where a, b, c, d a.re real integers and
ad-be= 1, forms a group (this group is said to be modular).
(2) Prove that if a, b, c and d a.re considered as complex
integers (that is, numbers of the form m+ni, where m and n a.re
real integers), which satisfy the condition ad-be= 1, then the
set of transformations of pa.rt (1) also forms a group (Picard'B group).
225. Find the fundamental domains of the groups generated
by the transformations :
T2 = -
ll1fl
(1) T(z) =en z (n a natural number);
32
:PROBLEMS ON COMPLEX ANALYSIS
27d
(2) T 1 (z) = enz, T 2 = 1/z;
(3) T(z) = z+w;
(4) T 1 (z) = z+w, T 2 (z) = -z;
(5) T 1 (z) = z+ro1 , T 2(z) = z+ro 2
(the
(Imro 2/ro1 =F 0)
doubly periodic group) ;
(6) T 1 (z) = z+ro1 , T 1 (z) = z+w1 , T 3 (z) = -z;
(7) T 1 (z) = z+w; T 2 (z) = iz;
(8) T 1 (z) = z+w, T 2 (z) = e9"'18 z;
(9) T 1 (z) = z+w, T 2 (z) = a21Cil&z.
226. Find the group of linear transformations corresponding
in stereographic projection to the rotation of the sphere:
(1) Round the vertical diameter;
(2) Round the diameter parallel to the real axis;
(3) Round the diameter parallel to the imaginary axis;
(4) Round the diameter, the stereographic projection of
one of the ends of which is the point a.
HINT. If z1 , z1 are the images of diametricaJly opposite points on the sphere,
then z1ii = -1 (see problem 48 of Chapter I).
227. (1) Prove that the group of linear transformations which
correspond to a rotation of the sphere, and transform the points
with stereographic projections a and b into one another, is defined
by the relation
w-b
z-a
---=e'•-l+bw
l+az ·
(2) Prove that the differential
ldzl
ds = 1+1z12
is invariant with respect to the transformations of this group and
represents the spherical length of the element of arc dz (that is, the
length of the image of this element on the sphere).
Linear transformations and non-Euclidean geometry
In the realisation of non.Euclidean geometry in the unit circle the part
of straight lines is played by arcs of circles in the unit circle and orthogonal
to it; the role of motion is played by linear transformations of the unit circle
onto itself, the role of distance between the points Zi and z1 by the quantity
!,>(Z1 , Z 2 )
=
1
2log(a, p,
z., z1 ),
where a and p are the points of intersection of the "straight lines" passing
through the points Zi and z2 , with the unit circle (the order of the points is
CONFORMAL MAPPINGS
33
a, z1 , z1 , fl), and (a, fl, z9 , z1 ) is the anharmonic ratio of the given points. The
angles are measured as in Euclidean geometry (see, for example, [I, Chapter II,
§ 4, section 8]).
228. Prove that Q(Z1, Z2) > 0 if Z1 i: Z2 and Q(Z, z) = 0.
229. Prove that Q(Z1, Z3) ~ Q(Z1, Z2)+Q(Z2, Z3) where the sign
of equality applies only when the point z3 lies on the "segment",
connecting the points z1 and z2 •
230. Prove that if one of the points 211 and z2 tends to a point of
the unit circle (or both of them to different points of the unit circle),
then the non-Euclidean distance e(z1 , z2) tends to infinity (that
is, the points of the unit circle correspond to the infinitely distant
points of the non-Euclidean plane).
231. Prove that the differential
ds=~
l-lzl2
(lzl <I)
is invariant with respect to the group of linear transformations which
transforms the circle lzl < 1 into itself and represents the non.
Euclidean length of the element of arc dz.
HINT.
<
lbl <
lzl
Write down the general form of the transformation of the disk
a into the point b (lal < 1,
1 into itself which transforms the point
1).
232. Indicate methods for the construction of the following
curves:
(I) The pencil of "straight lines" passing through the
point z0 ;
(2) "The straight line" passing through the points z1 and z2 ;
(3) The equidistance of a "straight line" (the geometrical
locus of points "equally distant" from the given "straight line");
(4) Limiting curves (curves orthogonal to the pencil of
"parallel straight lines").
233. (1) Prove that for the "rectilinear" triangle with angles
<Pv <{> 2 , </J3 the inequality </J 1 +</J 2 +</J3 < ;n; holds.
(2) Prove that the "rectilinear" triangle apart from
"motion" is defined by its angles </J 1 , </J 2 • </>3 • Construct the "rectilinear"
triangle from its angles.
§ 3. Rational and algebraic functions
The general mapping of a disk or half-plane onto a simply connected domain
of thew-plane is of the form w = <J>[l(z)] where t/>(z) is a particular mapping
and l is an arbitrary bilinear mapping of the disk or half-plane onto itself
(the inverse mapping is of the form z = l[V'(w)]). It is necessary to keep this
34
PROBLEMS ON COMPLEX ANALYSIS
remark in mind when finding a normed mapping, that is, a mapping which
satisfies certain supplementary conditions. If the condition of normalisation
is not given then one of the mapping functions is usually indicated in the
answer.
In the practical construction of conformal mappings an important part
is played by certain general principles (see, for example, [2, Chapter IX, § 5
and Chapter X, § 7] or [3, Chapter II, §§ 1 and 3].
The Riemann-Schwar7J principle of symmetry
Let the boundary of the domain D 1 contain an arc of the circle 0 (in particular a rectilinear segment), and let the function w =f1 (z) effect the conformal
mapping of this domain onto a domain D~ so that the arc 0 passes again into
an arc of a circle or rectilinear segment O*. Then the function J1 (z) which at
points symmetrical with respect to 0 assumes values symmetrical to the values
of f 1 (z) with respect to O*t, will be analytic in the domain D 1 symmetrical
to the domain D 1 with respect to 0 and will map it onto the domain n:, symmetrical to D~ with respect to O*.
The function
f1(z)
in D 1 ,
w = f1(z) = f 1(z) on 0,
J1 (z)
in D 1
effects the conformal mapping of the domain D 1 +0+D1 onto the domain
l
n~+o•+n:tt.
The principle of corresponding boundaries
Let D and D* be simply connected domains with boundaries 0 and o•
the domain D* being situated entirely in the finite part of the plane. If the
function w = f(z) is analytic in D, continuous on 15 and maps 0 onto O* one-one
with prese~ation of the direction of traversal then it effects the one·one and
conformal mapping of the domain D onto D*.
In the solutions of the problems of this and the following sections in cases
when the mapping is effected by a branch of a many-valued function, it is
recommended that the correspondence of points on the boundaries of the
mapped domain and its image should be traced out (this especially refers to the
mapping of domains with cuts).
234. By means of the function w = z2 and its inverse find the
conformal mapping of the following domains:
(I) The interior of the right hand branch of the equilateral
hyperbola x2-y2 = a 2 onto the upper half-plane;
(2) The exterior of the parabola y2 = 2px, p > 0 (that
is the domain bounded by this p~rabola and not containing its
focus) onto the right hand half-plane.
t If 0 and O* are segments of the real axis (it is always poSBible to arrange
this by making an additional bilinear transformation), then fz(z) = f 1 (z).
tt The mapping will be one-one if the domains D 1 and D 1 and also D~
and
do not intersect.
n:
35
CONFORMAL MAPPINGS
REIU.BX. For the mapping of domains bounded by curves of the second
order, see also problems 164, HS, 291, 212, 293, 328.
235. By means of the functions indicated in the preceding problem
map:
(1) The interior of the circle r = a cos cf> (a > 0) onto the
exterior of the cardioid e = ~ (1 +cos fJ);
(2) The interior of the same circle onto the interior of the
right hand branch of the lemniscate e = y (cos 20);
(3) The disk lzl < 1 onto the interior of the cardioid
e = A(l+cosO), A> 0 so that w (0) = A/8, w'(O) > O.
238. Find the domain onto which the disk lzl < 1 is mapped
by means of the function
R > 0,
0 ~ m ~ j.
Find the image of the polar net of the z-plane.
!37. Find the domain onto which the semicircle lzl < 1, Re z > 0
is mapped by the function
w=z+z•.
238. (1) Find the domain onto which the disk lzl < 1 is mapped
by the function
w = R(z+(zn/n)), R > 0,
w = R(z+mz•),
where n is an integer, n > 1.
(2) Find the domain onto which the exterior of the unit
disk lzl > 1 is mapped by the function w = R(z+ (1 /nzn)), R > 0, n
an integer, n > 1.
RE111A.:ax. For mappings involving the function
w
= B(z+ (1/z))
(Zhukovskii's function) see problem 261 et 11eq.
239. (1) Explain for what values of m the function w = R(z+mzn),
where n is a natural number, effects the conformal mapping of the
disk jzj <1 onto a certain domain, and find this domain.
(2) Explain the same questions for the mapping of the exterior
of the disk lzJ <1 by means of the function w = R(z+ (m/zn)) and of
the interior of the same disk by the function w = R((l/z)+mzn).
°'
240. (1) Map the angle 0 < argz < nx, (0 <
~ 2) onto the
upper half-plane.
(2) Map the angle -n/4<argz<n/2 onto the upper half.
plane so that w(l-i) = 2, w(i) = -1, w(O) = 0.
36
l'ROBLEMS ON COMPLEX ANALYSIS
241. Find the function w(z), which maps the semicircle lzl < 1,
Im z >0 onto the upper half-plane with the conditions:
(1) w(-1) = 0,
w(O) = 1, w(l) = oo;
(2) w(±l) ==fl, w(O) = oo;
(3) w(i/2) = i, argw'(i/2) = -7'/2.
242. Find the function w(z), which maps the semicircle lzl<l,
Im z>O onto the disk lwl<l with the conditions:
(1) w(±l) = ±1,
w(O) = -i;
(2) w(i/2) = 0, argw'(i/2) = 7'/2.
243. Find the function w(z), which maps the domain lzl >I,
Imz>O onto the upper half-plane.
244. Map onto the upper half-plane:
(1) The sector lzl <R,
0<argz<7'1X (O<ix ::;;;2);
(2) The domain lzl >R, O<argz<'J"Ccx (O<cx ~2).
In problems 245-250 map onto the upper half-plane the circular
lunes (two-angles).
245. lzl<l,
lz-il<l.
246. lzl<l,
lz-il>I.
247. lzl>l,
jz-ij<l.
248. lzl>l,
!z-ij>l.
249. lzl >2,
lz-Jf21 <J!2.
250. The exterior of the upper unit semicircle.
In problems 251-259 map the given domains onto the upper
half-plane.
251. The plane with a cut along the segment [-1, I].
252. The plane with a cut along the segment [-i, i].
253. The plane with a cut along the segment [z1' z1].
254. The plane with cuts along the rays (-oo, -R], [R, oo)
(R>O).
255. ';['he plane with a cut along the ray in the first quadrant
which issues from the point i and is parallel to the straight line
y=x.
256. The half-plane Imz>O with a cut along the segment [O, ih],
h>O.
257. The half-plane lmz>O with a cut from ih to oo along the
positive imaginary semi-axis (h>O).
258. The disk lzl<l with a cut along the radius [O, I].
259. The exterior of the unit circle with a cut along the ray [I, oo).
260. Find the mapping of the disk lzl <I onto the w-plane with
a cut along the ray (-oo, -1/4] subject to the conditions w(O) =0,
w'(O)>O.
37
OONFOBMAL M.Al'l'INGS
261. Find the transformation of the polar net lzl
by the function w
=
!
~
= R, arg z =et:
(z+ ).
282. Find the domain onto which the function w =
! !)
(z+
maps:
(1) The disk lzl<R<l;
(9) The domain lzl>l,
The domain lzl>R>l;
Imz>O;
The disk lail<l;
(10) The domain l<lzl<R,
The domain jzj>l;
Imz>O;
The half-plane Imz>O;
(11) The domain R<lzl<l,
The half-plane Im3<0;
Imai>O;
The semicircle jzj<l;
(12) The domain 1/R<
Imz>O;
<lzl<R, Imz>O;
(8) The semicircle lzl<l,
Rez>O;
Imz<O;
(13) The angle :n;/2-a<argz<:rc/2+rt. (O<rt.<:rc/2).
163. Find the transformation of the polar net using the functions:
(1) w = (z-1/z)/2;
(2) w = (z+a1 /z)/2
(a > 0);
(3) w = (z+c2/z)/2, e = jele17 (0 ~ I' <:re).
(2)
(3)
(4)
(5)
(6)
(7)
264:. Using the function w =
! !).
(z+
map:
(1) The exterior of the segment [-e, e], (e > 0) onto the
exterior of the unit circle with the condition that w(oo) = oo,
argw'(oo) =rt.;
(2) The exterior of the ellipse x2/a2 +y2/b2 = 1 onto the
exterior of the unit circle so that w(oo) = oo, arg w'(oo) = O;
(3) The upper half-plane with the semi-ellipse W-/a2 +y2/b2<. 1;
y > 0 removed, onto the upper half-plane.
285. Map the doubly connected domain bounded by the confocal
ellipses
onto a concentric circular ring with centre at the coordinate origin, and
find the modulus of the given doubly connected domain (seep. 29).
266. Find the domain onto which the function w =
! (z+-~-)
38
PROBLEMS ON COMPLEX ANALYSIS
maps the disk lzl < 1 with a cut along the segment [a,l] (-1
<a < 1). Consider the cases a > 0 and a < 0.
In problems 267-270 map the given domains onto the upper
half-plane.
267. The disk lzl < 1 with a out along the segment [1/2, l].
268. The disk lzl < 1 with cuts along the radius [- 1, OJ and
the segment [a,l] (0 < a < 1).
269. The upper half of the disk lzl < 1 with a cut along the segment
[O, exi] (0 < ex < 1).
270. The upper half of the disk lzl < 1 with a cut along the segment
[ixi, i] (0 < ex < 1).
271. Map the disk lzl < 1 with a cut along the segment [(l-k)e1•,
e'•J onto the unit disk of the w-plane.
272. Map the disk lzl < 1 with a cut along the segment [a, l],
0 <a< 1, onto the disk lwl < 1 80 that w(O) = 0, w'(O) > 0.
Find w'(O) and the length of the arc corresponding to the cut. For
what value of a does the cut pass into a semicircle 1
HINT.
disk lwl
It is an advantage first to map both the given domain and the
< 1 onto the exterior of the segment.
273. Map the disk lzl < 1 with outs along the segments [a, l],
[- l, -b] (0 <a < 1, 0 < b < 1) onto the disk lwl < 1 80 that
w(O) = 0, w' (0) > 0. Determine w' (0) and the length of the arc
corresponding to the cuts.
274. Represent Zhukovskii's function in the form
:+~ =(~+~r
and find:
(1) The image of the circle 0 which passes through the points
z = ± 1 at an angle ex (- n <IX < n) to the real axis at the point 1,
and the domain onto which the exterior of this circle is mapped;
(2) The image of the circle O' which passes through the point z = 1
at an angle ex to the real axis and contains within it the point -1,
and also the domain onto which the exterior of this circle is mapped.
275. (1) Find the images of the circles and domains of the z-plane
mentioned in problem 274 if the mapping function w(z) is given
by the equation
:+~=(~+~r-"
(0<«5<2,
w>O
for
z>l).
(2) In this mapping what is the image of the interior of the circle 01
39
CONFORMAL M.Al'PINGS
276. Map the exterior of the unit circle jzl > 1 onto the w-pla.ne
with a. cut a.long the a.re a.rg (w - l)/(w + 1) = fJ (0 < lfJI < :n) in
such a. way that w( ex>) = ex>, a.rg w' (ex>) =
In problems 277-280 find the domains obtained from the mapping
of the given domains by the functions indicated.
277. The disk lzl < 1; w = z/(z2 +1).
278. The semicircle lzl < 1, Imz > O; w = l/(z2 +1).
279. The angle 0 < a.rg z < :n/n; w = 1/2 (z"+l/z").
280. The sector - :n/n < a.rg z < :n/n, lzl < 1;
IX.
z
w=---2-
( w(z)
>0
for
z
>
0 )·
(l+z")"
IIINT. Represent the mapping function in the form w = F{/[t/>(z)]}, where
t
F(t) =
t/>(t) = t", J(t) = (l+t) 2 ,
j/t.
281. (1) Using the solution of problem 280 and the 'principle
of symmetry find the image of the unit circle in the mapping
z
w=---2-·
(l+z")"
(2) Find the function which maps the interior
exterior) of the unit circle onto the exterior of the "star" :
(and
2:n;lc
(I~= 0, 1, 2, ... , n - 1).
a.rgw = - n
282. Map onto the exterior of the unit circle:
(1) The whole plane with cuts a.long the segments [-1, l]
and [ - i, i] (the outside of a. cross);
(2) The whole plane with cuts a.long the rays (-ex>, -1 ],
[l, + ex>), (- i=, - i] and [i, + i=).
283. (l)* Using the function of problem 279, map the sector
lzl < 1, 0 < a.rg z < :n/n (n an integer) onto itself so that the radial
segments lzl <a, a.rg z = 0 and izl <a, a.rg z = :n/n (0 < a < 1)
pass into the corresponding radii.
(2) Map the exterior of the unit circle with cuts a.long the
a.rg z = 2lc:7t/n (le= 0, 1, 2, ... , n -1) onto the
lz!
segments 1
exterior of the unit circle.
jwj <l,
< <IX,
HINT. In the solution of (2) use the principle of symmetry.
40
PBOBLEMS ON COMPLEX ANALYSIS
284. Map onto the upper half-plane and onto the exterior of
the unit circle the exterior of the cross consisting of the segment
[-a, b] of the real axis and the segment [-ci, ci] of the imaginary
axis (a~ 0, b ~ 0, c ~ 0, a•+ b2 + c2 ¥:0).
Hnm Find the function which maps the upper half-plane with a cut
along the segment (0, ci] onto the upper half.plane and use the principle of
symmetry in virtue of which the exterior of the cross is mapped onto the whole
plane with a cut along a segment of the real axis.
285. Map the plane with outs along the ray [-a, + oo) (a ;;;;i: 0)
and along the segment [- ci, ci] (c > 0) onto the upper half-plane.
HINT. See the hint to problem 184.
286. Map onto the exterior of the unit circle the plane with outs
along the negative part of the imaginary axis and along the lower
half of the unit circle.
HINT. It is reduced to problem 182, (1) by a linear transformation.
287. Map the plane with outs along the segment [- ai, O] (a> 1)
and along the lower half of the unit circle (Fig. 2) onto the upper
half-plane.
HINT.
It is reduced to problem 284 by a linear transformation.
-t
eiA
-1
"
-ce'l
Fm. 2
FIG. 3
288. Map onto the upper half-plane the plane with outs along
the segment [-1, b] (b > -1) and along the circular arc with ends
at the points e:l:lai, which passes through the point z = -1 (Fig. 3).
OONFORMAL M.Al'l'INGS
41
289. Map onto the upper half-plane the exterior of the unit circle
with cuts along the segments: [i,bi], [-bi, - i], [l, a], [-a, -1]
(a> 1, b > 1).
HINT. Zhukovskii's function maps the domain considered onto the domain
of problem 284.
'
290*. Map onto the exterior of the unit circle the exterior of
the "star" represented on Fig. 4.
FIG. 4
291*. Map onto the upper half-plane the interior of the righthand branch of the hyperbola
~-1--1
coslla
sin2 oc - ·
292*. Map onto the upper half-plane the exterior of the rightha.nd branch of the hyperbola
x2
y2
-----=l.
cos2 ix
sin2 oc
293. Map onto the upper half-plane the domain included between
the branches of the hyperbola
x2
y2
aa--¥=1.
In problems 294-295 mappings onto many-sheeted domains (Riemann surfaces) a.re consideredt.
t Here only some of the simplest problems of this type a.re given. In
Chapter IX, § 2 is especially devoted to Riemann surfaces.
42
PROBLEMS ON COMPLEX ANALYSIS
294. Find the domains obtained on mapping by means of the
function w = z2:
(1) The part of the ring r 1 < !zl < r 2 , 0 < arg z < n+oc
(0
< oc ~ n);
(2) The domain lz2 - II <a (0 <a< ex>).
295. Find the domains obtained on mapping by means of Zhukov-
skii's function w =
! (z+ !):
(1) The disk !zl < R
(R >I).
It is advantageous first to consider the mapping of the disk
and the ring 1 < lzl < R (see problem 262).
HINT.
(2) The disk I:+! I< R
lzl <
1
(0 < R < ex>).
In problems 296-298 construct the Riemann surfaces of the
given functions .
• • (1) w =
297. (1) w
298. w
=
-V(:+!);
(2) w = v<z2-I).
y[z(z2+1)];
(2) w =
-v(
z2~ I ) .
= y(z9-1).
§ 4. Elementary transcendental functions
299. Explain what the given curves and domains are transformed
into by the mapping w = es:
(1) The rectangular net x = 0, y = O;
(2) The straight lines y = Tcx+b;
(3) The strip o: < y <fl (0 ~ ci <fl~ 2n);
(4) The strip between the straight lines y = x, y = x + 2n;
(5) The half-strip x < 0, 0 < y <IX~ 2n;
(6) The half-strip x > 0, 0 < y <IX::::;;; 2n;
(7) The rectangle IX < x < fl, y < y < o (o - y ~ 2n).
300. What is the original of the upper half-plane in the mapping
w = (l+z/n)11 i What is the limiting image of the upper half-plane
as n-+ ex>!
301. Explain what the following are transformed into by the
mapping w = log z:
(1) The polar net !zl = R, argz = O;
(2) The logarithmic spirals r = Ae"~ (A > 0);
43
CONFORM.AL MAPPINGS
(3) The angle 0 < arg z < oi ::;:; 2~;
(4) The sector lzl < I, 0 < arg z < oi ~ 2n;
(5) The ring r 1 <lzl <r2, with a cut along the segment [r1 , r 2].
The real and imaginary parts of the function
w
1'
•
a+z
= s-+i17
= 1og--,
a-z
are known as the bipolar coordinates of the point z = x + iy with
respect to the poles ± a (a > 0).
302. (I) Prove that the function w effects the schlicht mapping
of the whole z-plane with the cuts (-oo, -a] and [a, oo) onto the
strip -n ~ 17 ~ n of the w-plane: the upper limits of the cut corresponding to the straight line 17 = n, and the lower limits to the
straight line 17 = -n (Fig. 5).
a
a
Fm. 5
(2) Establish the correctness of the relations:
asinh~
x=
cosh~+cos17 '
y(x2+y2)
~ 17
=
y =
r =a-./(
JI
a sin 17
cosh~+cos17 '
cosh~-COS'YJ
cosh~+cos'YJ
)·
(3) Prove that the originals of the segments ~ = ~ 0 , -n ~
::;:; n are the circles of Apollonius with respect to the points ±a:
(x-acoth~o)2+y2 = ( sin~~o
(the original of the segment
axis) (Fig. 6).
~
= 0, -n
~
17
r
~
n is the ordinate
44
PROBLEMS ON COMPLEX .ANALYSIS
(4) Prove that the originals of the lines 'Y/ = 'Y/o are arcs of
circles passing through the points ±a,
a:l+(y+acot'Yj0 ) 2 =
2
(~)
,
Slll 'YJo
which lie in the upper half-plane for 'Y/o > 0 and in the lower half.
plane for 'Y/o < 0. The line 'Y/ = 0 corresponds to the segment [-a, a].
The arcs corresponding to the values 'YJ = 'Y/o and 'Y/ = 'Y/o-n ('l}o > 0),
supplement one another and make up a complete circle (Fig. 7).
!I
:r
FIG. 6
!I
z
acot'lo
.r
FIG. 7
(5) Find the magnitude of the segments b (see Fig. 6) and l
(see Fig. 7).
REMA.BE. A coordinate net constructed in this way in the z-plane is called
a bipolar nel.
OONFORMAL MAPPINGS
45
303. Explain what the mapping w = cosz transforms the following into:
(1) The rectangular net z = 0, 11 = O;
(2) The half-strip 0 < z < n, 11 < 0;
(3) The half-strip 0 < z < n/2, 11 > O;
(4) The half-strip -n/2 < z < n/2, y > O;
(5) The strip 0 < z < n;
(6) The rectangle 0 < z < n, -k < 11 < k, (k > 0).
304. Explain what the mapping w = sin-1 z transforms the following into;
(1) The upper half-plane;
(2) The plane with cuts along the real axis along the segments
(-<x:>, -1], [1, <x:>);
(3) The first quadrant;
(4) The half-plane z < 0 with a cut along the real axis along
the segment (-<X>, -1].
305. Explain what the mapping w = cosh z transforms the following into:
(1) The rectangular net z = 0, 11 = C;
(2) The strip 0<11 < n;
(3) The half-strip z > 0, 0 < 11 < n.
• · Explain what the mapping w = sinh-1z transforms the
following into:
(1) the plane with cuts along the imaginary axis along the
rays 1 ~ 11 < <X> and -<X> < 11 ~ -1;
(2) the first quadrant.
307. Explain what the mapping w = tanz transforms the following into:
(1) The rectangular net z == 0, 11 = O;
(2) The half-strip 0 < z < n, 11 > O;
(3) The strip 0 < z < n;
(4) The strip 0 < z < n/4;
(5) The strip -n/4 < z < n/4.
308. Explain what the mapping w = cothz transforms the following into:
(1) The strip 0 < 11 < n;
(2) The half-strip 0 < 11 < n, z > 0.
46
PROBLEMS ON COMPLEX ANALYSIS
In problems 309-314 map the given domains onto the upper
half-plane.
309. The strip bounded by the straight lines 11 = :i:, 11 = :i:+k.
310. The half-strip :i: < 1, 0 < y < k.
311. The circular lune bounded by the circles lzl = 2, lz-11=1.
312. The domain bounded by the circles lzl = 2, lz-31 = 1 (the
plane with the disks deleted).
313. The domain defined by the inequalities:
lz-11
> 1,
lz+ll
> 1,
Imz
>0
(the upper half-plane with two semicircles deleted).
314. The domain included between the confocal parabolas
ya= 4(:i:+l),
y" = 8(:i:+2).
HINT. See problem 234, (2).
315. Find the function w(21), which maps the domain bounded
by the circle lzl = 1 and the straight line Imz = 1 (the half-plane
Im z < 1 with the disk deleted):
(1) On the disk lwl < 1 with the normalisation
w(-3i) = 0, argw'(-3i) = n/3;
(2) On the disk lwl < 1 with the normalisation
w(-3i) = (-l+i)/2, argw'(-3i) = n/2;
(3) On the upper half-plane with the normalisation
w(-3i) = l+i, arg w'(-3i) = n.
316. Map on the upper half-plane:
(1) The strip 0 < :i: < 1 with a cut along the line :i: = f,
k ~y < oo;
(2) The strip 0 < :i: < 1 with cuts along the lines :i: = f,
hi ~ y < oo; and :i: = ·!. -oo < y ~ k2 (k2 < k1 ).
HINT. First map the strip 0 < a: < ! onto the upper half-plane. By the
principle of symmetry the mapping function will map the given domain on
the whole plane with certain cuts.
In problems 317-327 map the domains indicated onto the upper
half-plane.
317. The strip 0 < :i: < 1 with a cut along the segment 0 < :i: < k,
y = 0 (k < 1).
318. The strip 0 < :i: < 1 with cuts along the segments 0 ~ :i: ~ k1 ,
11 = 0 and 1-h.2 ~ :i: ~ 1, y = 0 (k1 +k 2 < 1).
CONFORMAL MAPPINGS
47
319. The half-strip 0 < x < n, y > 0 with a cut along the segment x = n/2, 0 ~ y ~ k.
320. The ha.If-strip 0 < x < n, y > 0 with a cut along the ray
x = n/2, k ~ y < oo (k > 0).
321. The half-strip 0 < x < n, y > 0 with cuts along the segment
x = n/2, 0 ~ y ~ k1 and along the ray x = n/2, k2 ~ y < oo
(k2 > k1)·
322. The domain bounded by the circles lz-11=1, lz+ll = 1
with a cut along the ray 2 ~ x < oo, y = 0.
323. The domain bounded by the circles lz-11=1, lz-21 = 2
with a cut along the segment y - 0, 2 ~ x ~ a (a < 4).
324. The domain bounded by the circles lz-11=1, lz-21=2
with cuts along the segments y = 0, 2 ~ x ~a and y = 0, b ~ x ~ 4
(a< b).
325. The domain bounded by the imaginary axis and the circle
1, with cuts along the segment y = 0, 2 ~ x ~a and along
the ray y = 0, b ~ x < oo (a < b).
326. The domain bounded by the circles lz-11=1, lz+ll = 1,
with a cut along the segment x = 0, -ix ~ y ~ fl (ix ~ 0, fl ~ 0).
327. The domain lz-11>1, lz+ll > 1, Im z > 0 (the upper
lz-11 =
half-plane with semicircles removed) with a cut along the segment
x=O, 0 ~y ~k.
328. Map the interior of the parabola y 2 = 4oi2 (x+oi2) onto the
upper half-plane and onto the unit disk.
HINT. Make a cut along the axis of symmetry of the parabola, map the
upper half of the parabola onto a half-strip (by means of the function yz)
and then onto the half-plane, and use the symmetry principle.
329*. Map the upper half-plane with cuts along the segments
0 ~ y ~a, x = n/2+kn (k = 0, ±1, ±2, + ... ) onto the upper
half-plane (Fig. 8).
II
Fm. 8
48
PROBLEMS ON COMPLEX ANALYSIS
330. Map the plane with parallel cuts -a ~ x ~ a, y = n/2+kn
(k = 0, ±1, ±2, ... ) onto the plane with cuts along the real axis
segments [kn-b, kn+b] (k = 0, ±1, ±2, ... ; 0 < b < n/2).
HINT. Make supplementary cuts along the imaginary axis, map one of
the domains formed onto the upper half-plane and use the principle of symmetry.
331. Map the plane with cuts along the rays (-oo, -n/2], [n/2,
+oo) and along the segments -a ~ y ~a, x = n/2+kn (k = 0,
± 1, ± 2, •.. }onto the exterior of the unit circle (Fig. 9).
FIG. 9
HINT. Map the function which gives the solution of problem 329 onto the
plane with cuts along the rays
]
l
( -co sin-1 sech a '
'
[sin-1 ~ch a' +oo] '
332. Map the plane with cuts along the rays (-oo,p], [q, +oo}
(-n/2 ~ p < q ~ n/2) and along the segments -a ~ y ~a,
x = n/2+kn (k = 0, ±1, ±2, ... ) onto the upper half-plane
(Fig. 10).
FIG. 10
333*. Map the plane with cuts along the rays 0 ~ y
x = kn/2 (k = 0, ±1, ±2, ... ) onto the upper half-plane.
< oo,
CONFORMAL MAPPINGS
49
In problems 334-337 the images are many-sheeted domains (see
the footnote on p. 41).
334. Find the domains onto which the following are mapped by
means of the function w = es:
(1) The rectangle 0 < x <a, 0 < y < b;
(2) The half-strip 0 < x < a, y > 0;
(3) The strip 0 < x < a.
335. Find the domains onto which the following are mapped by
means of the function w = cos z:
(1) The strip -n/2 < x < :i/2;
(2) The strip 0 < x < 2n.
336. Find the domain onto which the strip 0 < x < 2n is mapped
by means of the function w = tanz.
337. Construct the Riemann surface onto which the function
.!
w =es maps the z-plane.
§ 5. Boundaries of univalency, convexity and starlikeness
In problems 889-345 r 1 denotes the maximum radius of a circle with centre
at the coordinate origin within which the function w = f(z) is univalent;
r 1 denotes the maximum radius of a disk with centre at the coordinate origin
which the function w = f (z) maps schlicht onto a convex domain and r 3 the
maximum radius of a disk with centre at the coordinate origin mapped by
the function w = f(z) schlicht onto a domain which is sta.rlike with respect
to the point w = O. (A domain is atarlike with respect to a given point, if
any point of the domain can be joined to the given point by a rectilinear
segment lying wholly within the domain.) It is obvious that r 1 .;:;:;; r 3 .;:;:;; r 1•
338. Find r 1 , r 2, r 3 for the function w = z/(1-z) and construct the
images of the disks izl < rv izl < r 2, izl < r 3 •
339. Find r 1 for each of the following functions:
(1) w = z+z2;
(2) w = z+a.z2, (a is a real number);
z
(3) w = (l-z>9.
340. Prove that in the mapping w = f(z) the curvature of the
image of the circle izl = r is expressed by the formula
f"(z)]
l+Re [z7fz)
k=
jzj'(z)j
50
PROBLEMS ON COMPLEX ANALYSIS
341. Prove that an analytic function /(z) maps the circle Jzl = r
onto a convex curve when and only when
a [n
]= l+Re [zf"f'(z)(z)] ;;,:: O
a<P 2+</J+argf'(z)
for all <P (z = re14>)t.
342. Prove that the disk Jzl < r is mapped by the analytic function
f (z) onto a starlike domain when and only when
a~ arg/(z) =
Re[ z;;z(;>];;::: 0 for all
q, (z =re'"'>·
343. Find r 2 for each of the following functions:
(1) w = z+z2 ;
(2) w = 21+az2 (a is a real number);
(3) w = z/(1-z)D
344. Find r1 and r 2 for the function w =es-I.
345. Find r3 for each of the following functions:
(1) w = z+za;
(2) w = z+az2 (a is a real number);
(3) w = z/(1-z)D,
HINT. In the solution of problem 345, (3) it is more convenient to start·
directly from the inequality
!
[lfi+arg/'(z)];;:i: O.
t See, for example, [4, Chapter XIII, § 2].
CHAPTER
Ill
SUPPLEMENTARY G:mOMETRICAL
QUESTIONS. GENERALISED ANALYTIC
FUNCTIONS
§ 1. Some properties of domains and their boundaries. Mappings of
domains
A point of a set Eis said to be an interior point (of this set) if there exists
a neighbourhood of the point which is contained in E. A set E which consists
solely of interior points is called an open set, An open set is called a domain
jf any two of its points can be joined by a polygonal line all the points of which
are contained in the given set. Points which are limit points of an open set
E and do not belong to it are called boundary points; the set of boundary
points forms the boundary I' of the set E. Points not belonging to either E
or r are called 611lterior points.
346. (I) Prove that the boundary of a domain is a closed set.
(2) Prove that if the set of exterior points of an open set is
not empty then it is open.
(3) Prove that the boundary of a. domain is a set which does
not posseBB interior points.
347. (1) Give examples of domains posseBBing boundary points
but not poBSessing exterior points.
(2) Give examples of simply connected domains for which
the set of exterior points is not a domain.
runs through the set
848. Th~ number (! = inf iz-Z:I, where
E is known as the distance from the point z to the set E. Prove that
if Eis a closed set a point z' e E can be found such that iz-z'I = e.
349. The number (! =inf iz1 -z1 1, where z1 and za run through
the sets E 1, E1 , respectively is known as the distance between these
sets. Prove that if E 1 and Ea are closed and at least one of them is
bounded then there exist points z'e E 1 and z"e Ea such that iz' -z"I
= t!·
z;
350. Prove that every closed polygonal curve, that is, a continuous
curve consisting of a. finite number of segments, can be divided into
a finite number of aimple closed polygonal curves, that is not possessing multiple points.
51
52
PROBLEMS ON COMPLEX ANALYSIS
351. Prove that every polygon can be divided into a finite number
of convex polygons.
352. Prove that every simple closed polygonal line divides the
points of the plane not belonging to it into two domains, an interior
and an exterior, and is identical with the boundary of each of these
domains (a particular case of Jordan's theorem, which makes the
same assertion for any simple closed continuous curve).
HINT. Let us fix a direction (ray) which is not parallel to any side of the
polygon P considered and divide all the points not belonging to P into two
classes: we refer to one class those points for which a ray starting from them
and parallel to the given direction intersects P in an even number of points
(0, 2, 4, ... ), and to the other class the points for which the same ray gives
an odd number of intersections with P. A vertex of P is considered to be
a point of intersection if the two sides of P approaching it are situated on
different sides of the ray. Let us prove that two points of one class can be
connected by a polygonal line which does not have points in common with P,
and if two points belong to different classes then any polygonal line connecting
them has points in common with Pt.
353. Prove that in every simple polygon it is possible to draw
a diagonal, all the points of which excluding the ends lie within
the polygon.
354*. Prove that a Jordan curve does not contain interior points.
A set a S G is said to be open with respect to G if together with each of its
points z0 it contains all the points ze G lying in a sufficiently small neighbourhood Ua (Zo) of the point z0 (lz - z0 1 < e). The set e S E is said to be
olosed. with respect to E if it contains all its limit points belonging to E.
355. Let G be a domain in the z-plane and E =f(G) the image
of G in the w-plane in the mapping w = f(z).
Prove the equivalence of the following definitions of the continuity of the mapping:
(1) The mapping w = f (z) of the domain G is continuous at the
point Zoe G iffor every e-neighbourhood u.(wo) of the point Wo = f(zo)
it is possible to find a .5-neighbourhood U.,(:210) of the point z0 , situated
in G such that its image belongs to U8 (w0 ), that is, for ze U.,(z0 )
we have w = f(z)e U8 (w0). The mapping w = f(z) is said to continuous in the domain G, if it is continuous at every point of this
domain (Cauchy).
(2) The mapping w = f(z) of the domain G is continuous
if the original J-1 (e) of every set e c E, E = f(G), closed with
respect to E, is closed in G.
(3) The mapping w = f(z) of the domain G is continuous if the
originalf-1 (e) of every set e c E, open with respect to E is open in G.
t See R.
CoURANT
and H.
ROBBINS,
(1942). What iB mathematics?
SUPPLEMENTARY GEOMETRICAL QUESTIONS
53
(4) The mapping w = f(z) of the domain G is continuous
if for every set g s;; G the image of the closure of g with respect
to G is contained in the closure of f(g) with respect to E.
A mapping (single.valued) w = f(z) is said to be one-one or ~. if two
distinct points z always correspond to two distinct points to, that is f (z1 ) of: /(z 1)
if z1 of: z1• A one-one and mutually continuous mapping is said to be topolo·
gical or homeomorphic.
356. Construct a topological mapping of the unit disk lzl < 1 onto
the whole finite w-plane.
357. Is the following theorem correct~ [See, for example, 1, Chapter I, § 6.1.]
If w = f(z) is a function continuous in the domain G and maps
this domain one-one on some set E, then the latter is also a domain,
the mapping w = f(z) of the domain G on the domain E being topological (homeomorphic).
Using this theorem prove that if the function f(z) is continuous
in the closed region G and is one-one within G then the boundary
of the domain G is transformed into the boundary of the domain
f(G).
A path r is defined as a continuous image of a directed segment 0 ~ t ~ I,
z = z(t)
may have multiple points). The point z(O) is said to be the beginning
of the path, z(l) is its end. IfT = ~(t) is a homeomorphic mapping of the segment
0 ~ t ~ I on the segment 0 ~ T ~ I and t/>(O) = O, t/>(l) = 1, then the paths
z{t) and z[t/>{'r)l are considered identical. Two paths ,,0 , r 1 with common ends
a, b are said to be cominuoualy deformable into one another or homotopic in the
domain G, if there exists in G a family of paths
0 ~ .il ~ I, with ends
<r
r.,a
a, b, continuously varying with .il (this means that m:x lt/>A1(t) - ~A. (t)I -+ 0
as .ili -+ .il1) and such that the given paths ro· r1 correspond to the values
.il = 0, .il = I. The closed path r is said to be continuoualy deformable into
a point or homotopic to zero in the domain G if it is homotopic in the domain
G to a null path, that is, a path consisting of a single point.
358*. Prove the following assertions:
(1) Every path yin an arbitrary domain is homotopio to a polygonal path.
(2) Every closed path in a circle is homotopio to zero.
(3) Every closed path in a circular ring is either homotopic to
zero or homotopio to any boundary circle, traversed in a definite
direction one or several times.
359. Let y be an arbitrary path, closed or not. Prove the following assertions :
(1) arg (z-a), z e y, a y, are uniquely defined as continuous
functions of z, if arg (zti-a) is given for an arbitrary point z0 e y.
e
54
PROBLEMS ON COMPLEX ANALYSIS
(2) The quantity LI,. arg (z-a), equal to the increment of arg (z-a)
along y, does not depend on the choice of z0 or arg (z0 -a).
(3) H the path r lies within an angle with vertex at the point
a of magnitude IX (0 < IX < 2n) then LI.a arg (z-a) < IX.
(4) For a smooth (or even only rectifiable) path r the equality
Ll 7 arg (z-a) =
holds.
If y is a closed path and a
ey
n,a (a)
=
J,, darg(z-a).
then the number
I
2:n Ll 7 arg (z-a)
is said to be the indez of the point a with respect to y (or the winding number
of y with respect to the point a).
360. Prove the following assertions:
(1) In the homotopio deformation of the path r taking place
outside the point a the index n,.(a) is unchanged.
(2) In every closed path r it is possible to inscribe a closed polygonal curve n such that n 7 (a) = nn(a) (a is a given point outside y).
(3) If the points cii, a1 can be connected by a path which does
not encounter the path y, then n 7 (a1 ) = n 7 (a9).
(4) If the point a and the path rare separated by some straight
line then n 7 (a) = 0.
361. Prove that for a simple closed polygonal path n the index
n,.(a) = ± 1, if a lies within n (the sign depends on the direction
of traversal), and n,.(a) = 0 if a lies outside n.
362*. Let the function w = f(z) effect the topological mapping
of the domain G on the domain G'. Let Q be a small circle in G, r
a closed path in Q and a a point of Q which does not belong to y.
Let, finally, r' and a' be the images of r and a in G'. Prove that
n 7 ,(a') = n7 (a)c5,
where c5 equals either +1 or -1, independently of r and a. In particular prove that if the mapping is continuously differentiable
and the Jacobian
J
=
au !!!.. - ~ !!!.. =I= 0
ax oy oy ax '
then the sign of c5 is identical with the sign of J.
SUPPLEMENT.A.BY GEOMETRIC.AL QUESTIONS
55
363*. Let w = /(z) effect the continuous mapping of the closed
region G bounded by the Jordan curve y. Prove that if f(z) ¢a
on y and for the image y' = f(y) of the curve y the index n 7,(a) ¢ 0,
then within G, f(z) assumes the value a.
364. Prove the equivalence of the following three definitions of
a. simply connected domain:
(1) The domain G is said to be simply connected if every closed
pa.th lying in G is homotopic to zero in this domain.
(2) A domain G which does not contain the point at infinity is
simply connected if together with every simple polygon situated
in G it also contains its interior.
(3) A domain G different from the extended complex plane is
simply connected if it is bounded by a. single continuum or point.
RElllABX.
The extended complex plane is simply connected by the dret
definition.
§ 2. Quasi-conformal mappings. Generalised analytic functions
If in the relation
/(111, fl)
=
1(
z;i •
z~i )
= • (z, i)
z and i are considered as independent variables the derivatives with respect
to these variables are
-i- = ! (:Ill -
l
~).
! = ! (~ + i ~ ) .
In what follows the notations
au
8111 =
etc. will be used.
365. Prove the following relations:
(1) w11 =
(2) w11 =
!
!
[(ux+v,)+i(-u,+vx)];
[(ux-v1 )+i (u,-vx)];
(3) dw = w11 dz+WjdZ;
(4) ux =
! (w +wi+w.+w
11
11 );
"'x•
56
PROBLEMS ON COMPLEX ANALYSIS
-} ;
(5} u1 = 2i(Wz-Wi-wz+wr:
(6)
Vx
i
-
-
= -2(w.z-wi+wz-wr:};
(7) v1
- }.
= 2I (wr:+wz--wi-wr:
368. Prove that the Cauchy-Riemann equations are equivalent
tow;= 0.
387. Prove that Laplace's equation Liu= 0 can be written in
the form fJlu/fJz(}i = 0.
368. Prove that
wr: =wi, iOz =wr:, dw=dw
(the long bar over a symbol indicates passage to the conjugate
value after differentiation).
369. Prove that for the function z(w}, inverse to w(z},
dz=
lw.l~lw1l 1 dw+ lwr:l;:i~il1 dw,
370. Prove that the Jacobian of the transformation w(z} is given by
J w/r:
=
D(u,v}
D(:.i:, y}
' 1• .
= IWz 11 -,wi"
371. Prove the following equalities:
dw
(I} dz
=
wr:+wie-•1•,
maxi~:
(3) mini~:
(2)
I=
I=
where
IX=
argdz;
lw.l+lwsl;
lwr:l-lwil.
372. Prove that if IX= argdz, rt.'= argdw, then
(I} drt.' =
d O!
Jw1r:
(uxcosrt.+u, sinoc}1 + (vxcosrt.+v, sinoc}1
11
=
dz
Idw
I
I
Jr:fw;
Sm'l'LEMENTABY GEOMETRICAL QUESTIONS
da:'
(2) max da: = p,
57
min--=da:'
da:
1
p'
maxl*I
where
minl~~I
p=
-
lw.l+lw.I
liw.1-lwill
373. The 8et of monogeneity ID?. of the function w = u+iv = f (z)
at the point z is the set of derived numbers at this point, that is,
the set of all possible limiting values of the ratio LJw/LJz as LJz-+ 0.
Prove that if u and v are differentiable at the point z, then IDl. is
either a point or the circumference of a circle.
HINT. Use the relation 3'71 (1).
374. By the characteristics of an ellipse is meant the ratio p of
its semi-axes (p;;;:::: 1) and, if p #: 1, the angle 8 (0 ~ 8 < n), formed
by the major axis of the ellipse with the axis Oz. Show that the
equation of such an ellipse with centre at the coordinate origin and
semi-axis minor h can be written in the form
where
~=
pcos1 8+
~ sin•O,
r
fJ =
(p- ~)cos8
sin8,
= psin18+2-cos11 8.
p
REMARK. The quantities at, (J, y are also 1ialled characteristics of the ellipse.
They are connected by the relation
rx.y-(J•
'l'he function w
w-plane, if
l
== /(z) effects an affine
=
1.
(linear) mapping of the z-plane on the
bil
where LI= a1
of:.0.
a1 b1
375. Prove that in an affine transformation parallel straight lines
are transformed into parallel straight lines, and families of circles
into families of ellipses.
58
PROBLEMS ON COMPLEX ANALYSIS
What condition is it necessary to impose on the coefficients of the
transformation in order that circles should be transformed into
circles (see, for example, [3, Chapter II, § I]) 1
376. The characteristics of an affine transformation a.re the characteristics p, 8 or cc, fl, y of the ellipses which a.re transformed into
circles. Prove that:
(1) - " - -
-fl
__
cc _ _ _!_.
bf+bl - LI '
c4+a; - a1b1+a1b1 (2) p=K+y(K2 -I),
where K=
cc~_y-;
(3) ta.no= y-cc+y[(:pcc)1+4p•] .
377. By the characteristics of the continuously differentiable
mapping w = u(x, y)+iv(x, y) with Jacobian J > 0 is meant the
characteristics p(z), O(z) or cc(z), fl(z), y(z) of the affine transformation
du = uxdx+u1 dy,
d v = vxdx+v,.dy.
In this to the infinitely small circle
=de•
du11 +dv1
there corresponds the infinitely sma.11 ellipse
ydxl-2fldxdy+ccdyl = pdhl
(dh is the minor semi-a.xis).
Prove the relations:
SUPPLEMENT.ARY GEOMETRICAL QUESTIONS
59
378. A one-one continuously differentiable mapping w = u+itJ
with Jacobian J > 0 is called a quasi-conformal mapping with
characteristics p(z), O(z) or a(z), {J(z), y(z) if it transforms infinitely
small ellipses with these characteristics into infinitely small circles.
Prove that such a mapping satisfies the system of equations
au,.+flu1 = tJ1 ,
f3u,.+,,u,
= -v,..
which can also be written in the form
q(z)
HINT. Use the first relation for Gt,
nection between Gt, fJ, y and p, 8.
fJ,
=
p-l
- - - e119 •
p+l
y given in problem 877 and the con-
379. Prove that a one-one continuously differentiable mapping
w = u+itJ with positive Jacobian which transforms infinitely small
circles into infinitely small ellipses with the characteristics a(z),
{)(z), y(z), satisfies the system of equations
u,.+{Ju1
= av1 ,
-f3u,.+u1
= -ow,.,
which can be written in the form
p-l
q(z) = p+l e819 •
HINT. Use the relation
ydu1 -2/Jdud11+ctd111
= J(tlail+dy1).
380. Prove that the equation Ws = A (z)w. is invariant with respect
to analytic transformations of w, and the equation w, = B(z)ws is
invariant with respect to conformal transformations of z.
381. The one-one continuously differentiable mapping w = u(x, y)
+iv(x, y) with positive Jacobian is called a quasi-conforma.l mapping with two pairs of characteristics p, O; Pi• (Ji (or with two
triads of characteristics a, {J, y; ~./Ji, /'i), if it transforms infinitely
small ellipses with characteristics p, 6 into infinitely sma.11 ellipses
with characteristics Pi· Oi.
Prove that such a mapping satisfies the system of equations
au,.+(fJ+f3i)u1 = aiv1 ,
(fJ-f3i)u,.+,,u,
which can also be written in the form
w, = qi(z)w.+q (z)w•,
2
=
-~v,.,
60
PROBLEMS ON OOMPLEX ANALYSIS
where
()
q1 z = -
P1(pl-l)
el"'
(PP1+l)(p+P1)
HINT. Use the relation
y 1du1 -2P1dudt1+0idv1 = J(yda:l-2Pd:i:dy+ocdy1 ).
382. Let Pwt• be the characteristic p of a quasi-conformal mapping
w(z). Prove that
Pwt111 = P111tw
and that for the compound quasi-conformal mapping w[z(t)]
Pwtt ~ Pw1111P111tt
883. Show that for the quasi-conformal mappings
u = f (x), v = y (longitudinal stretching compression),
u = x, v = f(y) (transverse stretching compression),
e = r, (J = f(</>) (angular stretching compression)
the characteristic is
and for the mapping
e = f(r), 8 = "' (radial stretching compression)
f)
rf' rf'.
p=max ( j '
384. Construct the quasi-conformal mapping of the disk JzJ < R
onto itself, transforming the point z =a (Jal < R) into the origin
and leaving fixed the points of the circumference JzJ = R. Evaluate
the characteristic p.
385. Construct the quasi-conformal mapping of the oblique
half-strip x > 0, xtana < y < x tana+h onto the rectangular
half-strip u > 0, 0 < v < h without stretching on the base and
with constant stretching on the lateral side. Evaluate the characteristic p.
386*. Construct the quasi-conformal mapping of the domain,
consisting of a half-plane and a circular segment with central angle
61
SUPPLEMENTARY GEOMETRICAL QUESTIONS
2{J0 (Fig. 11), onto a half-plane with preservation of lengths on the
boundary. Evaluate the characteristic 'P·
FIG. 11
387. Reduce the quasi-linear equation
atu
atu
atu
(
au au)
A aXA +2B axay+o ay2 = F x, y, u, ax 'ay
of elliptical type (AO-B 2> 0) by mea.ns of the schlicht mapping
C= C(z) = E+i'YJ
to the canonical form
atu atu
(
au au)
aE2 + a'YJ2 =F1 E. 'Y/• u, aE , a'Y/
Prove that the mapping C(z) satisfies Beltre.mi's system of equations
AaE/ax+BaE/oy a'Y/
y(AO-B2) =" ay '
BaE/ax+oaE/ay
y(AO-B2)
= -
a'Y/
ax
and is a quasi-conformal mapping with the characteristics a, {J, y,
determined from the relations
r
(it is assumed
fJ
~
1
7f =Ii= A= y(AO-B2)
that A > 0).
A function w = u+iv, which satisfies the equation
wz-+Aw+Bw = F,
(I)
where ..4., B and F are functions of z, is known as a genemliaed analytic funo-
non.
62
PROBLEMS ON COMPLEX ANALYSIS
In problems 388-394 equations and systems of equations are
considered which reduce to the form (1) and also some properties
of their solutionst.
388. Show that Carlema.n's system of equations
u.¥ -v7 +au+bv =/,
u1
-
v.¥+cu+dv = g,
where a, b, c, d, J and g are continuous functions of the variables
z and y, can be written in the form (1)
tos+Aw+BW = F.
Express A, B and F in terms of the coefficients of the given system.
389. Show that the equation
w.-q2(z)wi+Aw+Bw = F
can be reduced to the form (1) by means of the "affine" transfor·
ma.tion
(2)
w = a(z)ro+b(z)w.
Find the general form of the transformation (2).
390. Show that the equation
w,-q1(z)w:r-i2(z)w1+Aw+.BW
=
F
can be reduced to the form
w.-q~ws+A'w+B'w =
F'
by means of the transformation of the preceding problem.
HINT. Apply the transformation considered to the given equation and
to the equation
w.-q1(z)w:r-f1(z}t.os+.A.W+iJ.w=F;
eliminate W:r and then select the coefficients a(z) and b(z).
391. Show that the equation
wi-q1(z)ws-qz(z)wi+Aw+Bw = F
t On this group of problems see the monograph: I. N.
VEXUA, G~
analyUo Junctiona (Obobahchennyye analitichukiye Junktayi), Chap. III. Fizmat-
giz (1958), Moscow, English translation published by Pergamon Press (1962).
63
SUl'PLEMENTABY GEOMET.RICAL QUESTIONS
oan be reduced to the form
wc-q:i»f+.Aw+Bw = F
by changing the independent variable z to the variable C connected with z by the relation Cs = ~ C111 • Find
and
and explain the geometrical meaning of the transformation C(z).
39!. Prove that an elliptic system of differential equations of
the form
qt
q:
et1v1 =
cxux+(/1+fl1)u,+au+bv+f, }
-at:1Vx = (/1-/11)ux+yu,+cu-t-dv+g
(the condition of ellipticity is here a{l-y 2 > 0; in addition at: > 0) can
be reduced to the form w.-q1(z)w111 -q2 (z)wi+.Aw+BW = F,
where lq1 (z)l+lq1 (z)I < 1, if at:1 >0, and jlq1 (z)l-lq2 (z)I\ > 1 if «1 <0.
HINT. See problem 181. The case at1 < 0 is reduced to the case at1 > 0
by the replacement of w ... u+w by iii = u-w.
393. Prove that if w(z) is a continuous differentiable solution
of the equation
o,
ios-q2(z)w• =
where q2 (z) is an analytic function of z and lq2 (z)I =f:. 1 then
w(z) = </>(z)+q2 (z) ~(z) ,
l-lq2(z)!
where <f>(z) is an arbitrary analytic function.
394. Prove that if w(z) is a twice continuously differentiable
solution of the equation
ios-q2(z)w.
=
where q2 (Z') is an analytic function of
w(z) = </>(z)+
o
z and
lq2 (z)I
.P 1, then
Jq2 (z)~sdz,
where <f>(z) is an arbitrary analytic function of z.
HINT. It is first necessary to prove that w(z) is the sum of an analytic
function of z and an analytic function of z.
OHAPTER IV
INTEGRALS AND POWER SERIES
In the problems of this and also of the following chapters if nothing is said
to the contrary, simple closed contours (that is, those without points of self
intersection) are traversed in the positive direction.
§ I. The integration of functions of a complex variable
395. By direct summation prove the equalities:
f
z,
Z1
(1)
dz = z1
-
z0 ;
(2)
J
zdz =
! M-zU .
z,
Zo
396. Let 0 be a simple closed contour bounding an area S. Prove
the following equalities:
(1)
fc zdz = iS;
fc ydz = -
(2)
S;
(3)
Ic zdz = 2iS.
397. Evaluate the integrals
11 =
Jzdz,
12 =
Jydz
along the following paths:
(1) Along the radius vector of the point z = 2+i;
(2) Along the semicircle lzi = 1, 0::::;;; arg z :s:;;;:n; (the commencement of the path is at the point z = 1);
(3) Along the circle lz - al = R.
398. Evaluate the integral lzldz along the following paths:
(1) Along the radius vector of the point z = 2-i;
(2) Along the semicircle lzl = 1, 0 ::::;;; arg z ::::;;; :n; (the
commencement of the path is at the point z = 1);
(3) Along the semicircle izl = 1, -:n;/2 ::::;;; arg z ::::;;; :n;/2 (the
commencement of the path is at the point z = -i);
(4) Along the circle lzl = R.
399. Evaluate the integral izlzdz, where 0 is the closed contour
c
consisting of the upper semicircle izl = 1 and the segment -1 ::::;;; z
::::;;; 1, y=O.
J
J
64
INTEGRALS AND POWER SERIES
J ~ dz,
400. Evaluate the integral
c z
65
where 0 is the boundary of
the ha.If ring represented on Fig. 12.
y
FIG. 12
J
401. Evaluate the integral (z-a)ndz (n is an integer):
(1) Along the semicircle lz-al = R, 0 ::::;;; a.rg (z-a) ~ :r:
(the commencement of the pa.th is at the point z = a+R);
(2) Along the circle lz-al = R;
(3) Along the perimeter of the square with centre at the
point a and sides parallel to the coordinate axes.
In problems 402-405 the branch of the many-valued function
which occurs as the integrand is selected by the specification of
its value at some point of the contour of integration. If the contour
is closed the initial point of the pa.th of integration is always considered to be that point at which the value of the integrand is specified
(it must be remembered that the value of the integral may depend
on the choice of this initial point).
402. Evaluate the integral
J ~:
a.long the following contours:
(1) Along the semicircle lzl = 1, y ~ 0, yl = 1;
(2) Along the semicircle lzl = 1, y ~ 0, yl = -1;
(3) Along the semicircle lzl = 1, y ~ 0, yl = 1;
(4) Along the circle lzl = 1, f.'l = 1;
(5) Along the circle izl = 1, y(- 1) = i.
403. Evaluate the integral Log z dz, where:
J
(1)
(2)
(3)
(4)
0
0
0
0
is
is
is
is
the
the
the
the
c
unit circle and Log 1 = 0;
unit circle and Logi = ni/2;
circle lzl = R and LogR =log R;
circle izl = R and LogR =log R+2ni.
66
PROBLEMS ON COMPLEX ANALYSIS
f
404. Evaluate the integral
and:
znLogzdz, where n is an integer
1zr=1
(2) Log (-1) = ni.
(1) Logl = O;
405. Evaluate the integral
J
~« dz, where
oi
is an arbitrary
lzl=l
complex number and 1« = 1.
406. Prove that for any choice of the initial value of the function ar
J a•dz = 0.
lzl=l
407. For what values of oi (0 ,,;;; oi
exist:
< 2n)
do the following integrals
(pis a natural number), taken along the radius vector of the point
z = e"'1
408. Prove that if lal #: R, then
f
lzl~R
jdzl
iz-all 21 +al
<
2nR
IR2-lal 2 I'
409. Prove the following assertions:
(1) If f(z) is continuous in the neighbourhood of the
coordinate origin then
2..
lim
Jf(re1~)d4' = 2n/(O).
, r-0 0
(2) If f (z) is continuous in the neighbourhood of the point
z=athen
lim
r-0
J
lz-al=r
f(z) dz = 2nif(a).
z-a
410. Prove the following assertions:
(1) If f(~) is continuous in the half-strip z ~ z 0 , 0,,;;; y,,;;; h
and the limit lim f(z+iy) =.A, not depending on y exists, then
lim
Jf(z)dz = i.Ah,
x-+00 Px
INTEGRALS AND POWER SERIES
67
where Px is a segment of the vertical straight line 0::::;;; y ~ h, traversed from below upwards.
(2) If f (z) is continuous in the sector 0 < lz - al ~ r0 ,
0 ~ arg (z - a) ~ cc, (0 < cc ~ 2n) and the limit
lim [(z-a)f(z)] = A
z_,.a
exists, then
lim
Jf(z) dz = iAcc ,
r...O 1,
where i'r is the arc of the circle lz-al < r which is present in the
given sector, traversed in the positive direction.
(3) If f(z) is continuous in the region !zl;;:?:: R 0 , 0 ~ arg z ~cc
(0 < cc ~ 2n) and the limit
lim zf(z)
=
A ,
Z-+00
exists, then
lim
Jf(z)dz =
iAcc,
R"""oorR
where JR. is the arc of the circle lzl = R, which lies in the given
region traversed in the positive direction with respect to the coordinate origin.
411. Prove the following theorems:
(1) If f(z) is continuous in the region lzl;;:?:: R 0 , Im z;;:?:: a (a is
a fixed real number) and lim /(z) = 0, then for any positive
number m
lim
Je',,,.f(z) dz = 0 ,
R....oorR
where I'R is the arc of the circle lzl
considered (Jordan's lemma).
= R,
which lies in the region
HINT. In the evaluation of the modulus of the integral along the semicircle
Im z > 0 make use of the inequality sin 6;;;;.: 28/n for 0 =s;;; 6 =s;;; n/2,
and in the evaluation along the a.res lying in the lower half-plane (in the case
a < 0), use the fact that the length of each of them tends to !al as R -+ co.
(2) If f(z) is continuous in the half-plane Re 21;;:?:: <1 (<1 is a fixed
lzl = R,
real number) and lim/(z) = 0, then for any negative number t
lim
Jff'f(z)dz = 0,
R-+oorR
68
PROBLEMS ON COMPLEX ANALYSIS
where I'R is the arc of the circle izl = R, Re z ";;!:;<1. If f(z) is continuous in the half plane Re z ~Cl, the assertion is true if tis positive
and I'R is the circular arc izi = R, Re z ~Cl.
REMABK.
The proof of both theorems Vi given, for example, in [3, Chapter V,
2, § 73].
§ 2. Cauchy's integral theoremt
412. Show that if the path does not pass through the coordinate
origin, then
Ji
z
= In r+i<P+2nik
1
where k is an integer indicating how many times the path of integration has encircled the coordinate origin (z =re'"'>·
413. Show that if the path does not pass through the points
± i, then
where k is an integer.
414. Show that if 0 is an arbitrary simple closed contour which
does not pass through the point a, and n is an integer, then
f
c
(z-a)ndz =
I
0, if n =F -1 ,
2ni, if n = -1, a is inside 0,
0, if n = -1, a is outside 0.
415. Cauchy's theorem holds in the following stronger form:
if f (z) is continuous in the closed region G, bounded by a simple
rectifiable contour 0, and analytic within G, then f (z)dz = 0.
Prove this for the case of a starlike contourtt.
Jc
HINT. Assuming 0 to be starlike with respect to the coordinate origin
consider the contour OJ,: C= .M (0 <A< 1, ze 0) and pass to the limit
as .A -+ 1 (see, for example, [2, Chapter V, sec. 8] or [3, Chapter I, § 4,
sec. 12]).
t The problems on the evaluation of integrals given in this and in the
following section are mainly of an illustrative nature. The majority of problems
of this type are located in Chapter VII on the calculus of residues.
tt A contour is said to be starlike with respect to some point if every ray
issuing from this point meets the contour in a single point.
69
INTEGRALS AND POWER SERIES
418. Prove the following asse1 tions:
(1) Ifj(z) is analytic in the strip 0
~
y:,;;;; k, liin f(x+iy) = 0
~:!:co
co
00
and the integral
Jf(x)dx exists, then the integral -coJf(x+ik)dx
-co
also exists and they are both equal.
(2) If f(z) is analytic in the angle 0 :,;;;; arg z ~ ex (0 < ex
~
2n),
00
limz f(z) = 0 and the integral
J f(x)dx exists,
then the integral
0
Z-+CO
Jf(z)dz along the
ray z = re1•, 0 ~ r
are equal to one another.
<
oo, also exists and they
HINT. Use the results of problem 410.
417. Prove that
co
J e-x"cos2bxdx =
0
yn e-b'.
2
HINT. Integrate the functionf(z) = e-zl along the boundary of the rectangle
l:i:I ~ R, 0 ~ 1J ~ b and use Laplace's integral
00
J e-t'dt=Jl':i.
-oo
418. Prove the equalities
j oosxldx= j sinxldz= t~v
0
(Fresnel's integrals).
0
HINT. Integrate the function j(z) = el•• along the boundary of the sector
0 ~ lzl ~ R, 0 ~ arg z ~ : snd use the result of problem 411 (1) (put
z• = C).
419. Prove that
co
J
0
sinx
n
---dx=x
2
(Dirichlet's integral).
HINT. Integrate the function J(z) = el• /z along the boundary of the
domain r ~ !zl ~ R, 0 ~ arg z ~ n and use the results of problems 410 and
'11.
70
:PROBLEMS ON COMPLEX .ANALYSIS
420. Prove that the following equations hold for 0
< 8 < 1:
00
(1)
J 11:"-lcosxdx = I'(8)cos 2
1'8
,
0
00
(2)
J X- sin xdx = I'(8) sin ~
8•
1
0
HINT.
Integrate the function f(z) = z&-1e-is along the boundary of the
domain r ~
lzl ~ R, - ; ~ arg z ~ O; use the results of problems 410 (2)
and 4ll (1) and the integral representation of the Gamma function:
00
I'Ce)
= J a:t-ie-tde.
0
§ 3. Cauchy's integral formula
Everywhere in the problems of this section 0 denotes a simple
closed rectifiable contour.
4!1. Calculate the integral
J
c
z•:9 , if:
(1) The point 3i lies inside the contour 0 and the point
-3i outside it;
(2) The point -3i lies inside the contour 0 and the point
3i outside it;
(3) The points ± 3i lie inside the contour 0.
4!2. Calculate all the possible values of the integral
J z(z~l)
c
for different positions of the contour 0. It is assumed that the contour
0 does not pass through any one of the points 0, 1 and -1.
4!3. How many different values can the integral
Jw~~z)
c
assume, where ron(z) = (z-z1)(z-z1) ... (z-zn) (z1 .P ZJ) and the
contour 0 does not pass through any of the points z1¥
424. Evaluate the integral
f
lz-•J- o
zdz
z4 - l '
a> 1.
71
INTEGRALS AND FOWER SERIES
1 -1.
·
,... E v aluate t h e mtegra
-5.
211:i
f
f
+
e•dz
z2 a11
c
contains within itself the circle izl ~ a.
1
.
426. Evaluate the mtegral - 2 .
11:i
within the contour 0.
HINT.
c
ze•dz
)3
(
z-a
if t h e
,
,
contour
O
if the point a lies
Use the formulae for the derivatives of Cauchy's integral.
J
(;•dz) 3 if:
427. Evaluate the integral - 21 .
11:i c z -z
(1) The point 0 lies inside, and the point 1 outside the
contour O;
(2) The point 1 lies inside and the point 0 outside the
contour O;
(3) The points 0 and 1 both lie inside the contour 0.
428. The function f(z) is analytic in the domain bounded by the
simple closed contour 0, which contains within it the coordinate
origin. Prove that for any choice of the branch of Log z
!i cf
2
J'(z)Logzdz =f(z0 )-f(O),
where z0 is the initial point of integration.
HINT.
Integrate by parts.
429. Evaluate the integral
-2 1 .
:rci
z+l dz,
f z Log-z- 1
2
c
if Log a = log a for a > 0, and for the contour 0 we have:
(1) The circle izl = 2;
(2) The circle iz-11=1 and the initia.l point of integration
is z = 1 i.
430. By Liouville's theorem a function f(z) which is analytic
and bounded throughout the whole plane is a constant. Prove this
theorem after calculating the integral
+
f
(!al
< R,
!bl
< R)
f(z)dz
(z-a)(21-b)
lzl=R
and taking its limit as R --. oo.
72
PROBLEMS ON COMPLEX: ANALYSIS
431. Let f M be analytic in the closed domain bounded by the
contour O; z1 , z2 , ••• ,Zn are arbitrary distinct points within 0 and
Wn (z) = (z-z1 )(z-z 2) ••• (z-zn)· Show that the integral
P(z)=-1-J f(C)
2:rti
c
wnm
Wn(C)-wn(z)dC
C-z
is a polynomial of degree (n - 1) which is equal to f (z) at the points
Z1' Z2, • • • ' Zn t •
432. Prove the following theorem (Cauchy's formula for an
infinite domain):
Let 0 be a simple closed contour bounding the finite domain D.
The function f(z) is analytic in the exterior of the domain D and
lim f(z) =A. Then
1
2"7
:n:i
J
1-f(z)+A, if the point z belongs to the exterior
f(C)
of the domain D,
-,-de=
A, if the point z belongs to the domain D.
c -z
The contour 0 is traversed in the positive direction with respect
to the domain D.
HINT. First consider the oase A = 0.
433. Let the function f (z) and the contour 0 satisfy the conditions
of the preceding problem.
Prove that if the coordinate origin belongs to the domain D,
then
1
2:rti
J
c
zf(C)
Cz-C 2
dC = {
0, if zeD,
f(z), if zeD.
§ 4. Numerical series
00
434. Prove that if the series
J; en converges and larg cnl ~ oc < :rt/2,
n=l
then the series converges absolutely.
00
00
435. Let the series
2
en and
2
n=l
n-1
c: . converge.
Prove that if
00
Re en ;?: 0, then the series
2
lcnl 2 also converges.
n=l
t The polynomial P(z) is known as Lagrange's interpolation polynomial.
73
INTEGRALS A.ND POWER SERIES
00
436. The series
2 en
possesses the property that the four parts
n=l
of it, each of which consists of the terms contained in one and the
same closed quadrant of the plane, converge. Prove that the given
series converges absolutely.
437. Prove the formula (Abel's transformation)
n
n-1
}; akbk = } ; 8k(bk-bk+ 1 )-S._ 1 b.+Snbn,
k-m
where 1 ::::;;; m
k=m
< n,
Sk
= ai+a2 + ... +ak
(le~
=
1), 8 0
0.
00
2 anbn where
bn > 0, it is sufficient that the partial sums of the series 2 an should
n=l
438. Prove that for the convergence of the series
n-1
00
be bounded and the sequence of numbers {bn} tend monotonically
to zero (Dirichlet's rule).
HINT. Use Abel's transformation.
00
439. Prove that for the convergence of the series
2
anbn where
n-1
00
the bn are real numbers, it is sufficient that the series
2
an should
n-1
converge and the sequence {bn} be monotonic and bounded (Abel's
test).
440. Let the sequence of real numbers {bn} satisfy the conditions:
00
(1) lim (yn)bn = O; (2) the series
n...oo
2 (yn)(bn-bn+i) converges.
n-1
Then, if the sequence { 8 n
yn
}
is bounded (Sn=
.f 1it), the series
k=l
00
2 akbk converges.
k=l
00
441. Let lim
Vlcnl
= q. Prove that the series
n-+oo
(absolutely), if q < 1, and diverges, if q
442. By the examples of the series
1 1 1
1+26+a-+7+ ...
2 en
n=l
>
I.
(1
<cc< p)
converges
74
PROBLEMS ON COMPLEX ANALYSIS
and
cc+pi+cca+p'+ ...
(0
< ex < p <
show that a series may converge even when lim
443. Prove that if lim
n...oo
I
vergence of the series
<
n-1
I
Cn+i
Cn
en
1),
Ic::
1
I > I.
I = 1, then for the absolute con-
it is sufficient that limn(I Cn+ 1
n-MX>
-1 (Ra.abe's test).
444. Prove Gauss's test: If
Ic~: 1
does not depend on n and a
absolutely.
<-
I = 1+
Cn
: +o ( ~),
l-1)
where
a
1, then the series converges
00
In problems 440-454 study the convergence of the series
445. Cn = n/2".
448. Cn = nl/nn.
449.
450.
2 Cn.
n=l
fJ,,. /n.
Cn
=
Cn
= eln /nl.
= e1".
451. en= e"''"/n.
= d"/n.
_ cx(cx+l) ... (ct+n-1)/1(/1+1) ... (/J+n-1)
447.
Cn
448.
Cn
,.,,,2
':l:u •
Cn -
----,---------------
nli'(i'+l) ... (i'+n-1)
(the hypergeometric series), Re (ct+/1-i')
453. c11 = (cosin)/2".
454. c11 = (nsinin)/3".
< 0.
§ 5. Power series
In problems 455-466 determine the radii of convergence of the
series.
00
455.
2z"
2z"
n.
n-o
-;;;·
00
457.
n-1
n•l
00
456.
-,•
2n"-a".
2n z.
00
418.
2"
n-1
n
75
INTEGRALS AND POWER SERIES
2 :! z".
ClO
ClO
459.
463.
2
ClO
ClO
znl,
464.
n=O
2
(cos in)z".
n=o
2 2nznl.
ClO
461.
[3+(-l)"]"z".
n-0
n=l
460.
2
ClO
465.
n=O
2
[n+an]z".
n=O
ClO
462 •
'1 z2" .
.L.J
n-0
ClO
466 . l+
'1 o:(o:+l) ... (ix+n-1),8(,8+1) ... (/J+n-l)z".
.L.J
n=l
nly(y+l) ... (y+n-1)
ClO
467. The radius of convergence of the series }; cnz" is equal to
n=O
R (O<R<oo). Determine the radii of convergence of the series:
ClO
ClO
(1)
2
nkrv,";
n=O
(4) }; n"cnz";
n=t
ClO
ClO
(2) }; (2"-l)cnz";
n=O
(5)
ClO
(3)
2 ~z";
n.
2 c~z";
n=O
ClO
(6) }; (l+z3)c4 z".
n=O
n=O
ClO
ClO
468. The radii of convergence of the series }; anz" and }; bnzn
n=O
n=O
are equal respectively to r1 and r 2• What can be said about the radii
of convergence of the series
ClO
(1)
2
n=O
ClO
ClO
(a4 ±bn)zn;
(2) }; anbnz";
n-o
(3)
2 :n z"?
n=O
n
76
PROBLEMS ON COMPLEX ANALYSIS
469. Sum the following series for
lzl <I:
co
(2 )
"1 z•
L.J n ;
00
(S)
"1
L.J
z1n+1
2n+ I
n=O
n-1
In problems 470-478 investigate the behaviour of the power
series on the boundary of the circle of convergence.
00
00
474. }; zPn/n (p is a natural number).
470.}; zn.
n=l
n=O
00
oo (
l)n
475. ~ ~z8n-l,
~ logn
471. }; z"/n.
n=l
00
00
472. }; zn/n".
478. }; znl /n•.
n=l
n-1
00
473. }; (-l)nzn/n.
n=l
00
By Abel's theorem if the series
2
c,, converges then
n=O
00
lim
r-+1
00
2 o,,r" = n=O
2
n=O
(0
Cn
< r<
1).
477. Show that the converse to Abel's theorem is not true, that
00
is, give an example of a divergent series
2
Cn,
for which there exists
n-0
00
the limit lim
2 c,, ,a.
r-+1 n-o
478. Using Abel's theorem and the solution of problem 489 prove
the following equalities:
=
-log,2sin11
I
2'
(o
< lc/>I
~ :n:):
77
INTEGRALS AND POWER SERIES
(0
< "' <
2:n;);
(o
<
l</>I
< :n:);
00
(4) ~ sin(2n+l)</>
=,L,;
2n+l
4
(O
< </> < :n;);
n-0
2
00
(5)
(-l)n+l co:n</> = log(2cos:)
(-:n;
< "' <
:n;);
n=l
2
00
(6)
(-l)n+i
sinnn</>
~
=
(-:n;
< "' <
:n;).
n-1
00
4'79*. Prove that the series
~
(-l)lt'n)zn
n-1
n
~
- converges, but condi-
tionally only at all the points of the circle of conv-ergence.
480. Prove that if the sequence of real positive numbers {a.} tends
monotonically to zero and the radius of convergence of the series
co
2 anz" is equal to 1, then this series converges everywhere on the
n=O
circle
izl
HINT.
= 1, excluding, perhaps, the point z = 1.
Use Dirichlet's test of convergence (see problem 488).
00
481. Prove that if the series
2 cnzn
n=O
converges at the point
C= Rfi-9 on the circumference of the circle of convergence then it
converges uniformly in every closed region G, belonging to the circle
of convergence lying in an angle between any two chords of the
circle lzl = R, issuing from the point C, and not containing any
points of the circle JzJ = R except the point C.
REMARx. This assertion is a more general form of Abel's theorem (see,
for example, [I, Chapter m, § 7, sec. 3)).
78
PROBLEMS ON COMPLEX .ANALYSIS
§ 6. The Taylor series
In problems 482-496 expand the given functions in the power
00
,L; cnzn
series
and find the radius of convergence.
n=O
482.
483.
484.
485.
486.
coshz.
sinhz.
sin1 z.
cosh1 z.
(a+z)'• (a• = e(Jl 10111).
487.
•f! (z+s)
. (·'·
f s=
1
488. az+b
(b
l+i)
y2 •
z•
490. (z+l)I .
l+z
491. log1- .
-z
492. ta.n-1 z (tan-1 0 = 0) •
493. sinh-1 z (sinh-10 = 0).
494. log (zl-3z+2).
..p 0).
z
496.
Js~z
dz.
0
In problems 497-502 expand the given functions in a series of
powers of (z-1) and find the radius of convergence.
497. z:2.
498.
z
z -2z+5
z•
499.
.
(z+1)1
1
500.
.
Vz (v1 = -l~iy3 ).
501. logz.
502. sin (2z-zl).
In problems 503-50'7 find the :first five terms of the expansion in
powers of z of the given functions.
503. e:r:a1u.
504. y(cosz)
(y(cosz) = 1 when z = 0).
505. (l+z)z = ez101(l+z).
506. er.
507. ezlog(l+z).
508. (1) Expand in a series of powers of z the function log(l+Ef)
(find the recurrence relation between the coefficients of the series).
INTEGRALS AND POWER SERIES
79
HINT. First 11.nd the expansion of the derivative of the given function.
(2) Prove that the only term of the expansion containing
an odd power of z, is z/2.
HINT. Use the identity
log(l+e•)-log(l+e-•)
= z.
In problems 509-013 use the multiplication of series and the substitution of a series into a series to expand the given functions in
powers of z.
509. [log(l-z)]D.
510. [Log.(l-z)]2
(Log I = 2ni).
511. (tan-1z)2 (tan-10 = 0).
512. tan-lz log (1 +z2)
(tan-10 = 0) .
•
513. e1-•.
514. Prove that if the expansion of the function I/cos z is written
in the form
00
_I_= '\1 (-l)11 Ea11 zB11
cosz
~
(2n!)
'
11=0
then the numbers E 211 (Euler's numbers) satisfy the relations:
Ea= 1, Eo+( 2;)E2+ ... +(::)E211 =0.
515. Prove that if the expansion of the function z/(e11 -l) in a series
of powers of z is written in the form
then the numbers B 11 (Bernoulli's numbers) satisfy the relations:
Ba= I,
(n!l )Bo+ (nil )B1+ ... +(n!l )B11 = 0.
516. Prove that all the Bernoulli numbers with odd subscripts,
except for B1, are equal to zero.
z
(-z)
HINT. Use the identity &11-l - e-•-l = -z.
80
PROBLEMS ON COMPLEX ANALYSIS
517. Expand in a series of powers of z the function zootz and
find the radius of convergence of the resulting series.
HINT. Use the equality which follows from Euler's formula
zcota
•
2iz
= u+ ell•- I·
518. Expand the given functions in series of powers of z and find
the radii of convergence of the resulting series:
sinz
z
(1) log--; (2) tanz; (3) logcosz; (4) -.-·
z
~z
519. Prove that the coefficients Cn of the expansion
l
2 e.z•
00
-:::----=- =
l-z-z•
n-o
satisfy the relation en= Cn-i+en-2 (n ~ 2). Find
of convergence of the series.
REMABB:,
en
and the radius
The numbers Cn are known as Fibonacci numbera.
520. In the expansion
find
e 0,
Cn-8
(n ~ 3).
o1,
e 8,
and also the recurrence relation between
If in some circle
!ti < R
e., Cn-1,
Cn-s•
the expansion
00
F(t, z)
=
2 In (z)t•
n=O
iB satisfied, then the function F(t, z) iB said to be a generating functidn. for
the sequence {/n(z)}. It is frequently possible to prove some of the properties
of the sequence of functions {/n(z)} from the properties of its generating
function.
521. The Bernoulli polynomials efJ.(z) are defined by the expansion
/"'-1
e'-1
00
=
~ _tPn(z_)_t•.
.L.J
n=l
nl
81
INTEGRALS AND FOWER SERIES
Prove the following properties of them:
(1) tf>11 (21+l)-t/>11 (z) = nz11-1;
(2) if m is a natural number then
t/>11 +1 \m) = 1+211 +311 + ... +(m-1) 11 ;
n+
11-I
(3) tf>11 (z)
=
.2 (;)
B"z11-", where B1c are the Bernoulli
k=O
numbers (see problem 515).
522. The function
y(l-~tz+t2 )
is the generating function of the
Legendre polynomials P 11 (z):
00
L
P11 (z)t".
=
l
2
y(l-2tz+t)
11 = 0
Prove the relations:
(1) (n+l)P..+i(z)-(2n+l)zP11 (z)+nP,,_1 (z) = O;
(2) P 11 (z) = P~+ 1 (z)-2zP~(z)+P~_ 1 (z);
(3) (2n+l)P11 (z) = P~+ 1 (z)-P~_ 1 (z).
HINT. Differentiate the generating function with respect to
respect to z.
e
and with
523. Using the integral formula for the coefficients of a Taylor
series prove that if -1 < s < 1 then
where 0 is a circle of radius R > 1 with centre at the point C= 0.
524. Prove that the function (4-t2)/(4-4tz+t2 ) is the generating
function of the Chebyshev polynomials:
T 11 (z) = 2111_ 1 cos (n cos-1 z).
Using the integral formulae for the coefficients of the Taylor series
establish that for n ;):. 2
4T11 + 1 (z)-4zT11 (z)+T11 _ 1 (z)
= 0.
82
PROBLEMS ON COMPLEX ANALYSIS
525. The Herm.ite-Chebyshev polynomials H.(z) a.re defined by
the expansion
e2t:r:-i•
"° Hn(z) en.
"1
=
L.;
n!
n=O
Prove the following relations:
(1) Hn+ 1(z)-2zH.(z)+2nHn_ 1(z) = 0
(2) H~(z) = 2nH._ 1 (z)
(n ~ 1);
(3) H~'(z)-2zH~(z)+2nHn(z)
(4) H (z) =
n
=
0
(n
~
l);
(n ~ O);
(-l)•e=·~·ce-"'') •
dzn
526. The Laguerre polynomials can be defined by the equation
Ln(z) = e=
d•(z•e-"')
dz•
•
Find a genera.ting function for the sequence {L.(z)} and by means
of it obtain the recurrence formula. connecting
Ln- 1 (z), Ln(z)
and
Ln+ 1 (z).
In problems 522-526 only a few of the particular properties
of the given systems of polynomials have been considered. For other important
properties of them which play an important part in the solution of various
problems of mathematical physics, see, for example, [3, Chapter VII, § 2]
or R. CoURANT and D. HILBERT, Methods of Mathematical Physica, (Vol. I,
Chapters II and VII, Interscience Publishers, New York).
REMARE.
In problems 52'7-529 find the solutions of the given differential
equations, which satisfy the conditions w(O) = 0, w' (0) = 1.
52'7. w"-z2 w = 3z2 -z4.
528. (1-z2)w"-2zw'+n(n+l)w = 0.
529. (l-z2)w"-4zw'-2w = 0.
530. Expand the function cos (m sin-1 z) (sin-1 0 = 0) in a series
00
of the form
2
CnZn
after finding a differential equation which is
n=O
satisfied by this function.
531. The differential equation
d2w
dw
z(l-z) dz2 +[c-(a+b+l)z] dz -abw = 0
is known as the hypergeometric equation.
INTEGRALS AND POWER SERIES
83
Find the solution w(z) of the hypergeometric equation analytic
at the point z = 0 which satisfies the condition w(O) = 1, assuming
that c is not equal to zero or a negative integer.
532. Prove that the general solution of the hypergeometric equation is of the form (c not equal to an integer)
w = 0 1 F(a, b, c, z)+09 z1 -cF(a+l-c, b+l-c, 2-c, z),
where F(a, b, c, z) is the function defined in the preceding problem
(the hypergeometric series).
583*. Prove that if c is not equal to zero or a negative integer,
then
dF(adb, c, z)
z
=
ab F(a+I,b+l,c+I,z).
c
§ 7. Some applications of Cauchy,s integral formula and power series
mu. Let the expansion of the function /(z) in the circle lzl < R
be of the form
00
/(z) =
2
.... o
011 z11 •
(l) Prove that
2ir
2~ Jj/(re'•)illd,P =
0
00
,21c 12rfln
11
(r < R).
m=O
(2) Prove that if
max
l.rl•r
lf(z)I
= M(r),
then the coefficients 011 satisfy the inequalities (Cauchy's inequalities)
Ic..I ~ ~r)
(r
< R) •
(3) Prove that if juE-t one of Cauchy's inequalities becomes an
equality, that is,
lctl =
.M(r) then the function /(z) is of the form
r1:
84
PROBLEMS ON COMPLEX ANALYSIS
00
HINT. Use the following inequality from section 1. }; lcn1•r1111~ [M(r)]'.
n=O
(4) Prove Liouville's theoremt.
00
535. The function /(z) =
2
n-0
Cnzn
2 c~~n
is analytic for lzl ~ r. Prove
00
that the series <f>(z) =
converges in the whole plane and
n=O
the estimates l</>(z)I
l!L
< Me r
,
i<f><k>(z)I
.M .!!!.
e r (M is a constant)
<7
hold for its sum.
00
536. Let the function /(z) =
2
CnZn be analytic in the disk jzj :,;;;; 1
n-o
and effect the schlicht mapping of this disk onto the domain G of
area 8. Prove that
00
S=n };nlcnl2 •
n-1
HINT. Write the formula for the calculation of the area S in polar coordi·
nates (see problem 152).
Rmt:ABJC. If we omit the condition that the mapping is schlicht the separate
parts of the domain G must be counted as many times as the corresponding
values of the function /(z) are ta.ken in the circle Ir.I .s;;; 1.
537. Prove that if in the conditions of the preceding problem the
function /(z) is analytic only in the open disk izl < 1 and if the
finite limit lim Sr = 8 then exists, where Br is the area of the open
r--.1
00
circle jzj :,;;;; r
<
1 then the series
2 n lcnl
n-1
2
converges and its sum
00
equals 8 /:n;. Prove also that if lim 8, =
<X>,
then the series
r-1
2 n lcnl2
n-1
diverges.
REMARE.
See, for example, [4, Chapter
xm.
§ l].
538. (1) Using the solution of problem 536 prove that ifj'(O) = 1
and the functionf(z) maps the disk lzl :,;;;; 1 conformally and one-one
t See problem 480 where another method of proof of this theorem is
given.
INTEGRALS AND l"OWER SERIES
85
onto some domain G, then the area of the domain G is not less than
the area of the mapped disk (the extremal property of mapping
onto a disk).
(2) Prove that of all the functions f(z) analytic in the disk
lzl ~ R and satisfying the condition
2n
f
if(Re1"')! 2 dcp = M
0
the linear function reali3es the mapping of the disk onto the domain
of least area. Find this area if f(O) = 0.
In problems 539-543 use the maximum modulus principle.
539. Prove that if the function f(z) is different from a constant,
analytic in the domain G and does not become zero, then the minimum
of lf(z)i cannot be attained inside the domain G.
540. Prove that within the domain bounded by a simple closed
level line of the modulus of the functionf(z) (that is, at all the points
of which lf(z)i = const) and contained together with its boundary
within the domain of analyticity of the function f(z), there is at
least one zero of this function (f(z) -:/= 0).
541. Prove that if P(z) is a polynomial of degree n, then the level
lines of its modulus IP(z)i = 0 (lemniscates) can be decomposed
into not more than n connected components.
542. Prove Schwarz's lemma:
if the function f(z) is analytic in the circle izl < 1, f(O) = 0 and
lf(z)I ~ 1, then in the entire circle
/f(z)I ~
lzl ·
Prove also that if even at a single interior point of the circle
lf(z)i = lz!, then f(z) = e11Zz (ix is a real number).
HINT. Consider the function f(z)/z and apply to it the principle of the
maximum modulus.
543. Prove that ifin the preceding problem the conditionf(O) = 0
is replaced by the condition
the inequality
/(IX) = 0 (lex! <
I
Z-IX
lf(z)i ~ 1-az
is satisfied.
I
1), then for !zl ~ 1
86
PROBLEMS ON COMPLEX .ANALYSIS
1-ii.z
HINT. Consider the function --/(z).
Z-13'
The point z0 is said to be a zero of order le of the function /(z), if
= /' (z0) = ... = /(k-1)(z0) = 0, /(k)(z0) -:/= 0. The point z0 is said to be
an A-point of order le of the function /(z) if it is a zero of order le of the
function /(z) - A.
/(z0)
544. Prove that the point z0 is a. zero of order k of the a.na.lytio
function /(z) when and only when the equality
/(z) = (z-Zo)"ef>(z),
holds in some neighbourhood of the point z0, where the function
</>(z) is ana.lytic at the point z0 and </>(z0) ..P 0.
543. Find the order of the zero a.t z = 0 for the functions:
(1) z1 (es"-l);
(2) 6 sinz8 +z8 (z8 -6);
(3) e81u-et•u.
546. The point z0 is a. zero of order k of the function/(z) and a zero
of order l of the function ef>(z). What is the point z0 for the following
functions:
(1) f(z) </>(z);
(2) /(z)+ef>(z);
(3) /(z)/t/>(z) ¥
In problems 54'7-561 find the order of all the zeros of the given
functions:
555. sin8 z.
lS4'7. z1 +9.
• 8
548. (z1 +9)/z4.
556 _ sm z •
lS49. z sin z.
z
550. (l-e•)(z•-4)B.
55'7. sinzB.
551. 1-cos z.
558. cos8 z.
559. cosz3 •
560. (yz-2) 3 •
1-cotz
561. (l-y(2-2 cos z))11 •
553. - - 554. etaas.
562. Prove that the distance of the nearest zero of the function
/(z) =
~o c.~ to the point z =
0 is not less than
i ~i~1 ·where e
is a.ny number not exceeding the radius of convergence of the series,
and M(e) =max l/(z)j.
1•1-e
87
INTEGRALS AND POWER SERIES
HINT. Establish that the function /(z) has no zeros in the domain where
f!(z) - c0 l<lo0 1: estimate l/(z)-c0 1, using Cauchy's inequality (see prob·
lem 134).
563. Does there exist a function analytic at the point z = 0 and
assuming .at the points z = ..!._ (n
n
... , 0,
1,
... '
1
... '
1, 0,
(2) 0,
1
1
1
2' 0, 4' 0, 6'· ... ' 0,
1
1,
values:
(1) O,
(S)
1, 0,
= 1, 2, ... ) the
1
1
1
1
1
2
3
4
5
6
1
2' 2' 4' 4' 6' 6' ... ' 2h '
1
<4> 2'
3' 4' 5' 6' 7' ... ,
2h~
1
2h , ... ;
_n_
~
n+l ' ... ·
584. Do functions exist which a.re analytic a.t the point z = 0
and sa.tisfy the conditions :
(1)
(2)
1(.!..)
=!(-.!..)n = ~.
n
n"'
1{~) =!(- ~) = ~8'
(n a na.tura.l number).
565. The function sin [1/(1-z] has an infinite sequence of zeros
converging to the point z = 1, hut nevertheless this function is not
a. constant. Does this contradict the uniqueness theorem~
566. Is it possible for the sequence of zeros (or in general A-points)
of a. function which is not identically a constant, and is analytic
in the whole finite plane, to ha.ve a. limit point 1
567*. The function f(z) = u (a:, y)+iv (a:, y) is analytic at the point
z0 = z0 +i110 a.nd f(z0 ) = c0• Prove that·
z+z-,~
z+zo) -c0 •
f(z)=2u (2
0
88
PROBLEMS ON COMPLEX ANALYSIS
568. Prove that with the conditions of the preceding problem
. ( z+z0 ~
21+z0)+e_0 •
f(z) = 2w - 2- ,
In problems 569-5'72 find the analytic function f(z) = u(x, y)
+iv(x, y) from its given real or imaginary part.
569. u(x, y) = xl-1'+2.
570. u(x, y) = ex(xcosy-ysiny)- z1!y• .
571. v (x, y) = :i:+y-3.
572. v(x, y) = cosxsinhy-sinhxsiny.
CHAPTER V
LAURENT SERIES, SINGULARITIES
OF SINGLE-VALUED FUNCTIONS.
INTEGRAL FUNCTIONS
§ 1. Laurent series
In problems 573-588 expand the given function in a Laurent
series either in the given ring or in the neighbourhood of the given
point. In the latter case determine the domain within which the
expansion holds.
573. l/(z-2) in the neighbourhood of the points z = 0 and z = oo.
574. l/(z-a)" (a ..P 0, k a natural number) in the neighbourhood
of the points z = 0 and z = oo.
575. l/(z(l-z)) in the neighbourhood of the points z = 0, z = I,
Z= oo.
576. l/((z-a) (z-b)) (0 <!al< !bl) in the neighbourhood of the
points z = 0, z =a, z = oo and in the ring !al< izl <lb!.
577. (z1L2z+5)/((z-2)(z11 +1)) in the neighbourhood of the point
z = 2 and in the ring I < izl < 2.
578. I/ (z1 +I )11 in the neighbourhood of the points z = iand z = oo.
579. y((z-a) (z-b)) (!bl ~ !al) in the neighbourhood of the point
z = oo (consider both branches of the function).
580.
/(z) =-VCz-l;(z- 2)) (Imf(i) >0)
in the ring
l<lz1<2.
1
581. z11 e• in the neighbourhood of the points z = 0 and z = oo.
1
582. e1=s" in the neighbourhood of the points z = 1 and z = oo.
1
583. es+• in the domain 0 < izl < oo.
584. sin z sin 1/z in the domain 0 < izl < oo.
585. sin (z/(1-z)) in the neighbourhood of the points z = 1 and
z = oo (in the latter case restrict yourself to the first four terms
of the series).
89
90
PROBLEMS ON OOMPLEX ANALYSIS
586. cot z in the neighbourhood of the point z = 0 and in the
ring n < !zl < 2n.
587. log z-ab in the neighbourhood of the point z = oo.
z-
588.
_!__2 Iog z+i· in the neighbourhood of the point z =
z-
z
oo and
i
in the ring 1 < izl < 2.
589. Can the given functions be expanded in a Laurent series
in the neighbourhood of the given points:
(1) cos l/z,
z = O;
(6) z1 cosecl/z,
z = O;
(2) cos l/z,
z = oo;
(7) logz,
z = O;
(3) sec l/(z-1) z = 1;
(8) log l/(z-1),
z = oo;
(4) ooh,
z = oo;
(9) log((z-1)/(z+i)), z = oo;
(5) tanhl/z,
z = O;
(10) Z11 = (e1110'•)
z = O!
590. Explain whether the given many-valued functions have
single-valued branches which can be expanded in a Laurent series
(in particular in a Taylor series) in the neighbourhood of the given
point:
(1) y(z),
Z=O;
(2) v(z(z-1)),
z = oo;
z = oo;
(11) Log ((z-l)(z-2)],
Z= oo;
Z= oo;
(4) V((z-l)(z-2)(z-3)),
Z= oo;
(12) Lo (z-a) (z-fJ)
g(z-y)(z-!5)'
Z= oo;
(13) sin-I z,
z = O;
(5) y(z(z-1)1),
z = oo;
(6) v(l+y(z))'
z = l;
(14) tan- 1 (l+z),
(15) sinh-1 (i+z),
(7) y(l+yz),
Z= O;
(16)
(8) y(z+y(z1 -l)), z = oo;
(9) y(z+y(z1 -l)), z = 1;
(17)
-i/(;
-i/(:
-sin-1
z=O;
z = O;
z), z = l;
-sin-1 z),
z =ly2.
91
LAURENT SERmS
Some properliea of univalent functionat
co
2 c,.z•,analyticintheringr:;:;:;;lzl~R,
n-=-00
591. (1) Thefunctionf(z) =
effects the schlicht mapping of this ring onto some domain D. Prove
that the area S of this domain is equal to
co
S = n _}; n lc11 12 (.R2•-r2•).
nc::.-oo
(2) Prove that the formula for the area S still applies when
f(z) is analytic only in the domain r < lzl < R; in this case both
sides of the equation may tend simultaneously to oo.
HINT. See problems 536 and 537.
592. The functionf(z) is univalent in the domain lzl
this domain can be expanded in the Laurent series
C-1
C-2
f( z) =z+-+2 +
z
z
>1
and in
...
Prove that
co
_}; n lc- 11 12 =::;; 1 ,
n=l
and explain the geometrical meaning of this inequality (the area
theorem).
HINT. Use the fact that the area S,, bounded by the image of the circle
lzl= r > I, is given by (f(:z:) =
u
+ w)
o~s, = J udv = J2" !+l
~ (of - o</Iaf)d•.
22io</I
Izj=r
0
593. Prove that if the function
J(z) =
z+c#+ ...
< r, then lc2 1 ~ 2.
1
function [F( ~)r ,
is schlioht in the disk lzl
HINT. Prove that the
where F(z)
=
y'(J(z1 )) is
sohlicht in the disk lzl > 1, and apply to it the inequality of the preceding
problem. (By y'(f{:z:2 )) is understood any definite branch of the root.)
t On the subject of this group of probleIIIJI on the properties of univalent
functions see, for example, [I, Chapter V, § 2]. where other references to this
subject are given.
92
PROBLEMS ON COMPLEX ANALYSIS
594. Prove that if under the conditions of the preceding problem
ICtl = 2, then the function f(z) is of the form
z
/(z) = (l+e'"z)•
(cc is a real number).
Explain onto what domain the unit disk 1211 < 1 is mapped by this
function.
595*. Prove that if the function f(z) is schlicht in the disk lzl < 1
andf(O) = O,J'(O) = 1, then the domain G onto which this function
maps the unit disk always contains the circle with centre at the
coordinate origin and of radius 1/4 (Kobe's theorem).
596. (1) Prove tha.t if the function f(z) satisfies the conditions
of the preceding problem and the points w1 and w1 lie on the boundary
of the domain G, where arg w1 = arg w1 +n, then
lw1 1+ lw1 1 ~ 1
(Szego's theorem).
HINT. Apply Kobe's theorem to the function
l-~~:~/wi .
(2) Prove that if the points w1 and w8 satisfy the cofulition
arg w 8 = argw1 +cc(O ~cc~ n), then
lw1l+lwsl
REMABK.
~sin~.
This result strengthens Kobe's theorem only for
°' >
n/3.
§ 2. Singular points of single-valued analytic functions
In problems 597-630 find the singular points of the functions,
explain their nature and investigate the behaviour of the function
at infinityt.
597.
1
--8.
Z-2'
z'
598. l+z'.
z&
599. (1-z)I •
t In the answers no distinction is ma.de between a removable and a
genuine singularity.
SINGULAR POINTS O.F SINGLE· V.ALUED FUNCTIONS
1
1
604. S-1
e- - -z.
as
605.---21(1-e-=)
1-e"
606. l+e=.
2
619. cotz--.
z
1
620. - - - sinz-sina
621.
1
cosz+cos a
.
. 1
622. sm-.
1 -z
607. tanhz.
608.
93
1
e-~.
1
623. cot-.
z
1
609. ze•.
1
1
z
z
624. cot- - -.
z
610. el-•.
:-.!.
625.
611. e "
sin..!:..+~.
z
z
1
ez-1
1
626. e-z cos-.
612. -.--.
e-1
z
1
1
627. e00
613. -.-.
smz
t•.
1
628. etan•.
614. co:z.
z
615. tanz.
616. tan1 z.
629.
sin(~).
Slll-
z
617. co~z .
z
1
1
630. s i n (l-)·
cos21
618. cotz--.
z
631. Let P,.(z) and Qm(z) be polynomials of the nth and mth
degree respectively. Describe the behaviour at infinity of the following functions:
(1) P,.(z)+Qm(z);
(2) P,.(21)/Qm(z);
(3) P 11 (21)Qm(z).
94
PROBLEMS ON COMPLEX ANALYSIS
632. Prove the equivalence of the following two definitions:
(1) The point x 0 is said to be a pole of order n of the function
/(z), if in the Laurent expansion of f(z) in the neighbourhood of z0
00
/(z) =
,2;
n-=
Cn(z-z0)n,
-oo
C_n =F 0,
C-(n+l) = C-(nH) = •.. = 0.
(2) The point z0 is said to be a pole of order n of the function
/(z), if in some neighbourhood of this point f(z) = </J(a)/(z-z0)n,
where the function </J(z) is analytic and </J(Zo) =F O.
633. Construct examples of functions which have in the extended
plane only the following singularities:
(1) A pole of the second order at infinity;
(2) A pole of the second order at the point z = 0 with the
principal pa.rt of the expansion c_ 8 /zl and a simple pole at infinity;
l•d
(3) Simple poles at the points zrc = w", where w = eI I
(k = 0, 1, 2, ... n-1).
634. Find the general form of the function which has in the extended plane only the following singula.rities :
(1) One simple pole;
(2) One pole of order n;
(3) A pole of the second order at the point z = 0 with principal
pa.rt of the expansion l/z1 ;
(4) A pole of order n at the point z = 0 and a pole of order
m at infinity;
(5) n poles of the first order.
635. Let f(z) be a single-valued function which has no singularities
other than poles in the domain G. Provethatthefunctionf'(z)/(f(z)-A)
(the logarithmic derivative of the functionf(z)-A) has simple poles
at all the poles of the function /(z) and at all the A-points of this
function and has no other singular points.
636. What singularity does the function F(z) =/[</J(z)] have at
the point z = 210 (the case z0 = oo is included), if the function </J(z)
is either analytic or has a pole at this point, and the point Co = </J(z0)
is a singularity of the following type for the function /(C):
(1) A removable singularity;
(2) A pole of order n;
(3) An essential singularity~
SINGULAR POINTS OF SINGLE-VALUED FUNOTIONS
95
837. The point z0 (the case z0 = oo is included) is an isolated
singularity of the function f(z), which maps the circular a.re (or
rectilinear segment) r on some circular arc (or rectilinear segment) y'.
What is the nature of the singularity of the function f(z) at the point
symmetrical to z0 with respect to r (the funotion/(z) is continued
a.cross y by the principle of symmetry), if the point z0 is for j(z):
(1) A pole of order n;
(2) An essentially singular point!
638. The Casorati-Weierstrass theorem asserts that if the point
z0 is an essential singularity of the function /(z), then whatever the
complex number A may be (including A = oo), a sequence of points
{zn} which converges to an essential singularity z0 exists, such that
lim f(z,.) =A. Prove thatthe theorem remains valid for a non-isolated
z:,
11-+0Q
singularity which is the limit of polest. (Sometimes singularities
of this kind are included among essential singularities.)
639. Find the limits:
1-; (3) lim
(1) lim ootlz; (2) lim -.Jl-+:!::OQ Sill Z
Jl-+:!::OQ
~h ;
X-+:!::OQ COS
Z
1
(4) lim
x-o
1
y-..o Bin-
z
Does the existence of these limits contradict the Casorati-Weierstrass theorem !
Pioard's theorem asserts that in the neighbourhood of an essential singularity
an analytic function assumes infinitely many times every finite value, with
the exception perhaps of one, which is said to be the Picard exceptional value.
If meromorphic functions are considered the possible number of exceptional
values (including oo) does not exceed two (see, for example, [I, Chapter VIII,
sec. 8.4]).
640. Verify Pioa.rd's theorem for the functions:
1
1
(1)
e=;
(2) e"'i";
(3) cos-;
z
(4) tanz;
(5) tantz.
Find the exceptional values for each of these functions and show
that these values (if they exist) a.re asymptotic, that is, that it is
possible to find not more than one line which :finishes at an essential
singularity along which the function tends to the exceptional value.
t It is assumed that the poles are the only singularities in the neighbourhood
of the point considered.
96
PROBLEMS ON COMPLEX ANALYSIS
Power series with singularities on tke boi11ru1ary of the circle
convergencet
of
00
641. Let the seriesf(z)
=
2
c.z• have the radius of convergence R
n=O
(0 < R < oo). Let us expand the function f(z) in a series of powers
of the difference (z-re111), where 0 < r < R:
00
f(z)
=}; bn (z-re
111)•.
n=O
Prove that if lim !JI 1
•~
r lb.I
=
R-r, then the point z = Re'11 is a sin-
!J/
gularity of the function f(z), and if lim
·~ r
a regular point.
Jb.I
> R-r then
it is
HINT. Use the fact that a power series always has at least one singular
point on the boundary of its circle of convergence.
642. Prove Pringsheim's theorem:
00
If the radius of convergence of the series
2
c.z• = f(z) is equal
n=O
to unity and all the en ;;;::: 0, then the point z = 1 is a singularity
of the funotionf(z) (the sum of the series).
HINT. Prove by means of the criterion of the preceding problem that if
it is assumed that the point z = l is regular then all the remaining points of
the unit circle will be regular (for the proof use the fact that lf<•l(z)I ~ 1/<•>(a:)I,
where a: > 0 and lzl = a:).
00
643. Prove that if the radius of convergence of the series
2 c.zn
n=O
= f(z) is equal to unity, where all the c. are real and the sequence
{s.} (s. =
i: c1c) tends to an infinity of a definite
sign as n -+ oo
k-0
(that is beginning with a certain n, all the s. have the same sign),
then the point z = 1 will be a singular point for the function f(z).
Prove, for example, that the assertion may not hold if we only
have lim ls.I = oo.
·~
t See [l, Chapter m, § 6]. See also G. P6LYA and G. Szoo6 (1957),
Aufgaben und Le1w8'i.tze a'U/1 der Analysia, 2nd ed. Chapter I, sec. 3,Chapter 5,
§ 3, or E. C. TITCHMABSH (1951), Theory of functiona, Chapter VII.
97
SINGULAR :POINTS OF SINGLE-VALUED FUNCTIONS
HINT. "Estimate lf(z)I for z real and close to I (z
identity
00
J; c
<
1), by means of the
00
11 z11
=
(1-z)
n-0
J; 1,,zn.
n=O
644. Prove that if on the circumference of the circle of convergence
00
of the series
2 c,,z!' = f(z)
there is just one pole of the function
n=O
f(z), then the series diverges at all the points of this circumference.
HINT. Using the fact that from the convergence of the series at just
one point of the circumference of the circle of convergence it follows tha.t
lim c,,R11 = 0 (R is the radius of convergence), prove that if (z'I = R then
n-+O
the radial limit lim (z' -z)/(z)
== O.
z-+-z 1
00
64lS. Prove that if the power series f(z) =
2 c,,z" has on the cirn=O
cumference of the circle of convergence only one singular point z0 ,
Anm-1
a. pole of order m, then c,, =
[I +<P (n)] where lim t/J(n) = 0,
11
Zo
n-o<>o
and A is a constant.
HINT. Expand in a series of powers of z the difference between the function
/(z) and the principal part of the expansion of /(z) in the neighbourhood of
the pole z..
00
846. Prove that ifthe radius of convergence of the series
2
c,,z"
n=O
= f(z) is equal to unity, and all the singular points of the function
f(z) on the circumference izl = I are poles of orders not exceeding m,
then lc,,I < Anm-1 , where A is a constant (if m = I, then ic,.I < A,
that is, the sequence of coefficients {c,,} is bounded).
847. Prove that if the only singular point z0 on the circumference
00
of the circle of convergence of the power series /(z) =
Hence, in particular, it follows that the formula R
c,,z" is
n-0
a pole, then
RElllARK.
2
=
c11
lim - n--.oo
0 n+l
for the radius of convergence always holds if the only singular point on the
circle of convergence is a pole.
98
l'ROBLEMS ON CO:Ml'LEX ANALYSIS
648. Prove that the assertion of the preceding problem remains
true if all the singular points of f(z) on the circumference of the circle
of convergence are poles, the order of one of them (z = z0) being
higher than that of all the others.
§ 3. Integral functionst
An analytic function is said to be an integral function if its only singularity
is the point at infinity. If the function /(z) is an integral function and the
point z = oo is a pole, then f(z) is a polynomial; if, however, the point z = oo
is an essential singularity then /(z) is called a tranacendental integral function.
Let M(r) =max l/(z)I· The number
lim loglogM(r)/logr is called
(! =
lzl=r
r-+oo
the order of the integral function f (z). If (! < oo, the function /(z) is said to
be a function of finite order; if(! = oo,/(z) is a function of infinite order. If
(!
> 0, the number a
= lim log M(r) is known as the flype of the function.
r-+oo
re
Ifa = 0, the funotion /(z) is said to be a function of minimal flype; if a= oo,
it is a function of maa:imal flype; if 0 < a < oo it is a function of normal flype.
649. Prove the following assertions:
(1) H (! ::fa oo and <1 ::fa oo are respectively the order and type
of the functionf(z), then for any e > 0 it is possible to find a number
R(e) such that the inequalities
M(r)
<
e«•+•)rll.
hold for r > R.
It is also possible to find sequences of numbers {rn} and {r~}
converging to infinity for which
,e-•
M(r0 )
> e'n
and
M(r~)
,e
> e<a-•>rn .
(2) If for some natural number le
.
O<hm
r-->OO
M(,r)
r
It
<oo,
then f(z) is a polynomial of degree le.
HINT.
Use Cauchy's inequalities for the coefficients of the power series
00
/(z) =
2
n=O
CnZ" •
t On the subject of this group of problems see, for example, [I, Chapter
VII§ I] or B. YA. LEVIN, The diatribution of the roots of integral functiona
(Raspredeleniye lcornei tselylch funktsii), Gostekhizdat, 1956.
INTEGRAL FUNOTIONS
99
(3) if /(z) is a transcendental integral function, then
fun logM(r) = oo.
logr
r-+OO
In problems 650-662 determine the order and type of the given
functions (n is a natural number).
655. e<2-l)z•.
650. Co'Zn+Cizn-1+ ... +en.
656. sinz.
651. e•zn (a > 0).
657. coshz.
652. znasz.
658. escosz.
653. zlel'Z-e3Z
659. cosy'z.
654. e621 -3e8st.
o
00
660*.
2 (2m~nn)
I
(m a natural number).
n=O
661. e•s.
1
662* •
JezPdt.
0
663. The integral function/(z) is of order(! and type a (0~0'~ oo).
Prove that the function P(z)/(z)+Q(z), where P(z) and Q(z) a.re
any polynomials is also of order (! and type a.
864. The integral functions f 1 (z) and f 2 (z) are of orders (!1 and (! 2
respectively, where (!1 ::p (!a· What can be said of the order e• of
the functions Ji (z) f 2(z) and f 1 (z) +f2 (z) 1
665. The integral functions f 1 (z) and f 2 (z) are of the same order
(! and are respectively of type& equal to a 1 and 0'2 , where 0'1 #: a 2.
What can be said of the order (!* and type a• of the functions:
(1) f 1(z)f2 (z);
(2) / 1 (z)+/2{z)1
666. The integral functions Ji {z) and f 2 (z) a.re of the same order
(! and of the same type a. What can be said of the order e• and type
a• of the functions :
(1) f1{z)f2(z);
(2) f1{z)+/2{z}1
667*. Prove that the order and type of an integral function a.re
not changed on differentiating the function.
Solve problems 668-675 by the use of the following theorem.
If the Taylor expansion of an integral f1mction is of the form
100
PROBLEMS ON OOMPLEX ANALYSIS
00
f(z) =
2 onzn, then the order and type of this function a.re given by
n•O
the formulae :
(O'ee)
.!.
11
-
= lim
( .!.
11
n
)
n
ylOnl
.
D-+00
688. Prove that the integral function
00
J(~) =
is of order
I
e = -a:
~
L.J
n-o
(Az)n
I'(o:n+I)
and type O'
HINT. Use Stirling's formula
I'(«n+I) = ( :"
(A >0, ct:>O)
.!
= A •.
t
y'(2n«n) [1+0
(~ )].
In problems 68M75 find the orders and types of the given
functions.
689. f(z) =
2 (:t
n=l
oo
670. f(z) =
2(
n
lo!n ) 0 zn
n=l
oo
(a> O).
n
671./(z)= 2(nl!gn)°zn
(a>O).
n=ll
2
00
672. f(z) =
e-n•zn.
n=O
673. f(z) =
~
L.J
n-1
zn
n,r.+n
(a> 0).
INTEGRAL FUNCTIONS
101
co
674. f(z) =
2
cos!tn zn.
n=O
co
~
-v
675. z Jv(z) =
(-l)nzlln
.L.J nlI'(n+,,+l)
(11
>
-1; Jv (z) is a Bessel
n-0
function of the 11th order).
If
A(•)
(!
is the order of the integral function /(z), then
= 1im"
r-+co
the function
logl/(re'">I is called the indicator funcwm of the function /(z).
rO
In problems 676-681 find the indicator functions of the given
functions.
676. es.
679. cosh z.
677. sin z.
680. ezn.
678. cosz.
681. e"+z1.
682. The integral function f(z) has the indicator function h(</>).
What is the indicator function h*(</>) of the function /(z)+P(z),
where P(z) is a polynomial~
CHAPTER VI
VARIOUS SERIES OF FUNCTIONS.
PARAMETRIC INTEGRALS.
INFINITE PRODUCTS
§ I. Series of functions
In problems 683-692 find the domains of convergence of the
given series.
00
'\1 (-l)"
z+n
L.J
n=O
688.
.
00
'\1
Zn
L,; 1-zn •
690.
n=l
2
2 s~nz.
00
00
686.
es ios n •
'\1
n-o
n-t
2
00
00
687.
.,.n
L,; 1 ~zll" •
691.
692
n-t
•
n-1
n
z
(4+z)(4+z2) ... (4+zn)
00
693*. Prove that if the series
J; a
11
converges then the series
n=l
00
n
.J: a,.zz"
,._1 1
converges everywhere where
102
izi :Fl; if, however, the series
VA.BIOUS SERIES OF FUNCTIONS
00
103
00
}; an diverges then the series }; 1anz;n converges in the circle of
n-1
00
n-1
2
anz" and diverges outside this circle.
n-1
694. (1) Expand in a sum of powers of z the sum of the series
convergence of the series
00
~a z"
,,c,,;
1 n z"; find the radius of convergence of the series obtained.
n-1
(2) Prove that for
lzl <
1
00
'1
L.J cf>(n)
n-1
z"
z
1-z" = (l-z) 2
'
where c/>(n) is the number of those positive integers less than and
coprime to n.
2
HINT. Use the relation
l/>(n) = m where n assumes in succeBBion the
value of each of the divisors of the number m, including 1 and m.
00
00
~.
e-s 1o1n (Riemann's
n=l
n-1
Zeta function) in a Taylor series in the neighbourhood of the point
z = 2 and find its radius of convergence.
695. Expand the function C(z) =
,2 =};
In problems 696-699 find the sum of the given series.
696.
.f(l~zn -l+~n-1)
(jzj.,t:l).
n-1
00
697 ·
2
n-1
(1-z");;_zn+l)
(izl ¥: 1).
HINT. Multiply numerator and denominator by (1 - z).
2
2"
00
698.
n=l
n Z
fl (l+zs")
k-0
•
104
PROBLEMS ON COMPLEX ANALYSIS
00
The series
2
ft(z) is said to be uniformly co1wergent on the set E if for
k=l
any
B
> 0 it is possible to find a number N(e) such that the inequality
00
12 /rc(z)I < B is satisfied for all n >
k=n
N(e) and for all points z of the set E.
700. Prove the propositions:
00
(1) For the uniform convergence of the series
2
In (z) on
n-1
the set E it is necessary and sufficient that for any e > 0 a number
N = N(e) should exist such that for all n > N, all z e E and any
natural number p the inequality
should be satisfied.
00
(2) From the uniform convergence of the series
2 lfn(z)I
n=l
on the set E there follows the uniform convergence on the same
00
set of the series
2 fn(z).
n-1
701. Find the sets on which the given seq\'J.ences converge
uniformly:
(l) {
l~zn };
(a) {
sn:
nz } .
702. Prove:
In order that the sequence of continuous functions {!n(z)} should
converge uniformly on the bounded closed set E, it is necessary
and sufficient that this sequence should converge at all the points
of this set and that it should converge continuously at all the limit
points of the set E, that is, that for every sequence of points Zn,
belonging to the set E and converging to the point z0 ,
lim In (zn) = l(zo) •
n-+00
In problems 703-707 find the sets on which the given series
converge uniformly.
VARIOUS SERIES OF FUNCTIONS
00
703.
00
2 ~8 (zn+ ~ ).
2
2
706.
n-1
2
2
(sin nz)/n1 •
n-1
00
704.
105
00
707.
e-ns.
n-0
(sinnz)/n.
n~l
00
705.
e-slosn •
n=l
L :. converges uniformly in the closed
n
oo
708. Prove that the series
n-1
circle Iz I : : ; ; 1. Does the series obtained by term-by-term differentiation converge uniformly in the circle Iz I < 1 ~
oo
709. Prove that the series
,l:
(-l)n-1
z+n
converges uniformly in
n=l
any finite pa.rt of the plane, from which there has been removed
disks of arbitrarily small radius e with centres at the points z = 0,
-1, -2, ...
Prove that this series does not converge absolutely at a single
point.
n
oo
710. Prove that the series }; :
converges uniformly in the
n=l
interval (-1, 0), and the series }; 1
1'
~I
n=l
converges in the same in-
I
.L:-:
oo
terval, but not uniformly. (Hence the series
n
cannot be ma.jorised
n=l
in the interval (-1, 0) by a convergent series of numbers.)
REMABJC. This example shows that Weierstrass's sufficient test for uniformity of convergence is not necessary.
00
711. (1) The series
,L: (I ~z2f converges absolutely for izl ;;:,: 0,
n=O
largzi ::::;;; n/4 (these values of z do not exhaust the whole of the domain
of absolute convergence which, as is easily seen, consists of the point
z = 0 and the outside of the lemniscate I I+z2 1 =I). Prove that
the series converges non-uniformly in the given domain.
106
PROBLEMS ON COMPLEX ANALYSIS
REMA.RX. This shows that uniform convergence does not follow even
from the absolute convergence of a series in a closed region.
(lO
(2) Prove that in the same region the series ,L; c\-l~:;n
con+
n=O
verges uniformly and absolutely, but not absolutely uniformly
(that is, the series of moduli does not converge uniformly).
(lO
712. Prove that if the series 21/n(z)i converges uniformly in
n-1
(lO
every closed region inside the domain G, then the series 21/~(z)i
n=l
also possesses the same property.
§ 2. Dirichlet seriest
(lO
Series of the form
2
ane -A,,z, where the an are complex coefficients
n-1
and the An are real non-negative numbers, satisfying the conditions
A1 < A1 < ... and lim An
=
oo,
II-co
are known as Diriokke Bef'iea.
713. Prove that if a Dirichlet series converges at the point
z0 = :r:0 +iy0 , then it converges at all the points of the half-plane
Re z > Re z0 , the convergence being uniform in every angle
Iarg(z-z0 )1 ~ () < n/2.
HINT.
Apply Abel's transformation to the sum
q
~
L.J ane
q
..
)
-A,,z _ ~
-Anzo -AnCz-z0
- L.J ane
e
11-=p
n==p
and use the inequality (a< b, te =a:+ iy)
b
je-az-e-11•1
= jz J e-l:tdtl~~(e-u-e-llx).
a
a;
714. Prove that if a Dirichlet series converges absolutely at the
point z = z0 , then it converges absolutely and uniformly in the
half-plane Re z ~Re z0 •
From the theorems formulated in problems '718-'714 it follows that the
domain of convergence of a Dirichlet series (if it exists) is a half.plane
Re z > ll:c (a:c ~ - oo), and the domain of absolute convergence (if it exists)
t On the subject of this group of problems see, for example, [l, Chapter IV,
§ I].
107
VARIOUS SERIES OF FUNCTIONS
is a half-plane Re z > a:0 (a:a;;;a. - oo), the series either converging absolutely
on the whole of the straight line Re z = a:0 , or not converging absolutely at
a single point of this straight line. The numbers a:c and a:0 are called respectively
the abBciaBa of oonoorgenoe and the absciaaa of abaolul.e cont1ergence of the Dirichlet
series.
In problems 715-721 find the abscissas of convergence (zc) and
the abscissas of absolute convergence (z0 ) of the given series.
2
00
715.
e-n•e-zn•
719.
2(
00
00
nl)n e-"'lOglOg n,
720.
e-slogn.
n=l
n=2
2 (Y~n
2
00
00
717.
(-l)ne-slogn •
n=l
n=O
716.
2
2 ~.
2
00
e-slollocn.
721.
n-2
e"" e-sn•.
n=O
00
718.
(-l)ne-Zlo1101n,
n-2
722. Prove that if lim (log n) /An :-- 0, then
723. Prove that if lim (logn)/An
=
Z, then
Z 4 -Zc ~
Z.
In problems 724-728 investigate the convergence of the Dirichlet
series on the boundary of the half-plane of convergence.
2
2 ~3
2~
00
724.
725.
727.
n!)n e-sn •
n=l
n=l
00
00
e-s 101 n •
n=l
00
726.
2(
2(
00
(-l)ne-"' 101fn.
e-zn.
n=l
HINT.
See problem 4'79.
728.
n=l
-I)lvnJ
e-"'n.
n
108
PROBLEMS ON COMPLEX ANALYSIS
The series
<: ... ; lim A.a
Ioo
ane
-A z
n , where tJ.e An are complex numbers
n-1
(ll1 1< IA.I
<
= oo ), is known as a generalised Dirichlee series.
n-+00
729. Let the numbers An satisfy the conditions
lim logn = 0
n-oo
lim log Ian! = k
IA.al
and
An
<
oo.
n-oo
Prove that if oc ~ arg An~ p, then the generalised Dirichlet
series converges absolutely at every point z = :t + iy for which
the inequality :t cos c/J-y sin <f>-k > 0, is satisfied for all q, of [oc, {1],
and diverges at a point for which for all </> of [oc, PJ
x cos <f>-y sin <f>-k < 0.
00
730. An arbitrary generalised Dirichlet series
L; a.,. e-Ans is given.
n=l
_ log!antl
Let k(<f>, oc) = lim - ,
1 -and k(</>) = lim k(c/J, oc), where {nt} is
1
k-oo
«~
Anrc
the sequence of all the suffixes for which <f>-oc::::;;; arg Ant::::;;; <f>+a (if
no such subsequence {n,,,} exists for which lim arg Anm = <f>, then it
m-oo
is necessary to put k(</>) = -oo).
Prove that if lim (logn)/An = 0, then the series converges absolutely
-
within the domain G, the points z = x+iy of which satisfy fo1·
any <f> the condition zcosef>-ysin<f>-k(c/J) >0, and diverges at
every point lying outside G.
§ 3. Parametric integrals
731. Prove the theorem:
Let 0 be a simple contour (closed or not), possessing a finite
length, and f(T, z) a function, analytic with respect to the variable
z and continuous with respect to T for every z of some domain D and
for all the points T, belonging to the contour 0. Then the function
represented by the integral
F(z) =
Jf(r:, z)dr:,
c
is an analytic function of the variable z and
F'(z) =
JJ;(r:, z)dr:.
c
INTEGRALS DEFENDING ON A l'ARAMETER
If the integral
Jf (T, z)dT is
c
109
improper, that is, if the integrand has dis·
continuities at some isolated values TE 0 or the contour of integration includes
the point at infinity, then the definitions of convergence and uniform conver·
gence of such an integral are exactly similar to the corresponding definitions
given in books on mathematical analysis.
732. Prove that for the uniform convergence of the integral
f(T, z)dT on the set E with respect to some point To#= oo of the
J
c
contour 0 it is necessary and sufficient that for any e > 0 there
should exist a number o(e) such that
IJf(T, z)dTJ < e
Co
for all the points z of the set E and for every arc 0 a of the contour 0,
lying in a a-neighbourhood of the point To and not containing this
point either as an interior or as an end point.
733. Formulate and prove a similar criterion for uniform convergence of the integral if To = oo. Consider the cases when the contour
0 is not bounded at one end, or at both ends.
734. Prove that if lf(T, z)j ~ IVJ(T)I for all the points z of the set
E and if IVJ(T)ldT converges then the integral f(T, z)dT converges
f
f
c
c
uniformly on the set E.
735. Let f(T, z) be a function analytic with respect to z and continuous with respect to T for all the points z, belonging to some
domain D, and points T, belonging to the contour 0, with the
exception of some isolated points of it where the conditions imposed
on the function f(T, z) are not satisfied either for all points z, or
only for some of them.
Prove that if the improper integral
F(z) =
Jf(T, z)dT
c
converges locally uniformly in the domain D (that is, in every closed
sub-region of the domain D), the function F(z) is analytic and
F' (z) =
Jazaf
c
dT '
the latter integral converging uniformly within D.
110
PROBLEMS ON COMPLEX ANALYSIS
In problems 736-743 find the sets on which the given integrals
converge uniformly.
Jts00
736. I'(z) =
w-1 = e<•-l)lolt).
1 e-'dt
0
J
J t•
J c;t
J
c+loo
00
737.
e-•t'dt.
741.
0
c-loo
c+loo
00
738.
sint dt
742.
.
0
c
c+loo
00
739.
dt.
743.
0
c-loo
00
740.
e•'
-dt
(c:FO).
-e•'
dt
(c:F 0).
f
J
J z'
t
t
-dt
t
(c ::/: 0, z'=etlOIS).
-sintz
, - dt.
0
An
integral of the form
00
J e-t•j(t)dt,
(1)
0
where the function /(t) is integrable on the segment [O, a] for any positive
a< oo is called a Laplace integral.
744. Prove the following propositions:
(1) If the integral (1) converges at the point z = z0 then it
converges in the half-plane Re z >Re z0 , the convergence being
uniform in the angle la.rg (z-z0)1 :,; ; (J < n/2.
(2) If the integral (1) converges absolutely for z = z0 then it
converges absolutely and uniformly in the half-plane Re z ~Re z0 •
(3) If lim loglf(t)I
t-+OO
t
= p, then the integral (1) converges
absolutely in the half-plane Re z >fl and uniformly in every half-plane
Rez ~ fl+e (e > 0) (constructa.n example of a La.place integral which
converges absolutely in the whole plane and for which fl= oo).
(4) If lim log lf(t)I =ex, then the integral (1) does not cont-.oo
t
verge absolutely at a single point of the half-plane Re z < oc.
111
INFINITE PRODUCTS
It follows from the theorem formulated in problem '744 that the domains
of convergence and of absolute convergence of a Laplace integral (if they
exist) are the half-planes Re z > a:c and Re z > a:0 ; the number a:c is called
the absciasa of convergence, and a:0 is called the abacisaa of absolute convergence
of the Laplace integral.
co
In problems 745-751 find
Xe
and x 0 for the integral [ e_.' /(t) d.t,
where f(t) is the given function.
746. /(t) = e-t•.
747. /(t) = e'".
745. /(t) = 1.
748. f(t) = e-t• for 0 < t < log log 3 and
log log 2le t <log log (2le+l) (le= 2, 3, ... ) and f(t) = -e-t• for
log log (2le+l) t <log log (2le+2) (le= 1, 2, ... ).
749. /(t) = eie' for 0 t < log log 3 and
log log 2le t <log log (2le+l) (le= 2, 3, ... ) and f(t) = -eie' for
log log (2le+l) < t <log log (2le+2) (le= 1, 2, ... ).
750. /(t) = e8 ' for 0 < t < log log 3 and
log log 2le ~ t < log log(2le+ 1) (le = 2, 3, ... ) and f(t) = -ee' for
log log (2le+l) ~ t <log log (2le+2) (le= 1, 2, ... ).
751. /(t) = e' for log (2le-l) ~ t <log 2le (le= 1, 2, ... ) and
f(t) = -e' for log 2le
t <log (2le+l) (le= 1, 2, ... ).
In problems 752-755 investigate the convergence of the Laplace
<
<
<
<
<
00
integrals
J e..,.'f(t)d.t on the boundary of the half-plane of conver-
o
gence.
752. /(t) = 1.
753. /(t) = 0 for 0 ~ t ~ 1, /(t) = l/t2 for t > 1.
754. /(t) = 0 for 0 t 1, f(t) = l/t for t > 1.
755. f(t) = 0 for t = 0, /(t) =--= 1 for 0 < t ~ 1, and for t > 1
/(t) is defined as follows:
.f(t+l) =f(t)+l, if (2le-l)ll < t+l "'.; (2le)2
and
/(t+l) =/(t)-1, if (2le)2 < t+l ~ (2le+l)2 (le= I, 2.... ) .
< <
§ 4. Infinite products
In problems 756-762 prove the given equalities.
n(
00
756.
•
1- :. ) =
•=2
!.
112
PROBLEMS ON COMPLEX ANALYSIS
00
757. n(1+
n
n=l
00
758.
n
n
00
n(n~2r)=2.
760.
n 8- l
ns+1
2
=a·
n=2
1
n2_4
n2-l = 4.
oo [
761.
=
n
l+ (-1)•+1]
l.
n=l
n=S
00
759.
n(1- n(n~l)) ! .
=
n=2
oo
1
elil
762.JJ
n=l
]
+-n
n
=ecwhere0=!~(2!-logn)
constant.
n
00
763. Prove that
n=l
HINT.
is Euler's
k=l
•
sm o:
o:
oos 2n = -o:- ·
First prove the identity
.
sin
k
<[>
=
21:·</>n,,,
cos 2
sm liJ'
n- •
n=l
764. Using the solution of problem 763 prove that
'J'C
2
2
2
... = - .
2
y2 v' (2+ y2 y[2+v' (2+J1'2)]
765.* Prove Wallis's formula
'J'C
2
=
J::OI (
n=l
2n )
2n
2n-l 2n+l ·
766. Prove that if, as usual, we take
--'J'C
< argpn ~ 'J'C, then
00
the infinite product
fl Pn
converges and diverges together with the
n=l
00
series
l; logpn.
n=l
767. Examine whether the assertion of the preceding problem
is still true if it is assumed that:
(2) o: < argpn o:+2n (o: < 0).
(l) 0 ~ argpn < 2'J'C;
<
113
INFINITE PRODUOTS
768. Prove that for the absolute convergence of the infinite
n (1 +an) (that is, for the absolute C!>nvergence of the
00
product
n-1
00
series
21 log (!_+an))
it is necessary and sufficient that the series
11=
00
2 an
should converge absolutely.
11=1
00
769. The infinite products
fl p
OCI
11
and
n=l
fl qn converge. Investigate
n=l
the convergence of the infinite products:
n
00
(1)
n
00
(Pn+qn); (2)
n-1
n
00
(pn-qn);
(3)
11=1
00
p 11 q11 ;
(4) nPn.
n=l qn
n=l
In problems 770-774 investigate the convergence and absolute
convergence of the given infinite products.
710.
fJ[1+ <-;n+l].
112.
n=l
n
n
n<-u".
773.
n-1
fJ(1+~,,)
(p>O).
n=l
00
774,
(p>O).
11=1
00
771.
fJ[1+ <-~;a+l]
00
COS Zn, if it is known that the series
n=l
,l11z jll converges.
11
11-l
775. Prove that inside the unit circle
n
00
11
1
1-z
(l+zll ) = - - - ,
n=O
the infinite product converging absolutely.
In problems 776-784 find the domain of convergence of the infinite
products.
00
776.
11
11=1
(1-zn).
777.
fJ(1--;:).
11-1
114
:PROBLEMS ON OOMPLEX ANALYSIS
00
778. f/(1-::).
779. fJ[ 1-(1- ! z-·J.
780. f/[1+(1+ !rznJ
781. llcos:.
n=l
r·
n=l
n=2
n=l
n
00
784.
00
(l+cnz), if it is known that the series
2: lcnl converges.
n=l
n=l
785. Prove that the infinite product
converges in the half-plane Re z > 1/2 and converges absolutely
in the half-plane Re z > 1.
786. {fn(z)} is a sequence of functions analytic in the domain G,
and all these functions, with the exception of a finite number of
them, do not become zero in the domain G. Prove that if !fn(z)I ~ ocn
00
for all z e G, where ocn is independent of z, and the series
2
rxn
n=l
converges, then the function
n
00
.fi'(z) =
[l+fn(z)]
n=l
is analytic in the domain G.
In problems 787-790 some of the.properties of the Gamma function
which follow from its definition as the limit of an infinite product
a.re explained (see, for example, [l, Chapter VII, § 4] or [3, Chapter VII, § l]).
787. Prove that the infinite product
I'(z-!--1) =
n-- -00
n=l
n
z+n
(
n+l
n
)%
115
INFINITE PRODUCTS
converges absolutely in the whole plane, except for negative integral
values of z, and represents a function analytic in the whole plane
except at the points z = - 1, - 2, ...
788. Prove Euler's formula
n! n"
.
r(z) = lun
(n·=e
•
r:lo1n)
n-+oo z(z+l)(z+2) ... (z+n)
and show that:
(1)
(2)
I'(z + 1) = zI'(z);
r (m + 1) =ml, if m is a natural number.
789. Prove that
n
00
n-1
n(a+/J+n)
_ I'(a+l)I'(/J+l) _
(a+nH/J+n) -
(a,p #: -1, -2, ... ).
I'(a+/J+l)
790. Prove Weierstrass's formula
z)
1
noo (l-1-- e _!..n,
-=---=ec"
I'(z+l)
n=l
' n
where 0 is Euler's constant.
HINT. Use the solution of problem '762.
791*. Let p 1 , p 2, ···•Pn• ... be the sequence of all the prime
00
numbers (p1
= 2,
p8
= 3,
p3
= 5, ... ).
C(a)
= 2 n-•(n-" = e-• 1o1n)
n=l
(Riemann's Zeta function) is analytic in the half-plane Re a> 1
(see problem 705). Prove that:
(1) CM = _oo_1 __
n c1-p;•)
n=l
(2) The function C(a) has no zeros in the half-plane
Rea> 1.
REHAB.JC. An extensive literature is devoted to Riemann's Zeta function.
See, for example, the monograph: E. C. TITOBMARSH, The zeta-function,
Cambridge Tracts, No. 26.
00
792*. Prove that the series
.L; _!_,
n=l
of prime numbers, diverges.
Pn
where {Pn} is the sequence
CHAPTER Vll
RESIDUES AND THEIR APPLICATIONS
§ 1. The calculus of residues
In problems 793-813 it is required to find the residues of the
given functions at all their isolated singularities and at infinity
(if it is not the limit of singularities).
793.
1
-8--6.
z -z
z2
794. (z•+ l)ll .
zan
795. - - -
(l+z)n
1
796.
I
z(l-z)
zll+z-1
797. zl(z-1) .
(n is a natural number).
sin2z
798. (z+l)a .
ez
799 • z11 (z1 +9)
800. tan z.
1
801. - . - .
smz
802. cot11 z.
803. cot8 z.
1 .
804. (1) cos-•
z- 2
1
(2) z& cos - - ·
z-2
1
z+805. e z
.
. 1
806 • smzsm-·
z
•
z
z+l
807 am--·
.
808. cos
z11 +4z-l
z+S ·
1
809. z(l-e-hz)
810. zn sin l/z
(h.~O).
(n is an integer).
1
811.--.
. 1
sm-
z
812. Jiz/sin Jiz.
813. (tan z)/zn (n is a natural
number).
116
RESIDUES AND TREIB APPLICATIONS
117
In problems 814-821 it is required to find the residues of each
of the single-valued branches of the corresponding many-valued
functions at the given points.
814. lyz at the point z = 1.
-z
815.
l+y~ 2 -zf
z•
(z 0
l-..-z
816. - - .1-
at the point z = 1.
= e•Loss-)
at the point z = 1.
817. v(<z-a)(z-b)) at the point z = oo.
z-Cl
818. (1) Log--{J at the point z = oo;
z-
z-Cl
(2) as- Log --{J at the point z = oo.
z-
819. (1) Logzsin_.!_
at the point z = 1.
z- 1
(2) Log z cos __!_l at the point z = 1.
z-
tan-1z
820. - - at the points z = 0 and z = oo.
z
821. z"Log z-{Jel (n is an integer) at the points z = 0 and z = oo
z-
(in calculating the residue at the point z = 0 it is assumed that
a;'= 0, {J;o!= 0).
822. The expansion of a function in the neighbourhood of infinity
is of the form
f(z)
C1
=co+-+
...
z
Find res [{f(z)}2]:-=oo·
823. Find res[<f>(z)/(z)]z=a• if <f>(z) is analytic at the point a, and
/(z) has at this point:
(1) A simple pole with residue A;
(2) A pole of order Ii with principal part ~~ + ...
z-a
+ -( z-a
c_1: )" .
us
PROBLEMS ON COMPLEX ANALYSIS
824. Find
res[~(~i l=a
if:
a is a zero of order n of the function f(z);
(2) a is a pole of order n of the function f(z).
(1)
825. Find res [ <f>(z)
~(~i l=a
if <f>(z) is analytic at the point a and:
(1) a is a zero of order n of the function f(z);
(2) a is a pole of order n of the function f(z).
826. Find res{f[</>(z)]}s=a if the function <f>(z) is analytic at the
point a and </>'(a)¥= 0, andf(C) has a simplepoleatthepoint C= <f>(a)
with residue A.
82'7. At the point a the function <f>(z) has a simple pole with
residue A, and J(C) has at infinity a simple pole with principal part
BC. Find
res{f[<f>(z)]}z=a.
828. The function f(z), which assumes real values on the arc l
of the circle Iz-a I = R, is continued analytically through this arc
by the symmetry principle. Let the point z = fJ ({J ¥= a) be for /(z)
a pole of order k with principal part
k
'\,
LJ
11=1
c_.
(z-fJ)•.
Find res [f(z)]z=/J*• where {J* is the point symmetrical to z = {J
with respect to l.
§ 2. The evaluation of integrals
In problems 829-83'7 evaluate the integrals, assuming that the
closed contours are traversed in the positive direction.
829.
830.
831.
r
J(z-:)~-2)1 ,
c
J(z-S~~6 -l),
c
'dzl ' where 0 is the circle xl+y1 =2x •
c.. z +
where 0 is the circle lz-21 =
where 0 is the circle lzl = 2.
!.
119
RESIDUES AND THEm APPLICATIONS
HINT. Use the fact that the sum of the residues at all the singularities
(including the point at infinity) is equal to zero.
832.
f 2~:1
f z•(z~9)
f
c
:n, cf !
!;, cJz•e~dz,
• where
a is
the circle lzl =I.
c
883.
dz, where
c
a is the
834. 21 .
sin2:..dz, where
836.
sin'
ni
z
88'7. 21 .
a is the circle
dz, where
836. 2
circle lzl =I.
a is
lzl = r.
the circle Iz I = r •
wherenisaninteger, and 0 is the circle lzl = r.
f /(z)(dz) 'if the simple closed contour a is the boundary
me zgz
of a domain G, containing the point z = 0, the function/(z) is analytic
in the closed region ii, the function l/g(z) is analytic on 0, and in
the domain G has no singularities in addition to the simple poles
a,., a1, ••• , a. (a11 =#= 0 for any k).
In problems 838-841 evaluate the given integrals.
!;, j
838. 2
y'(z'!:+l), where 0 is the circle lzl = r
=#=
I.
1 j
dz
889. 2n;, c (z'+l)y'(z•+l) (y'l = 1), where 0 is the pa1·abola
'If"= x, traversed in the direction of increasing y.
-J
840 - 1
dz
(a•= e•.1°• 0 ), wherel a> 0, and 0 is the
• 2ni c a• sin nz
straight line x = IX, 0 < IX < 1, traversed from below upwards.
HINT.
Consider _l_
J
ck
, where the contour y is indicated in
!:iii y a=sinnz
Fig. 13, and pass to the limit as fJ
J
-+
oo.
841. 21 .
e=dz , where the contour of integration 0 is indicated
nt c cosz
in Fig. 14.
120
PROBLEMS ON OOMPLEX ANALYSIS
If the function f(a:) becomes infinite at a:
b
principal value of the integral
=
c (a
Jf(a:) dz is defined
< c < b), then the Cauchy
by
a
lim [7°/(a:)dz+
•-+0
a
j f (a:)dz]·
c+•
This definition is generalised to the case of a curvilinear integral in an
obvious manner.
g
-r
tt+t
c
D
z
I
ft
A
tZ-i,8
-
Z+tt
-
:&
z
t
8
!/
-c
FIG. 13
-ot
FIG. 14
If the functionf(a:) is continuous on the whole of the real axis, the principal
oo
J f(a:) dz is defined
value of the integral
-oo
N
as lim
J f(a:) dz.
N-+oo -N
In problems 842-849 find the definite integrals. If an integral
is improper and diverges, find its principal value (if it exists).
f
In
842.
def>
a+coscf>
0
(a> 1).
.,.
HINT. Put eld> = z.
843.
f
def>
(a+b COS c/>)1
0
844
,.
def>
J
•
(a+b cos
.,.
2
0
845.
J
cf>)8
1 _ 2a:!cf>+as
0
(a> b > 0).
(a
> 0, b > 0).
(a is a complex number and
a~±l).
RESIDUES AND THEIR .APPLICATIONS
J
121
2n
846
•
cos1 3</> def>
l-2acoscf>+a1
0
(a is a complex number and a#: ±1) .
2n
84'7.
J
..
J
J
e008 ~cos (ncf>-sin </>)def>
(n is an integer).
0
848.
tan (x+ia)cb;
(a is a real number).
0
2n
849.
cot (x+a)cb;
(a is a complex number and Im a#: 0).
0
850. Prove that if b
>a> -
1
y
FIG. 15
where 0 is the contour represented in Fig. 15, and let the radii of the arcs
of the small circles tend to zero. In calculating the integral along the vertical
segment divide it into two and by corresponding substitutions reduce them
122
PROBLEMS ON OOMPLEX ANALYSIS
to Eulerian integrals of the first kind; also use the known relation
B(p, q)
= i;![)I'(~)
p+q
and the formula I'(p)I'(l-p)
= ~.
smnp
In problems 851-855 evaluate the integrals with infinite limits.
00
852.
zldx
(zl+a11) 1
f
f
J (zl+a•~za+bl)
0
(a> 0).
co
853.
(n is a natural number).
(zl:l)"
0
co
854.
(a > 0, b > 0).
-oo
00
855.
f
zl+l
x'+l dx.
0
f
856. Prove that
1
dT
ilc-n-l(n+lc-2) I
2ni T"r = (2k)n+lc-l(lc-l)! (n-l)! (na.ndlca.rena.tura.lnumbers),
c
where 0 is a straight line parallel to the real a.xis which cuts off on
the imaginary a.xis a segment equal to k (k > 0).
857. Evaluate the integral
1
2ni
f
dT
T"(T-z)
(le is a natural number),
c
where 0 is the contour of the preceding problem.
In problems 858-861, using Jorda.n's lemma. (see problem 411),
evaluate the given integrals.
f
co
00
858 • (l)
-oo
zcoszdx
z 11_2z+IO '
(2)
f
co
zsinzdx
zl-2z+10.
123
RESIDUES AND THEIR APPLIOATIONS
f
f
f
00
859.
-oo
xsinxdx
x8+4x+20 ·
00
860.
cosa:x:
(a and b are real numbers).
xB+b• dx
·-00
00
861
.
x sin a:x: d
x8+b1
(a and b are real numbers).
x
-oo
862. Let f(z) = e1m"F(z), where m > 0 and the function F(z) possesses the following properties :
(1) In the upper half-plane it has a finite number of singularities a 1 , a8 , ••• , a,. ;
(2) It is analytic at all the points of the real axis, except
the points x1 , x2 , ••• , Xm, which are simple poles;
(3) F(z)-+ 0 if z-+ oo and Imz ~ 0.
Prove that
oo
J
f(x) dx =
-oo
n
m
k-1
k=l
2:n:i{2 resU(z)]:r=ak + ! }; resU(z)]:r-~t},
where the integrals are understood in the sense of the principal
value (with respect to all the points xk and oo).
In problems 863-866 find the principal values of the given
integrals.
co
f
f
00
863.
J
1· x -5x+6
e''~
_ 00
- d x (t is a real
x
number).
865.
-oo
00
864 •
xcosxdx.
2
sinxdx
(x1 +4)(x-1)
'
00
866.
-oo
cosax
l+x3 dx (a~ 0).
-co
In problems 867-872 evaluate the given integrals (a and b are
real numbers).
f
00
868 ·
0
sina:x:dx
x(x1 +b8) '
124
PROBLEMS ON OOMPLEX ANALYSIS
co
8'71.
f
s:x ch.
0
J
co
S'70.
cos2aa:;;cos2bx dx.
0
!I
FIG. 16.
BniT. Use the integral Jells-l dz, where the contour 0 is indicated
C
z•
in Fig. 16.
co
f fi:8
sin8 x
8'72.
ch .
0
JlniT, Use the integral
f e81s-3e1•+2
zS
dz, where the contour 0 is indi-
C
cated in Fig. 16.
In problems 8'73--8'76 evaluate the integrals, considering that
xP > 0 for x > 0 (this condition is retained in all the following problems).
00
8'73. (1)
JzP--
1
cos aa:ch (a
> 0, 0 < 'P <
1);
0
co
(2)
Jzl'- sin aa:ch (a> 0, -1<p<1).
1
0
HINT.
Use the integral
in Fig. 17.
JzP-le-asdz, where the contour 0
c
is indicated
RESIDUES AND THEIR APPLICATIONS
125
g
FIG. 17.
co
874.
f cos x"dx
(p
> 1).
0
co
875.
f sin x"clx
(IPI >
1).
0
877. Let the rational function f(z) have no poles on the positive
part of the real axis and
lim [z"f(z)] = lim [z"f(z)] = 0.
z-0
Z-+CO
Prove that
CO
11
J
o
:where
zP-1f(x)dx
ix1 , ix1 , ... ,
=
~
.2
smpn k~l
res [(-z)"-1 f(z)]z=et1:•
°'n are the poles of the function f(z), pis an integer.
IIINT. Consider the integral
J(-z)P-1/(z)dz
( (-z)P-1
c
= e(P-1)101(-z)),
where 0 is the contour represented in Fig. 18.
co
878. Evaluate
J
x"(:l)
(0<p<1).
0
n
879. Prove that I'(a)I'(l-a) = - . - Blll
na
(0 < a < 1).
126
PROBLEMS ON COMPLEX ANALYSIS
FIG. 18
HI:NT. Use the following known relation between the Beta and Gamma
functions:
I'(a)I'(b)
== I'(a+b)B(a, b)
and in the integral defining the Beta funotion
I
B(a,b) =
J:i;0-1(1-z)ll-ldz,
0
ohange the variable, putting a:= y/(l+y).
RmfA.IUC, The relation proved in the problem only for real numbers a
contained in the interval (0,1), holds for all complex numbers. For z = -n
where n is a natural number both parts of the equality become infinite. See,
for example, (1, Chapter VII, § 4) or (3, Chapter VIt, §1].
In problems 880-882 evaluate the given integrals.
CCI
880.
f
f
f
x"dx
l+xl
(-l<p<l).
0
CCI
881.
x"dx
(I+x9)1 • (-1
<p<
3).
0
CCI
882.
0
x-Pdx
1
A
)
xl+2xcosii.+I (-l<p< ,-n< <n.
RESIDUES .AND THEIR .APPLIOATIONS
127
883. On the positive real axis let the rational function /(z) have
only poles of the first order and let
lim [z"f(z)] = lim [z"/(z)] = 0.
z...0
z~oo
Prove that
m
-:Ji
cot pn ~ {Jf""1 res [f(z)J11.pt,
k=l
where the /31; are poles of the function /(z) situated on the positive
real axis, the «11: are other poles of /(z) and the integral is taken in
the sense of its principal value. (p is an integer.)
HINT. See problem 8'1'1.
In problems 884, 885 evaluate the principal values of the integrals.
00
Jr W:/(1-a:)
885. f e",.cb/(1-tr)
884.
1
co < p < I).
0
00
(0
<p<
1).
-co
In problems 886-890 evaluate the given integrals.
l
886.
J
x1-P(l-z)P
(l+x)a cb
(-1
< p < 2).
0
!I
-
FIG. 19
128
PROBLEMS ON OOMPI.EX ANALYSIS
Jzl-P(l-z)P
(I +z)'
HINT. ConBI"der
..>-
~.
h
0 JS
• t he contour m
• dio
.
w ere
ated m
c
Fig. 19 bounding a doubly connected domain, and pass to the limit as R -+ co.
1
887.
f
xl-P(l-x)"
1+xl
dx
HINT. Prove
circle
lzl
that llin
I
R-+oocR
+z
.
dz = 2me-.rmJ, where
a
is the
= R traversed in the positive direction.
1
888.
(-l<p<2).
. Jzl-P(l-z)P
I
0
f
xl-P(l-x)"
(l+x)• dx
(-1<p<2).
0
1
f (l+x)l -P(l-x)"
xi
dx
1
889.
f v
1
890.
dx
0
(-1<p<2).
+
-1
(x+l)
(x1 (1-x))
.
J
l
891. Evaluate the integral
dx
(x-a)y'(l-xl)' where y'(l-xl)>O
when -1<x<1, a is a complex number and a =F ±1. In particular, find the values of the integral for a=± e"' (0 <ix< n),
a= iy, a< -1 and - l <a< 1 (the principal value).
1
892. Evaluate the integral
J xP-1(1-x)-P
b-x
dx, where 0 < p < 1,
0
b is a complex number and b =F 0, b =F 1.
In problems 893-895 find the given integrals.
1
893. In=
f
x2ndx
(l+xl)y'(l-xl)
(n = 0, 1, 2, ... ).
0
HINT. Put z = l/t.
1
894. In=
f
0
v
xi"
(x(l-x1))
(n = 0, 1, 2, ... ).
129
RESIDUES AND THEm APPLIOATIONS
I
895. In =
fV
0
(n = 2, 3, ... ).
dx
(1-z")
f
dz
where a is the contour COD•
y(l+zn)
sisting of segments along the radius vectors of the points 1, co, co•, .. ., con-l,
HINT. Consider the integral
n
C
....
where
co= en, and along the circle lzl = R.
In problems 896-901 evaluate the integrals (a .P 0 is a real number).
00
896
•
f
log xdx •
xl+a1
0
...._,_ e use or t he mte
• gral
HINT...........
f logzdz
zl+a• , wh ere t he contour a·
.
1s given
c
in Fig. 20.
g
FIG. 20
J
f
f
I
00
897 •
0
log1 xdx •
xl+a1
900. J1og(x+
0
00
898.
logxdx
(xl+al)I VX
0
•
901.
~) l~x1·
J ez+1) x
oo
I
(
dx
log - - - --·
eX-1
00
logxdx
899. o (x+l)• yx.
902. Let f(z) be a rational function which has no poles on the
positive part of the real axis or at the point z = 0, and is such that
130
PROBLEMS ON COMPLEX ANALYSIS
f(z) =
o( ~)
as z-+ oo. Prove that
J
f(x)dx
-logl xl+n1 =
oo
0
~
L.J
k-1
[
f(z)
res Log z-ni
J
z-ak
where iii= -1, and a2, aa •... ,an are poles of the function f(z)
different from -1, and Log z =log lzl+i arg z, 0::::;;; arg z < 2n.
·
al
HINT. Consider the mtegr
1
~
J
:inc
-"L f(z) • wo
ogz-:in
~here
the contour 0 IS
·
given in Fig. 21.
g
FIG. 21
In problems 903-905 evaluate the integrals assuming that a
and n is a natural number.
00
903. (1)
f
0
j'
dx
(x+a) (loglx+31:11 )
00
904
•
0
HINT.
r
----,---,,----dx_ _ __
(xl+a1)[log• x+ (2n+ 1)131;1]
Make use of the integral
1 [
I
• z1 +a1 Log z- (2n+ l)ni
c
1
+ Log z- (2n-l)ni
+ ··· + Logz+(~n-l)m]m
>0
131
RESIDUES AND THEIR APPLICATIONS
where the contour 0 is given in Fig. 21, and the branch of Log z is chosen as
in problem 902.
co
905 •
J
dz
(xl+a1)([log x]1 +4n1 xl) ·
0
HINT. Make use of the integral
Jz•+a•
I
c
[
I
I
Logz-2mri+ Logz-(2n-2):ni + ···
+ Log z+
(~-2):ni] dz,
where the contour 0 is given in Fig. 22, and the branch of Log z is chosen just
as in problem 902.
-
FIG. 22
FIG. 23
906. Let f(z) be a rational function which has no poles on the open
contour 0, the first point of which is a and last b. Prove that
J
[
J [
~ res /(z) Log z-b
c f(z)dz = ,L,;
z-a +res /(z) Log z-b
z-a
J
z=oo
where the summation extends over all the poles of the function f(z)
132
PROBLEMS ON COMPLEX ANALYSIS
different from oo (the choice of the branch of the logarithm singlevalued outside 0 is arbitrary).
HINT. Consider
f
t(z) Log z-b dz, where the contour
z-a
r
r,
bounding the
doubly connected domain is given in Fig. 23.
In problems 907-910 find the given integrals, for real values of a.
f
co
907 •
sinaxdx
sinhx •
0
HINT. Make use of the integral
J- . - - where the contour 0 is given
c sinhz
e•lsdz
in Fig. 24.
'
-
--c
FIG. 24
co
908.
f
co
xcosaxdx
sinhx
909.
•
0
of the rectangle
J"'
0
J
0
.
al
HINT. Make use of the mtegr
910.
f
-or.~ Re
cosh a.=_dx
coshnz
J-col$hns
- - , where 0
c
e"dz
(-n <a < n).
is the perimeter
z ~ or., 0 ~ Im z ~ 1.
zsinzdx
l+a1 -2a coax
(a> 0).
HINT. Make use of the integral
Ja-e-
zdz 1 , where 0 is the perimeter
c
11
of the rectangle -n ~ Re z ~ n, 0 ~ Im z ~ h, and pass to the limit as
h-+ co.
RESIDUES AND THEIR APPLIOATIONS
133
From here to the end of this section it is assumed that t > 0, 0 1
is the straight line Re z = IX > 0, going from below upwards, IX being
chosen so that all the singularities of the integrand are situated on
the left of 0 1t.
911. Prove that if f(z) -+ 0 as Im z -+ ±oo, IX1 <Re z < IX 2 and
the function f(z) is analytic in the strip tX1 < Re z < IX1 , then the
integral f(z)dz, where 0 is the straight line Re z = tX, does not
J
c
depend on the choice of IX, if IX1 < IX < tX1 •
912. Let the analytic function <f>(z) have not more than a finite
number of singularities au a1 , ••• ,an on the left of 0 1 and let <f>(z)-+ 0
as z-+ oo and Rez <IX.
Prove that
2~i
n
J
e"t<f>(z)dz =
c,
2
res[e"t</>(z)]z,.ak.
k=l
In problems 913-919 find the integrals (n is a natural number).
913. (1) _l_
2ni
f
c,
e"'dz ;
zn+l
f
Jaztaz
2m
f
914. _1_
2ni
e"t dz
c,
1
915._
c,
916. _l_
2ni
c,
(ai-a)n+l
.
(2) _l_
2ni
f
c,
zn+l
f
f
2m
917. _l_
2ni
--·
z1 +1
918. _l_
ze"' d.:_ .
919. _l_
z1+1
(F = Eflost).
F dz
c,
c,
2ni
e"'
dz
zl(z•+ 1) ·
e"tdz
.
(z-a) (z-b) (z-c)
f z(z+l)t"dz... (z+n) .
c,
920. Using the identity I'(z)I'(l-z) = ~ (see the remark to
smnz
t The problems of this section are mainly connected with the inversion
formula of the Laplace integral which plays an important part in the operational calculus. See, for example, (3, Chapter VI] or V. A. Ditkin and P. I.
Kuznetsov, Te.dboolo on the Operational Oakulus (Spraoochnilo po operatsionnomu iaehialeniyu), Gostekhizdat, 1951, (in this book there are also given E>xten ·
sive tables or formulae on the operational calculus).
134
PROBLEMS ON COMPLEX ANALYSIS
problem 879), prove that if Re"
J~Y+l
I
<0
ezdz
2ni
I
= I'('1+1)'
)'
where the contour y is given in Fig. 25.
FIG. 25
-J as
dz conver""•
~ 0 also, it
zv+l
e-- for Re v ~
R:mMARx. As the integral - 1
2m
)'
1
continues the function I'(v+ I) analytically to the whole plane.
921. Prove that for Re"
I
2ni
f
>
-1
eztdz
tv
zY+l = I'(,,+I) ·
c,
In problems 922-933 find the given integrals.
922. (I) _I_
2ni
923. _I_
2:n:i
I
924. 2ni
927. _ I
2:n:i
J-
c,
f
f
c,
c,
f
c,
eztdz
(2)
;
y(I+z)
e ..tdz
2ni
I
c,
ez'dz
.
y(z+i)
.
e zt~'~
""'
(z+I)y(z+2)
.
e•tdz
(a> 0).
z(I-e- 0 ")
f etz-Jtlfz dz
c,
f
zy(I+z)
HINT. Make use of the expansion
928. 2:n:i
_!__
z
(x
> 0).
1
1-e oz
=
I+ e-oz
+ e-llloz +
...
135
RESIDUES AND TREIB APPLICATIONS
HmT. Replace 0 1 by the contour shown in Fig. 26.
26
FIG.
929 . _1_
2ni
930• ~
2m
J
fl' sinh ry21 dz
c,
J
c,
~ffl'log(z+l)dz.
2ni
z-1
931.
rz sinhayz
c,
(a >r >0).
co
lo~(l+z)as' dz.
z
ot
J Je:~•
dt
932.
dz (a >0).
c,
0
HINT. Change the order of integration.
2~i c,J~ dz Je-•scosudx (a> 0, b a real number).
00
933.
0
HINT.
Make use of the fact that
e-•
J-z
-dz
=
0 for u
> 0.
c,
934. From the series expansion of the Bessel function
~
J,,(z) =
L.J
(-l)t
klI'(k+v+l)
k=O
(z2 )v+lt '
deduce the following integral representations ( y is the contour given
in problem 920):
1
(I) 2ni
f c•+i.. dC (z)
2 •
eC-it
=
J,(z);
136
PROBLEMS ON COMPLEX .ANALYSIS
r..
2~i c,f ~~~ dC = (:
(2)
J,(z} (Re,, > -1).
HINT. Expand the function e - '' into a power series and use the solu·
tions of problems 910 and HI.
-n
n
FIG. 27
935. Prove that if Re z > 0
f
J,(z} = __!__ eliratnt-fvtdC,
2:n; II
where ll is the contour given in Fig. 27, and hence obtain that for
evecy integer n
J.(z} =
~
.
J
cos(zsinC-nC}dC.
0
In problems 938-938 find the integrals which involve Bessel functions.
J
00
936.
e-ztJn(t}dt (Rez
> 0, n
is an integer}.
0
HINT. Use the integral representations of the preceding problem and
change the order of integration.
J
00
J 0 (at} cosbtdt;
937. (1)
J
00
0
(2)
J 0 (at} sinbtdt
0
(a and b are real numbers}.
J
00
938.
cos bx
0
sin t JI (xi- a2)
V(xD-aB)
dx (t > lbl}.
RESIDUES A.ND THEm APPLIOATIONS
137
HINT. Make use of the fact that
sinut
u
=
y/(!!!...)J
2u
u•t•
(ut)
l
=
y'(nt) _l_
2
2ni
Jez-~ dz
c,
?!la
(see problem 93'), and change the order of integration.
The asymptotic behaviour of integral,at
939. Let the analytic function </>(z) have on the left of 0 1 only
a finite number of singularities, all of them being poles, and let
<f>(z) -+ 0 as z -+ oo and Re z :,;;;; ct. Let us put
f(t)
=
2~i c,Je%'</>(z)dz
Find limf(t). Consider various cases of the distribution of the
t-+00
poles with respect to the imaginary axis.
HINT. Use Jordan's lemma (see problem 4ll).
940. Let the analytic function </>(z) have on the left of 0 1 a finite
number of singularities, and let </>(z) -+ 0 as z-+ oo 11.nd Re 21 :,;;;; ct.
Prove that the asymptotic equality
J
2~ic, e%'</>(z)dz"' .2res [e%'<f>(z)],
holds for large values of t, the summation extending over all the
singularities of </>(z) with negative real part.
REMA.Rx. The functions f (t) and F (t) are asymptotically equal as t
(f(t)-F(t)), if lim f(t)/F(t) = 1.
-+
co
f-+00
941. Investigate the asymptotic behaviour as t-+ oo of the
function
l
e:r:'dz
(Rea> 0).
f(t) = 2ni 2!2(z+a)3
J
c,
t On the subject of this group of problems and also on the question of
the application of asymptotic estimates and other methods for obtaining them
see, for example, (3, Chapter V, § 3]; B. A. FucHs and V. I. LEvlN, FuootionB of a oomple3: variable and some of their applications (Funktaii komplekmogo
peremennogo i nekotoryye ikh prilozheniya), Chap. IV, Gostekhizdat, 1951
English translation published by Pergamon Press (1961); M. A. EvGRAFOV
Asymptotic e8timatu and integral Junetions (Asimptoticheskiye otaenki i taelyyl!i
Junktaii), Gostekhizdat, (1957).
138
PROBLEMS ON COMPLEX ANALYSIS
942. Find a.n asymptotic expression as t-+ oo for the function
I
t =-
f
zen-"(z•+aa)
(z-wi)y(z2+2az)
2ni
f()
dz
(w
> 0, a> 0).
C1
where
y(z +2az) > 0
>
for z
2
0.
HINT. Repla.ce the contour 0 1 by lihe contour represented in Fig. 2·0,
and prove that the integrals along the arc of the circle and along the negative
part of the real axis tend to zero as I-+ oo.
00
The series
1:
_<'.!_ is said to be the asymptotic expansion of the function
n=O z11
f (z) as z -+ oo,
if
lim zl: [j(z)-
f ;: ]
=
(k
0
=
0, 1, 2, ... ).
n=-0
Z-+-00
(From this it does not even follow that the series converges!)
Asymptotic expansions of a more general form are also frequently considered. Let {q11 (z)} be a.n arbitrary sequence of functions such that lim q11+1( ()z) = O,
Z-+00
and {µ 11 (z)} a sequence satisfying the conditions:
lim 1'11+1(z)
q11(z)
= 0,
Z-+00
The series
lim ,. µ11(z)
q11(Z)
Z-+00
I>
qll
z
0.
1:"" c11µ 11 (z) is said to be a.n asymptotic expansion of the function
n=O
00
j(z),....
2 c11µ 11 (z),
n=O
if
i; ~µ11(z)] = 0
lim - 1 -[J(z)q1:(z)
n=O
z-+oo
(n = O, 1, 2, ... ).
Frequently the sequence { z!n } is chosen, as the sequence {µ 11 (z)} where
the
ot,s
are positive real numbers monotonically tending to oo.
943. Prove that for :x:
J
>
0
00
0
e-xt
I+t2
I
2!
4!
dt.....,-;-7+7- ...
+(-l)
11
(2n)!
:x:Dn+i
+ ...
139
RESIDUES AND THEIR APPLICATIONS
Make use of the expansion
HINT.
I
-=
I+t•
(-I)n+lt2•+2
l-tD+t'- ... +(-I)nt2n+-'---'---I+t•
and estimate the remainder term.
944. Prove that for :x: > 0
f -,-dt,....,-;-7+-;a-- ...
00
1
e"-t
2
1
n!
11
+(-1) z11+1
+ ...
h
Integrate by parts and estimate the remainder.
HINT.
945*. Prove that for :x:
>
0
00
f -t-dt-.i- -;+-;a+7+ ... +
e-ir-t
(
1
1
2
(n-1)1
x"
)
+ ... ,
-x
where the integral is understood as its principal value.
946. Prove that for real values of x
f
00
-co
00
e-'"dt
t-x
~ (2n) !
,...,-Ji:n; .L.J
1
22n.n! x2n+l'
n-0
where the integral is understood as its principal value.
947. Prove that for real values of :x:
where for :x: > 0 the integral is understood as its principal value.
948*. Prove that
the signs + or - being taken according as Re z > 0 or Re z < 0.
If Re z = 0 the term in front of the brackets must be omitted.
140
PROBLEMS ON COMPLEX ANALYSIS
949. Find the asymptotic expansion of the function
f(t) =
2~i c,Je:r~~~
(ro
> 0).
Find also the expansion of f(t) for small t.
HINT.
Replace 0 1 by the contour given in Fig. 28. In order to obtain the
co
asymptotic expansion of the integral
J e-~tyaicb
w+w•
use the hint to problem
0
HS. For small t it is necessary to choose 0 1 such that
and expand l/(11 +w8) in a series.
at
is greater than co,
y
FIG.
28
950. Prove that
1
2ni
J
pJt
• (
y(z)(z1 +l)dz-sm
C1
n)
2
~
t-4,..., yn .Lt (-l)
n
(4n)I ( 1
(2n)I 2yt
)'n+l •
n=O
951. Find the asymptotic expansion of the function
1
f(t) = 23ii
J
endz
!.
c, z(l+z1 )
8
(i
>0 for z >0).
Obtain also an approximate formula for f(t) for small t.
HINT. In order to obtain the asymptotic expansion replace 0 1 by the
I
iy3
I
iy8
contour shown in Fig. 29, where z1 = - 2 + - 2-, z, = -2- - 2- •
141
RESIDUES AND THEm Al'l'LIOATIONS
For small I the abscissa of the straight line 0 1 muat be taken greater than
unity.
FIG. 29
§ 3. The distribution of zeros. The inversion of series
In problems 952-954 using Rouche's theorem find the number of
roots of the given equations lying within the circle lzl < 1.
952. z9 -2z&+z•-Sz-2 = 0.
953. 2z6 -z3 +sz•-z+s = O.
954. z7 -5z'+z2 -2 = 0.
955. Prove that if the inequality
la1:zil > lau+a1 z+ ... +a1:_ 1 zt-1 +a1:+1zi+i+ ... +anz"I,
is satisfied at all the points of a contour 0, then the polynomial
a0 +12iz+ ... +anzn
has le zeros within the contour 0, if the point z = 0 lies within this
contour, and has no zeros if it lies outside the contour 0.
956. How many roots of the equation
z'-5z+l =0
a.re situated inside the circle lzl < 1 ¥ In the ring 1
957. How many roots of the equation
< lzl < 21
z'-Sz+IO = 0
a.re situated inside the circle lzl < 11 In the ring 1 <
lz! <
31
142
PROBLEMS ON COMPLEX ANALYSIS
958. How many roots has the equation z = </>(z) in the circle
lzl < 1, if for izl ~ 1 the function </>(z) is analytic and satisfies the
inequality l</>(z)I < 1?
959. How many roots has the equation ez = az" in the cii'clti Jzj < R
(n a natural number), if !al> eR/R"?
960. Prove that in the right half-plane the equation
z = A.-e-z
(A> 1)
has a unique (and hence real) root.
961*. Prove that no matter how small is
great n all the zeros of the function
I
f,.(z) =I+ z
1
+ 2!z2
-1 ··•
(!
> 0,
for sufficiently
I
+ n!z".
are situated inside the circle izl < (!·
962. Prove that if (! < I then the polynomial
P,.(z)
= I+2z+3z2+ •.• +nz"-1
for sufficiently great n has no zeros in the circle izi
<
(!.
HINT. Use the method of solution of problem 961.
963. The function </>(z) is meromorphic in the domain G and analytic on its boundary 0. Prove the following assertions:
(I) If l</>(z)i < 1 on 0, then the number of roots of the equation </>(z) = I situated inside the domain G is equal to the number
of poles of the function </>(z) in the domain G.
(2) If l</>(z)i >I on 0, then the number of roots of the equation <f>(z) =I situated in the domain G is equal to the number of
zeros of the function t/>(z) in the domain G.
(3) The assertions (I) and (2) remain true if the equation
</>(z) = I is replaced by the equation </>(z) =a, where Jal ?- I in case
(1) and 1~! < l in case (2).
964. Let none of the zeros of the polynomial
P,.(z)
= z"+a1 z"-1 + ... +a,.
lie on the imaginary axis.
Prove that when the point z traverses the imaginary axis from
below upwards the increase of the argument of P,.(z) equals kn,
where k is an integer of the same parity as n, and !kl ~ n.
143
RESIDUES AND THEm APPLICATIONS
Prove that the polynomial P.(z) then has (n+k)/2 zeros in the
right half-plane.
lhNT. Represent Pn(Z) in the form
Pn(Z)
= zn ( l +
: 1 + ..•
+ :: )
and apply the principle of the argument to the semicircle
for a sufficiently large R.
lzl < R,
Res > O
96lJ. Find the number of zeros of the polynomial
z8 +z6 +6z4+1W+Sz1 +4z+1
in the right half-plane.
966. Find the number of roots of the equation
z'+2zs+sz•+z+2
= 0
in the right half-plane and in the first quadrant.
967. How many roots in each quadrant has the equation
2z4-3z3 +3z1 -z+ l = 0?
968. In which quadrants are the roots of the following equation
to be found1
z4+zs+4z1 +2z+3 = 0.
969. Prove that the number of roots of the equation
z11n+°'z••-1+p• =
0
(°' and p are real numbers, °' ¥= 0, p ¥= O; n is a natural number),
which have positive real part is equal to n if n is even. However,
if n is odd, the number of them is n-1if°'>0 and n+l if°'< 0.
lhNT. Consider the increase of arg (zll•+oczln-1+/JI) when the point
describes the boundary of a right semicircle of large radius.
11.
If the coefficients of the polynomial
Pn(Z)
= z•+0iz•-l+
... +an-1.::-i an
depend continuously on the real parameters a:, {J, then in order to find the
relation between the number of zeros of Pn(z) situated in the right half-plane
and the parameters it is possible to proceed as follows (commencing with the
fact that each zero depends continuously on the coefficients of the polynomial):
144
PROBLEMS ON COMPLEX ANALYSIS
In the a.{J•plane, construct the lines Pn(i-r) = 0 (1' is a real parameter).
that is, lines for the points of which among the roots of the polynomial there
are purely imaginary (or zero) roots. These lines divide the a.fJ·plane into domains in each of which the number of zeros of Pn(Z) with positive real part
is constant. This number can be found by taking an arbitrary point of the
corresponding domain and applying to it, for example, the method of problem 96'.
In problems 970-972 determine the domains of the ix{J-plane in
which the number of zeros of the corresponding polynomial Pn(z)
with positive real part is constant; find this number m for each domain.
970. P(z) = z8 +az1 +cxz+f3.
971. P(z) = z8 -l-az1 +f3z+I.
972. P(z) = z8 +(ix+{J)z1 +(ix-{3)z+ix.
973. Let f(z) = Pn(z)+Qm(z)e-", where T > O; Pn(Z) and Qm(z)
are polynomials prime to one another, where n > m and f(z) has
no zeros on the imaginary axis. N is the number of zeros of the polynomial Pn(z) in the right half-plane. Prove that in order that the
function f(z) should have no zeros in the right half-plane it is neces-
sary and sufficient that the point w = -
~:::: e-"•
should go
round the point w = 1 in the positive direction N times while the
point z traverses the whole imaginary axis from below upwards
(if Pn(z) has zeros on the imaginary axis it is necessary in the motion
of the point z along this axis to avoid the zeros of P nCz> on the right
by semicircles of sufficiently small radii).
In problems 974-976 it is necessary to find the domains in the
space of the coefficients a, b (that is in the ab-plane) for which all
the zeros of the corresponding function lie in the left half-plane.
a and b are real numbers, and T > 0.
The use is recommended of the theorem of 973 and the method
given on page 143.
974. z+a+be-'"•
976. z•+ (az+b)e-"•.
975. z2 +az+be-r:
977. Prove by means of Rouche's theorem that if the function
f (z) has the expansion
w=
f(z)
=
w0 +crc(z-z0 )"+...
(ere ;i' 0, k ~ 1)
RESIDUES AND THEm Al'PLIOATION'S
145
in the neighbourhood of the point z0 then, for sufficiently small r > 0,
there exists a (! > 0 such that any value w =F w0 of the circle lw-w0 1
<(!is assumed precisely k times in the circle lz-z0 1<rand moreover at distinct points.
978. Using the result of the preceding problem prove that an
analytic function has the property of mapping domains onto domains.
979. Using the result of the preceding problem prove the maximum
modulus principle for analytic functions. Prove that this principle
is valid for arbitrary continuous mappings w = f(z) which have
the property described in 978.
980. Prove that if in the conditions of problem 977 we have k = 1,
that is, J'(Zo) =F 0 then the function f(z) establishes a one-one and
conformal correspondence between some simply connected neighbourhood of the point z0 and the small circle lw-w0 1< !!·
HINT. Consider the function z = /-1 (w) in the small circle lw-w0 1 < (!.
981. Prove that if k >I in the conditions of problem 977 then
the function w = f(z) maps one-one some simply connected neighbourhood of the point z0 onto a k-sheeted circle with centre at the
point Wo·
982. Extend the theorems proved in problems 980 and 981 to the
oases when the point z0 is a simple or multiple pole of the function
f(z).
983. Prove that if the expansion of the function f(z) in the neighbourhood of infinity is of the form
Ai
A1
Ak
f(z) = Ao-l-z-+7+ ... +7+ ...
then some neighbourhood of infinity can be mapped one-one and
conformally onto a single-sheeted circle, if Ai =F 0, and onto a
k-sheeted circle, if Ai = A 2 = ... = Ak-i = 0 and Ak =F 0.
RRMARX.
On the subject of this group of problems see, for example, [2,
Chapter X].
984. Let F(z) = z-a-wf(z), the function f(z) being analytic
at the point z = a. Using Rouohe's theorem prove that for sufficiently
small lwl there exists a circle K with centre at the point z = a, in
which the function F(21) has only one (simple) zero. Show also that
if f(a) =F 0, then for an appropriate choice of the value of w any
point of some neighbourhood of the point z = a can become a zero
of the function F(z).
146
PROBLEMS ON COMPLEX ANALYSIS
985. Let z = z(w) be a single-valued function, defined for sufficiently small !w! by the equation z-a-wf(z) = 0, the function f(z)
analytic at the point z = a and /(a) '# 0. Prove that for every
function g>(z) analytic at the point z = a, for sufficiently small
lwl there holds the expansion
00
g>(z)
'\1 w" d"
n
1-wf'(z) = g>(a)+ L.;
da" {g>(a)[/(a)] } .
nr
n-1
HINT. If we denote by 0 the circumference of the circle K, in which the
equation z-a-wf(z) = 0 has only one root (see problem 984), then
lfl(z)
= _1_
1-wf'(z)
2ni
f
«P(C)
c C-a-wf(C)
dC.
Now expand the integrand into a series of powers of w and estimate the
remainder term.
986. Using the notation of the preceding problem prove Lagrange's
formula
00
g>(z) = g>(a)
'\1 w"
dn-1
+L.; nr dan-1 {g>' (a)[/(a)]"}.
n=l
Hence, in particular, obtain the expansion in a Taylor series
of the function z = z(w) itself.
HINT.
Apply to the function </l(z)[l-wf'(z)] the solution of the preceding
problem.
987. Expand in a series of powers of w each of the branches of
the function z(w), defined by the equation w = 2z z1 (for one
branch z(O) = 0, for the other z(O) = - 2).
988. Expand in a series of powers of w the branch of the function
z = z(w) defined by the eqttation w = 2(z-a)/(z1 -l) for which
z(O) =a.
989. Starting from the definition of the Legendre polynomials
P 11 (z) in terms of the generating function l/y'(l-2zt+t2) (see problem
1 d"
522), prove that Pn(z) = 2"nl dz" [(z8 -l)"].
+
HINT. In the conditions of problem 988 apply Lagrange's formula to
the function
1
l/l(z) =
/ (l---2aw +w2)
l-z 2
= z2 -2az+ l ·
RESIDUES A.ND THEm APPLICATIONS
147
• · In the neighbourhood of the point w = 0 the function
z = z (w) is defined by the equation w = ze-•. Expand in a series
of powers of w:
(I) z(w);
(2) e11•<w> •
991. Expand in powers of w the function z = z (w), defined in
the neighbourhood of the point w = 0 by Kepler's equation
z-a=wsinz (a=F 0, ±n, ±2n, ... ).
992*. Determine the radius of convergence obtained in the preceding problem for the expansion of z(w) in the case when a = n/2.
993. Prove the following generalisation of La.grange's theorem.
Let f(z) and ef>(z) be functions analytic in the neighbourhood of
the point a, 0 a circle with centre at the point a and radius r such
that at all its points
Joif(z)+Pef>(z)J < r.
If ~(C) is an analytic function of a single root of the equation
z-a-af(z)-Pt/J(z) = 0
then
"'1 oi111{Jn
d111+n-l
~(C) =~(a)+ L.J mini c1a111+n-1 {~'(a)[f(a)]•[ef>(a)ln}'
where the summation extends to all m and n except m = n = 0.
§ 4.. Pariia.J. fraction and infinite product expansions.
The summation of series
9M. Let f(z) be a meromorphic function with simple poles at
the points ai,ai.,. . .,an,. .. , where O<lail ~lai.J~ ... and 11->00
liman
= oo. Let us denote by An the residue of the function f(z) at the
pole an (n = I, 2, ... ). Let us assume that there exists a sequence
of closed contours 0 111 satisfying the following conditions:
(I) 0 111 does not pass through even one of the points an;
(2) Every contour Om is contained within the contour Om+i:
(3) The minimum distance of the contour 0., from the origin
of coordinates (let us denote it by Rm) increases without limit as
m-+ oo;
(4) The ratio of the length L111 of the contour 0., to R.,
remains bounded, that is, L., = O(R111 ).
(5) max lf(z)J = o(R111 ) (condition 5 is obviously satisfied if
zecm
the function f(z) is bounded on all the contours Om) •
PROBLEMS ON COMPLEX ANALYSIS
148
Prove that subject to these conditions the partial fraction expansion
00
f(z) =f(O)+
1-+_!_),
~ An(z-a,.
an
~
n-1
of f(z) is valid, the convergence of the series being uniform in every
closed region, not containing points an, if terms referring to poles
contained between Om and Om+l (m = 1, 2, ... ) are grouped under
the summation sign.
HINT.
Apply the residue theorem to the integral
I
2m
f C<C-z>
J(C)dC
cm
and pass t.o the limit as m-+ oo.
REMARK. For various generalisations of this theorem see, for example,
[2, Chapter VIII, sec. 4].
In problems 995-1002 prove the validity of the expansions.
= -+ 2
2
00
995. cotz
2z
z1 -nBn• •
1
z
n-1
00
1
· sin z
1
(-l)n2z
z
z" -n11i"
998 - = - +
n=l
.
00
~
2z
2
99'7. tanz = ~
[
-l
(2n-l)3t ]
I
n- - - -z
2
00
1
9118. oosz
~
(-l)n(2n-l)
~"ft "-[ (2" 2 l)n
J.
00
~
999. tanhz = ~
_
n-
1
2z
..
z·+
[
(2n-1)3t ] 11
2
•
RESIDUES AND THEIR Al'PLIOATIONS
149
co
1
~
1
1002. -.-=
sm1 z
(z-nn)1
.
n=--oo
1003. Let f(z) be an integral function with simple zeros at the
points tli' as' ... ' an' ... ' where
O< ltiil ~ lasl
lim a.= oo.
< ... ,
Let us assume that there exists a system of contours {o.}, satisfying
conditions 1-4 of problem 994, and on which
max lf'(z)
zECm
/(z)
Prove that the expansion
/'(0)
j(z) =f(O)e/CO>r:
I=
o(Rm).
n
co (
1-
:n
)
"
ell,,
n-1
is valid in the whole plane.
In problems 1004-1010 prove- the validity of the expansions.
co
1004. sin
Z=Zn(
1- n::I)'
n-1
1005. cosz =fl{1-[ (2n~~l)n Jl
n=O
co
1006. sinhz =Zn (1 + n~:I )·
n=l
1007. cosh z
=fl
{1 + [
n=O
..!.nco (
1008. e"-1 = ze 2
n=l
( 2n~l)n J} ·
1 + 4n•n2 ·
zl
)
150
PROBLEMS ON COMPLEX ANALYSIS
co
1009. e"-eb• = (a-b)zeHa+b>zfl[1 + (a-b)•z•J.
4n1 n 1
n=l
n(+
00
1010. cosh z-cos z = z1
1
4:.·n') •
n-1
1011. Let f(z) be a meromorphic function with a finite number
of poles: ~.a., ... , a,,., not coincident with even one of the points
z = 0, ± 1, ± 2, .... Prove that if there exists a sequence of contours
{On}, tending to the point at infinity, and
lim
Jf(z) cot nzdz = 0
(1)
n-+00cn
then
m
co
}; f(n) = -n}; res [f(z) cotnz]z-arc.
n=-OO
k=l
1012. Prove that if in the conditions of the preceding problem
the requirement (1) is replaced by the condition
lim
n-+00
Jf ~z)
en
dz = 0 '
smnz
then
In problems 1018-1019 find the sum of the series, assuming the
number a to be such that not one of the denominators becomes
zero.
00
1013.
~
.L.J
n==-co
00
1 .
(a+n)1
1016.
00
~
1014. ,L.,;
00
(-l)n
(a+n)• ·
1017. ~i (-l)n •
.L.J n1+a1
n-o
n--oo
00
1015.
2 (2n~l)ll.
n=O
2 n•!a• ·
n-o
00
~
1018 •
.L.J
n-o
(-l)n
(2n+l)B •
RESIDUES AND THEm APPLIOATIONS
151
(-:n<b<:n).
HINT. Make use of the integral
J(a -z
zellr:dz
c
1
1)
sin nz'
1020. Prove that
LJ
J
rata
co
"1
n
(n•-3)y(4n•-3) =
n=l
o
xoot:nx
l
(3-xl)y(3-4x8) rlx+6oot[:n(2-y(3)].
!I
FIG. 30
HINT. Ut1e the integral
nz dz
J.-(z zcot
8)y'(4z•- 3 ) •
c
1_
where 0 is the contour
bounding the doubly connected domain shown in Fig. 30.
vm
OHAl'TEB
INTEGRALS OF CAUCHY TYPE.
THE INTEGRAL FORMULAE OF POISSON AND
SCHWARZ.
SINGULAR INTEGRALS
§ 1. Integrals of Cauchy type
An integral of the form
_1_J
'1(CldC,
2ni C
C-z
where 0 .is a smooth contourt (closed or open) and '1(C) is a function, con•
tinuous on the contour 0, with the exception perhaps of a finite number of
points where it has an integrable discontinuity, is known as an integral of
Cauchy typs. The function '1<Cl is known as the density and l/(C-z) as the
kernel. An integral of Cauchy type represents a function .F(z) which is analytic
in every domain not containing points of the cont.our O. Hence
F<n>(z)
= ~J
'1(CldC
2ni C (C-z)n+l
.
On the contour 0 let •<Cl satisfy a Lipschitz condition of order or: (0
(for brevity, '1(C) e Lip or:), that is,
l•<Cil-tfi(Call
<
(1)
<
or;~
1
Tc!C1-Cal 11•
where the points C1 and Ca belong to the contour 0, and le is a constant. Then
if a point of the contour Co is not an end point of it, there exists the singula
W6grGl
.F(Col
= ~ j•<CldC
2ni C
C-Co
defined as the principal value of· the integral of Cauchy type.
t By a smooth contour we understand a simple (that is, without points
of self-int.ersection) curve with continuously varying tangent and without
cusps. However, the conditions imposed on the contour 0 can be significantly
extended. See I. I. Privalov, (1950), Boundary PropBrtiea of Analytic Functiona
(GranichnyyB 8"0iahia anaUtichealcilch Junlctaii) 2nd ed., Chap. m. It is also
poBSible to consider compound contours consisting of a finite number of contours of the given type.
152
153
INTEGRALS OF OAUOHY TYPE
This principal value can be expressed in terms of ordinary improper integrals
by the formula
.F(Co) =
~J
2m c
l/l(C}-1/l(Co) dC +~(Co>+ l/l(C~) Log b-Co
C-Co
2
2m
a-C0
•
(2)
where the points a and b are the ends of the contour 0, if it is open. The singlevalued branch of Log is chosen so that for a closed contour (a = b) the term
with the logarithm vanishes and the formula assumes the form
.FCCo> =
2~ cf lfi(C~=~:Co>
dC+
~ lfi(Co>·
(3)
If we denote by .F+(C0 ) and .F-(C0 ) the limiting values of the Cauchy type integral .F(z) as z-+ Co on the left of a and on the right of a respectively, then
by Sokhotskii's formulae
•"«·> - J'(C.l+ : •<C.>
.F-(Co)
I
(4)
= .F(Co)-21/l(Co)
or
.F"o)
l
=2
[.F+(Co>+.F-(Co)], .F+(Co)-.F-(Co) = l/l(Co) ·
(5)
If the contour 0 is closed and is traversed in the usual direction, then .F+(C)
is the limiting value of the function .F+(z), determined within the contour
(the domain D+), and .F-(C) is the function determined outside the contour
(the domain D-)t. (See, for example, [l, Chapter III,§ 3] or [3, Chapter m, § 3]).
1021. Prove the formula (1) for n = 1, find the limit of the difference
Jc/J(C)df_
F(z+k)-F(z) __1_
k
2ni c
(C-z) 2
as k-+ 0.
1022. Prove formula (1) for any n by the method of mathematical
induction.
1023. Prove that if 0 is a closed contour and the density of the
Cauchy type integral
<P(z) =
~
2ni
f </>(C)
dC
C-z
c
can be represented in the form
c/J(C) = cfJ+(C)+c/J-(C)
t The case when the contour is an infinite curve dividing the plane into
two domains is also referred to this one.
154
PROBLEMS ON COMPLEX .ANALYSIS
where cf>+(C) and cf>-(C) are the boundary values offunctions analytic
respectively inside and outside the contour 0, then
REMARK. If in the conditions of the problem one of the functions If>- or
tf>+ is identically equal to zero the Cauchy type integral becomes the Cauchy
integral of the interior or exterior dome.in respectively.
1024. Let 0 be a closed contour. Find F+(z) and F-(z), if the density
of the Cauchy type integral is the function (n is a natural number):
(1) c/>(C) = (C-a)n;
1
(2) c/>(C) = (C-a)n (a is inside 0);
1
(3) c/>(C) = (C-a)n (a is outside 0).
1025. Find F+(z) and F-(z), if:
(1) The function c/>(C) is the boundary value of a function
analytic in D+, with the exception of a finite number of points ak,
where it has poles;
(2) The function cf> (C) is the boundary value of a function
analytic in D-, with the exception of a finite number of points
ak, where it has poles (among the points ak there may also occur
the point z = oo ).
1026. Find F+(z) and F-(z), if
= CS+iC2-C+4i
C"-3'2-4
circle !Cl = 3/2.
c/>(C)
+ log[(C-2)/(C-3n
'2-4
and 0 is the
1027. Find F+(z) and F-(z), if c/>(C) = cot C and 0 is the circle
ICl=5.
1028. Find F+(z) and F-(z), if c/>(C) = C2/(C2+1) and 0 is the real
axis traversed from left to right.
REMABK.
By an integral of Cauchy type taken along the real axis
F(z)
=
J--dT,
l
oo l/>(T)
-:n:i
2
•
T-Z
-00
is to be understood its principal value if it diverges in the usual sense.
INTEGRALS OF CAUCHY TYPE
155
1029. Find F+(z) and F-(z), and also the limiting values of
F::1:<C> on the contour of integration 0, if 0 is the circle !Cl= R, and
00
ef>(C) =a;+
.2
(ancosnO+bnsinnO)
n=l
is the uniformly convergent Fourier series of the real function
tp(O) = ef>(Re18 ).
1030. (1) Let 0 be the circle !Cl = n/2 and/(C) a function analytic
in the circle IC I <, n /2. Find the functions defined by the integrals
J
f
11 (z) = 21 .
f(C,) cot(C-z)dC,
ni c
1
f(C)dC
J 9 (z) =2 -.
---;---(,.-),
ni c sm .,-z
in domains, the points z of which possess the property that not
one of the points z+lcn(lc is an integer) lies on 0.
(2) Solve the problems formulated in part (1) on the assumption
that 0 is the circle !Cl = n.
1031. Let 0 be the segment [-1, l], traversed from left to right,
and ef>(C) 1. Find F(z) outside 0, the limiting values F::l:(C) and
the principal value of F(C) on 0. Evaluate in particular F(±i),F::1:(0)
and F(O).
1032. Let 0 be the semicircle ICI = R, 0 < arg C< n (commencing
at the point R) and ef>(C) 1. Find F(z) outside 0, the limiting
values F::l:(C) and the principal value of F(C) on 0. Evaluate, in
particular, F(O), F=(iR) and F(iR). Find also F'(O).
1033. Let 0 be the semicircle IC! = R, -n < arg C< 0 (commencing at the point R) and ef>(C) = 1. Find F(z) outside 0, the
limiting values of F1=(C) on 0, F(O) and F'(O).
1034. Let the density of a Cauchy type integral ef>(C) = 1gn. Find
F(z) outside 0, if the contour 0 is given by:
(1) The boundary of the ring r < !z! < R;
(2) The straight line Im C= n, traversed from left to right;
(3) The boundary of the strip [Imz[ < n;
(4) The semicircle !Cl = R, 0 < arg C< n (commencing at
the point R);
(5) The semicircle !Cl = R, -n < arg C< 0 (commencing at
the point R).
==
=
156
PROBLEMS ON OOMPLEX ANALYSIS
In parts (4) and (5) find the limiting values of F*(C) on 0 and
evaluate F(O).
In problems 1035-1040 find F(z) outside 0, assuming that the
contour 0 is an arc connecting the points a and b, and c/>(C) is the
given function.
1035. efJ(C) = I.
1036. t/J(C) = c.
1037. </J(C) = C".
00
J; c.. C", an integral function.
1038.
t/J(C) =
1039.
1040.
t/J(C) = l/(C-z0 ) (zoe 0).
t/J(C) = l/(C-z0)" (z0 e0). Evaluate, in particular,
n=O
F(z0 ).
HINT. Use the equality
<t>(C) = <t>(C)-<f>(z)
+ <t>(z)
•
C-z
C-z
C-z
1041. Find F+(z), F-(z) and the limiting values F*(C) if 0 is
the circle ICI = R and t/J(C) is the logarithmic function defined by
the conditions:
(1) c/>(C) = log C= log R + it/J, -n < tfJ ~ n;
(2) t/J(C) = Log C=log R + it/J,
O :,;;;; t/J < 2n.
HINT. Consider the contour consisting of the circle lzl = R with a cut
along the radius [-R, O] in the first case and [0, R] in the second.
REMARK. If the branch of the logarithm is not fixed in advance and its
continuous prolongation along the contour of integration is required, then
the integral will depend on the choice of the initial point of integration. See
also the examples given on page 65.
1042. Find F+(z), F-(z) and the limiting values F:l:(C) on 0, if
efJ(C) =log (C/(C-1)), and the contour 0 is:
(1) The circle ICI = R (R > l);
(2) The straight line Im C= 1, traversed from left to right.
1043. Find F(z) and the limiting values F± CC) on 0, if cf>(C)
=log (C/(C - 1)), and the contour 0 is the semicircle ICI = R(R > 1),
0 ~ arg C ~ n (commencing at the point R).
1044. Find F+(z) and F-(z), if t/J(C) = y (C) (0 :,;;;; arg yC < n)
and 0 is the circle ICI =I.
In problems 1045-1047 find F±(z), if 0 is a closed contour, the
points a and b lie within it and efJ(C) is a single-valued branch of
a many-valued function defined outside the cut which connects the
points a and b and lies within the contour 0.
1045. c/>(C) =Log ((C-a)/(C-b)) (Log 1 = 0).
157
INTEGRALS OF CAUCHY TYPE
1046. t/J(C) = y[(C-a)/(C-b)] (t/J(oo)=-'.-1).
HINT.
In order to find the integral along the contour surrounding the
out, expand
"CC) in a series of powers of _!_.
C-z
1047. <P<CJ = <C-aJ.1(C-bJ 1-A
C
(<f>(C)l
= 1)·
C-b c=oo
1048. Find F:l:(z) if the contour 0 is closed, the point a belongs
to the domain D+, the point b to the domain D- and efJ(C)
=Log [(C-a)/(C-b)] (Log 1=0) is the single-valued branch defined
outside the out connecting the points a and b and intersecting the
contour 0 at one point Co·
HINT. Join to 0 the cut along the arc C0 a, if z is in the domain D+, and
that along Cob if z is in the domain D-.
1049. Prove the validity of the equalities (0
< A<
1):
(ze [O, l]);
HINT. In order to obtain the first formula consider the Cauchy integral
l
23ii
where
f (c-1C )ii C-z
dC
'
c
(-C-)il
c-1 ill a function single-valued in the plane with the straight line
cut (0,1] and equal to lat oo, and 0 is the boundary of the doubly connected
domain: the disk !Cl < R (R > 1) with a cut along the segment [0,1]. The
second formula is obtained from the first by means of Sokhotskii's formulae.
1000. Evaluate the singular integral
1
_!_
n
J--JII( 1-t) t2+a dt
-1
I+t
t-x
<-1<x<1).
1051. Find the integrals:
1
(1)
J
I-t dt
log --t- t-z
0
(ze[O,l]);
158
PROBLEMS ON' COMPLEX ANALYSIS
1
(2)
f
1-t dt
log -t- t-•
(i:e (0, 1)).
0
HI"T· Use the method recommended in problem 1049 11ud the limit
( _t
-)..t-1
1-t
lim-----
A
..t-.0
t
= log-.
1-t
1052. Consider the singular integral
1
F(zt)=0
f
di:- -
(y = rx+i{J, 0 ~a < 1),
c
where 0 is an arc connecting the points a and b, t0 a point of thit:1
arc and (•- t 0 )Y is a single-valued branch in the plane with a cut
joining the points t0 and oo. If the point t0 coincides with one of
the ends of the arc 0, we shall consider that the cut extends along
the whole of the contour 0; if, however, the point t0 is internal, the
cut extends along the arc (t0 , b) of the contour 0.
Prove the following assertions:
(1) In the neighbourhood of the point a
'
23ti
(T-t0)11 (i:-z)
(zeO),
(t E 0),
where F 1 (z) is a function analytic in the neighbourhood of the
point a.
HINT. Consider the difference
I
F1(z)
= 2m
J
c
d-r
(-r-a)Y(-r-z) -
ell'"
2i sin y:ll' (z-a)-Y
and using Sokhotskii's formulae prove that F 1 (z) is an analytic function in
the neighbourhood of the point a.
(2) In the neighbourhood of the point b
e-11"'
F(z, b) = -- .,. .
-(z-b)-Y+FQ(z)
-ism 7'3t
~
(zeO),
159
INTEGRALS OF OAUOHY TYPE
Jl'(t,b) = -
co:ri
(t-b)-11 +F1 (t)
(teO),
whert• 1'2 (z) is a function analytic in the neighbourhood of the point b.
(3) In the neighbourhood of an interior point t0 of the
contour 0
F(z, t0) = (z-t0)-11 +F8 (z) on the left of 0,
on the right of 0,
F(t, t0 ) =
!
(t-t0 )-11 +F0 (z) if t e 0,
where F 3 (z), F,(z) and F 6 (z) are analytic in the neighbourhood of
the point t0 •
1053. Explain the behaviour of the Cauchy type integral
J
_1_
log-C- _dC __
2ni c
C-l C-z
close to the pointR z = -R and z = R if 0 is the semicircle ICI = R
(R > 1), in the upper half-plane (commencing at the point R).
HINT. See problem 1043.
For the behaviour of integrals of Cauchy type close to a singular
line see N. I. MusBKELIRHVILI, (1951), Singular Integral Equations (Bingidyarfll/l!/6 integral'nyye uratmemya) Chap. I, Gostekhizdat, or F. D. GAKHOV, (1958).
Boundary Problems (Krat'IJ1fY6 wdachi) Chap. I, Fizmatgiz.
REMARK.
§ 2. Some integral relations and double integrals
In the problems of this section the following notation is used: G is the
domain bounded by the contour C,
dz= tl:i:-idy = dz ,
(see page 55)
IOMt. Using Green's formula prove the following relations:
t For problems 1054-1057 see the supplement by M. Schiffer to the book:
R. CoUBANT, Dirichlet's Principle, Conformal Mapping and Minimal Burfaoea.
Interscience, New York, 1950.
160
PROBLEMS ON COMPLEX ANALYSIS
(1)
ff~£ dxdy =
G
(2)
j
CIZ
1
2i.
fdC +udf = 2i
(3) f(z) =
~
J
2m c
f
fdC;
C
£! (;{ -~~
JJ
_!_
f(C)dC _
C-z
)dxdy;
1'
G
of(£) ~d7J
oC C-z
(C = ;+iri).
aJ<Cl C-1 z = a ( 1m
)
C-z) , and apply
formula (1) to the domain G excluding the small circle IC-zl < e (fl-+ 0).
1055. Assuming that the functions f and g are continuously
differentiable in the closed region G, prove that the expression
fdz+gd.Z is the total differential of some function when and only
when
oc
llINT. Make use of the fact that
oc (
of
au
oz =-az·
1056. Prove that if f and g are functions analytic in the domain
G and continuously differentiable on 0, then
ff
f'udxdy =
G
! f fUd°E,
C
in particular
JJ
f'g'dxdy =
G
i
J
2C
fdg.
1057. Prove that ifthe functionf(z) conforma.lly maps the domain
G onto a domain G', bounded by a contour O' =f(O), then the
integral
1= 2i
Jfdf
c
is equal to the area 8 of the dome.in G'; if, however,f(z) conformally
maps the domain G onto the exterior of the contour O', then
l=-8.
161
INTEGRALS OF CAUCHY TYPE
IOOSt. Prove that if the function cf>(C) is defined and continuous
in the bounded closed region G, then the function
F(z)
=ff cf>(C)d~d-YJ
C-z
(C =
~+ifJ);
G
(1) Is analytic outside G and F(oo) = 0.
(2) Is continuous in the whole plane and
jF(z2)-F(z1)1
< Mlz2 --z1i llog IX iz2-ai1ll,
where M and IX a.re constants, and Zi and ai2 a.re arbitrary points
of the region if.
HINT. In (2), to estimate the integral
JJIC-~~~-Zil
make use of
G
the fact that this integral is obviously less than
ff
d~d17
IC - Z1~Z11 < R
IC-zal IC-z1I
where R is the radius of the circle with centre at the point (z1 +:i:a}/2, which
contains within it the closed region G. By a simple substitution we arrive
at the last integral in the form
ff
~<R
~d17
IC-al IC+al
2
=
ff
d~d17
~<R
IC-al
where a
lf+ai'
=
lzz-Z1I.
2
l>O
From geometrical considerations it is clear that
ff IC-ald~d17IC+al + a<IC-aJ<Rt'l
ff IC-ald~d17IC+al
ff ~'!L__ < IC-aJ<a
ICl<R IC-al IC+al
1>0
E>O
Noting that in the first integral IC+al > a and in the second IC+al >IC-al,
and passing to polar coordinates (C-a = rel<fl), we obtain the required
estimate.
1059. Let G be the circle izl
c/>(C) =
e2'1J
-1
< R '< 1
and
CC= ee18 , z =
re1"').
log-
e
t For problems 1058-1062 see I. N. VEKUA, GsnBralisBd .Analytic Functions
(ObobshchennyyB anaJitioh68kiys funktaii), Fizma.tgiz, 1959. English translation
published by Pergamon Press (1962).
162
PROBLEMS ON COMPLEX ANALYSIS
Find the function
F(z)
= -
_!_fj'
c/J(C)d~d'f/
:n:
C-z
(z E (})
G
and its partial deriva.tes oF/ox, oF /oy, oF/oz, oF /oz for z =F 0.
Show that of the given derivatives only oF/oz exists and is continuous at the coordinate origin.
1060. Prove that if the function f(z) is continuously differentiable,
then
I
of
. 2f jfdC
-=hm---•
Oz
ff d~d17
G
when the domain G shrinks to the point z.
HINT. Use the relation (1) of problem 1054.
1061. Let the function ef>(C) be continuous in the closed region
G. Prove that the function
F(z)
I
= --;-
ff
ef>(C)d~d'YJ
C-z
G
satisfies the equation
oF
oz
= ef>(z)
in the domain G if the derivative oF /o'i is defined by the formula.
of problem 1060.
1062. Prove that with the conditions of problem 1061 the genera.I
solution of the equation
of
oz= ef>(z)
in the domain G can be represented in the form
f(z) = ~
2:n:i
f f(C)dC
_ _!_ff ef>(C)~d'f/
C-z
:n:
C-z
C
G
HINT. Use the relation (3) of problem 1054.
(Pompeius's formula.).
163
INTEGRALS OF CAUCHY TYl'E
§ 3. Dirichlet,s integral, harmonic functions, the logarithmic potential
and Green,s function
The integral
D(u)
= JJ (u!+u~)da; dy
G
is known as Dirichlet's integral, and the integral
D(u, v)
= JJ (tixVx tu1v1 )da; d1t
G
is its corresponding bilinear form.
1063. Prove the following properties of Dirichlet's integral and
its corresponding bilinear form :
·
(1) D(u)
!
= £J (u:+ u~)dxdy
(z = x+iy
= re11fi);
(2) D(u) and D(u, v) are invariants in conformal mappings
of the domain G;
(3) .D2(u,v) ~ D(u)D(v) (the sign of equality holds only
for the case u/v = constant);
(4) If the function f(z) = u+iv is analytic in the domain G,
then
D(u) = D(v) = D(f)
=ff if'l dxdy.
2
G
In this case what is the geometrical meaning of D(f)?
1064. Using Green's formula
J J vLludxdy+D(u, v) = J v ~= ds
G
C
(n is the outward normal; LI is Laplace's operator), prove the
following properties of the harmonic functions u, u 1 , u 2 (v is the
function conjugate to u):
(I) D(u)
(2)
=
Ju ~u ds = { udv;
un
c
c•'
J:: = cJ
ds
c
dv = 0;
=
(3) If u1 = u 2 on 0, then u 1 u 2 in G;
(4) If OU1/on = a~/on on 0, then U1-U2 =constant in G.
164
:PROBLEMS ON COMPLEX ANALYSIS
1065. Prove that if u is harmonic in the circle izl
tinuous in the closed circle lzl ~ R, then
an
u(O)
<R
and con-
2~ Ju(Re' )d0 = n~s lzlJJu(re ")rdrd<f>.
=
9
1
<R
0
HINT. It follows from the equality
J:: dB = 0 that the integrallzl-rJ~
C
does not depend on ,. ; find its value by passing to the limit as ,. -+ O, and
then pass to the limit as ,. -+ R.
1066. Integrals of the form
JJe<C>
log IC 1 zl
G
J
c
~ d17,
a
1
P(C)Tnlog-IC-zl ldCI
(C = ~+i17, n is the normal to 0) are known respectively as the
logarithmic potential, the logarithmic potential of a simple layer and
the logarithmic potential of a double layer.
Verify the following equalities:
l_
(I) 2n
J
_ _l _
-!log 1! 1 , if lzl
log IC-zl dO1
o
loglf, if lz!
l
<2 > 7&R 2
2:r
fJ
log IC-zl
ICl<R
(3)
l
J a:
c
P(C)
d~d1J =
log IC 1 zj ldCI =
>R
<R
/log l!I , if lz!
<4 >
Ja
-a
> R,
1
1[
{lzl ) 2]
loglf+2 1- 1f
J
1
{ </>, if Im z > 0 ,
a'YJ log IC-zl ag = -</>, if Im z < o,
,
if izl <R;
P(C)d arg (C-z)
c
(in traversing 0 the normal n is taken on the left);
a
(C = Re19);
INTEGRALS OF CAUCHY TYI'E
165
where <P (0 < <P < 3t) is the angle subtended by the segment
(-a, a) at the point z (a > 0).
Gt'een'B junction g(a:, y, ~. 11> of the domain G (abbreviated to g(z, CJ,
z = a:+iy, C= ~+i1J) is defined as a harmonic function of both pairs ofvariables a:, y and~. fl• which is equal to zero on the boundary of the domain G.
and has a singularity at z = C, where
g(z,
Cl
lz~CI +
= log
(a harmonic function).
Green's function is symmetrical with respect to its arguments, that is, g(z,
g(C, z) (see, for example, (1, Chapter VI, § I]).
C>
=
1067. Formulate the Dirichlet problem for harmonic functions
which is equivalent to the finding of Green's function g(z, C).
1068. Let the function w = f(z, C) conformally map the simply
connected Jordan domain G onto the disk jwj < 1, so that f(C, C) = 0
(Ce G).
Prove the relations
g(z, C) = - log lf(z, C)j,
(1)
f(z, C) = e-<•+lh),
(2)
where h(21, C) is the harmonic function conjugate to g(z, C).
1069. Using the relation (1) of problem 1068 find Green's function
g(z, C) for the following domains:
(1) For the disk izl < R;
(2) For the half-plane Imz>O;
(3) For the strip 0 < Im z < 1.
1070. Prove the following assertions (n is the inward normal.
Yr is the circle Jz-CI = r):
(1) If u(z) is continuous near to z = C, then
lim
r....O
J
u(r) og(az, n_ds = 23tu(z) .
n
Yr
(2) If u(z) is continuously differentiable near to z = C, then
lim
r-+0
f
g(z, C) ou(z) ds
on
Yr
= o.
(3) If u(z) is harmonic in G and continuously differentiabfo
on 0, then
u(C) = - 1-
J
23t c
u(z)
og~, C) ds
c·n
(.;: e G).
166
PROBLEMS ON OOMPL'IDX .ANALYSIS
HINT, In the formula
f ( u~-gou)da
on on = J(u~-go")da
an on
C
l'r
pass to the limit as r -+ 0.
10'71. The kernel functions of tke domain
2 o"fl(z, C)
L(z. ') = -. n -Tz"&.--
are defined in terms of Green's function g(z, C).
Prove the following assertions :
(1) For the disk izl < R
1
L(z, C> = n(z-C)'.
(2) K(z, f> is a function of 21, [ analytic everywhere in D
and L(z, C) is a function of z, Canalytic everywhere in D except at
the point z = C, at which it has a pole of the second order,
1
L(z, C) = n(z-C)• +l(z, C) ,
where l(z, C) is a function of z, C which is analytic everywhere.
(3) K(z, C) = K(C, z), L(z, C) = L(C,z), l(z, C) = Z(C, z) (the
symmetry of Green's function g(z, C) = g(C, z) is considered known).
(4) Ifj(z) is an arbitrary analytic function in G, continuous
in the closed region G, then
(J(z), K(z, C))=
Jf K(C, z)f(z)dxdy =f(C)
G
(the "multiplicative property" of the kernel K(z, E}).
(5) With the same conditions
(f(z), L(z, C>) = - 2
n
I Ja_"g~z,_C>
G
ozoC
f(z) dxdy = 0
(the "orthogonality property" of the kernel L(z, C>)·
RBMAB:S:.
See the literature indicated in the remark to problem 1054.
167
THE INTEGRAL FORMULAE OF :POISSON AND SCHWARZ
§ 4. Poisson's integral, Schwarz's formula, harmonic measure
If the real function u(C) = u(R, 0) is defined and piecewise continuous on
the circumference C= RellJ (0 ~ 0 < 2n), then the PotaBOfl inlegral.
Ju(R,
2n
I
u(z) = u(r, rf>) = .2n
R•-r'
0) R•-.2Rr cos (0-4>)+r8 dO
(1)
0
defines in the disk lzl < R (z = rel</>) a harmonic function having at the points
of continuity of u(Cl boundary values equal to u(C):
limu(z) = u(C)
z-+C
(z-+ Calong any non•tangentia.I. path). The corresponding function/(z)
analytic in the disk lzl < R, is defined by Bchwarz's formula;
=u+w,
2n
/(z)
= .2~
J
u(C)
~: :
dO+iv(O)
0
(v(O) is an arbitrary real number).
1072. Prove the following assertions :
J
In
I
(I) 2n
R"-r8
Jll-2Rrcos(8-4>)+r8dr=l;
0
(2) u(r, cf>)-u(R, 00)
f
In
I
= 2n
R1 -r8
[u(R,O)-u(R,Oo)] Jll-2Rrcos (0-cf>)+rl d(J;
0
(3) If lu(R, 0)-u(R, 00 )1
9
I
J
.-n 111-0,1 <er
<e
for 18-00 1 < ct, then
Jll-r8
iu(R, 8)-u(R, 00)1 R 2_ 2R
(0-.1.) +-•- dO
r cos
'I'
r-
< e;
ct
(4) If 10-80 1 >ct and lcf>-0 0 1 < 9 , then
-
B1 -2Rroos (0-cf>)+rl > 4Rrsin2 :
(5) If lcf>-00 1 <
;
;
and the condition of pa.rt (3) is satisfied,
168
PROBLEMS ON COMPLEX ANALYSIS
then
iu(r,,P)-u(R,0 0)1
< e+
M(Jll-r8)
23tA -
where
f
2•
A= 43tRrsin2 : ,
M =
lu(R, 0 )-u(R, 00)id0.
0
1073. Prove that if
C+z
C-z
C=Re'', z = re1•, then
=
(R2-rl)+i2Rrsin (t/>-0)
R 8 -2Rrcos (O-,P)+r8
Using this identity obtain the following expansions:
r
2 (;r
co
u(z) = u(O)+
2 (;
(a. cos n,P+bn sin n,P),
n=l
co
v(z) = v(O)+
(-b,. cos ncf>+an sin ncf>),
n=l
co
f (z) = f(O)+
2 o.z•,
f (0) = u(O)+iv(O),
n=I
where
I•
a.=
!J
u(R, 0) cosnOdO,
bn =
! J••
u(R, 8) sin nOdO,
0
0
In
- - a.-ibn
a) -'"'dav.
OnR-- _I_f
3tRu (R ,ve
0
1074. Prove that for a harmonic function u(z) the Dirichlet
integral is given by
D(u) =
JJ(u~+u~)dxdy =
~<R
co
3t
~ n(a:+~)
n-1
THE INTEGRAL FORMULAE OF POISSON AND SOHW.A.BZ
169
(the coefficients a,, and b,. are defined in the preceding problem).
It is possible that both sides of the equation are equal to infinity.
HINT. Pass to polar coordinates (r, ~) and prove that for r
<R
00
Dr(u)
=
JJ(u~+u~)c:Wdy = 2; n(; f"c~+bl}.
:n;
izl<r
1t-I
1075. Prove that for the continuous function
the Dirichlet integral determines a function u(z) harmonic in the
circle lzl < R with boundary values u(R, 0) and an infinite Dirichlet
integral D(u) = oo.
1076. By means of Poisson's integral solve the exterior Dirichlet
problem for the circle lzl < R: find the function u(z) harmonic
in the domain lzl > R, regular at infinity and possessing given
boundary values u(C) on the circle !Cl = R. Determine the value of
u(oo).
HINT. Make the substitution Zi =
.Z:z
1077. Prove that for lzl > R, Schwarz's formula (see the introduction to § 4) defines an analytic function / 1 (z) = u 1 (z) + iv1 (z) regular
at infinity and that the following expansions hold:
--
/ 1 (z)
=
00
-!( ~) =
-/(0)-
2 c;:,
c_,, = R'"c,,,
1t=I
00
Ui<z> = -{uco> + 2( :)" (a,,cosn<P+b,,sinnef>+
n=I
00
v1 (z) = -v(O)+
2( !)"
(-b,,cosnef>+a.sinncf>),
n-1
where a,,, b,,, c,. are defined just as in problem 1073.
Here Re/1 (C) = - Ref(C) = -u(C), Imf1 (C) = Im/(C).
170
PROBLEMS ON' COMPLEX .ANALYSIS
lzl > R make the substitution z = ~
HINT. In Schwarz's formula for
/f on the
and make use of the fact that C = R 8
Z1
ICI = R.
In problems 1078-1083 find the functions f(z) (!zl < R) and
/ 1 (z) (lzl > R) defined by Schwarz's formula if u(C) is the given
function.
1078. (I) u(C) = Re [cp(C)+vi(C)],
(2) u(C) = Im [cp(C)+vi(C)],
where cp(z) is a function analytic for !zl :( R, and VJ(z) is a function
analytic for !zl ;;;:i: R.
1079. u(C) =Re
1080. u(C) = Re I/en.
circumference
en.
1081. u(C)
=
Re log CC 1 (R
>
1).
1082. u(C)=Re-i/(c C1 ){forC>l,-V(c C1 )>0, R>l)·
1083. u(~) = Re log C.
1084. Prove that Schwarz's formula can be written in the form
f(z) = _.;.
ni
J
ICl=R
u(C)dC -f(O).
C-z
HINT. Use the equality
C+z
2
C<C-zl = C-z
1
--r·
1085. Obtain formulae similar to Poisson's integral and Schwarz's
formula for the upper half-plane Im z > 0, that is, express the
harmonic function u(z) and the analytic function f(z) = u(z)+iv(z)
in terms of u(t) (-oo < t < oo).
HINT. Make use of the conformal mapping of the half-plane onto the
circle.
1086. Derive Schwarz's formula for the strip 0
HINT.
< Im z <
1.
Use the conformal mapping of the strip onto the half-plane.
The harmonic measurs w(z, oi:, G) of the boundary arc oi: at the point z with
respect to the domain G is defined as a bounded function, harmonic in G, which
is equal to 1 at the interior points of the a.re oi: and to 0 at the interior points
of the remaining pa.rt of the boundary. The harmonic measure w(z, oi:, G)
is an invariant in conformal mappings.
TH'E INTEGRAL FORMULAE OF POISSON AND SCHWARZ
171
In problems 1087-1090 the domain G is the circle lzl < 1 and
ro(z, (Ji, 02 ) is the harmonic measure of the arc IX= (0 1 , 02 ), w = e19,
81::::;;; (J::::;;; 82.
1087. Using Poisson's integral prove that
o,
1
l-r2
ru(z,0 1• 82)= 2n
l-2rcos(O-<f>)+r2d0,
o,
and in particular, that c.o(O, 01 , 02 ) = (08 -01)/2n.
1088. Find the level lines of the function dc.o (z, 01 , 0) /d(J for a fixed
value of 6 (z = re'• is a variable point).
J
HINT. Prove that
~: = 2~ I::=: I·
where w' is the end of the chord from w which passes through z.
1089. Let us denote by w' the end of the chord from w which
passes through the point z. Let IX be the arc (0 1 , 61), and IX'(z)
the arc described by the point w', when w traverses the arc IX. Prove
that the length of the arc IX'(z) equals 2nc.o(z, Ov 02).
1090. Find the level lines of the harmonic measure c.o(z, 01 , 01 )
of the arc (01 , 02). Using this prove that the integral defining the
harmonic measure (see problem 1087), actually has the limiting
values 1 on (61 , 02 ) and 0 on the complement (interior points of the
arc are considered).
1091. For the half-plane Im z > 0 determine the harmonic measure
c.o(z, a, b) of the segment (a, b), of the ray ( -oo, b) and of the ray
(a, oo). What is the geometrical meaning of these harmonic measures 1
1092. Find the harmonic measures of the boundary rays of the
angular sector.
O<argz<y.
1093. For the semicircle lzl < R, Im z > 0 find the harmonic
measures of the diameter LI and of the semicircu]ar arc I', and also
the level lines of these harmonic measures.
1094. Find the harmonic measure of the bounding semicircular
arc I' of the domain lzl > R, Im z > 0.
1095. Find the harmonic measure of the boundary circular arc
I' of the domain lzl > R, 0 < arg z < 2n.
1096. Find the harmonic measure of the boundary circumferences
of the annulus r < lzl < R.
172
PROBLEMS ON COMPLEX ANALYSIS
In problems 109'7-1101 the domain G is bounded by a compound
contour I', consisting of n simple, smooth contours I',.(v = 1, 2, ... , n).
The contour is traversed in the positive direction with respect to
the domain G; the norm.al n is inward with respect to the domain G.
The period with respect to I',. of a function analytic in G is defined
as the integral
Jdf(z)
r,,
(see page 16).
109'7t. Prove that if the harmonic function u(z) is single valued
in G, then the period of the analytic function /(z) = u(z) +w(z)
along
is equal to
r.
-f~:da.
r,.
1098. Prove that for the complex Green function g + ik of the
domain G (g(z, C) is the Green function of the domain G, k(z, C)
its harmonic conjugate) the period along
equals 2mo,.(z), where
m,,(z) is the harmonic measure of I',. with respect to the domain G.
Prove that
r.
2" w,.(C) = 1.
HINT. The function u(C) harmonic in G can be represented in the form
u(C) = - 1-
Ju (z) (Jg(z, t') da
2nr
an
(llff problem 1070).
l•.
Express in terms of w,,(z) the function u(z) which is bounded,
harmonic in G and has constant values c,, on the I',,('1' = 1, 2, ... , n).
1100. Prove that for a function v(z) which is conjugate to a function
u(z), single valued and harmonic in G, the periods p .. along
can be represented in the form
r.
p,, = -
f ()aw..
r
(z)
..3-
u z - 0- w s .
n
t For problems 1097-1104 see the references indicated in the remarks
to problem 1014.
THE INTEGRAL FORMULAE OF POISSON AND SOHWARZ
173
1101. Let w,.(z) be the harmonic function conjugate to ro.(z)
and p"., the periods of the function w,.(z) = ro,.(z)+iw.,(z) along I'".
(1) Prove that
(µ, 71 = 1, 2, ••• 1 11.) •
'P/JY ='PY/I
HINT. Use the representation
p"., =
__1_
J
wp
2n r
o(J)y da.
0n
(2) Prove that
"
2
•-I
tppy
(µ = 1, 2, ... , n) •
= 0
1102. Let C.(71 = 1, 2, ... , n) be arbitrary real numbers. Prove
that if p." are the numbers defined in problem 1101, then the quadratic form
2"
... "
p.,pC.Op
~0
1
the case of equality occurring only when allthe "• are equal.
llDrT. Apply the formula
D(w) = -
f
OJ :
da
r
to the harmonic function w(z) ...
I" C,,OJy (1)
(see problem 1084; the afgn is
..-1
ravened because n is now the inward normal).
1103. Prove that the quadratic form
is positive definite, that is, positive for all sets of values {c.}, excluding 0i = 0 11 = ••• = o...1 = 0.
116'. Prove that the system of equations
11-I
2 p.,".A" = B"
.. -1
(µ = 1, 2, ... , n - 1)
174
PROBLEMS ON COMPLEX ANALYt!!IS
(the A,, a.re unknowns) has a. unique solution for any B,.. Using this,
prove that for any function u(z) harmonic in G, generally speaking
not single-valued, it is possible to select constants A 1 , A 2 , ... , A 11_ 1
in such a. way that the harmonic function
11-l
u 1 (z) = u(z)+}; A.w.(z)
v-1
is single-valued in G.
§ 5. Some singular integralst
Let h(of>) be an integrable function with period 231; I.Jo, IJn, b11 (n = 1, 2, ... )
its Fourier coefficients; J(z) = u(z)+it1(z) and J 1 (z) = u 1 (z)+it11 (z) functions
analytic respectively in the disk lzl < 1 and in the domain lzl > 1, defined
by means of Bchwa.rz's integral
an
J
..n
elt+z
1
c;-
h(t)-1-,- d t
e -z
(z
= rellf>).
(1)
0
The function h(•), defined by means of a singular integral with the Hilbert
kernel - 1- cot !-t ·
2n
2 ·
~
h(of>) = T[h(of>)]
J...
1
~ !n
J + JIn) ,
cf>-t
l
(.__.
h(t) cot2 -dt = !n ~
0
0
(2)
·~
is said to be conjugate to the function h(cf>). The series
00
}; (-b11COSf'lcf>-l-a11 sinnof>)
(3)
11=1
is said to be conjugate to the Fourier series
-i +};
00
(a11cosnof>+b11 sinncf>)
(4)
n-1
of the function h(cf>).
1105. Prove the following assertions:
(1) The principal value of Schwa.rz's integral (1) at the
point e1•, if it exists, is equal to ih(</J).
(2) If k(</J) = const, then h(</J) = 0.
t Problems 1119-1118 and 1142-1143 on the approximate evaluation of
singular integrals were composed by B. A. Vertgeim.
175
SINGULAR INTEGRALS
(3) If k(4') eLip1X(O <IX~ 1) (see page 152), then h(tP)
exists and is continuous. In addition, if IX < 1, then k(4') e Lip ex.
HINT.
Write h(</I) in the form
IK
b(+)
M
=
1
2n
f [l•(l)-h(</l)]cot-2-dl = - 2 f [h(<fl+t)-h(</l-t)]cot 2' dl.
n
<fl-t
1n
0
0
1106. Let k(tP) be continuous and have a piecewise continuous
derivative k'(q,). Prove the following assertions:
(1) h(tP) = limf(re1"') = lim f 1 (re1"') (the definition of the
r-1
r-+1
functions f(z) and f 1 (z) are given at the beginning of the section).
HrNT. Show that
00
(2) The function f(z)
= a2° + 2
cnz• (en
n=l
= an-ibn)
is e.on-
tinuous in the closed disk lzl ,_;;; 1 and f(e'"') = k(tP)+ik(tP).
HINT. With the conditions of the problem an
00
=
0 ( :. } , b.,.
=
0 ( :. ) •
-
(3) The function f 1 (z) = -- } ; :: is continuous in the
domain lzl ~ 1 and / 1 (e'"')
n-1
= -
k(tP) +
ik (<fa).
00
(4) h(tP)
verges uniformly.
=}; (-bn cos n<P+an sin n<P) and the series
con-
n-1
Bn
(5)
f h(tP)dtP =
0.
0
1107. Let 1J1(Z) be an analytic function in the disk lzJ ~ 1 and
1J11 (z) an analytic function in the domain izl ~ 1. Prove the following
properties of the transformation (2) :
(1) T[Re 1JI (e1"')] = Im 1Jl(e1<f>) - Im 1J1(0);
(2) T[Im 1JI (e1<fl)] = -Re 1Jl(e1"')+Re 1J1(0);
(3) T[Re 1J11 (e1"')] = - Im 1J11 (e'"')+Im 1J11 (oo);
(4) T[Im 1J11 (e1"')] =Re 1J11 (e'"')-Re 1J11 (oo).
176
l'ROBLEMS ON COMl'LEX ANALYSIS
In problems 1108-1116 find the function k(<f>) from the given function k(<f>).
1108. cos n<f>.
1109. sin n<f>.
1-acos q,
1110. 1- 2acos <P+ a1
(-l<a<l).
1111. log (l-2 a cos q, + a 1) (-1<a<1).
1112. cos•• <P (m is a natural number).
1113.
1114.
1115.
1116.
1117.
sin (n+i><P
.
2 sin l<P (n is a natural number).
'"'' (-:n: ~ "' ~ :n:).
"' (-:n: ~ "' ~ :n:).
"' (0 ~ "' ~ 2:n:).
Prove the following assertions:
(1) If k(<f>) is an even (odd) function, then k(<f>) is an odd
(even) function.
(2) If k(<f>+ro) = k(<f>), then k(<f>+ ro) = k(<f>).
(3) If k(<f>+ro) = -k(<f>), then k(t/>+ro) = -k(<f>).
1118. Let F(t, </>), k(<f>) and kt(<f>) be periodic functions (F(t, t/>)
with respect to both variables) with period 2:n:, belonging to the
class Lip ct (0 <ct ~ 1).
Prove the following assertions:
llll
(1)
llll
I•
JJ
F(t, <f>) cot q, 2 t dt =
d<f>
0
0
In
JJ
2
•-I
F(I, <P> cot
dt
0
0
00
= 2:n:8
where
A.=
~
(A 8 -B11) ,
I• I•
JJFcosntsinn<f>dtd<f>,
0 0
B 11 =
~1
••••
ff
0 0
F sin nt cos n<f>dt dq,;
t/> 2 t dtf>
177
SINGULAR INTEGRALS
In
2's
J{2~ J
f { 2~ f
(2)
k(t) cot </> 2
0
~
=
t dl}k,.(</>)def>
0
~
~
k1 (</>) cot </> 2 t d<f>}k(t)dl
0
=n
0
2 (anb~1)_bna~1)),
n=l
where (an, bn) and (a~1>, b~1>) are the Fourier coefficients of the
functions k(</>) and k1 (</>);
(3) 21n
I
1
"{
1
2n
0
fl>I k(t)cotT-t
</>-T
-dl} cot-dT
2
2
0
= -k(</>)+
or, if
k=
2~
111:
f
k(tf>) def>
0
T(k),
In
T[T(k)] = T(i) = -k(ef>)+
2~ f k(t/>)dtf>.
0
1119. Applying the rectangle formula for the approximate evaluation of the singular integral
obtain the formula
(5)
where
k1:,n = k
nk) • -k1:,n = k-(2nk)
(2n
7
(k=O,
±1, ±2, ... ).
In particular, if n = 5 and
k1;
=k(7~).
- -(nk)
k1;=k10,
178
PROBLEMS ON' OOMPLEX ANALYSIS
then
-
I[
hk = -- 10 (h11:+i-hk
1)
3n
n
cot 20 +(krc+s-hk-a) cot 00
5n
7n
+(hrc+6 -k1 _ 6) cot 00 +(h1:+7-h1:_7 ) cot 00
+ (h1:+u-hrc_ 9) cot
:~ J·
(6)
1
t
1120. Let (l>(t) = - 231: [h(cf>+t)-h(cf>-t)]cot 2 •
Assuming that h(</>) can be continuously differentiated three times,
prove that
where
1}Irc = max jh,<l:>(cf>)I
(k = 1, 3).
o.;;q,.;;;n
Using this and the estimate of error Rn of the rectangle formula
for the calculation of k(c/>):
Rn= (b-ar (l>"(E),
24n
a,.,_;; E ~ b
(a=
o, b = n),
obtain the estimate
IRnl
~ 2~:1 [Mi+(;+ :)Ma]·
(7)
In particular, if n = 5
IR!< l ·65X l0- 2{M1 +7 ·6U·f3 ).
(8)
HINT. Expand 4'(t) by Taylor's formula with remainder term in Lagrange'11
form (with h'") and use the inequalities
a: cot a: .:;;; I,
and
a:Bcosa: ~ 1 fi
sins a: """
or
1-a: cot a:
.
sm1 a:
~
1
179
SINGULAR INTEGRALS
1121. Prove that if k(<f>+n) = k(<f>), then formula. (6) (see problem
1119) acquires the form
~ + (k1c+ -k1c-a) cot ~~].
k1c = - 110 [ (k1c+ 1 -k1c- 1 )2 cot
and if k(<f>+n) =
(9)
8
-k(</>), then
In problems 1122-1128 calculate the values of h1c to five significant
figures by one of the formulae (6), (9) or (10) and compare with
the exact values. Also estimate the errors by formula. (8) and compare
with the actual errors.
1125. k(</>) = sin <f>.
1122. 'k(</>) = cos 2<f>.
1126. k(</>) = sin 3<f>.
1123. k(</>) = cos 3</>.
1124. k(</>) = cos 4cp.
1127. k(</>) = <1>( 1</>,
1128. k(</>)
= { _,/,.
n '""
~ ).
0
if 0 ~
if /2
n
~ </> ~ n
</>
~
and k(-</>)
n/2,
,/,. ./'
=
-k(</>).
k(-</>) = -k(</>).
~"'""'n,
REMARK. In problems 1H'7, 1128 compare the result with calculations,
based on the expansion of h(</I) in a Fourier series.
In problems 1129, 1130 singular integrals a.re considered of the
form
f
00
h(x) = T[k(x)] = _
_!...
n
-oo
k(t)dt •
t-x
(11)
connected with Schwa.rz's integral for the half-plane (see problem
1085)
f
00
__!_
ni
-oo
k(t)dt •
t-z
(12)
180
PROBLEMS ON COMPLEX ANAL"!SlS
If the integral in the usual sense diverges, its principal value must
be considered. The function h = T(k) is said to be conjugate to
the function k.
1129. Prove the following assertions:
(1) The principal value of the Schwarz integral (12) at the
point x, if it exists, is equal to ih(x).
(2) If k(x) = const, then h(x)
0, if the principal value
of the integral (11) is considered not only at the point x, but also
at oo;
=
oo
.x-e
R
f /'x =eJJ + J)=
-oo
-o
(3) If k(x) e LiplX (0
.x+•
-R
<IX : : ; ; 1)
0·
on any finite interval
and the integral (7) converges on ±oo (e.g., if lk(x)I =
o(
!xii')'
> 0, for large !xi), then h(x) exists and can be expressed by the
formulae
{J
J
00
h(x)
= _ _!_
n
-oo
00
k(t)-k(x) dt =
t-x
_ _!_J
n
0
k(x+t)-k(x-t) dt. (IS)
t
(4) Prove that with the conditions of section (3) h(x) is continuous and in addition, if IX< 1, h(x) e LiplX.
(5) If k(-x) = ± k(x), then h(-x) = =Fk(x).
(6) If k(x+ro) = ± k(x), then h(x+ro) = =F k(x).
1130. Let /(z) and fi(z) be analytic functions defined by the
Schwarz integral (12) in the half-planes Im z > 0 and Im z < 0
respectively. Prove that if k (x)e LiplX (0 <IX ~ 1) uniformly
on the x-axis and lk(x)I = 0 ( l:l' ).
(1)
fJ > 0,
at oo, then
f(Z) (Im z < 0);
limf(x+iy) = k(x)+ih(x)
,,_o
f 1 (z) = -
(2) /(x)
/ 1 (x)
=
= limf1 (x+iy) = 7-+0
k(x)+ik(x)
(y
> 0),
(y
< 0).
181
SINGULAR INTEGRALS
HINT. In order to estimate the difference
use the fact that
Re/(:i:+iy)-h(:i:) as y-. 0
co
and divide the
x-1
x+I
J=
integral
Re/(:i:+iy)-h(:i:)
-co
co
f +f+f.
-co
x-1 x+ I
To
into
the
integrals
co
estimate
the
J=
integral
co
I
J+ J
(see (13)) subdivide it into the integrals
0
Im/(:i:+iy)-h(:i:)
0
and in the first of these
I
make the substitution e = wy.
Rmu.Rx:. The assertion of the problem oa.n be significantly strengthened.
1131. (1) Let <f>(z) and 'J'(Z) be analytic functions in the half-planes
Im z > 0 and Im z < 0 respectively, bounded at oo and continuous
in the closed half-planes. Prove that if <f>(x) and 'l'(x) e Lip ix (0 < oi~l)
and there exist transformations T[</>(x)], T['J'(z)] (T(ki, +ik8)
= T(k1 )+iT("9)) and the limits
n
lim
R-+00
_!_
f
no
n
</>(Re11)d0 =
</>"',
lim
R-+00
_!_
f
no
'J'(Re- 11)d0 = "'"',
then
T[<f>(x)] = -i[<f>(x)-</>"'],
HINT. Let </l(z)
= u(z)+iv(z)
T['J'(x)] = i['J'(Z)-'J'"'].
and "'"'
of problems 1085 and 1129, we have (z
= "'"'+"'"''
= :i:+iy)
Applying the results
J u(e)-u(:i:)
-00
.
.
v(:i:) = lim[~(s)-u(:i:)] =Jim
{
I
,....0
y-+11
:Tr -oo
}
de +""''
·-·
J u(t)-u(:i:)
00
!l'[u(:i:)]
= _ _!_
:Tr
-oo
de.
t-:i:
Estimating the difference of integrals it is possible to obtain
J u(t)dt
e-:i: = v(:i:)-v"''
00
!l'[u(:i:)]
=
--nI
-co
which reduces to the first proved relation. The remaining one is proved similarly.
182
PROBLEMS ON OOMPLEX .ANALYSIS
(2) In the conditions of part (1) let efJ(z) and 1jl(z) have
respectively the isolated singularities a (Im a > 0) and b (Im b < 0).
Find T[efJ(x)] and T[1j1(x)].
In problems 1132-1137 find h(x), /(z) and / 1 (z) for the following
functions k(x) (the parameters A., a, b are real numbers, A. > 0,
b > 0, n is a natural number):
1132. COB Ax.
1
1135. l+x' ·
1133. sin A.a:.
x
1136. Re l/[x-(a+ib)]•.
1137. Re 1/[x-(a-ib)]•.
1134. xl+bl .
1138. Prove that
00
-!J
dt
tA(t-x) =
1_
0
cotA.n
---;r-'
1
l:tl"sinm'
HINT, Consider the integral of l/[C"(C-111)) along the boundary of the
annulus r < !Cl < R (deleting the circle IC-a:I < e. If a:> O) with a. cut
along the re.dial segment r <Re C= t < R, Im C= 0 (or the remaining
pa.rt of it if a: > 0) and pass to the limits e.s r -+ 0, R -+ oo (and e -+ O, if
a: > 0). On the out CA = tA above and C" = e.\e8'1U below.
1139. Prove that
- : Joo
,.
(:t ¥= a, a
a
> 0,
dt
t"(t-x)
= _1
log 1~ -11 +-1 ~-1 (~)k
nx8
a
L.J n-k x
'JW,n
k-l
n is a natural number). For
be represented in the form -
i
!xi < a the integral can
~ k-o ___..!:_k
(~)k. Both expansions
n+ a
'j(,Q,
are also valid for x = -a.
HINT,
Use the expansion
I
"(I-a:)
=-
1 ( 1
a:"
:r
a:n-1 )
1
,-+,.+ ... +-en- + :i:"{e-:r.) •
183
SINGULAR INTEGRALS
1140. Prove that
co
_ cot«n _ _
1_
:if-
'j(,Q,fl.
~-1-(~)k
.L.i k-a.
x
•
if x>a,
k-l
if
~-1- ~),
na« .L.i k+cc a
co
1--
(
k
x< -a,
if lxl<a
k-0
(a> 0, cc> 0 are not integers). Both the latter expansions a.re
valid also for x = -a, where the function k(x) defined by the integral
standing on the left remains continuous.
HINT. For la:!> a consider the integral of l/[Co'(C-111)) along the bound·
ary of the ring a < IC! < R with a out along the re.dial segment a < Re C
= t < R, Im C= 0 (and deleting the circle IC-a:! < (/, if a: > a), and pass
to the limit as R-+ oo (and (!-+ 0, if a:> a). For la:I <a expand the
integrand in powers of a:/t. The latter assertion is established by means of
Abel's theorem on power series (for the latter expansion) and by means of
the expansion of I/sin or:n in partial fractions (see problem 998).
RllMABK, If 0 < or: < I, then for la:I > a it is possible to use the integral
•
J
•
"o
of problem 1111, from which it is necessary to subtract the integral _..!._
evaluated by expanding the integrand into a series of powers of t/a:. The
resulting expression retains its meaning (analyticity) also for or: ;;;i. I, and
hence by analytic continuation, it is valid also for all or; > O.
1141. Prove that if k(x) = 0 outside the interval (-a, a), then
for !xi> a
a
h(x) =
_..!_
n
n
j' k(t)dt
= ~ Ak +Rn(x),
t-x
.L.i x
-a
k-l
where
a
"
Ai=
! f et-
1 k(t)dt,
-a
IR
n
I<
1
1
lxl~(lxl-a) :re
f
ltlnlk(x)I dt.
-a
Find k(x) in explicit form, if k(x) = 0 outside the interval (-1,1)
andintheinterval (-1,1) itequalsrespectively 1, 1-lxl, signx, and
compare the result with the expansion in powers of l/x given above.
184
PROBLEMS ON OOMPLEX ANALYSIS
For an approximate ca.lcula.tion of h(a:) for sma.11 va.lues of a:, la:I < a, the
integrals a.long the infinite intervals (- oo, -b), (b, oo) and a.long the interva.1
(- b, b), where b >a a.re considered separately. For the interva.1 (-b, b)
the formula
J e-a:lk = _..:_ .L..J'1
n-1
b
_ _!_
h(e)dt
:ri
:ri
i=-n
-b
h 81+1
+Rak
2i-21c+I
can be obtained from the rectangle formula. (here Xk
= Tc:,,,;
'Tc
=
0,
± 1,
=
hk; Rak is the remainder). A simila.r formula. is obtained
for the points Xak+i· To ca.lcule.te hk = k(a:k) we take la:kl ~a. The integra.1
a.long the infinite interva.1 (b, oo) is eva.lue.ted by means of the series expansion
±2, •.• , ±2"'; h(a:k)
where
!J
00
ok = -
e-k-1 h(e)dt.
b
The integral a.long the infinite interva.1 (- oo, -b) is treated simila.rly.
1142. Evaluate h(x) for k(x) = e--¥' at the points x = l, 2, 3, 4, 5.
Take a = 5, b = 6, n = 15. The calculation is to be taken to five
significant figures. Show that the integrals along the infinite intervals can be neglected. For comparison evaluate h(x) at the points
x = 2, 3, 4, 5, by the asymptotic expansion given in problem 946
(the terms of the series are to be used as long as they decrease).
1143. Evaluate h(x) for k(x) = l/(l+x2) at the points x = 1,
x = 5, taking b =IO, n = 25. Compare the results with the exact
values.
In problems 1144-1147 singular integrals are considered of the
form
k(T) = T[k(T)] = - 2In
N
J
-
t-T
k(t) coth - 2 - dt
-+
t-T
t-TJ
= - 2ln Joo [ k(t)coth2--k(t+ni)tanh2 - dt,
-00
(14)
SINGULAR INTEGRALS
185
connected with Schwarz's integral for the strip 0 <Im z < n (see
problem 1086), and
1
2ni
(~denotes
J
t-z
h(t) coth ~ dt,
-
....
(15)
the boundary of the strip).
The function h = T(h) is said to be conjugate to the function k.
1144. Prove the following assertions:
(1) The principal value of the integral (15) at the boundary
point T of the strip 0 <Im z < n, if it exists, equals ih(-r).
const, then h 0.
(2) If h
(3) If h(t) e Lip a (0 < a ::=:;; 1) on any finite interval and
the integral (9) converges to ± oo (for example, if k(t) is bounded
=
=
00
and
J lh(t)-h(t+ni)jdt <
-oo
~
= - 21n
h(-r)
oo), then h(-r) exists:
f
-....
t-T
[h(t)-h(-r)] coth2- dt
and is continuous (the stronger assertion: h(-r) e Lip a, if 0 < a < 1).
(4) If h(t+ni) = ± h(t), then h(t+ni) = =i= h(t).
(5) If h(t+w) = ± h(t), then h(t+w) = ± h(t) (w a real.
number).
1145. Let f(z) and f 1 (z) be analytic functions defined by Schwarz's
integral (15) in the strips 0 < Im z < n and -n < Im z < 0 respectively.
Prove that:
(1) f 1 (z) = -f(z) (-n < Imz < 0).
(2) Schwarz's integral (15) defines in each of the strips
kn< Im z < (k+l)n (k = 0, ± 1, ± 2, ... )an analytic function such
that if F(z) is the value of the integral (15) at the point z and
kn< Im z < (k+l)n, then F(z) = f(z-kni), if k is even, and
F(z)=f[2-(k+l)ni], if k is odd.
(3) With the additional assumptions that the function
h(t) is bounded, belongs to Lip a (0 <a::=:;; 1) uniformly on the
boundary of the strip and
00
J lh(t)-h(t+ni)j dt < oo:
-oo
186
PROBLEMS ON OOMPLEX .ANALYSIS
we have
f(t) = limf(z) = h.(t)+ih(t),
=
(Im t
0, Im t
= n) .
1146. (1) Let cf>(z) and 1Jl(z) be functions analytic respectively in
the strips 0 < Im z < n and -n < Im z < 0, bounded at ± oo
and continuous up to and including the edges of the strips. Prove,
that if cf>(t) and 1/'(t) (Im t = 0, Im t = n) belong to Lip cc (0 <cc ~ 1)
and there exist the transformations T[cf>(t)], T[1j1(t)] and the limits
n
n
fun ..!..Jc/>(x-t-iy)dy = cf>±a>,
X-+:1::00 1/: O
Jim ..!..J'l'(x-iy)dy = 1/'±rr,,,
n
X-+:1::00
O
then
T[cf>(t)]
= -{c/>(t)-
c/><Xl~c/>-a>].
T[1J1(t)] =
{1/'~)- 1/'oo~'l'-- 00 }
(2) Find T[c/>(t)] and T[1J1(~] (Im t = 0, Im t = n), if the
functions cf>(z) and 1/'(z) have poles at the point a (0 < Im a < n)
and at the point b (-n < Im b < 0) respectively and elsewhere
satisfy the conditions of part (1).
1147. Find functions defined by Schwarz's integral (15) in the
strip 0 < Im z < n for the following functions k(t):
(1) k(t) = Re cf>(t), cf>(z) is from problem 1146 (1);
(2) k(t) = Re 1/'(t), 1J1(z) is from problem 1146 (1);
(3) k(t) = Re cf>(t), cf>(z) is from problem 1146 (2);
(4) k(t) =Re 1f'(t), 1/'(z) is from problem 1146 (2);
(5) k(t) =Re 1/(t-a), 0 <Im a< n;
(6) k(t) = Re coth (t-a), 0 < Im a < n.
We consider below the composition of singular integrals with the Cauchy
kernel l/(l;-z). A fundamental position is occupied by the Poincar6-Bertrand
formula:
f ~ cf
c
11-7:
F(l;, 11) dl;
l;-a
=
-nl F(T T)
' +
f f
c
dl;
c
F(l;, 11) d11
(a-T) (~-a)'
187
SINGULAR INTEGRALS
where 0 is a smooth contour, closed or open (in the J.a.tter case a point Te 0
differs from an end point), the function F(C, a) e Lip et with respect to C and
a and if 0 goes off to infinity, decreases sufficiently rapidly at infinity (see the
book: N. I. MusKHELISBVILI, Singular Integral Equations (Singulyamwe
integral'nwe uravneniya), § 23, Gostekhizdat, 1946 or I<'. D. GAKBOV, Boundary
Probkma (Kraevyye zadachi) Chapter I, § 7, Fizmatgiz, 1958).
In problems 1148-1164 verify the Poinca.rlS-Bertra.nd formula.
for the given functions F(l;, a) (l; = x+iy) by the actual evaluation
of both sides of it. In these problems: 0 1 is the closed contour
enclosing the origin of coordinates, 0 2 is the horizontal a.xis, 0 3
is the segment [O, 1] and 0 4 is the semicircle ICI = R, Im l; ~ 0,
n = 0, 1, 2, ...
1148. t;n, the contour 0 1 •
1149. 1/t;n, the contour 0 1 •
1150. an, the contour f!11.
1151. 1/an, the contour 01"
1152. l;"a", the contour 0 1 (p, q = 0, ± 1, ± 2, ... ).
1153. 1/(x+bi), the contour 0 2 •
1154. cos A.x cos µa (A. > 0, µ ~ 0), the contour 0 2 •
cos A.x
1155. --b-. (A. > 0), the contour 0 2 •
a+
i
1156. x, the contour 0 3 •
llmT. Make use of the formula
1
I
J
0
1-z dz
1 (
t -) - x1
log--- - =log.
x
x--t
2
1-t
2
See problem 1051 (1).
1157. xa, the contour 0 3 •
1158. xn, the contour 03.
In problems 1159-1164 make use of the integrals:
f ~=logR+r
R-T
f
c,
• O'-T
C:,
R+a
log R-a da = 0,
f
c,
=logjR+rj+i~,
R-T
1
2
R+a
:i21
-log--da= - - ,
a
R-a
2
R+a
1 [log
j •lo g
- -da
- -_- R-a a-T
2
2
R+T
R-T
2]
---~
c,
(for the last equality see the hint to problem 1051).
188
PBOBLEMS ON COMPLEX ANALYSIS
1162. C/u, the contour 0 4•
1163. ufC, the contour
1164. l/<18, the contour
1159. 1, the contour 0 4 •
1160. C, the contour 0 4 •
1161. Cl, the contour
a,.
a,.
a,.
Singular integrals of the form
•w=-~ff°'m~~
:ri
(C-z)•
<~
E
are considered below, where C= e+lf'/, extending to the entire C-plane
and defined as the principal value
Sw(z) = lim e.....O
~
:ri
1~
f£
•-Z >I!
w(C) d;df'/ •
<C-z)•
(17)
The transformation SOJ is closely connected with the integral transformation
Tw(z)
= -~ff °'(C)~df'/.
:ri
(18)
C-z
E
See problems 1058-1062.
1165. Prove that if ro(C) e Lip o: (0 < o: ~ 1) and the integral
f Iro (C) IdE dfJ < oo, then the integral (16) exists and
J
JJ
E
Sro(z) = _
_!_
:n;
IC-zJ >R
HINT. Prove that
ro(C)dEd17 _
(" -z)•
_!_
:n;
C.
ff
e<IC-zl<R
d;df'/
(C-z)•
ff
ro(C)-ro(z) dEdfJ.
IC-zl <R
(" -z)•
C.
= 0.
1166. Let ro(z) e og (twice continuously differentiable and equal
to zero outside a compact set). Prove that
(Tro),, = Sro, (Tro)s = ro
(see problem 1061), where the partial derivatives with respect to
z and denote as usual.
z
1167. The functions w(z) which effect the quasi-conformal mapping
transforming the infinitesimal ellipses
Edx2+2Fdxdy+Gdy2 = &•
189
SINGULAR INTEGRALS
or, in a different notation,
l&+q(z)dii 2 = ds8,
lq(z)i
<
1,
into infinitesimal circles, satisfy the equation
W~= q(~)W:r:,
and conversely (see problem 378). Show that if a solution of this
equation is sought in the form
Tc.o(z)+~(~),
w(z) =
where ~(z) is an arbitrary integral function, and c.o (z) is an unknown
function, then for the definition of c.o (z) there is obtained the singular
integral equation
c.o-qSc.o = q~'.
REMARK. For the development of this idea for the actual construction
of w(z) see the book: I. N. VEKUA, Generalised Analytic Functions (Obobakckenwyye analitioheaki11e Junlctrii), Fizma.tgiz, 1959, English translation
published by Pergamon Press (1962). References to the literature on this
problem a.re given there.
1168. Prove that the transformation
is isometric, that is,
llSc.oll =lie.oil.
HINT. First prove the formula. for w e Gn by integrating !1Bwll1 by parts
and using the relations of problem 1166. Then prove the passage to V from
which is everywhere dense in v.
m
1169. Prove that if c.o e Gg, then
(19)
where
.,.
Mec.o(C)
S
eC.O
= 2~
( ) = _ _!__
z
:ii
.r c.o(C+ee' )d0,
9
0
JI
IC-zi>e
c.o(C)d~d.,,
(C-z)2 '
190
PROBLEMS ON COMPLEX ANALYSIS
and Sew converges in mean to Sw, that is llSew-Swll -+ oo as
0.
(!-+
HINT. Putting t = ee'' and changing the order of' integration in Bew
with the substitution ·(;1 = C+ee", use the fa.ct that
_l_
2ni
= { ((; ~z)I' if IC1-zl > (},
-~-t((; -z-t)z
f
1
1
ltl=e
o,
if
ICs-zl < e.
For the proof of the second assertion use the fact that (prove this)
llMe a>-roll
~
0 as
e ~ 0.
REMARK. The relation (19) also holds almost everywhere for roeL 2 • This
enables us to define for co e L 2 the transformation Bro as the limit in mean
of the transformation Be co.
1170. Let H« = L 2 n Lip IX, that is, H« is the space of functions
which belong simultaneously to L 2 and Lip IX, Prove that all functions
g(C) e H« satisfy the inequality
ISg(l)-Sg(-1)1 ::;:;; A«lul",
where
HINT. Subtract from the difference which is being estinia.ted multiplied
by -1/n the quantity
g(l)
ff
ff
4(;Md11
((;1-1)2 +g(-1)
4(;dEd11
((;•-1)1
ReC>O
ReC>O
(prove that both sides a.re equal to zero!).
1171. Let w{C) e H« and
w(C) =
w( u;v C+ u~v).
Prove that
-- 2-"lu-v'«lwl
'rol oc-
I
I
Gt
and using this, that
!Sw(u)-Sw(v)I ::::;; 2 "Aat'w'.oc:u-v1°'.
191
SINGULAR INTEGRALS
REJirAB:s:, From this result it follows that 8 is a bounded linear operator
in H •.
1172. Let g(C) e H 11 and
I
1- _ _!.)g(C)dEd11
_!_ff
(-C-z
n
C
T 0 g(z) = _
B
(in comparison with Tg the passage to T 0 g ensures integrability at
infinity). Prove that
1--_!.)dEd11 =
_..!...ff
(-c-1
n
C
l
(20)
B
and using this fact, that
ITog(l)-g(O)-Sg(O)I ~
where
1
B" =-;;
Bcx!Y!cx,
ff IC-II
1ci11-1
dEd'fJ.
B
HINT. Multiply g(O) by the left; side of (20).
1178. Let w(C) e H 11• Putting
ro(C) = w[u+C(u-v)],
prove that
IToro(u)-Toro(v)-w(v)(ii-ii)-Sw(v)(u-v)I ~ B 11 \wl 11 \u-1J[H 11,
and deduce from this the relation:
(T0 co(z)k= w(z), (T0 w(z))s = Sw(z).
1174. Show that the search for the quasi-conformal mapping
w(z) of the whole w-plane onto itself, which satisfies the equation
(q e Ha.)
WW= q(z)w:
by means of the substitution
w(z) = z+T0w(z)
leads to the singular integral equation
w-qSw
(see the remark on problem 1167).
=
q,
OHAPTEB IX
ANALYTIC CONTINUATION.
SINGULARITIES OF MANY-VALUED
CHARACTER.
RIEMANN SURFACES
§ I. Analytic continuation
00
1175. The function /(z) =
2 t' is expanded into a Taylor
series
n-o
in the neighbourhood of the point z =a (lal < 1). For what values
of a does this expansion permit the function f (z) to be continued
analytically ~
oo
n
2 !.__ is
n=l n
1176. The sum of tlie power series /(z) =
expanded
into a Taylor series in the neighbourhood of the point z = -1/2.
What is the domain into which the function/(z) is continued by this!
oo
1177. Prove that the function /(z) =
n
2 (- l)n+i!._n
n=I
can be
continued into a larger domain by means of the series
1-z
(1-z)2
log 2--2-- 2.22
(1-z)B
3.28
1178. The power series
fi(z) =
l
00
n=l
l
zn
-
n
.2
00
and
/ 2 (z) = i:n:+
(-l)R
(z~ 2 )n
n-1
have no common domain of convergence. Prove that the functions
/ 1 (z) and f 2(z) are nevertheless analytic continuations of one another.
1179. Prove that the functions defined by the series
1
(1-a)z
(1-a)2z2
1+az+a1z2+ ... and - - ... '
1-z
(l-z)2
(1-z) 3
are analytic continuations of one another.
+
192
i93
ANA.LYTIO OONTINUATION
1180. Let the power series
f(z) = a0 +a1z+ ... +anzn+ ...
have the radius of convergence R = 1. By carrying out the change
of variable z = Z/(i+Z) let us transform it into the form
f(z) =J[Z/(i+zn = F(Z)
= ea+eiz+ ... +c.Z"+ ...
Denoting bye. the radius of convergence of the series obtained
prove the following assertions:
(i)
e~
!'
function /(z), then
(2) If
and if the point z = - i is a singularity of the
!.
e=
! < e<
i, then the equality
/(z) =
F(Z) = F( i z
z)
permits the analytic continuation of the function /(z) into a
domain exterior to the disk lzl < i and interior to the circle of Apollonius
lz zii= e·
(3) If e = i, the equality given in part 2 analytically
continues the function /(z) into the half-plane Re z < i/2.
(4) If e > i, then the function f(z) is analytically conti-
I I
nuable into a domain exterior to the circle of Apollonius z z i =
e·
00
1181. Prove that the power series f(z) =
2 z n represents a func2
•-O
tion, analytic in the disk lzl < i and having the circle lzl = i as
its natural boundary (that is, f(z) is a function which is not continuable beyond the unit circle).
HINT. Using the identity
/(z)
= z1 +z'+
... +zaA:+J(z•t),
prove that for any point of the form
< t < I).
J(tC> -+ oo as t -+ 1
co
C= ·vt1
(le is a natural number)
In problems 1182-1184 prove that the functions represented
by the given power series are not continuable beyond the unit circle.
194
PROBLEMS ON COM"PLEX .ANALYSIS
00
~-, ~2n
1182. f(z) =- L.J --~,,-
·
n=O
HINT. Use Pringsheim's theorem (see problem 642).
00
1183. /(z) = ,}; znl .
n=O
HINT. If p and q are coprime integers and
n;;;;;, q, then
(
2pnl
re q
)nl
= ,.n !.
00
~1
1184. /(z) =
znl
L_; rii"'
n=O
REMARK. In problems
1182-1184
particular cases are
considered of
Hadamard's general gap theorem:
00
If the indices of non-zero coefficients of the power series /(z)
= I a,,zn
n=O
form a sequence nu n 2 , •• ., in which n1:+ 1 > (I+ix)n11; where IX> 0, then
the boundary of the circle of convergence of the series is a natural
boundary of the function /(z).
1185. Prove that the series
.i:(l -zl
n=l
l n)
n 1 1-
+
-z
in the domains
[z[ < 1 and [z[ > 1 represents two analytic functions which are
not analytic continuations of one another (see also problems 696-699).
1186. Let /(z) and </>(z) be arbitrary integral functions, and
S(z)
oo (--------1-zn
l-zn-1) . Prove
=,};
n=l I+zn
I-j-zn-1
that the expression
1
1
1j!(z) = 2 [f(z)+</>(z)]+2S(z)[f(z)-<f>(z)]
represents in the domain JzJ < 1 the function /(z) and in the domain
[z[ > 1 the function <f>(z).
1187. (1) Prove that if oi is a real irrational number the series
represents in the domains JzJ < 1 and Jzj > 1 analytic functions
for each of which the circle [z! = 1 is a natural boundary.
195
ANALYTIC CONTINUATION
HINT. Prove that the sum of the series increasel'I without limit as z __,. e•imtzn
along a. radill8 vector.
REMARK. This problem is a. parlicula.r case of the following general
theorem:
Let L be a curve, closed or open, which has at every point a definite radius
00
of curvature. If the series
l;
converges absolutel;v· and the
Cn
points
n-1
a1' a 1 , ... , an, ... all lie on the curve L and are distributed on it BO that.
any finite aro of the curve L always contains an infinite 110t of them, then
the series
2 ---"-·
an-z
00
.F(z)
c
=
n-1
represents a. function analytic in any domain not containing points of the
curve L, and for which this curve is a. singular line (see GoUltSAT, A Oourae
in Mathematical Analysis, Vol. 2, Chapter XVI).
(2) Prove that if oi is a real rational number, then the
series of part (1) represents a rational function.
00
1188. Prove that the series
J;-,,-1n=l
n +n
(11"
=- e" 101 ") converges for
Re z > 1 and its sum has the straight line Re z
boundary.
=
1 as its natural
00
1189. Prove that the functionf(z) = J; e-mlz is analytic for Re z>O
m=o
and has the straight line Re z = 0 as its natural boundary.
1190. Prove that the functionf(2!) defined in the half-plane Re z > 0
by the Dirichlet series
00
f(z) =
L
ane-Anz
n=l
where an= (-l)n+i, A.21c_ 1 = 2k, A.21c = 2lc+e- 2" (le= 1, 2, ... ), can
be continued analytically into the half-plane Re z > - l.
HINT. Write /(z) in the form
00
/(z)
=
J;(1-e-ze- 2")e--llcz
k-1
and prove that in any finite domain belonging to the half-plane Re z ··. --1,
11-e-ze-llcl <Me-lie,
where M is a constant for the domain considered.
196
PROBLEMS ON COMPLEX .ANALYSIS
REMA.BE. The given problem shows that on the straight line bounding
the half-plane of convergence the sum of a Dirichlet series may not have
singular points.
1191 The function /(a) defined in the half-plane Re a > 0 by
*.
means of the Laplace integral
00
/(a)
=
Je-•e' sine' dt,
0
is to be continued analytically into the half-plane Re 8 > -1.
1192. Euler's Gamma function is defined in the half-plane Re z
by means of the integral
>0
00
I'(z)
=
Je-'F-ldt
(F-1 = e<s-1) Joi:') •
0
Integrating by parts the right-hand side of this equation, show
that the function I'(z) is continued analytically to the whole plane
as a meromorphic function with the simple poles 0, -1, -2, ... ,
-n, ... , the residue at the pole -n being equal to
(-l)n.
--;r-·
1193. Show that it is possible to continue the Gamma function
analytically by means of the formula
00
00
+f e-'F-ldt,
.L.J nl(z+n)
I'(z) = ~ (-l)n
n-o
1
1
HnrT. In the integral
into a power series.
Je-tt•-ldt replace the function e-r by its expansion
0
1194. In problem 420 it was proved that for 0
<x<
1
00
f
F-1 sin t dt = I'(x) sin
~
·
0
In what domains of the z-plane are the given formulae valid!
1195. Prove that I'(z) can be continued to the whole of its domain
of existence by means of the formula
I'(z)
=
-.-i-J
2smnz
c
e-w(-w)•-ldw
((-w)•-1 = e<•-..1•01(-w>)'
197
ANA.LYTIO OONTINUATION
where the contour 0 consists of a cut a.long the positive part of the
real axis, the detour round the coordinate origin being traversed
anticlockwise.
1196. Let C(z) be Riemann's Zeta function (see problem 695)
co
~ _!_
L.J
n:r:
C(z) =
(Rez >I).
n=l
Prove that for Re 21
>I
J wco
C(z)
I
= I'(z)
o
1 dw
(ew-1)-
and obtain from this the analytic continuation of the function
C(z) to the whole plane, excluding the point 21 = I; explain the
character of the singularity of the function C(z) at the point z = I.
HINT. For the analytic continuation consider the integral
J(-w):r:~i dw
C
eW-
Where 0 is the Contour of problem 1195.
1197*. Let the function /(z) be expanded into the power series
co
/(z) =
2 anzn
which has the radius of convergence R = I. Let us
n=O
co
denote by efJ(z) the sum of the series <fo(z) =
J; a~~
n
(the function
n=O
t/J(z) is the Borel transform of the function /(z) - it is an integral
function, see problem 535).
Prove that for izl <I the equality
co
Je~t q, (zt) dt = J(z)
0
co
is satisfied. Prove also, that the functionf e-'q,(zt)dt effects the anao
lytic continuation of the function /(z) into the domain G defined
as follows: A straight line is drawn through every singularity of
the function /(z), perpendicular to the segment connecting this
singularity to the coordinate origin: G is the convex domain containing the circle izl <I, the boundary of which consists of points
of these straight lines; if the number of these straight lines is finite,
then G is a polygon (Borel's method of continuation).
198
PBOBLEMS ON OOMPLEX ANALYSIS
1198. Verify Borel's method of continuation for the following
serieR:
00
(3)
2z'".
11-0
1199. In the integral of Cauchy type
F(z) = ~f ,P(C)dC
2ni
C-z
c
let 0 be a. simple closed contour and ,P(C) a function oontinuou•
a.long 0. Prove that in order that one of the functions F+(z) and
F-(z) (see page 153) should be the analytic continuation of the
other through an arc ye 0, it is necessary and sufficient thatf(z) = 0
on the arc y.
1200. Prove that if the function ,P(C) is not analytiot a.t even
one point of the simple open a.re 0, then all the points of the arc 0
are singularities of the integra.l of Cauchy type
_I_f ,P(C)dC
2:n:i
c
C-z .
HINT, Commence with Sokhotskii'a formulae for the limiting valuea
of a Cauohy type integral.
1201. Let y be a simple closed contour tra.versed in the positin
direction, consisting of arcs y 1 , y 8 with common end points Zi and z1
(Fig. 31), let G+ be the domain within y, G- the domain outside y.
rz
z,
Zt
tr
FIG, 31
t That is, there does not exist any analytic function identical with • (C)
on any arc belonging to O.
SINGULABITIBS OF MANY•VALUED ClHABA.CTEB
In addition, let .F(z) =
,.J"'i~:c
where
199
t/>(C) =a if Ce Yv
and t/J(C) = b, if Ce y 8 (a and b a.re complex constants).
Find the functions .F+(z) and .F-(z) and continue analytically the
function .F-(z) into the domain G+:
(a) through the a.re y1 ; (b) through the arc y11 •
1202. Let G be a doubly connected domain bounded by an interior
contour y and an exterior contour I', and efJ(z) a function analytic
in the closed region G+y+I'.
Prove that the function
t/J(z) -
~f t/>(C)
2m
r
dC
C-z
is analytically continuable to the whole of the exterior of the contour
y, and the function
t/J(z)+~f t/J(C)
2ni ,. C-z
dC
to the whole of the interior of the contour
a.re traversed anticlockwise.
r. The contours ,, and r
§ 2. Singularities of many-valued character. Riemann surfacest
An isolated branch point z -= a of order k-1 of the function w(z) (k is
a natural number. k ;;;i: 2) is characterised by the fact that there exists a branch
of w(z). which in the neighbourhood of the point z .... a admits of the represen•
tation
oo
n
w = L;on(z-a) 1-co
(a 7' oo)
01'
oo
w
n
= L; CnZ-T
(a= oo).
-co
If only a finite number of the coeftl.cients On with negative subscripts differ
from zero the point z = a (or z = oo) is called an algel>raio branch point. In
the contrary case the corresponding point is called a ~ branch
f)Oint (an essential singularity of many-valued character).
Above one and the same point of the z-plane the function w(z) may have
not more than an enumerable set of different algebraic and transcendental
t For tbis section see [l, Chapter VUI]; V. V. GoLUBEV Lecffwea on eke
Analytic Theory of DiffereMal Eqwtiona, (Lektrii po analiUchealcoJ teorU rllfferenttal'nylch. twa11716nU). Gostekhizdat, 1951; R. NEvANLINNA, Umformien.mg,
Berlin, 19153.
200
l'BOBLEMS ON OO:Ml'LEX ANALYSIS
branch points, regular points and singularities of single-valued character.
On the Riemann surface of the function w(z) above the z-plane such points
have neighbourhoods which do not intersect.
In every such neighbourhood w is a single-valued function of the para·
meter t:
00
w
2
=
0ntn,
n=-N
where
(a of: oo),
(a= oo).
A point z = a or z = oo is said to be a logarithmic branch point if it is such
that any branch w(z) admits of unbounded analytic continuation in the domain
0 < lz-al < r (or R < lzl < oo) and is infinitely many valued there. Such
a branch w(z) in the neighbourhood of a logarithmic branch point becomes
a single-valued analytic function on passing to the parameter t = Log (z - a),
Re t < e (or t = Log z, Re t >(!)·It must be remembered that on the Riemann
surface above one and the same point z, in addition to various logarithmic
branch points there may also be found other points of single-valued and
many-valued character.
1203. Explain for which values of z the values of w(z) a.re identical
on all the sheets of its Riemann surface above the z-pla.ne, if:
(1) w= (z2-9)y(z);
(2) w =sin z + (z2 + 4) Log z;
(3) w = sin z + (z• + 4)• Log z.
Are the values of w' (z) identical at the same points 1
_.1204. For ea.oh of the functions:
(1) w = zyz.
(2) w = z1 Log z, w(O) = 0,
verify that at the point z = 0 there exists a first derivative which
is the same for all the branches and that a finite second derivative
does not exist.
In problems 1205-1212 expand ea.oh of the given functions w(z)
into a series of powers of the local para.meter t in the neighbourhood of all the points of its Riemann surface situated above the
given z points; indicate the domains of convergence of the resulting
series.
1
1205. w= l+y( 2 -z), z = 1, z = 2.
1206. w
=
y[y(z-1)-2], z
=
1, z = 5, z = oo.
SINGULARITIES OF MANY-VALUED CHARACTER
201
1207. W= y[l+J/(z -1)], z = 1.
1208. w = y[(-i/z-a)(yz-b)], z = oo (a#= b).
1
= 0.
1209. w = eTz, z
sinJl'z
1210. w = --2-, z = 0.
z
1211. w= cotVz, z = 0.
1212. w = JI'sin z, z = o.
If the function w = /(z) is single valued and the function z(w) inverse
it is many valued, for the determination of the algebraic branch points
of the function z(w) it is neceBBary to find the zeros of /'(z), the multiple poles
of /(z) and investigate the behaviour of /(z) at infinity. In this, a point z0 #: oo
corresponds oo an algebraic branch point of order k-1 of the function z(w),
if in the neighbourhood of Zo the Laurent expansion of the function /(z) has
the form
oo
00
= w0 + 2
/(z)
o.(z-z0)n
(01c
#:
0)
n=k
or
00
/(z)
If Zo
=
=
2
en(z-z.)n
n=-k
(o-1c
#: 0).
oo, then the given expansions must be of the form
(ore #: 0)
or
-00
/(z)
= Wo+
2
enzn
n=-k
(ll-Tc
#: 0).
In problems 1213-1219 determine the singularities of z(w) if
w is the given function.
1213. w = z(l-z).
1214. w = z3-3z.
ums.
w=
1216. w =
1217.
(l~z)2
(-z
)
1-z
2
•
z-a
W=~l
z-
(0<a<1).
1218. w = Pn(z) (a polynomial
of degree n).
1219. w = R(z) (a rational function).
202
PBOBLEMS ON COMPLEX ANALYSIS
In problems 1220-1228 investigate the mapping effected by the
function w(z), constmct the Riemann surface B over the w-plane
and subdivide the z-plane into domains corresponding to the sheets
or half-sheets of B.
1220. w = (I + :
1221.
W= ( :
1222. w =
1224. w =
:
r
r
Consider the limiting case n -+ oo.
! (z + !)·
! z~
(zn+
1223. w =
(l~z)• •
). Find also the group of linear transfor-
mations, with respect to which the function w is invariant and
explain which transformations of the Riemann surface correspond
to the transformations of the group.
z
I
~
1225. w= (I+zn)• •
1227. w=z-+-;-·
zn
1226. w = z--·
n
I
1228. w = Tn(z) = 2n-l cos (n cos-1 z), n ~ I (Tn(z) are Chebyshev polynomials).
In problems 1229-1232 find the singularities of the functions
inverse to those given.
1
w= e"i",
t{.r-.!)z
1230. w = e
1229.
(t is a complex number).
1231. w = cos z+sin z.
sinz
1232, W = - - •
z
In problems 1233-1241 construct the Riemann surfaces above
the w-plane.
1233. w = cos z.
1236. w = cot z.
1234. w = sin z.
1237. w = cosh z.
1235. w = tan z.
1238. w = sinh 21.
SINGULARITIES OF MANY-VALUED Cil.ARAOTEB
203
1239. w = ta.nh z.
1241. w = z+e=.
1240. w = cothz.
1242. Assuming that the Riemann surface above the w-plane
of the rational function w = R(C) is known, construct the Riemann
surfaces of the functions z(w), if:
(1) w = R (e11);
(2) w = R (sin z).
In problems 1243-1248 construct the Riemann surfaces of the
given functions (above the z-pla.ne).
1243. (1) w = y[(z-a)(z-b)];
(2) w = y[(z-a)(z-b)(z-c)];
(3) w =
VCft
(z-a")) (consider separately the
even and odd n).
1244. (1) w=y (z-a);
(2)
w=
of
f/[(z-a)(z-b)];
j/ [(z-a)(z-b)(z-c)];
(3) w =
ca~s
4) w =
VLU.
(z-a,,,)}
n>3.
V[(z-a) (z-b)(z-c)], n > 3.
1246. w =
(:=~2 ) +v (z-c).
1245. w =
-,I(
1247. w = y[y(z)-1].
1248. w=y(sinz).
In problems 1249-1254 investigate the mappings and construct
the Riemann surfaces for the algebraic functions w(z) and z(w).
1249.
w2+z1 =
z
1.
1250. w• = (l-z)• ·
HINT, In problem 1250, use the solution of problem 1228.
1251.
w+z8-3wz =
0.
HINT. Use the parametric representation
3t
z = I+t•'
St•
w
= I+t8 '
zS
1252. w8 = - - ·
1-z
HINT. Use the parametric representation
11
= t1 /(l+t1),
w = t8 /(I+e1 ).
204
PROBLEMS ON COMPLEX ANALYSIS
l+z
1253. w8 = z2 - 1-z
•
HINT, Put
1254. uJ'
=
zn
(1 +zn)I
HINT, Consider separately the cases of even and odd n.
If z =a is a singular point of one of the branches J(z) of a function w(z)
then the Bet of indeterminaten68B of f(z) at the point z = a is the set of the
limiting values of f(z) obtained from its values in the neighbourhood of z = a
above lz - al < r as r -+ O. For algebraic branch points and poles the set
of indeterminateneBB consists of a single point. If the function is single valued
and the point a is an isolated eBBential singularity, then by the Ce.sore.tiWeierstrass theorem the set of indeterminateness covers the whole plane
(see also problem 688). For transcendental and logarithmic branch points,
and also for non-isolated essential singularities the set of indeterminateneBB
may have a more complicated structure (see the book: V. V. GoLUBEV, LectureB
on the Analytic Theory of Differ<mtiaZ EquatioM (Lektrii po anaZiticheBkoi teoril
dijferentai,al'nylch uravnenii), Chapter I, § 7, Gostekhizdat, 1951.
In problems 1255-1266 determine the singularities of the functions and
also find the sets of indeterminateness at the transcendental and logarithmic
branch points.
1255. w =
(y z)
(y z)n
4•
1256. w =
(n, m are natural numbers).
1257. w = eit<vcz>-1>
1
1
1263. W=-Log--•
1258. w = (sin yz)/yz.
z
1-z
1-yz
1259. w =tan l+yz •
1264. w = z+Logz.
1265. w = .!..sin-lz.
1260. w = e<Los :z)n (n is an integer).
z
1261. w = z1 = e1 Losz.
1
1266. w = 1 +cot-lz,
1262. w = sin Log z.
z
In problems 1267-1270 construct the Riemann surfaces of the
given functions.
1267. w = z" (a is a complex number).
1268. w =Log [(z-a) (z-b)].
1269. w =Log [(z-a) (z-b) (z-c)].
1270. w = Log sin z.
SINGULARITIES OF MANY-VALUED CHARACTER
205
1271. Let C= <f>(z) be a single-valued or many-valued analytic
function, R,., ~ its Riemann surface above the z-plane and w = f(C)
a single-valued analytic function with set of indeterminateness Gi;.
In what domains on Rr:, ~ does each of the given expressions:
(1) w(z) = f[<f>(z)],
(2) w(z) = f(z)+<f>(z),
(3) w(z) = f(z)<f>(z)
determine a unique analytic function i
In particular, consider the case when </> is algebraic or inverse
to meromorphic, and f is a rational or transcendental meromorphic
function.
In problems 1272-1302 explain which of the given functions
w(z) can be resolved into distinct algebraic functions and which not;
determine also their singularities and where indicated construct
their Riemann surfaces (n and m are natural numbers).
1272. w = f.(~) (compare with (tz) 2).
1273. w = fl (z') (compare with
z)4).
1274. w = y(z)m (compare with (yzr).
r:
1275. W= 1!!
t' e.
1276. w = y(sin z).
1277. W =Log zn.
1278. w=Loge"'.
(f/
1279. w=Log(z-
!)·
1280. w =Log (e"-1). Construct the Riemann surface.
1281. w = Log sin z. Construct the Riemann surface.
1282. w = Log tan z. Construct the Riemann surface.
1283. w = cos-1 (cosz) (compare with cos(cos-lz)).
1284. w = tan-1 (tan z) (compare with tan (tan-1 z)).
1285. (1) w = (z'•)'• (r1 , r 2 are rational numbers). Compare with
(z'•)'•. Consider, in particular, r 1 = 2/3, r 2 = 3/2;
(2) w = z'•z'•;
(3) w = z'•+z'•.
1286. w =
z j/ (1-z).
y
1287. w =
1288,
W=
y[j/(z)-1].
yz ,
v(z)+l
y
1289. w =
(Log z). Construct the Riemann surface.
1290. w = Log Log z. Construct the Riemann surface.
1291. w = [Log (z - 1)]1•
206
PBOBLEMS ON OOMPLEX .ANALYSIS
y
1292. w = (sin-1 z). Construct the Riemann surface.
1293. w = sin-1 Log z. Construct the Riemann surface.
1294. w =Log (j/(z)-1). Construct the Riemann surface.
1295. w =Log [(1-yz)/(l+Jlz).
1296. w = sin-1 [yz/(l+z)].
1297. w = Log z• (o: is a real number).
1298. w = y (z)+Log z. Construct the Riemann surface.
1299. w = Logz Logz.
1300. w = sin-lz + cos-1 z .
1301. w = ta.n-lz + cot-lz.
1302. w = ta.nh-1z - coth-lz.
1303. Construct the Riemann surface of the function w = (Log z) 1
and investigate the set of limiting values of w for one logarithmic
branch point above the point z = 0, obtained for: (1) r-+ 0,
</> = const; (2) r-+ 0, o: < q, < {J;
(3) r = const, q,-+ ± oo;
(4) r-+ 0, q, -+ ± oo.
1804. Let x(z) be a single-valued analytic function in the circle
lzl < 1, which cannot be continued beyond the circle lzl = 1.
Explain for what values of a the given fV.n,ctions resolve into
distinct analytic functions, and for what values they do not.
(1) w = x(z)+v (z-a);
(2) w = x(z) r.og (z-a):
(3) w = x(a+z");
(4) w = x(a+e11 ).
1305. Explain the problem of the factorisation of the functions:
(1) w = V(C-a) (2) w = Log(C-a), where = x-1(z) and
xis the function of problem 1804.
1306. Investigate the behaviour of the separate analytic functions
determined by the equations:
(1) w = x(z)(Log z)1; (2) w = x(z)[Log (z-1)]1, where x(z)
is the function of problem 1304.
Find, in particular, the set of indeterminateness in the neighbourhood of the logarithmic branch point.
+
c
HINT. Use the solution of problem 1808.
1307. Let f(z) be an integral function.
Construct the Riemann surfaces of the functions:
(1) w= y(f(z)); (2)w=Logf(z); (3)w=[/(z)J1 (0: is an
irrational number).
1308. Construct the Riemann surface of the function
SINGULARITIES OF MANY-VALUED CHABAOTEB
207
1309. Let
Construct the Riemann surfaces of the functions:
(1) w = y(f(z));
(2) w =
(3)
Logf(z);
w =Log 1(;)+Logf{:z)·
HINT. First prove that the function /(z) is sohlioht in the disk lzl
and has the oirole fzf = I as its natural boundary.
<
I
§ 3. Some classes of analytic functions with non-isolated singularitiest
Let E be a closed set of points with connected complement !J in the extended
z-plane. Let AB, AD and AO be classes of single-valued analytic functions
in !J, respectively: bounded (the class AB), having a bounded Dirichlet integral
with respect to !J (the class AD), continuous in the extended z-plane (the
class AO). If the set Eis in some respect small it can be expected that from
the analyticity of the function in !J and the additional limitations connected
with the class of function, it would follow that the function in E is analytic.
Then the function is a constant in virtue of Liouville's theorem. In this case
the set E is said to be a nul aet of the class of functions considered.
The classes of nul sets corresponding to the given classes of functions are
denoted by NAB• N AD• N AC• and the class of their complements in the
domain !J by 0 AB• 0 AD• 0 AC•
A set Eis said to be .AB-removable in the domain G, containing E, if from
the analyticity and boundedness in G-E there follows analyticity everywhere
in G (if G is identical with the extended z-plane, then AB-removableness in
Gisequivalent to the fact that EeNAB)· AD-removableness and .AO-remova·
bleness of a set E in the domain G are similarly defined (in the case of AO·
removableness, functions are considered of the class AO in G, that is, analytic
in G-E and continuous everywhere in G). For brevity F denotes any of the
classes AB, AD, AO, if it is a question of some property common to them.
1310. Prove that the set E with connected complement !J, which
is the set of singularities of a single-valued function analytic in Q,
is closed.
1311. Prove that the isolated singular points of functions of the
classes .AO, .AB, .AD are removable and the sets of singularities
of the functions of these classes are perfect.
t For this section see L. Alu.FOBS, A. BEtJBLING, (1950), Conformal invariants
and function theoretic null-sets, Acta Math. vol. 83, 100-129.
208
PROBLEMS ON COMPLEX .ANALYSIS
HINT. For /e.AD consider a.Bf?-+0 the Dirichlet integral
fJ l/'l8 rdrd•
C1<fzi<C11
(a singularity at zero), starting from the Laurent expansion
.f(z)
"" Cnzn •
= .J:
-00
1312. Let G be an arbitrary domain in the extended z-plane
and Ek(k = I, 2, ... , q) bounded closed sets with connected complements Dk, located in G, no two of which have points in common:
(i#=j).
Prove that the function /(z), single-valued and analytic in G-E
q
E
=
LJ E", can be represented uniquely in the form
k=l
q
f(z) = ,P(z)+ 21J11i(z),
k=l
where ,P(z) is analytic in G, 1Jlk(z) is analytic in D" and 1J11i(oo) = 0.
In particular, if G is the extended z-plane, then
q
J(z) =f(oo)+ L1Pk(z)
k=l
HINT. Separate the sets Et in (} by contours "" (la = I, 2, ..• , q) and in
the domain cut off from the variable domain Gn, converging to G, by the contours Yk• represent /(z) by Cauchy's formula..
REMARK, 'Pt(z) has on Ek the same singularities a.s /(z) and is the analogue
of the principal pa.rt of the Laurent expansion in the neighbourhood of an
isolated singularity.
1313. Prove that if Ee NF, then E is .F-removable in any domain
G => E (removableness of nul sets).
HINT,
Make use of the result of the preceding problem.
1314. Prove that if Eke NF (k = I, 2, ... , q) and E 1 n EJ = 0
q
(i =F j), then E =
LJ Ek c: NF
k=l
(the property of analyticity of nul
sets).
1315. Prove that if E c: NF• then E does not contain interior
points.
SINGULARITIES OF MANY-VALUED CHARACTER
HINT. Assuming that J!J contains the circle
function
/(z)
=
{ a+~.
_z-_a
if
a+z-a,
rz -
lz-al >
209
al<(], consider the
(!,
if lz-al< fl·
1316. Prove that if E e NF, then the plane measure mE = 0
(a stronger form of the result of problem 1313).
HINT. On the assumption that E is bounded (this is not essential) and
mE > 0, consider the function (see problem 1058)
/(z)
=
ff d~d1/
E
CC = ~+i11).
(;-z
1317. Prove that if E forms a rectifiable arc, then EeNA.c·
HINT. Make use of the generalised theorem of Cauchy (if /(z) is continuous
in the closed region <J, bounded by a rectifiable contour
and is analytic
inside G, then f(z)dz = 0) and Morera's theorem.
r,
J
r
A set E is said to be everywhere discontinuous, if in an arbitrarily small
neighbourhood of every point z e E it is possible to draw a closed Jordan curve
y, y E = 0, surrounding the point z. A bounded, perfect, everywhere discontinuous set can be included in a finite number of closed rectifiable Jordan
curves { Yt} (k = I, 2, ... , q) with diameters less than an arbitrarily specified
number e > 0, lying outside one another and not having points in common
with E. The quantity
n
q
l(E) =inf}; Z(y1c),
k=l
where Z(y1c) is the length of Yt and inf is applied to all possible systems {Y1c},
is lmown as the length Z(E) of the set E.
1318. Prove the following assertions:
(I) If Eis a perfect everywhere discontinuous set of finite
length Z(E) < oo, then E e N A.c.
HINT. For the domain G=>E and e> 0 there exists,d(e)> 0 such that
for z', z" E G, lz' -z"I < d we have lf(z')-/(z")I < e. For the system of
contours { Yt} with diameters less than d, containing E and having the sum
of lengths l:Z(y1c) < l(E) +I, the expansion
k
/(z)
= lf>(z)+ 2 'Pk(z)
k
210
PROBLEMS ON COMPLEX ANALYSIS
(Ctc is inside ;'le•
of problem 1312 is constructed and by means of the relations
z is in G outside {;'le}),
"'"
(z)
= _1_ Jf<C>dC = _1_ Jf<C>-J(Cle) d,.
2m
2m
C-z
C-z
"" leads to the ""equality J(z) =
If 'Pk (z)j is estimated and this
'
If> (z).
(2) If a perfect set E is situated on a rectifiable a.re, then
(3) There exist bounded, perfect, everywhere discontinuous
sets Ee N AC which have infinite length.
(4) There exist bounded, perfect, everywhere discontinuous
sets E of infinite length and plane measure zero, l(E) = oo, mE = 0,
not belonging to N AC.
HINT. See V. V. GoLUBEV, Smgls-mlued. analytio functions f.Dith. perfect
asta of aingular pointa, (Od.nomaoh.nyye analitiohukiys funktaii a aovsr11hennym
mnozheatoom oaobylch. toohelc), Saratov, 1916.
1319. Prove the following assertions on the class of null sets
NAB·
(1) If the set E contains a continuum, then EeN.tB·
(2) NABc.NAc• correspondingly OAcc.OAB (the inclusion is
strict).
(3) If E is a perfect everywhere discontinuous set and the
length l(E) = 0, then E e N .AB·
HINT. As above (see the preceding hint) use the expansion f(z)
+ 2 '1'1c(z), but with 2 Z(y1c) <
"
k
proving the equality /(z) =
e and estimate
= lf>(z)
j2 '1'1c(z)j for the purpose of
k
If> (z).
(4) If Eis a perfect set, situated on a straight line, then in
order that E e N .AB• it is necessary and sufficient that the linear
measure mes E = 0.
HINT. Assuming that mes E > 0, consider the function
J(z)-
·J~.
t-z
E
1320. Prove that if Ee N .AD• then every function f(z), which is
schlicht and conformally maps the domain D which is the complement
of E, is linear.
HINT. Use the AD-removableness of E. To prove that f(z) is schlioht
on E use the principle of corresponding boundaries, after enclosing E by some
contour.
CHAPTER X
CONFORMAL MAPPINGS
(CONTINUATION)
§ I. The Schwarz-Christofl'el formulat
Let us denote by P a polygon in the w-plane, by Ak(la = I, 2, ... , n) its
vertices, arranged in the order of positive traversal of P with respect to its
n
interior, and by CCkn its internal angles (
2
k=I
ar:k
= n-2). The function w = /(z)
which maps the upper half-plane Im z > 0 onto the interior of the polygon
P, is determined by the Schwarz-Christoffel formula:
(1)
where - oo < IJi < a 1 < .. . < an < oo a.re points on the m-e.xis which correspond to the vertices Air A 1 , ... , An of the polygon P; 0 and 0 1 a.re complex
constants.
The mapping of the upper half-plane onto the exterior of the same polygon
ia effected by the function
f n(z-ai/kz
F(z) = 0
n
1
o k-1
dz
_ +Ou
(2)
(z-b)•(z-b)•
where b is· the point of the upper half-plane corresponding to the point at
infinity of the w-pla.ne; ale a.re points which correspond to the vertices Ak
of the polygon (now oo > a~> ~ > ... > ak > - oo); the fJkn a.re the
n
external angles of the polygon (/Jk
= 2-or:t, 2
k=I
/Jk
= n+2). If it is
a ques-
tion of the mapping of the exterior of a polygon some of the points Ak
may be geometrically coincident.
1321. Prove that in formula (1), of then points aa:, three can be
assigned arbitrarily, and the remaining n-3 points and the cont For thia section see [I, Chapter VIII, § 7], [3, Chapter II, § 3]; B. A. FucHs
and B. V. SHA.BAT, l!'imotWna of a Oomple:a Variable, (l!'unlatsii laomplelamogo
peremennogo), hd ed., Chap. VIII, Fizmatgiz, 1959. English translation
published by Pergamon Press (1962).
211
212
PROBLEMS ON COMPLEX ANALYSIS
stants 0 and 0 1 (in all n+I real parameters) are determined from
the relations
0 1+1
J lf'(x)ldx =
~
°1+1
101
J
~
n
n
lx-atj«t-ldx = l1
t-1
a1
(i=I,2, .. .,n;an+l=a1, andj
an
oo
a1
=J + J),
an
-oo
where the l 1 are the lengths of the sides A 1A 1+1 (An+i = A1), 0 1 = f(a 1)
= w 0 (w0 is the affix of the vertex A 1) and arg 0 = () (() is the
angle of inclination of the side AnA1 of the polygon to the real axis),
The given relations exceed the unknown parameters in number,
however: they determine these parameters uniquely. Explain this
and show which of the relations can be discarded.
1322. Prove that if one of the vertices of the polygon is the image
of the point at infinity, for example an= oo, then the mapping
function is of the form
z n-1
f(z) = 0
Jn (z-at) t-ldz+01.
11
0
t=I
HINT. Make the transformation C= -1/z, if none of the at equals 0
and C= - - 1 - , where a at, if one of the points at= 0 (k ... 1, 2, ... , n-1).
z-a
+
1323. Prove that if one or several of the vertices A1; is at oo,
then formula (I) remains in force, if by 1Xt31: is understood the angle
between the corresponding rays at the finite point of their
intersection, taken with a negative sign.
A.y
2)
FIG.
32
213
CONFORMAL MAPPINGS (CONTINUATION)
HINT. Let At= oo. If OCA: < 1, then consider the polygon P', cut off from
P by the segment Ai,At' where Ai, and At' lie sufficiently far from the sides
At-iAk and AtHAt (Fig. 32 (1)) and in formula (1) for P' pass to the limit as
Al: -+ oo, A/: -+ oo. If, however, oc1c ;;;i:: 1, then join A_tA/i' in P by a polygonal
l.irie (see Fig. 32(2)) and by enlarging it similarly extend it to infinity.
1324. Determine the quantities °'1:, which enter into formula
(I) for the infinitely distant vertex of the "polygon" formed from
parallel rays represented in Fig. 33.
A,~~Az
A
A,=======
A 1 ~Aa
1)
2)
A,-A3
A
z
AJ
3}
Az
A,=====~
A*
A+
----
A,=======;:
Az
6)
5)
A,
A,c::=======!
A1
A,
As======;;
A•
11.
A3=======?Az
~ ,_A•
'\
tZ/(
Aa
--
A
A~'
8)
7)
FIG.
33
HINT. Pass to the limit as recommended in the hint to problem 1323.
1325. (I) Prove that in the mapping of the unit circle !z! <I
onto the polygon P, situated in the finite part of the plane, the
mapping function is of the form
f n(z-ak)
z
/(z) = 0
n
0 k=l
•1:-l
dz+ 01 '
214
PBOBLEMS 0111' COMPLEX .ANALYSIS
"'t (
where arc == e1 </>1 < </> 1 < ... < <f>n) are the points on the circle
lzl == 1, corresponding to the vertices At, traversed in the positive
direction and the IXtn are the internal angles of the polygon P.
(2) Prove that the function which maps the unit disk izl < 1
onto the exterior of the same polygon P, subject to the condition
that the point z == 0 passes into the point w == oo, is of the form
/(z)
== 0
j fl (z-<t,/k-i: •
1
k-1
where ai == e11fii: (</>~ > </>~ > ... > </>i), and the f3tn are the external
angles of P.
1326. Find all the cases of single-valued transformation of the
Schwarz-Christoffel formula. (1), that is, explain for which polygons P
the inverse function z == z(w) is defined and single valued in the
whole w-plane.
HINT. Polygons obtained from P by any number of reflections in the
sides, must without omissions or overlapping cover the whole 1.0-plane.
In problems 132'7-1332 map the upper half-plane Im z > 0 onto
the given domain P, located in the w-plane with the given correspondence of the vertices of P and the points of the real axis. Determine also the period or periods of the inverse function z(w), the
group G of its invariant linear transformations of the w-plane and
the fundamental domain B of this group (see page 30).
132'7. (1) P is the strip 0 <" < k; w(- oo, oo) -+Z (0, oo)t.
(2) P is the strip 0 <" < k; w(- oo, oo)-+ z(-1, 1).
1328. P is the half-strip 0 < u < n; v < O; w(O, n, - i oo) -+Z
(1, -1, oo).
1329. (1) P is the right-angled triangle with acute angles n/3,
n/6;
w(o, co, co+ ~~ )-+ z(O, 1, oo),
(2) P is the right-angled isosceles triangle;
w(O, co, co + ico) -+ z(O, 1, oo).
(3) P is the equilateral triangle;
w(o, co, co 1+1(3>)-+ z(O, 1, oo).
t Here and in what follows by the symbol w(A, B, ... ) -+ s(a, b, ... ) is
meant the correspondence of points of thew- and•· planes A.+--+a, ~.+--+b, ...
CONFORMAL MAl'PINGS (ClO:NTINUATION)
215
1880. Find the domain of the to-plane onto which the function
z
to(z)
= Jt«-l(l-t)ll-1 dt
(a.rg w'(l)
= 0)
0
maps the upper half-plane Im z > 0, if:
(1) 0 <ex~ 2, 0 < {J ~ 2; consider the oases:
(a) ex+{J < 1, (b) ex+{J = 1, (o) ex+{J > 1, in particular ex+{J = 2
and ex = {J = 3/2
(2) 1 ~ex~ 2, -1 ~ {J ~ 0, ex+{J ~ 1; consider the oases:
(a) ex= 1, (b) ex+{J = 1, (o) ex = 2, (d) ex = 2, {J = -!, (e) ex= 2,
{J = -1.
1)
2)
3)
FIG. 34
1331. (1) Find the domains of the to-plane onto which the function
z
w(z)
=
f
1-).
y[l(l-1)] de
0
= y[z(z-1)]+(1-2).)(log [y(z)-y(z-1))-tin)
maps the upper half-plane. Consider the oases: ). < 0, 0 <). < !.
). =
i < ). < 1 and ). > 1.
t.
216
l'ROBLEMS ON' COl\£.PLEX ANALYSIS
(2) Find the domains of thew-plane onto which the function
1
w(z) = 2ni
Jz -i/[t(t-1)]
1-A.
, /(z-1)]
tdt = ni1 [ log[y(z)+y(z-l)]-A.JI
-z1
maps the upper half-plane. Consider the oases: A. < 0, 0 <A. < 1
and A.> 1.
1332. Map the upper half-plane Im z > 0 onto the domains of
the w-plane indicated in Fig. 34 with the given correspondence
of points.
(1) w (A= 0, B = 1, 0 = oo)-+ z (0, 1, oo);
(2) w (A = 0, B = 1, 0 = oo) -+ z (0, 1, oo);
(3) w (A= 0, B = oo, 0 = oo)-+ z (0, 1, oo);
(4) w (A = 0, B = oo, 0 = oo) -+ z (0, 1, oo);
(5) w (A = 0, B = ia, 0 = oo) -+ z (0, 1, oo).
1333. Map the upper half-plane Im z > 0 onto the domains in
the w-plane indicated in Fig. 35 (0 < 0 < 1).
(1) w (A, B = 0, 0 = oo)-+ z (0, 1, oo).
(2) w (A, B = 0, 0 = oo) -+ z (0, 1, oo).
C
B
~
ti
C
A
Fm. 35
In the case when 0 is a rational number (0 = p/q), express the
integrals obtained in terms of elementary functions.
1334. Map the upper half-plane onto the domain indicated in
Fig. 36 (the arc AO is a semioir.ole).
w(A
= a,,
B
=
oo, 0 = 0) -+ z (0, 1, oo).
liINT. This is reduced to a particular case of problem 1333, (2) by means
of the mapping C= a/11•.
217
CONFORM.AL MAPPINGS (CONTINUATION)
FIG. 36
1335. Map the upper half-plane Im z > 0 onto the domain of
the w-plane indicated in Fig. 37, with the condition
w(A
=
= oo,
-h, B
0
= h,
D
=
oo)- z (-1,0, 1, oo).
(The possibility of this mapping follows from the symmetryprinciple.)
1336. Map the upper half-plane Im z > 0 onto the lozenge in
the w-plane with angle nrt at the vertex A and side d (Fig. 38).
The corresponding points are given by
w(A
= 0,
B
= d,
0
= d(l+e1""),
D
= de1"")-+ z (0,
1, oo, -1).
Justify the possibility of this mapping.
FIG. 37
Fm. 38
1337. Map the upper half-plane Im z > 0 onto the domain in
the w-plane indicated in Fig. 39. The parameters a, b (a> O,
218
PROBLEMS ON OOMPLEX .ANALYSIS
b > 0) are the images of the corresponding vertices, they cannot
be specified arbitrarily and they must be found.
(1) w (A= 0, B = oo, 0 = H +ik, D = oo)-+ z (0, 1, a, oo).
(2) w (B = oo,O = H+ik, D = oo, O' = -H+ik,
B' = oo)-+ z (1, a, oo, -a, -1).
HINT. Make an additional cut a.long the imaginary axis and use the sym·
metry principle.
-
FIG.
39
(3) w (A = 0, B = oo, 0 = oo, D = -ha-i(k2 -k1,),
JiJ = oo)-+ z (0, 1, a, oo, -b).
1338. (1) Find the function w(z) which maps the disk lzl < 1 onto
the interior of the regular n-gon with centre at the coordinate origin
and one of its vertices at the point w = 1, subject to the conditions
that w (0) = 0, w' (0) > 0.
(2) Map the disk lzl < 1 onto the exterior of the same
n-gon subject to the conditions that w(O) = oo, w(z) > 0 (0 < z
<
1).
Determine c_1 in the expansion w(z) = c_1
z
+....
1339. (1) Find the function w(z) which maps the disk lzl < 1
onto the interior of the regular five pointed star with centre at
219
OONFORMAL MAPPINGS (OONTINUATION)
the coordinate origin and one of the vertices at the point w = 1;
the condition of normalisation: w (0) = 0, w' (0) > 0.
(2) Map the disk lzl < 1 onto the exterior of the same star;
w(O) = oo, w(x) > 0 (0 <a:< l); determine c_ 1 in the expansion
( )
wz
C-1
=--+
....
z
1340. Find the domain onto which the function
w(z) = Jz (l+tn):dt , w'(O)
> 0,
-1<A<1-~
o (l-tn)n-+A
n
maps the unit disk lzl < 1.
1341. Map the upper half-plane Im z > 0 onto the regular n-gon
in the w-pla.ne with centre at the coordinate origin and one of the
vertices at the point w = 1; the normalisation is w(i) = 0, w(O) = 1.
1342. Let the domain P in the w-pla.ne be the exterior of the
"star" consisting of n segments issmng from the point w = 0
(Fig. 40). Let .A1; be the vertices of P at the coordinate origin and
n
rx1;11: the corresponding angles (
2 IX1; =
2), B1; the vertices of P a.t
k-1
the ends of the segments of the star (.A1, B 1, .A 2, B 1 ,
•••
a.re arranged
FIG. 40
in the order of positive traversal of P), l1; = .A1;B,. the lengths of
the segments of the star. Prove that the function w =f(z),f(O) = oo
which maps the unit disk 1zl < 1 onto Pis of the form
n
n
f(z) =
~
(z-a1;)rx,.'
k=l
220
PROBLEMS ON OOMPLEX ANALYSIS
where 0 is a. complex constant and tit = e14'1c a.re the points on the
circle lzl =I which correspond to the A1c. The points b1c = e1"1c which
correspond to the vertices B1c a.re the roots of the equation
n
~~-_!__=0.
k=I
z-a"
How a.re the para.meters 0,
a",
(3)
z
b1c determined 1
HINT. In continuation by the symmetry principle the function f(s) is
multiplied by a constant factor, hence the function
4>(z)
=
f'(z)
f(z)
is single valued in the z-ple.ne.
REMARK. This formula immediately gives the solution of a number of
problems of Chapter II, for example, problems 281, (2), 282(1), 284, 290.
It is recommended that these problems should be solved again e.nd the
constants entering into the general formula determined.
1343. Map the exterior of the double angle (Fig. 41) onto the
exterior of the circle lzl >I subject to the conditions: w(oo) = oo,
w'(oo) > 0.
1344. Prove that the function w = f(z), which maps the upper
half-plane Im z > 0 onto the exterior of the star of problem 1342
is of the form
n
f(z) = 0
n
(z-aa:)11"
k-1
-
(z-z0 ) (z-z0 )
'
where z0 is the point of the upper half-plane which is transformed
into oo.
0
FIG.
41
1345. Let the domain P in the w-plane be bounded by the ray
[O, oo), by n - I segments, going from w = 0 to the points B 1
CONFORMAL MAPPINGS (CONTINUATION)
221
(i = 1, 2, ... , n - 1), and m - 1 rays going from the points D.
(8 = 1, 2, ... , m -1) to oo, such that their continuations in the
opposite direction pass through the coordinate origin (Fig. 42). Let
At (le = 1, 2, ... , n) be the vertices of P at the origin, rtt'lt the cor-
42
FIG.
responding angles, 0 1(j = 1, 2, ... , m) the vertices of P at oo, i'J""
the corresponding angles, Ai Bi A 2 ••• An Oi Di ... Om a positive
traversal of the boundary of P. Prove that the function w = f(z),
which maps the upper half-plane Im z > 0 onto P, is of the form
n
0
n
n
f (z) =
(z-ak)«a,
(4)
k=l
m
(z-e1)'JIJ
j=l
where ak, e1 are the points on the x-axis which correspond to the
vertices At, 01. The points b,, a. on the x-axis which correspond
to the vertices B,, D. are the roots of the equation
2,
2
- 'JIJ
-=0
k=l
j=l
J
n
m
-rtk- Z-!Zt
(5)
Z-C·
a.
How are the parameters 0, ak' bi. Cj, determined 1 What is changed
if one of the parameters ak, b,, Cj, a8 is equal to 001
HINT. Prove that
n
=
n
n-1
n
/'(z)
/(z)
m
'\'"1 IXk
'\'"1 "/J
= L.J z-ak - L.J z-e1
k-1
J=l
i-1
(z-ds)
s=l
n
k=l
n
n
m-1
(z-bi)
m
(z-ak)
1-1
(Z-Cj)
222
PROBLEMS ON COMPLEX ANAJ,YSIS
1346. Prove that formula. (4) of problem 1840 still applies to
the case where the domain P, contains two rays [O, oo) (Fig. 43).
Here the vertex at the coordinate origin is taken into account also
in the case when there issue from it only two rays forming a single
straight line.
JJ,
/'
81
c,
In problems 134'7-1351 map onto the upper half-plane Im z > 0
the domains of the w-plane given in the corresponding diagrams
subject to the given conditions, and determine the parameters
a and b (a > 0, b > 0).
c,[
~
c
c,
FIG.44
A,1~
FIG.
c,
46
134'7. The domain indicated in Fig. 44,
w (.A 1 = 0, B = ke'-., .A 1 , 0 = oo) -+ z (-1, b, 1, oo).
1348. The domain indicated in Fig. 45,
w (.A 1
= 0,
B
= ik,
.A 1 , 0 1
= oo, D = iH,
01
= oo)
-.z(-a,O,a,
!
,oo,-!)·
223
CONFORMAL MAPPINGS (CONTINUATION)
1849. The dome.in indicated in Fig. 46,
w (..4.1 = 0, B1 =
-
he,_, ..4. 8, B 8 =he*, ..4.3 , 0 = oo)
-+ z (-1, -b, 0, b, 1, oo).
1350. The dome.in indicated in Fig. 47,
w (A1 = 0, B =he'••, A11 , 0 1 = oo, D =He,.•, 0 1 = oo)
-+
8,~Bz
C
z(-a
-1
I
0, b,
.!..,
b
.!..).
oo '
a
c
FIG. 46
I
I
'
FIG. 47
'\\
/
\
I
1)
2)
FIG. 48
1351. (1) The dome.in indicated in Fig. 48, (1),
w (oo) = oo, w (± 1) =
± 1.
(2) The dome.in indicated in Fig. 48, (2) (the angles between
the cuts are TC/n; the extreme segments form with the corresponding
rays of the real axis angles n/2n),
w(oo) = oo,
w(A 1 , An) -+
z (-
1, 1).
1352. Let the domain P in the w-plane be the horizontal strip
of width H with cuts going left to oo from the points B 1 (i = 1,
224
PROBLEMS ON' OOMPLEX ANALYSIS
2, ... , n - 1) and right from the points Ds (8 = 1, 2, ... , m - 1)
{Fig. 49). Prove that the function w = f(z), which maps the upper
half-plane Im z > 0 onto the domain P is of the form
n
f(z) =
2~
m
log (z-a1c)-
k=l
2 ~log
(z-c1)+0,
(6)
J=l
where a1c, c1 are the points on the x-axis corresponding to the vertices
A1c and 0 1 of the domain P; the h1c are the distances between the
cuts on the left, and the l1 are the corresponding distances on the
right. The points b1(i = 1, 2, ... , n - 1) and d8 (8 = 1, 2, ... , m - 1)
on the x-axis corresponding to the vertices B 1 and D,, are the roots of
the equation
n
'1 h1c
.L.J z-a1c
k-1
m
-
'1
.L.J
J=l
l1 - 0
z-c1 - •
(7)
How are the parameters 0, a1c, b1o c1, ds determined 1 What is changed
if one of the parameters a1c, b1o c1,
becomes equal to 001
a.
FIG. 49
HINT. The problem reduces to problem 13'6. It is also possible to find
the functionf'(z) which is single valued or to derive directly from the SchwarzChristoff'el formula.
1353. Let the domain P be the upper half-plane Im w > 0 with
horizontal cuts going leftwards to oo from the points B 1 (i = 1,
2, ... , n) and to oo on the right from the points D. (8 = 1, 2, ... , m)
(Fig. 50).
CONFORMAL MAl'PINGS (CONTINUATION)
225
Prove that the function w = /(z) which maps the upper ha.If-plane
Im z > 0 onto the domain P is of the form
n
m
f (z) = L.J
~!!I!..
log (z-a1;)- ~!:!..log (z-c·)+Az+B,
1'
L.J 1'
J
k=l
(8)
J=l
where ~. cJ are points on the x-axis corresponding to the vertices A1:
and OJ of the strips of the domain P, and k", Z1 a.re the distances
FIG. 50
between the cuts going in one direction. The point A 0 is transformed
m
>0
into oo, A
and Im B =
2
l1 • The points b1(i = 1, 2, ... , n)
J=l
and d. (8 = 1, 2, ... , m) on the x-axis corresponding to the vertices
B1 and D. a.re roots of the equation
n
m
~~- ~-l1 -+An=0.
L.J ·z-a1; L.J z-c1
k=l
How a.re the para.meters A, B, a", b1, cj,
if A 0 passes into a. point a0 #= oo, then
/(z) =
(9)
J=l
n
m
1:-1
J=l
a. determined 1 Show that
2 ~log (z-~)-2 ~log (z-c
1)
L-H
A
n
z-a0
+--log (z-a0)+--+B,
(10)
226
PROBLEMS ON OOMPLEX ANALYSIS
where
n
H= };h1" L=
k=l
m
};zi, A <0 and ImB =
L.
J-1
If the parameter a1c or cJ is equal to oo, then in formula (10) the
corresponding term is absent.
HINT. Use the Schwarz-Christoffel formula. To determine the coefficients
of the logarithmic terms compare the increments of w on going round the
points a 0 , a1c, CJ (along a semicircle) calculated geometrically and by the formula
fo.r /(z).
FIG. 61
1354. Let the domain P be the w-plane with horizontal cuts
going from the points B 1(i = I, 2, .. ., n) to oo on the left (Fig. 51).
Prove that the function w = f(z), which maps the upper half-plane
Im z > 0 onto the domain P, is of the form
n-1
f(z) = -Az2 +Bz+O+
,2 ~ log (z-a1c),
k=l
where the a1c are points on the x-axis corresponding to the vertices
A1c of the strips of the domain P, and k1c are the distances between
the cuts. The point A 0 maps into oo, A> 0 and Bis a real number.
The points b1 (i = 1, 2, .. ., n) corresponding to the vertices B 1
are zeros of the derivative J'(z). How are the parameters A, B,
0, ~. b1 determined 1
OONFOBMAL MAl'l'INGS (OONTINUATION)
227
HINT. See the hint to problem 1353.
1355. Let the domain P be the w-plane with horizontal outs
going to oo on the left and on the right (Fig. 52). Prove that the
function
FIG. 52
w = /(z), which maps the upper half-plane Im z > 0 onto the domain
P, is of the form
n-1
/(z) =
2
k-0
2l1
m-1
·k"
-log(z-arc):n;
J-0
A
0
)+--+--+B,
z-a
z-eo
(ll)
-log(z-c1
:n;
0
where A > 0, 0 > 0, k 0 = Im (Drn - B1), l0 = Re (D1 - Bn), and
the remaining parameters (including b1 and d11 for the vertices B 1
and D.) have the same values as in problem 1353. How are the
parameters A, B, 0, arc. b,, CJ, a. determined 1 Show that if ao = oo,
then
n-1
/(z) =
2n
k-1
2l1n
m-1
k"
-log
(z-arc)-
J-0
0
-log (z-c1 )---+Az+B,
z-c0
(12)
228
PROBLEMS ON OOMPLEX ANALYSIS
where A > 0 and 0 > 0. If the parameter ak(Tc ¥= 0) or c1 (j ¥= 0)
is equal to oo, then in formula (12) the corresponding term is absent.
HINT.
See the hint to problem 1353.
1356. Map the upper half-plane Im z > 0 onto the domains in the
w-plane indicated in Fig. 53 (all the dimensions are given on the
corresponding diagrams) for the given conditions; find a and b
(a> 0, b > 0).
FIG. 53
(1) (A 1 , B, A 2 , 0)-+ (-1, b, 1, oo);
(2) (A 1 , B 1 , A 2, B 2 , A 3 , 0)-+ (-1, -b, 0, b, 1, oo);
(3) (E, 0, D,)-+ (-1, 1, oo);
(4) (E, 0, D)-+ (-1, 1, oo);
(5) (E, 0, 0)-+ (-1, 1, oo);
(6) (A 0 ,B,A 1 )-+ (oo,O, l);
(7) (A 1 , 0, 0)-+ (-1, 0, l);
(8) (A 0 , B 1 , Av B 2)-+ (oo, -(1 +a), -1, O); d = Re(B2 -B1 );
(9) (A,B,O,D)-+ (oo,-l,a,l);d=Re(D-B);h=Im(B-D).
229
OONFORMAL MAPPINGS (OONTINUATION}
§ 2. Conformal mappings involving the use of elliptic functionst
The integral
z
u
=
J
dz
y[(l-z1) (l-lc1 :z:2)]
0
.p
=
d.P
y(l-lc•sin•.p)
J
=
{l)
F(,P, le)
0
is known as the normal elliptic integral of the first kind in Legendre'a form. The
inverse function
z = sn (u, le)
(2)
is one of the fundamental elliptic functiona of Jacobi, known as the sn function
of Jacobi. Two other elliptic functions of Jacobi a.re connected with it:
en (u, le)= y[l-sn1 (u, le)],
dn (u, le) = y[l-lc1 sn1 (u, le)],
(3)
called respectively the en- and dn-functiona of Jacobi. If there is no necessity to
indicate the modulus le, then they a.re simply written sn u, en u _and dn u.
It follows from (1)-(3), that
dsnu
du
- - - = cnudnu,
dcnu
du
(4)
- - - = -snudnu,
ddnu
du
=
-lc1 snucnu.
In what follows, if nothing is said to the contrary, it is assumed that 0
The integrals
<
le
< I.
(5)
1
K' = Jy[(I-:z:•)dtl-Tc'•zs)] = F( ~,le')
0
are known as complete elliptic intll{l1"ala of the jirat lcind, and le'
as the complementary parameter.
= y (l -
lc2 )
t For this section see: N. I. AxHIYEZER, Elementa of the Theory of Elliptic
Functiona (Elementy teorii elliptichealcilch funlcta-ii), Gostekhizda.t, 1948; Yu.
S. Sm:oBSXII, Elementa of the Theory of Elliptic Functiona with Applicationa
to Mechanica (Elementy teorii elliptichealcilch funlctBii B prilozheniyami le melchanilce), Gostekhizdat, 1936; A. M. ZHURAvsxrr, Textbook on Elliptic Functiona,
(Spraooohnik po elliptichealcim funlctaiyam), Gostekhizdat, 1941; A. BETZ,
Konforme Abbildung, Berlin, 1948.
230
PROBLEMS ON COMPLEX ANALYSIS
1357. Investigate the mapping of the z-pla.ne and the ,P-pla.ne
by means of the normal elliptic integral of the first kind (1). Continuing
this mapping by the symmetry principle, prove that sn u is a doubly
periodic function with periods 4K and 2iK'.
HINT. Consider the mapping of the upper half-plane Im 111 > 0 by means
of the principle of corresponding boundaries. Trace the variation of the argument of du/dz when 111 is varied along the :i:-u:is.
1358. Map the upper half-plane Im~> 0 onto the recta.ngle in
the w-plane with vertices ±a, ±a±ib (a > 0, b > 0).
1359. Investigate the mapping of the z-plane by means of the
elliptic integral
I
w=
f J"[(l-z•)~'•+klz•)]'
z
Find the inverse function and prove that it is doubly periodic
with periods 4K, 2K + 2iK'.
HINT.
Make the substitution
111
== y(l-P).
1380. Investigate the mapping of the z-pla.ne by means of the
elliptic integral
Find the inverse function and prove that it is doubly periodic
with periods 2K, 4iK'.
HINT.
Make the substitution
111
== y'(l-k't1).
1361. Verify Table 1 and Table 2 (apart from the la.st column).
The dash (') denotes passage to the complementary parameter k'.
HINT. Table 2 is obtained from Table 1. Let B be the rectangle with
vertices (0, K, K+iK', tK') in the u-plane and B 1 its image in the u,,-plane.
To establish the relationship between 1111 = sn (a,,, Ji) and 111 - sn (u, Ji) map
the image B onto the image B 1, preserving the correspondence of vertices.
This applies to all the rows of Table I except the last where as R there is taken
the rectangle with vertices (0, K', K'+tK, tK) and a correspondence is esta.
blished between 1111 and 111' = sn' u = sn (u, k').
1362. Consider the mapping of the z1-plane onto the Ui·pla.ne
by means of the elliptic integral of the first kind Ui = F(,P1, ii)
((z1 =sin <P1 = sn (Ui, k1 )) for the following values of ii:
~·
k',
~',
231
CONFORM.AL MAPPINGS (CONTINtT.A.TION)
TABLl!I l
U1
sn(u10 k)
-u
-snu
onu
-snu
-onu
dnu
sn ti
-onu
-dnu
onu
-k'~
dnu
k'-1-
u+2K
u+2iK'
u+K
on (Ui. k)
dnu
idnu
I
u+iK'
iu
ksnu
-1snt:i""
. sn'u
on'u
on'u
l
t---
TABLJ!I
u
I
K+i11
dn' 1/
-K+i71
;+iK'
ik' sn' 1/
dn' 1/
k' on' 1/
1
kmE
dnu
k' on' 1/
dn' 1/
l
-
idn;
ksnE
I
idnE
ksn;
ksnE
im'71
on' 1/
on' 1/
l
dnu
dnu
-·-. onu
snu
dn'u
on'u
2
tk' sn' 1/
- - dn' 1/
- dn'71
E-iK'
i71
onu
snu
dn(u10 k)
dn' 1/
Z(u)+ dn u on u_
snu
-•{z'<11>+
2i~'}
__!E_}
_ · {z'<11>+
•
2KK'
-·-,on;
snE
Z(E)-~
2K
ion;
snE
zm+ 2K
dn'71
on'71
-i{Z'(71)+ dn'11cn'11
sn' 1/
in
n71 }
+2KK'
l
±iK'
'--
k
i
=t=k
=Fi
Residues
232
ik
PROBLEMS ON OOMPLEX ANALYSIS
ik'
k' and verify Table 3, where the
K 1 = K(ki), Ki= K(lci) (lei= y(I-kf)) is
k' ,
notation
used.
HINT. To establish the connection between z1 and z = sn (u, Tc) compare
the mappings of the first quadrants of the zrplane and the z-plane by means
of the functions F(•1 , Tc1 ) and F(•, Tc) (in the case of k1 = Tc' take the second
quadrant of the z1-plane). The necessary relationships are given in the third
column.
TABLE
3
K'l
I
k
Tc'
I
v
ilc
v
ilc'
T
1cu
Tc sn (u, Tc)
iu
I
-
dn (u, Tc)
en (u, Tc)
Tc(K+iK')
TcK'
• sn (u, Tc)
• on (u, Tc)
I
en (u, Tc)
dn(u, Tc)
en (u, Tc)
K'
K
ilc'u
ilc' m (u, Tc)
en (u, Tc)
dn(u, Tc)
on (u, Tc)
I
en (u, Tc)
Tc'(K'+iK)
Tc'K
lc'u
Tc' sn (u, Tc)
dn (u, Tc)
on(u, k)
dn(u, Tc)
I
dn(u, Tc)
Tc'K
Tc'(K'+iK)
ilcu
ilc sn (u, Tc)
dn (u, Tc)
l
dn(u, Tc)
en (u, Tc)
dn (u, Tc)
TcK'
Tc(K+iK')
I
The integral
j Ji( l;~;:•)dz
0
= jy'(I-Tc•sins•)d• = E(•,Tc)
(6)
0
is known as the normal elliptic integral of the second kind. By the substitution
z = sn u it is reduced to the form
jy(1;~::1 )dz=
0
j
dn1 udu=E(u).
(7)
0
Two functions of Jacobi, Z(u) and O(u) are connected with this integral:
Z(u)
E
O'(u)
== E(u)--u
=-,
K
O(u)
(8)
233
CONFORMAL MAPPINGS (CONTINUATION')
where
(9)
is the complete elUptic integral of the second kind. The quantities K, E and K'
E', obtained by p&BBing to the complementary parameter Tc', are connected
by the relations
EK' +E'K-KK' = n/2.
(10)
We slian also use the notation
f
1
D _ K-IC _
- ~-
0
z1 dz
f[(l-zl)(l-lclz 1 )]
(11)
•
Ma.king the substitution lclzl+Tc'1 t1 = 1, it is poBBible to establish the formulae
(12)
j
1
v( 1;__k;:9)dz
+i
= i(K'-JC')
ft
dt
)
(ft,JII( l-Tc'•t•)
l-t• dty[(l-t )(1-Tc' P] •
1
0
1
(13)
0
Ma.king use of the notation for the elliptic integrals given by (1) and (7)
these formulae can be written briefly as:
u-K = i(K' -Ui),
where
In particular, it follows from formula (13) that
1
k
v
f (1 ;~:• )
1
dz
= i(K' -JC') .
234
PROBLEMS ON COMPLEX ANALYSIS
The integral
z
J
0
(l+sizl)y'[(l
~z9)(1-klz1 )]
4'
=J
d4'
(J +11sin1 4') y'(l-k8 sin1 4')
0
= ll(4', 11' k)
(14)
is called the normal elliptic mtegral of the third kitld. By the substitution
z = sn u it is reduced to the form
II
J
du
1+11sn•u'
(l 5 )
0
and is expressed in terms of the 8-function of Jacobi. The quantity
1
ll1 = ll1(S1, k) =
n(; '"· k) = J (l +n•) y'[(l~zl)(l-klzl)]
(16)
0
is known as the oomplet.e eUiptio integral of the third kitld. The quantity ll1
can be expressed in terms of elliptic integrals of the first and second kindst.
1363. Consider the mapping of the z1-plane onto the w1-plane
by means of the elliptic integral of the second kind w1 = E(</J1, lei)
(z1
=
k1
:
~
sin </J1
•
k',
=
sn u1) for the following values of the para.meter
~' •
; •
~~
(0<k<1) and verify Table 4 (page 235).
HINT. See the hint to problem 1362.
1364*. Map the upper half-plane Im z > 0 onto the upper ha.lfplane Im w > 0 with two vertical cuts a.long the segments Re w
= ± a, 0 < Im w < k with the normalisation: w(O) = 0, w( oo) = oo,
w'(oo) > 0.
HINT, Use the Schwarz-Christoffel formula in the form
z
J
w(z) =Oki
b1 -z1
y'[(l-zl)(l-klz•)] dz
0
=
0
z
[Jz-.JIf( l-k
1-z•
1 1 )
0
dz-(1-klb•)
Jz y'[(l-z•)(l-klz•)]
dz
]
;
0
t For the calculation of elliptic integrals of the third kind and of related
quantities see ERDELYI, MA.GNUB, OBERHlllTTINGER, Tm:com, (1953), Higher,
Tramllllndemal Funotiom, vol. II. Note that for "e (-oo, -1), (-1, -k1 ),
(-k•, 0),(0, oo) the formulae for the calculation of (15) and (16) have different
forms.
I
- [E(u)-k'2u]
k
I ku I
I
! iu
I ik'u l ~[k'2u-E(u)+
k'
l/k
k'
l/k'
ik
ik'
T
I iku
. I
v I k'u I
~ [ E'(u)-k'2 sn:i:~:' u]
:, [ E(u)-k• sn~c= u]
sn u dn u]
cnu
4u-E(u)+ snudnu]
cnu
E(u1 , k1 )
I Ui I
k'
I
I
I
I
I
I
Ei
(l/k)E'
(I/k')E
~, [(E'-k2K')-i(E-k'2K)]
E'
_!_ [(E-k'2K)-i(E'-k2K')]
k
TABLE 4
I
I
I
I
I
_!_[(E-k' 2 K)-i(E'-k2K')]
k
~, [(E' -k2K')-i(E-k'2K')]
(l/k')E
E
(l/k)E'
E'1
z
0
~
i:..:I
Ot
~
~
'"3
....
0
z
'"3
z
0
rJl
14'.l
....
~
Is:
~
I'd
~
~
!;lj
z
0
0
236
PROBLEMS ON OOMPLEX .ANALYSIS
for the determination of the constants use the values of a and h with the equation
w{ ~ )-w(l) =
0 (it determines b in terms of k) and the condition
w'(oo)> O.
1365. Verify the last column of Table 2, (see problem 1361) and,
using it, consider the mapping of the rectangle R: !El < K, 1111 < K',
in the u-pla.ne (u = E+ifJ), by means of the function
dnuonu w=Z(u)+·
snu
HINT. Use the equations
J
0 en• t
II
'-'I
If;
dt
'-'• u- E(u+
)
--=If;
Bn
u dn u ,
en u
II
k'•J~
=E(u)-kl snuenu'
1
0
dn t
dnu
easily verified by differentiation (see formulae (4)).
1366. Investigate the mapping of the z-pla.ne by means of the
normal elliptic integral of the third kind (14) for 0 < k < 1. Consider
separately the oases when v belongs to the intervals:
(1) (- oo, -1); (2) (-1, -k•); (3) (-k1, O); (4) (0, oo). Consider
also the oases: v = - 1 and,,= - k1 •
llmT. Investigate the mapping of the first quadrant of the z-plane.
The integral
with discriminant Lf = gf-27 gl, #: 0 (with this condition no two of Iii• 8t• e1
are equal) is known as the normal elUpeio integral ofthefir1t kind in Weierltra81'1
form, and the function
z = p(w)
(18)
is known as the "p"-function of Weier1Wa11. This is one of the fundamental
elliptic functions with periods 2<.o, 2ro' (Im !!t__ ofi 0). It satisfies the differential
(.()
equation
(19)
CONFORMAL MAPPINGS (CONTINUATION)
237
The fwiction p(w) is even, two-sheeted in the parallelogram of periods
(Fig. 54), having there a pole of the second order at zero and the double points
(p' = O)co, co+co', co':
e1 = p(co), llz = p(co+co'), e3 = p(co').
(20)
FIG. 54
It follows from (19) that
(21)
1367*. Investigate the mapping of the z-plane by means of the
normal elliptic integral of the first kind in Weierstrass's form (17)
for real g2 , g3 and LI > 0. Consider the cases g3 > 0, g3 < 0, g3 = 0.
Find the periods of p(w).
HINT. Consider the mapping of the upper half-plane Im z > 0 by the
principle of corresponding bowidaries.
1368*. Investigate the mapping of the z-plane by means of the
normal elliptic integral of the first kind (17) for real g2 , g3 and LI < 0.
In particular, consider the case g2 = 0. Find the periods of p(w).
238
PROBLEMS ON OOMPLEX ANALYSIS
HINT. Since .d < 0, two of the quantities 8 1, 8 8, 8 8, are complex conjugates,
and one is real. Let 8 8 be the real quantity, 8 1 = a.+ip, 8a = a.-ip <P > 0).
Consider the mapping of the semicircle lz-e.I = le1 - 81 1, Im z > 0 by means
of the principle of corresponding boundaries and continue this mapping by
the symmetry principle.
1369. Find the mapping onto the upper half-plane Im w
of the triangle ABO with the given conditions:
(1) (A= 0, B = ro > 0, 0 = ro(l+i))-+ (oo, -1, O);
>0
in)
ay3
(2) ( A=O, B=a>O, 0 = 2 -es -+(oo,-1,1);
!'!!)
ay3-es -+(oo,-1,0).
(3) ( A=O,B=a>0,0=2
HINT, Use the solutions of problem 136'7 (the case g8 = 0) and 1368 (the
case 91 = 0).
REMARK. For this problem see also problems 1330-1332.
1370. Map the doubly connected domains 1-15 of the z-plane shown
in Fig. 55 onto the circular ring (h < lwl < es and determine the
modulus µ = e.le1 (see page 29).
In problems 1371-1373 map the given domains onto the unit
circle ltl < 1.
1371. The rectangle Re luj < K, Im jitl < K' (0 < Ii < 1). Find
the positions of the vertices in the mapping.
1372. The interior of the ellipse jz - lj + jz + 1J = 2a (a> 1)
with the cuts [-a, - 1], [1, a].
1373. The interior of the ellipse jz - lj + jz + lj = 2a (a> 1).
Find the positions of the foci in the mapping.
1374. Map the exterior of the unit circle jtj > 1 onto the domains 1 - 3, of the z-plane indicated in Fig. 56.
1375f. Let R be the rectangle IEI < K, l'l'/I < K' in the plane
of u = E+in. Using Table 2 (see problem 1361) and the principle
of corresponding boundaries, prove that the functions
z
dnu
-l = sn u-J-ig en u '
.:_ = Z(u)+ dn u (en u+~ sn u+ik) +mu
l
snu-f-igcnu
(all the para.meters are real, / 1 g• < 1, m > 0, 0 <Ii< 1) map
the rectangle R onto the exterior of a cross, the dimensions of which
+
t For this problem see DAB.WIN, (1950), Some conformal transformations
involving elliptic functions, Phil. Mag., '7, vol. 41, No. 312, 1-11.
11) '
FIG. 55
2)
FIG. 56
240
PROBLEMS ON COMPLEX .ANALYSIS
are determined by the parameters l, f, g, k, or onto the exterior
of a rectangle with four branches-continuations of the sides, the
dimensions of which are determined by the parameters l, f, g, k,
m, k. In particular, the functions
(1) dn u/sn u,
(2) Z(u) + dn u on u/sn u,
(3)
dnuonu
snu
Z (u)+---- +mu,
(4 ) Z(u)+ dnuonu_ + ikdnu,
snu
snu
(5 ) Z(u)+-dnuonu,
snu-/
(a) Z(u)+ dnuonu +
snu
ikdnu +mu,
snu
(k = .~2 )
(7) Z(u)+ dn u on u_ + ik dn u + __::__
snu
snu
4.KB
y
map R onto the exterior of the diagrams 1-7 in Fig. 57.
R:m:MABx. Ee.ch of the functions (1)-(7) corresponds to the figures of a
horizontal row of Fig. 57. The figures of one such row illustrate the effect
(qualitatively) of the variation of some parameter, namely:
In the cases (I). (2) the para.meter k increases (with motion to the right).
In case (3) the para.meter m increases. For m = ;
obtained, in particular, if k = : 2 (then m =
figure of the series is obtained form=
-k'• a rectangle is
4~) •it is a square. The last
2;K' .
In the case (4) the parameter h increases from zero. For h =
one pair of branches disappears.
In case (5) the para.meter
f increases. For f =
!(
1- k'1
k~' ( ; -
k' 1 )
!}
one pair
of branches disappears.
Case (6) differs from case (4) by the presence of the term with m, which
causes thickening.
Case (7) is the particular case of (6) when k = l/y2, m = n/4K1 ; a "wind·
mill" is obtained.
CONFORMAL MAPPINGS (CONTINUATION)
11+
+
+
~III
~nrro ~~H
~ILLi-i~IJJ ~ ~
·Jn 111Ln n
~-µ_
FIG 57
24
242
PROBLEMS ON COMPLEX ANALYSIS
On the application of elliptic functions to the problems of the mapping of
the upper half-plane onto the exterior of an ellipse, hyperbola. and para.bola.,
and also on the mapping of the exterior of two arbitrarily situated rectilinear
segments or of two concentric a.res onto a circular ring see, for example, L. I.
SEDOV, Plane Problsma of Hydrodynamios and Asrodynamios, (Ploakiys zadachi
giclrodinamiki i asrodinamiki), Chapter VI,§ 2, Gostekhizda.t, 1950; W. KoPPENFELS, F. STALLMAN, Praa:is d!lr konJormsn Abbildung, Berlin, 1959 (in
this book there is also a catalogue of mappings of different types of polygons).
OH.APTER XI
APPLICATIONS TO MECHANICS
AND PHYSJCSt
§ 1. Applications to hydrodynamics:
Steady flows
The established plane irrotational flow of an incompressible fluid is described
by the analytic function
(1)
w(z) = t/l(a:, y)+iV'(a:, y)
known as the complea: potential or chamcteriBticfunction of the flow; t/I is known
as the potential function, V' as the atream function. The curves t/I = const
are the equipotential linea, the curves V' = oonst a.re the stream linea. The
stream velocity V is connected with w(z) by the relations:
V
V
=
=
Vel« = Vx+iV1 = w'(z)}
lw'(z)I, oi: = - arg w'(z),
(2)
V=gradt/I.
Let 0 be a closed contour traversed in the positive direqtion (the contour
0 may also consist of the two sides of an arc traversed in opposite directions).
The quantity
I'= f Vada= f Vxda:+ V1 dy
c
c
=f
c
dt/I
(3)
is known as the circulation of the vector V with respect to the contour O.
The quantity
Q= fVnda= f<-V1 d11+Vxd11)= fd'P
c
c
(4)
c
(n is the outward normal to the closed contour 0, traversed in the positive
direction) is called the flow of the vector V through the contour O. Similarly
the flow of the vector V through the arc AB is defined as the integral
Vn dB
J
(the direction of the normal n must be indicated).
Combining formulae (3) and (4), WA obtain
I'+iQ
=
Jw'(z)dz.
AB
(5)
c
t For this chapter see [3, Chapter III] and the references given there.
243
244
PROBLEMS ON COMPLEX ANALYSIS
If to'(z) is defined within 0 and has a finite number of singularities there,
then
r+iQ = 2ni};resw'(z).
If a is a pole of the function w' (z), then w(z) possesses in the neighbourhood
of a an expansion of the form
w(:z:)
O-n
p
l
I'+iQ
= -(--)-n
0 +c 1 (z-a)+ ...
2n --+2-.-log(z-a)+o
z-a + ... +z-a
ni
It is said that the term I'~Q log (z-a)(I', Q, real numbers) defines a vortear
BOUrce Of strength Q and intensity
I', denoted by (a; Q, I')t, the term 2P _l_
nz-a
is a doublet with moment p, denoted by (a; p) (p is a complex number;
the argument of p determines the direction of the axis of the doublet passing
through the point a in the direction of a stream line), the remaining terms
(0-k)/(z-a)li define at the point a multipletB of order 2k.
Correspondingly, if at oo
w(z)
=
CnZ
n
P
I'+iQ
IJ-1
+ ... +2- z + - 2 -.-logz+c0 + - - + ... ,
:n;
m
z
I'+iQ
2:n;i log z defines at oo a vortea: BOuroe of strength Q and
intensity
the term pz/2:n; defines a doublet with moment p (the direction
of the stream line at infinity coincides with the argument of p), the remaining
terms o,.zk define multipletB of order 2k.
~oints at which V = 0, and consequently w'(z) = 0, are known as /ltag·
nation pointB of the fl.ow; from these points stream lines and equipotential
lines issue alternately. If a stagnation point is a zero of order (n-1),
these curves enclose an angle :n;/n. This branching of the curves may also
occur at infinity.
then the term
r,
.In problems 1376-1389 it is required from the given complex
potential of the fl.ow to construct the equipotential lines and stream
lines, to determine V, the singularities and stagnation points, the
strength and intensity of the vortex sources, the moments of the
doublets and also to investigate the behaviour of the fl.ow at oo.
1376. W := CZ (c = IX + i{J).
1377. w = zn (in particular, n = 2, 3).
r+iQ
1378. w = - 2-.- log z. Consider, in particular, the cases I'= 0
:n;s
and Q = O.
I'+iQ
z-a
1379. w =~log z-a ·
t If Q
= O,
then we have a oortea: (a; I'). If I'= 0, we have a aouroe
< O, it is often described as a ainlc.
(a; Q). If the intensity of the source Q
245
APPLICATIONS TO MECHANICS AND PHYSICS
1380. w = l/z. Determine also the velocity at the points 2 ± i.
R2
1381. (1) w = z+-;
(2) w =
z
R2
z--·
z
1382. w = l/z2 •
1383. w =log (z2 -a2) (a> 0). Determine also the velocity at
the points ±ia.
z2-a2
1384. w =log~+
2
z
(a> 0).
a
1385. w =!?.__log
2:n;
(z- _!_)z ·
1386. w=log(1+ : 2 )
1387. w =log (z2
•
+ :2 )
Q
1388. w = az+ 2:n; log z
•
(a > 0, Q > 0).
I'
(a >0, I'>O).
1389. w = az+2:n;i. log z
1390. Investigate the character of the flow in the region lzl
w
= a(z+
~2 ) + 2~i log z
~
R, if
(a> O, I'> 0).
Consider the cases: I'< 4naR, I'= 4naR, I'> 4naR.
1391. Find the complex potential w(z) of the flow in the whole
plane formed by the vortex sources {(ak;Qk,I'k)} (k=l,2, ... ,n)
and having at infinity the given velocity V00 = Ve1«.
1392. Is it possible for stream lines to emerge from a point hav:ing:
(1) a vortex, (2) a doublet, (3) a vortex and a doublet together~
1393. Find the law of variation of a vortex source, doublet
and multiplet, situated at a point a or a.t oo, for the following
schlicht conformal mappings of the neighbourhood of these points
(Ci =fi 0, C-1 =fi 0):
(1) C= a+Ci.(z-a)+ ... ;
246
PBOBLEMS ON COMFLEX .ANALYSIS
(2)
Ca: X + c_l
(3)
+co+ ... ;
C= eiz+eo+ ...
(4)
z
C=
+ ... ;
C-1
z-a
1394. Find the law of variation of the vortex source for the
n-sheeted mappings:
C= ix+c
11
(z-a) 11 +
... ,
c_ ..
.,,. =oc+-+
... ,
z"
In problem 1395 there is established the 81Jmmett'y principle, which together
with conformal mapping is widely used for the construction of flows (see
problems 139'7-1405).
1395. Prove that the fl.ow can be continued by the symmetry
principle a.cross a rectilinear or circular segment of a stream line
or equipotential line, a vortex source passing into a vortex source,
a doublet into a doublet, a multiplet, generally speaking, into
a set of multiplets of the same order. Find the stl't"ngth and intensity
of the reflected vortex source and the moment of the reflected doublet.
Rmt:ABK. It follows from the symmetry principle that if there are recti•
linear or circular segments on the stream line or equipotential line the fl.ow
muat be symmetrical with respect to this curve. This imposes a definite
restriction not only on the singularities of the fl.ow outside the given lines
but also on them or at their end points (if any).
1396. A fl.ow in the z. plane is formed of a finite number of sources,
vortices and doublets.
(1) Find the necessary and sufficient condition for the circle
lzl = R to be a stream line, if the sources, vortices, doublets:
(a.) a.re not situated on this circle;
(b) a.re all situated on it;
(c) some a.re on it and some not.
(2) With the same assumptions find the conditions for the
circle lzl = R to be an equipotential line.
1397. Find the complex potentials of the flows in the upper half.
plane Im z > 0 with the given singularities and the velocity Voci·
(1) The velocity V00 = V.
247
APPLICATIONS TO MECHANICS AND PHYSICS
(2) The vortex (a; I') and velocity V00 = O.
(3) The source (a; Q) and velocity Vco = 0.
(4) The doublet (a;p) and velocity Vco = 0.
(5) The vortex sources {(a1c;Q1c,I'1c)} (k=l,2, ... , n), the
doublet (a; p) and velocity Vco = V. What can be said of the
behaviour of the fl.ow at 001
(6) The vortex source (O; Q, I') and doublet (O; p); Vco =0.
What values can the moment .P of the doublet assume 1 Is the :ftow
always possible if r #= 01
1398. In the circle izl < R construct flows, having respectively:
(1) a vortex (a; I'); (2) a doublet (a; p).
1399. Find conditions for the possibility of constructing flows
in the circle lzl < R, if:
(1) There exist only the sources {(a1c; Q1c)} (k = 1, 2, ... , n,)
situated inside the circle;
(2) In addition to the sources of part (1) there are sources
{(a~; Q~)} (k = 1, 2, ... , m), situated on the circumference izl = R.
In both cases find the complex potentials of the flows.
1400. In the domain Jzl > R construct flows having respectively:
(1) The vortex (a; I'), the velocity Vco = 0 and circulation
at infinity co = 0;
(2) The doublet (a; p), the velocity Vco= 0 and circulation
r
l"'co=O;
(3) The velocity Vco = Ve111 and circulation I'co = 0;
(4) The velocity Vco = Ve'11 and circulation I' round the
circle
izl = R.
REMAIUC. The last two examples of problem 1400 give the streamlining
ofa circle with given velocity at infinity, without circulation and with circulation (see, for example, [3, Chapter III, § 49]).
In examples 1401-1405, using the symmetry principle construct
flows with given singularities (at infinity and at angular points
the velocity is equal to zero).
1401. In the domain Jzl > 1, Im z > 0, with the vortex (ia; I'),
a>O.
, !!.
1402. In the angle O<arg z <n/3, with the source (ae a ;Q),a>O.
1403. In the first quadrant Re z > 0, Im z > 0, with the source
(l;Q).
1404. In the first quadrant Re z > 0, Im z > 0, with the source
(1 ;Q) and the sink (i; -Q).
248
PROBLEMS ON COMPLEX ANALYSIS
1405. In the first quadrant Re z > 0, Im z > 0, with the source
(l+i; Q) and the sink (O; -Q).
1406. Construct a fl.ow in the whole z-plane, if it is known that
in the upper half-plane Im z > 0 there are vortex sources { (ak; Qk,
I'k)} (k = 1, 2, .. .,n) and a doublet (a; p), the x-axis is an equipotential line and the velocity V00 = Vei". Is such a flow always possible?
1407. Construct a fl.ow in the whole z-plane if it is known that
in the circle [z[ < R there are vortex sources {(ak;Qk,I'k)}(k=l,
2, ... , n) and a doublet (a; p), the circle [z[ = R is an equipotential line and the velocity V00 = Ve1". Is such a flow always possible?
1408. In the simply connected domain D, bounded by the contour 0, construct a fl.ow with streamline 0, having the vortex sources
{(ak; Qk, I'k)} (k = 1, 2, ... , n). Is such a fl.ow always possible~
1409. In the domain D, bounded by the contour 0 and containing the point at infinity, construct a flow with stream line 0, which
has the vortex sources {(ak; Qk, I'k)} (k = 1, 2, .. ., n) and the given
·
velocity V00 = Ve1". Is such a flow always possible?
In problems 1410-1417 the streamlining of bounded and unbounded contours (they must be stream lines) is considered. The problems are solved by means of conformal mapping onto the exterior
of a circle, the upper half-plane and a rectilinear strip.
1410. Construct the streamlining of a bounded contour 0 with
given circulation I' and velocity V00 = Ve1". What is the mapping
effected by the complex potential w(z) in the case I'= 01
1411. Construct the streamlining of the ellipse x 2 /a2
y2 /b 2 = 1:
(1) With given velocity V00 , without circulation;
(2) With given velocity V00 and circulation I'.
1412. Construct the streamlining of a flat plate [xi ~ 0, y = 0:
(I) With given velocity V00 , without circulation;
(2) With given velocity V00 and circulation I', defined
by the condition that one of the ends of the plate is a point of
departure of the flow (the Zhukovskii-Chaplygin postulate).
1413. Construct the streamlining of a Zhukovskiit profile with
given velocity V00 and circulation I', defined by the Zhukovskii-
+
t See problem 274.
.Al'PLICATIONS TO MECHANICS AND PHYSICS
249
Chaplygin postulate (the sharp end of the profile must be a point
of departure).
In problems 1414-1417 construct the streamlining of the given
contours.
1414. The parabolas y2 = 2px (from the inside and from the
outside).
1415. The right branch of the hyperbola x1 /a8 -y1/b• = 1 (from
the inside, and from the outside with the velocity V00 = 0).
1416. The half lines:- oo < x < - 1, y = ± n.
1417. The half lines: 1 < [x[ < oo, y = 0.
In problems 1418-1422 periodic flows {V(z+co) = V(z)} and
flows in curvilinear strips (channels) are considered. For the construction of these flows the curvilinear strips must be conformally
mapped onto a rectilinear strip, and then the fl.ow continued by
the symmetry principle and use made of the expansion of a meromorphic function into a series of partial fractions.
In problems 1418, 1419 investigate the singularities, construct
schematically stream lines and equipotential lines and determine
the velocity at infinity in the strip of periods for periodic flows
with the given complex potentials.
'· 1418. (1) w
1419. w =
=
2Qn log sin z; (2) w
=
r.
2ni log sin z.
in cot z (0 < arg p < ; )·
1420. In the rectilinear strip 8: 0 < x < co of the z-plane construct the fl.ow formed by the vortex source (a; Q, I'), a e 8, having
the given velocities V (x+ioo) = V1 , V(x-i oo) = V1 • Is such
a fl.ow always possible 1
Construct schematically stream lines and equipotential lines
if I' = 0 or Q = 0.
HINT. Continue the fl.ow by the symmetry principle and use the result
of problem 1418.
1421. In a rectilinear strip of the z-plane 8: 0 < x < co
construct the flow formed by the doublet (a;p), a es, having
the given velocity V(x ± i oo) = iV. Construct schematically
the stream lines and equipotential lines.
1422. In a curvilinear strip of the z-plane 8, bounded by
contours 0 1 , 0 1 , construct the fl.ow streamlining 0 1 , 0 1 , having
250
PROBLEMS ON OOMPLEX ANALYSIS
given vortex sources, and doublets in S and given velocities
Vv V2 at the points at infinity !J1 , !J 2 of the strip S. Indicate
sufficient conditions for the existence of such a flow.
A fl.ow is said to be doubly pM'iodic if its velocity w'(z) is an elliptic function.
Elliptic functions are doubly periodic meromorphic functions, possessing
periods 2<.o and 2<.o', where Im (ro'/ro) ,,P 0 (in what follows it is assumed
that Im (<.o' /<.o) > 0). It follows from this definition that
f(z+2mro+2nro')
= j(z),
where m and n are any integers or zero.
The parallelogram with vertices z0 , z0 +2co, z0 +2ro', z0 +2co+2ro' (z0 is
an arbitrary point) is lmown as the parallelogram of periods.
If j(z) is an elliptic function different from a constant, then it possesses
the following properties (Liouville's theorem):
(1) j(z) has at least one pole in the parallelogram of periods;
(2) The sum of the residues of the function j(z) at all the poles situated
in the period parallelogram is equal to zero;
(3) In the period parallelogram the equationj(z) =a has the same number
of roots for any complex number a, finite or infinite (this number of roots
is known as the order of the elliptic function);
(4) The difference between the sum of all the zeros and the sum of all the
poles of the function j(z), situated in the period parallelogram is equal to
some period of it, that is,
2 cx11:- 2 Pt = 2µro+2vro'
(µ and
v are integers).
The Sigma junction of Weierstrass is defined as the integral function
O'(Z) = Z
n'(
z)
..!.+~
1-Q e!J
2.QI,
(6)
where D = 2nro+2mco' and the product extends to all D different from
zero. The function a(z) is odd.
The zeta junction of Weierstrass is the meromorphic function
a'(z)
I
2'(--+-+-,
I
I
z )
C<z>=-=-+
a(z)
z
z-D
D
D•
(7)
where the summation extends to all D different from zero. The function
C(z) is odd.
The Weierstrass function p(:z:) with periods 2co and 2<.o' (see pa.ge 236)
is connected with C(z) by the relation p(:z:) = - C'(:z:).
Since
[C(z+2<.o)-C(:z:)]' == p(:z:)-p(z+2ro) = 0,
it follows that
APPLICATIONS TO MECHANICS AND PHYSICS
251
and similarly
where '1 and 71' are constants. Using the oddness of the function CCz), it is easy
to prove that '1/ = CC<») and 71' = CC<»').
The quantities 71, 71', <» and <»' are connected by Legendre's relation
11<»'-11'<» = :n.i/2.
Let us use the notation
11 = 11u 11' = 71a and 11+11' = 11••
<» = <»u <»' = <»a and <»+<»' = <»2·
The functions aa:Cz) are determined by the relations
flt,"a(z-<»1:)
<>"kCz) = -e
a(<»t)
Ck= I, 2, 3).
(8)
Correspondingly
C C) k
alcCz)
z - <>"tCz) •
C9)
The functions <>"tCz) are connected wit~ Weierstrass's pCz) and the functions
sn z, en z, dn z of Jacobi by the following formula.et.
vr pCz)-e11:] =
a11:Cz)
aCz) •
(10)
a(z)
a1 (z)
snu=y(ei-e8 ) - - , cnu=--, dnu = <>"2Cz) ,
(11)
<>"a(z)
<>"a(z)
<>"aCz)
where u = zy(e1-e3 ).
It is possible to express any elliptic function in terms of a(z) and C(z).
If f(z) has only the simple poles b11: with residues At (k = I, 2, ••. , n), t·hen
n
f(z)
= _2 AtC(z-bkl+O.
CI2)
k-1
If f(z) has the zeros a11: and poles bt (of any multiplicity), then
f(z)
=
oa(z-a1)a(z-a1) ... a(z-an) '
a(z-b1) a(z-b 2 ) ••• a(z-b:)
where
b: =
t ea:
= pC<»•l· See page 236.
n
n-1
2aa:-2bk.
k=l
k=1
(13)
252
PROBLEMS ON OOMPLEX ANALYSIS
The Theta functions of Jacobi are the functions 81M (i
by the relations:
fli(v) = i
00
1, 2, 3, 4), defined
1)"
(
.J;
=
(-l)•q n-2 e(lt-l)nlu
n--oo
01 (v)
= 01 ( v+
! ),
0,(v)
=
= q~ e"1uo1 ( v+
81 (11)
~ + ~),
-iqt emuo.( v+ ; ) •
(lo)
em-r, T
where q =
= (J) 1 /(J).
The Theta functions are connected with the Sigma functions by relations
of the form
(16)
al( ~~)
.,... 0
-- elma
(
~
81+1(0)
("i
= l ' 2, 3) ,
(17)
where z = 2°'1v.
The advantage of the Theta functions is the rapidity of the convergence
of the series defining them. Using formula (16), the representation (13)
can be written in the form
(z-a1)o1(z-a•)
(z-an)
2
... 01 2
°'
°'
°' .
z;:1 )o1(z;:•) ... z;;:)
By the use of formulae (11), (16) and (17) it
possible to write the
expression for the Jacobi functions sn z, on z,
z
in terms
the Theta
functions.
f(z)
=0
01 2
(18)
81 (
81 (
is
dn
of
°'
In what follows it is a88Uilled that
is a real number, °'' is purely imaginary, that is. the period parallelograms are rectangles.
1423. Show that the function /(u) =
~ C(u-ix)+Ou is the
complex potential of a doubly periodic fl.ow with one doublet
(ix; M) (Mis the moment of the doublet) in the period parallelogram.
In particular, consider the cases:
( 1) ix = 0 and the lines Im u = ± Im ro' are stream lines;
sketch the stream lines and equipotentials; investigate the conformal mapping effected by the function t = f(u).
APPLICATIONS TO MECHANICS AND PHYSICS
253
+
(2) f(u
2w) = f(u); sketch the stream lines and
equipotentials and investigate the mapping t = f(u).
1424. Show that the flows defined by the complex potentials
Cdu) (k = 2, 3, 4), reduce to the flows of problem 1423 (for
0 = 0) by means of displacements in the u and C-planes.
HINT.
Make use of formulae (8) and (9).
1425. Show that the flows defined by the complex potentials
(k
=
2, 3, 4)
and
Z(u)
(see page 232), are reduced to the flows of problem 1423, (2)
by means of linear transformations.
1426. Show that the fl.ow defined by the complex potential
E(u) (see page 232) is reduced to the fl.ow of problem 1423, (1)
by means of linear transformations.
\HINT.
As a preliminary prove the relations
E
K 1 = (e 1 -e3 )ro• and - =
K
e
ro
--+-1].
e -e
K
1
2
3
2
1427. Find the complex potential f(u) of the doubly periodic
fl.ow with two doublets (a;M), ({J;N) in the period parallelogram.
Explain the conditions for f(u) to be an elliptic function
and for the curves Im u = ± Im w ', Re u = ± w to be stream
lines and equipotentials (or conversely); sketch the stream
lines and equipotentials.
In problems 1428-1430 investigate the doubly periodic flows
defined by the given complex potentials f(u).
1428. sn u.
1429. en u.
1430. dn u.
1431. Find the complex potential f(u) of the doubly periodic
fl.ow with two vortex sources (a; Q, I'), ({J;-Q, -I') in the
period parallelogram. In particular, consider the case a = 0,
a = w, a = w + w' and f3 = w'.
Find the form of the function f(u) which satisfies the condition f(u+2w) = f(u).
In problems 1432-1434 investigate the flows defined by the
given complex potentials f(u).
254
l'ROBLEMS ON COMl'LEX .ANALYSIS
1432. (l)logsnu; (2) logcnu; (3) logdnu.
1433 . .p(u).
u
1434. log 61cM (v = 200 , k = 1, 2, 3, 4).
For the construction of a complex potential f (z) in a doubly connected
domain D it is usually conformally mapped onto a circular ring R: f! < !ti < I
(µ = l/o is the modulus of D) ; in its turn the ring R with the radial cut
.
...,,
[(!, I] is mapped by means of the function t = e "' onto the rectangle with
vertices 0, 2<.u, 2<»+<»', <»' in the u-plane in such a way that the edges
of the cut become the vertical sides and T = <»'/<» = i/nlog I/e. The char.
acteristics of the flow in the rectangle are determined by the method of solu·
tion of problem 1393. Since the bases of the rectangle are stream lines,
the flow is continued through them by the symmetry principle (see problem
1395), after which the complex potential !Z>(u) of the resulting doubly periodic
flow with periods 2<.u, 2<.u' is determined; then f(z) = !Z>[u(z)] (see the book
by L. I. Sedov mentioned on page 242).
1435. Find the complex potential of the flow:
(1) In the circular ring R: r 1 < jzj < r 2 with circulations
a.long the boundary circles;
(2) In the arbitrary bounded doubly connected domain Dt with circulation I' a.long the boundary contours;
(3) In the exterior of two circles lying outside one
another with circulations± r on the boundary circles and
with the condition Vco = 0;
(4) In the doubly connected domain D, containing the
point at infinity, with the circulations ±
along the boundary contours and with the condition V00 = 0.
1436. Construct the fl.ow in the circular ring R: (! < jzj < 1,
formed by the doublet (a;p) ((! <a < 1) and streamlining
without circulation the boundary contours. Investigate the
mapping t = f(z) and sketch the positions of the stream lines.
r
r
HINT. Make use of the solutions of problems 1393,(1) and 1427.
1437. Construct the fl.ow in the doubly connected domain
D, containing the point at infinity, which streamlines without
circulation the boundary contours and has at infinity the given
velocity Vco = Veirr..
t Here and in what follows it is assumed that functions mapping the
domain D onto the ring are known.
APPLIOATIONS TO MECHANICS AND PHYSICS
255
1438. Construct the fl.ow formed in the circular ring R:
1, by a doublet and quadruplet located at the point
z = 1 and streamlining the boundary circles without circulation. Sketch the distribution of stream lines and investigate
the mapping t = f(z). In particular, consider the case of a doublet only.
e < lzl <
HINT.
Write the complex potential /{z) in the form
c_, + (z-1)
C-1
/(z) = (z-1)•
(
l
+co+e1 z- )+ ...
and explain what values of c_1 and c_ 1 are possible.
1439. (1) In the circular ring R:e < lzl < 1 construct the fl.ow
formed by the vortex (a; I')(e <a< 1) and streamlining the
boundary circles with circulations I'1 (along the circle lzl = 1)
and I'2 (along the circle lzl = e). Is it possible to specify I',
I'1 , I'2 arbitrarily~ Consider, in particular, the cases I'2 = 0
and
= -I'l"
Investigate the mapping effected by the functions t = f(z),
r,_
lnl t
-4n; 1
.8 = er in the first case and by the functions t = f(z), .8 = e r
4m/Jo
and s = y'(.8-3 0 ) (3 0 = -e----r-, where VJo is the value of the
stream function at the stagnation point) in the second case.
Sketch the stream lines and equipotentials in the u-plane.
(2) In the doubly connected domain D, containing the
point at infinity, construct the fl.ow streamlining the boundary
contours with given circulations I'v
and having at 00 the
velocityV00 = Ve111•
r,_
HINT. The solutions of problems 1487 and 1439, (1) are superposed by
means of conformal mapping.
Unsteady vortex flowst
Let w(z) be a complex potential with the isolated singularities {ak}
(k = 1, 2, ... ).If they are fixed we have a steady flow; if however, they are
movable we have an unsteady flow in which the velocity at each singularity ak
is determined by the formula
da11:
d.t
=
[dw1c(z)]
dz
z=ak
t For this section see: N.E. KooHIN, I.A. KlBEL' and N.V. RoZE, TheO'l"lltfuaJ,
hydromechanics (Teoreticheakaya gidromekhanika), Chapter V, Gostekhizdat,
1948. For problems 1448-1450 see: F. OBERHETTINGER and w. MAGNUS,
..i.lnwendungen der elliptiBohen Funotionen in Physik und Teohnik, Chapt.er IV,
1949.
256
PROBLEMS ON COMPLEX ANALYSIS
where w11:(z) = w(z) - S(z, a1:), S(z, a11:) is the principal part of the expansion
of w(z) close to a,., the logarithmic term also being included in this if a11: is
a vortex source, that is, the velocity of the point ak is determined by the
action on it of all the remaining singularities. In this case ak = a11:(t) and
the complex potential w(z) depends on the time.
1440. Prove that two vortices (z1 = x1+iy1; I'1), (z 2 = x2
+iy2 ; I'2), acting on one another rotate round a "centre of
.
t• "
I'1X1 +I'2X2
I'1Y1 +I'2Y2 .f r +r .J. 0
d
mer 18. Xe= r1+r2 ' Ye= --I'1+r--;:-· l
1
2 r
an
move steadily perpendicular to the straight line connecting
them if I'1+r2 = O.
1441. Along the real a.xis there is an impenetrable wall.
At a distance h from it there is located a vortex of intensity
I'. Prove that it moves parallel to the x-axis with velocity
I'
4nh·
HINT.
Apply the symmetry principle.
1442. Prove that a vortex located in a circle with an impenetrable boundary circumference, moves a.long a concentric
circumference.
1443. Prove that n vortices of the same intensity I'; arranged
at a certain instant at the vertices of a regular n-gon inscribed
in a given circle of radius R, move along this circle with conI' (n- l)
stant speed 4 nR ·
1444. Prove that a vortex located in the first quadrant
x > 0, y > 0 with impenetrable walls moves a.long the curve
1
1
-+-=0.
xs
y2
1445. The fl.ow of a fluid is ca.used by a vortex cha.in consisting of an infinite number of vortices of identical intensity
I' situated at a certain instant at the points Zt = z0 +kl (k
= 0, ± 1, ± 2, ... ). Prove that the fl.ow is steady with complex potential
w(z) =
~
±
log sin
7
(z-z0 ).
It is assumed that the velocity V(m+iy) remains bounded as
oo. See problem 1420.
REMARK.
y
~i
APPLICATIONS TO MECHANICS AND PHYSICS
257
1446. Prove that a vortex of intensity I', located in a rectilinear channel with impenetrable walls moves parallel to these
walls with constant velocity -
~
tan
~a
(l is the width of the
channel, a is the distance from the vortex to the centre line
of the channel). What is the fl.ow if a = 01
144'7. The fl.ow of a fluid is caused by two vortex chains
consisting of vortices of intensity I'1 , which at a certain instant
are situated at the points z11> = z~1>+kl, and vortices of intensity I'2 situated at the points 42> = z~2>+kl (k = 0, ± l, ± 2, ... ) .
Prove that the chains move with the same velocity when
and only when I'2 = -I'1 , this motion taking place parallel
to the x-axis when and only when zf,8>-41>= ik, the symme-
trical arrangement or
z~2> - 41> =
!+
ik the chessboard arrange-
ment (k is an arbitrary real number); the speeds of motion in
.
. 1y b y 2f
I' cot h -l:nk an d 2f
I' tan h -z:nk
t h ese cases are given
respective
(I' is the magnitude of the intensity of the vortex).
REMABX. It is assumed that as y-+ ± oo the velocity V(m+iy)
remains
bounded.
1448. The fl.ow of a fluid in a rectilinear channel with impenetrable walls a:= ±k is caused by a vortex chain consisting
of vortices of intensity I', situated at a certain instant at the
points a+ikl (k = 0, ± 1, ± 2, .. ., ; -k <a< k). Prove that
the complex potential of the fl.ow is given by
I'
81
(z 4k a)
w(z) = - 2 . log-(---'---..,..-),
m
i-2k+a
01 -4-.,,,-
where for the function (;Ji we have 2ro = 4k and 2ro' = il, and
that the vortex chain moves steadily parallel to the walls of
the channel with constant speed
What is the fl.ow if a = 01
..!:___
[C(2k-2a) - !l (k-a)] ·
2:n;
k
258
l'BOBLEMS ON OOMl'LEX ANALYSIS
At every instant the flow is doubly periodic with periods
4h, 2<.u' = il.
REMA.RX.
2<.u
=
1449. The fl.ow of a. fluid in a rectilinear channel with impenetrable walls X = ± h is caused by two vortex cha.ins consisting respectively of vortices of intensity I', and -I', situated
at a. certain instant at the points a+ilcl, and
-a+i(~ +1cz)
(le = 0, ± 1, ± 2, ... ; -h < a < h), respectively, in chessboard
order in the channel. Prove that the complex potential of the
fl.ow is given by
o,(~)
r [
w(z) = -2:ni.
log
o,
(•+:{) ]
( z-2h+a ) - log
(
il. )
01 - - - - z+2h-a- 4h
0
2
1
4h
and that the vortex chains move steadily parallel to the walls
of the channel with the constant velocity
r [2
-- 2:n
a
1
C< 2a)- 217 k"+2
.p'(2a)
.p(2a)-e1
What flow is obtained in the limit as h
1
+2
-+
.p'(2a)
.p(2a)-e3
J
•
001
In order to obtain the final expression for the speed of motion
of the vortex chains use the addition formula of the C-function:
HINT.
C<u+v>
= CM+CM+
! t>;~~>=~<~> ·
1450*. Find the motion of a. vortex of intensity I', located
in a rectangle with vertices 0, 2a, 2bi, 2a+2bi and with impenetrable walls. In particular, consider the case when at a.
certain instant the vortex is at the centre of the rectangle or
close to it.
§ 2. Applications to electrostatics
A plane electrostatic field with intensity E = E,.+iE1 = Eei« is characterised by the analytic function w(z) = u+iv, known as the complex potential; v is known as the potential function (it is always single valued), and u is
the force function. The curves v = const are the equipotential ZineB, and the
APPLICATIONS TO MECHANICS AND PHYSICS
curves u
= const
are the lines of force of the field. We have
E
Ex
259
grad11 = -iw'(z)
= -
E = lw'(z)I, a: = -n/2- arg w'(z),
= -avfa:i: = au/CJy, E 1 = -a11/8y = -au/ax.
In all the problems of this section where it is a question of electrostatic
fields in domains bounded by one or several boundary contours it is assumed
that the potential function is constant along each simple contour (that is,
every such contour is a conductor).
If a is a pole of w' (z) and close to a the function w has the expansion
w(z) =
C-n
(z-a)n + ... +
pi
"l
l
z=a
+2qi og z-a
then the term 2qi log - 1-
z-a
+ c0 +c1 (z-a) + ... ,
defines at the point a a plane point charge of
magnitude (! = 2q, denoted by (a; 2q) (on unit length of a rectilinear conductor perpendicular to the z-plane at the point a, there occurs the charge
q); the term pi/(z-a) defines at the point a a dipole of moment p, denoted by
(a;p) (pis a complex number; its argument defines the direction of the
axis of the dipole); the remaining terms
(z".::)k
(k
=
2, 3, ... , n) define at
the point a a multipole of order 2k.
Correspondingly, if at oo
• 2qi"log z+c0 +-zC-1
w (z) = cnzn + ... +piz+
+ ... ,
then the term 2qi log z defines at infinity a plane point charge of magnitude
(! = 2q, and the term piz a dipole of moment p.
If the function w = u+i11 is considered as the complex potential of an
electrostatic field E = -iw'(z) and at the same time as the fl.ow of a fluid
with velocity V = w' (z), then this leads to the following electrohydro·
dynamical analogy:
Fluid fl.ow
Electrostatic field
u
Potential function
Force function
u = const
Equipotential lines
Lines of force
Stream function (possibly
many valued)
Potential function
(always single valued)
11
11
= const
118 -11i
f
du
Stream lines
Equipotential lines
Outflow of fluid
Potential difference
Circulation I'=
f
VadB
Flux N
=
f
End II
260
PROBLEMS ON COMPLEX ANALYSIS
Fluid flow
Electrostatic field
Point charge (a; 2q)
Vortex (a;I')
r
q=-
4n
Source
Doublet with moment p
Dipole with moment 2p .
ni
Streamlining with given
vortices and doublets
Field with given charges,
dipoles and equipotential
boundary curves
In problems 1451-1458 it is required, from the given complex
potentials, to determine the force and potential functions, the
intensity of the field, the nature of the singularities (including
that at oo ), and also to sketch the families of lines of force
and equipotential lines (q is a real number). Compare with
the solutions of problems 1376-1389.
1451. w = cz (c = cx+ip).
1452. w = 2qi log_!__ ·
z
1453. w
=
z-a
z-b
2qi log--·
1454. w = 2qi log (zB-a2) (a
1455. w = pi/z (p = jpjeicc).
1456. w = z± R2/z.
> O).
1457. w = piz+2qi log..!:.. (p
> 0, q > 0).
z
n
1458. w = piz+2i
.2
k=l
qk log-1-
z-a1c
(p
> 0,
q,.
> 0,
al <as< ... <an).
1459. Find the law of variation of the point charge (a; 2q)
and of the dipole (a; p):
(1) In a single-valued conformal mapping;
(2) In continuation by the symmetry principle across
a rectilinear or circular segment of an equipotential line.
APPLICATIONS TO MECHANICS AND PHYSICS
261
1460. Show that the complex potential of the electrostatic
field formed by the point charge (a; 2q) in the arbitrary singleva.lued domain D is determined by the formula
w=2qilog /(!,a) +c,
where /(z,a) is the function which conformally maps the domain
D onto the unit circle in such a way that/(a,a) = 0, and c is a real
constant.
Establish the connection between the potential function v(z)
and the Green's function of the domain D (see problem 1068).
In problems 1461-1467, using the results of problem 1460
or the symmetry principle, find the complex potentials of the
electrostatic fields formed by the given point charges in the
domains indicated.
1461. In the upper half-plane Im z > 0, by the charge (z 0 ; 2q).
1462. (1) In the circle [z[ < R, by the charge (z 0 ; 2q);
(2) In the exterior of the circle [zj > R, by the charge
(z 0 ; 2q).
1463. In the exterior of the ellipse x2/a2 +y2 /b2 = 1, by the
charge ( oo; 2q).
1464. In the exterior of the segment jxj < R, y = 0, by the
charge ( oo; 2q).
1465. In the exterior of the square !xi <a, IYI < d, by the
charge ( oo; 2q).
1466. In the rectangle jxj <a, IYI < b, by the charge (O; 2q).
1467. In the rectangle 0 < x < 2a, 0 < y < 2b, by the
charge (z 0 ; 2q).
In problems 1468-1472 construct the electrostatic fiE>lds
formed by the given dipoles.
1468. In the circle jzj < R, by the dipole (a; p).
1469. In the exterior of the circle Jzl > R, by the dipole
(a; p).
1470. In the exterior of the segment jxj < R, y = O. by tlw
dipole ( oo; p).
1471. In the exterior of the ellipse x 2 /a 2 +y2 /b 2 = 1, by the
dipole ( oo; p).
1472. In the rectangle jxj < a, IYI < b, by the dipole
(O; p) (pi = ee111 ).
1173. Prove that the electrostatic field formed by the dipole
(a; p) in the arbitrary simply connected domain D, is deter-
262
PROBLEMS ON OOMPLEX ANALYSIS
mined by the complex potential w = /(z), where the function
/(z) maps the domain D onto the exterior of a horizontal segment in such a way that f(a) = oo, and the principal part
of /(z) at the point a equals pi/(z-a) if a ..p oo, and equals
piz if a= oo.
Find f(z) if a known function t(z) maps the domain D:
(1) Onto the interior of the unit circle, if a ..p oo, where
t(a) = 0, t'(a) > O;
(2) Onto the exterior of the circle !ti > R, if a= oo, where
t(oo) = oo and t'(oo) = 1.
1474. Construct the electrostatic field formed by the point
charges {(a1:; 2q1:)} (k = 1, 2, ... , n) and the dipoles (a; p) in
the simply connected domain D.
Let g(C,z) be the Green's function of the domain D (see page 165), the
boundary of which consists of the piecewise smooth simple contours I'u
... ' I'n; also let ti be the inward normal to and let be traversed in the
positive direction with respect to D. If u(z) is a function harmonic in the
domain D and continuous on I', then it follows from Green's formula that
r
r
r,,
r
J
u(z) = - 1u(C) 8g(C, z) 'da
2nr
an.
or
u(z)
=
J[u<C> ana log IC-zl
l
1
aum]
-log IC-zl --a,;-
l
2nr
da.
If the domain D contains the point at infinity and the function v(z) is
harmonic there, u (oo) must be added to the right hand sides of the given for·
mulae. Then in the neighbourhood of the point at infinity the Green's func·
tion g (z,oo) can be represented in the form
g(z, oo) =log lzl+Y+o(
i}.
The quantity
/' = lim [g(z, oo)--log lzl]
is known as the Roben's constant of the closed set representing the com·
plement of D on the z-plane; the quantity e-7 is known as the capacity of
this set.
1475. Prove the following assertions (n is the inward normal):
.
1
1
(1) g(z, a)= log-1 - -1 - 2-
z-a..
n
f ag(C,
a)
1
.
.
--!l-log-1, . - -1 ds if a#: oo,
F
MZ.
.,-z '
263
Al'PLICATIONS TO MECHANICS AND PHYSICS
(2)
g(z,
oo) =
I
r- 2n
J
ag(C,oo)
an
r
log
I
IC-zl
ds,
if zeD and the domain D contains the point at infinity;
1
ag(C,oo)
1
.
(3) 2n
an log IC-zl
ds = ,... If z e D and the domain
r
D contains the point at infinity;
J
(4) ;nf
ag~~,a)
ds =I, if a#: ooor ifa = ooeD.
r
HINT. In part (1) use the symmetry property of the Green function
g(C, z) = g(z, C) and the integral representation of the harmonic function in
terms of its boundary values. In part (2) use the integral representation of
the function log!C-zf+g(C,z)-g(C,oo), which is harmonic in D, pass to the
limit and use the symmetry property g{oo,z) ... g(z,oo). In part (3) proceed
in the same way but start with the function log IC-zf-g(C,oo)t.
The function
l
tJ0 (z) -2qlog-1- 1
z-a
is known as the logarithmic potential of the point charge (a; 2q). In the
extended z-pla.ne tJ0 (z) represents the logarithmic potential of the two point
charges: (a, 2q) and (oo; -2q).
Let the contour I' satisfy the conditions indicated on page 262, and let
and •<C> be real and continuous on
The integral
am
r.
"(z)
=
J
(!{C) log
r
IC~zl
da
am
is known as the logarithmic potential of the simple layer wUh density
(in
three dimensions there corresponds to it the potential of a charged cylindri·
cal surface with base r and surface density of charge : • that is bearing the
charge
~ da8 on the element of area cfal).
The function tJ(z) is continuous in the finite z-plane and is harmonic everywhere outside I', except at the point z = oo, where it has a logarithmic
singularity
"(z)
=
-2q log l•l+o{
i),
2q
=
f
e(C)da
r
t See R. N:mvANLINNA, Eindeutige analytiache Funkf.ionen, Chapter V, § 2,
Springer, Berlin, 1936,
264
PROBLEMS ON COMPLEX ANALYSIS
(this means that the potential "(z) corresponds to a charge (oo; -2q)).
The integral
tJi.{Z)=
J
r
8
11m
1
log IC-zi ds
an
r
is known as the logarithmic potential of the double layer with density 11((:) (if
is the boundary of a domain there is on I' a distribution of dipoles with axes
directed along the inward normal to I'; 11((:) is the density distribution of
dipole moments). If 8((:,z) is the angle between n and a vector going from
Cto z, and d<J> ((:, z) is the angle subtended by the element of arc ds at z, then
111(21) =
J
r
11((:)
cos 8((:, z)
IC-zl
J
ds =
11((:)d<1>((:, z).
r
In particular, for a closed contour I' and 11 ((:) = 1
f :n
l'
I
2n, if z is inside
log
IC~zl
ds
=
z
r,
r,
n,
if
0,
if z is outside
is on
r
(see also problem 1066).
The Green function g(z, a) of the domain D can be considered as the po·
tential of the electrostatic field formed by the point charge (a; 1) ifthe boundary I' of the domain D is earthed. Problem 1475, (1) shows that in the
case a
00 the earthing of r is e~uivalent to the placing on r of a charge
*
of linear densityg((:)
= - 2~ ag~~ a)
• In this case, by part (4) of problem
1475, the total quantity of charge is equal to -1. In the case a = oo the
point charge (oo; 1) and the earthing of rare together equivalent to the
placing on
r
of a charge with density given by
em= - 2~ au(~~oo)
of total value -1 and to the addition of a field with constant potential y (see
1475, (2)). In these oases the given distributions on r are said to be induced
by the charge (a; 1).
In problems 1476-1479 find the density e(C, a) of the distribution induced by the charge (a; 1) on the contour
and
the corresponding potential v(z,a) of this simple layer for the
domains bounded by the contour
1476. I' is the real axis, Im a > 0.
1477. (1) I' is the circle lzl = R, lal < R;
(2) I' is the circle lzl = R, lal > R (consider, in particular, the case a = oo).
1478. I' is the real-axis segment lxl
R, y = 0, a= oo.
1479. r is the ellipse x"/a.2+y2/(32 = 1, a= 00.
1480. Considering the Green function of the domain D
containing the point z = oo to be known, solve Boben' s problem:
r
r.
<
APPLICATIONS TO MECHANICS AND PHYSICS
265
Find the density of the distribution e(C) on the boundary I' of
the domain D due to unit charge creating outside D and on
I' a constant potentialt.
HINT. See problem 1475, (3) and (4).
In problems 1481-1483 solve Roben's problem for the given
domains D.
1481. D is the exterior of the circle jzj > R.
1482. D is the exterior of the segment jxl < R, y = O.
1483. D is the exterior of the ellipse x2/as+y2/b2 = I.
In problems 1484-1487 find the capacity (see page 262)
of the closed sets.
1484. jzj < R.
1485. jxj
R, y = 0.
1486. x2/a2+y2/b2
I.
1487. jz2-a2[ a2 (a > 0).
1488. On the simple closed contour I' let there be given
a real function c/>(C) continuous and differentiable along the
contour.
Prove that the real part of the Cauchy type integral
-1
c/>(C) dC is the logarithmic potential of a double layer
23ii r C-z
with density c/>(C) and its imaginary part is the logarithmic
<
<
<
-J
:!: ·
1
2 31:
1489. Prove that the function v(z), bounded and harmonic in the
upper half-plane Im z > 0, can be represented as the logarithmic
potential of a double layer:
potential of a simple layer with density -
00
v(z) =
~
f
v(t)
a~ log
It I zj dt.
-oo
If however v(z) is regular at infinity, then it can also be represented
as the logarithmic potential of a simple layer:
f
00
1
v(z) = v(oo)-31;
-00
ov(t)
I dt.
- 0- l o g 1 1
n
1t-z
t For the general Roben's problem which requires a non-negative distribution of unit charge on a given set E to be found in order that the corresponding logarithmic potential should assume one and the same value at every
point of the set E, see the book by R. Nevanlinna mentioned on page 263.
266
PROBLEMS ON COMPLEX ANALYSIS
1490. In the upper half-plane Im z > 0 find the complex potential of an electrostatic field if its potential v(z) assumes on the real
axis given piecewise continuous values. Write the potential function
in terms of the harmonic measures of the corresponding segments
of the real axis (see page 170):
(1) cf> in the interval (-oo, a), 0 in the interval (a, oo);
(2) cf> in the interval (a, b), 0 in the intervals (-oo, a),
(b, oo);
(3) c/>1 , c/> 2 , ••• , cf>n respectively in the intervals (-oo, a 1 ),
(a1 , a 2), ••• , (an-l• an) and 0 in the interval (an, oo) (here a1 <a,<
... <a,,);
(4) c/> 0 in the interval (an, oo), c/>1 , c/> 2 , ••• , cf>n respectively
in the intervals (-oo, a 1 ), (ai, a 2) and so on.
HINT. In part (1) use conformal mapping onto a strip; in the remaining
parts use the method of superposition (it is possible also to use Schwarz's
integral formula for the half-plane, see page 170).
1491. Find the complex potentials w(z) and the potentials v(z) in
the following doubly connected domains with a given difference
d = v2 -v1 of potentials v1 , v2 on the boundary contours:
(1) In the circular ring r 1 < lzl < r2;
(2) In an arbitrary doubly connected domain D.
1492. Prove that if D is an arbitrary doubly connected domain
and on each of the contours bounding this domain the potential
function assumes constant values (v1 and v2 ), then
w(z)
=
i(;2-v1 ) logt(z)+e+iv1 , v(z) = v12-v1 log lt(z)!+v1 •
ogµ
ogµ
where t(z) conformally maps D onto the ring 1 < !ti < µ (µ is the
modulus of D) and the boundary contour with potential v1 passes
into the circle ltl = 1; e is a real number.
1493. Find the complex potentials in the given doubly connected
domains (the potentials v1 and v2 on the boundary contours are
constants).
(1) In the exterior of the circles lz± al = R (a> R) (v1
is the potential on the circumference of the circle on the left).
(2) In the exterior of the circles lzl = r 1 (potential v1 ) and
lz - al = r 2 (a> r 1 +r2 ).
(3) In the non-concentric circular ring, bounded by the
circles lzl = R (the potential v1) and lz-aj = r (0 < a < R - r).
(4) In the ellipse x'l-/a2+y2/b" < 1 with a cut along the
segment joining the foci (the potential on the ellipse is v1 ).
267
APPLIOATIONS TO MEOHANIOS AND PHYSIOS
(5) In the exterior of the segments 1
<
)x)
<
!,
JI= 0
(0 < le< 1). On the segment on the left the potential is v1 •
(6) In the exterior of the segments !xi < 1, y = ± n. On
the upper segment the potential is vp
1494. Let D be a multiply connected domain with boundary I',
consisting of n piecewise smooth contours I',, (le= 1, 2, ... , n), and
let ro,,(z) be the harmonic measure of I'" (see page 170). If the domain
D is bounded we shall consider I'n to be the exterior boundary contour. Prove the following assertions:
(1) If the domain D is bounded, then
1
<.Ot(Z) = - 211:
f
ow,,(C)
on
r
Wn(z)
=
1
log IC-zl da
f
1
1- 2n
OWn(C)
r
on
(le= 1 '2, ... 'n-1)'
1
log IC-zl d8.
If the domain D contains the point at infinity, then
1
Wt(z) = ro,,(oo)- 21'
f --a;ow"(C)
1
log IC-zl da
(le= 1, 2, ... , n).
r
(2) For points z not belonging to the domain D the right
sides ofthe equations given in part (1) assume the value 1 in the
domain complementary to D bounded by I't (or I'n), and 0 in the
domain complementary to D, bounded by the I'1 (i =F le).
REMARK. By part (1) the functions <»t(z) represent in the domain D
potentials identical with the distributions induced by the charges distributed
onI',withdensitiesofthelayer!?t(C) = - -1- a<»at(C) .Inthecaseofa bound-
2n
n
ed domain D the quantities ro 1 (z), ... , <»n- 1 (z) exactly coincide with the
logarithmic potentials of the given induced layers on
The values of the charges of the layer Plk• induced by the potential <»t(z)
on the contour
(i, k = 1, 2, •••• n), that is,
r.
r,
P11<
=-
_1_
2n
f
r1
a<»t(C) dB
an
= -
_1_
2n
f
r
<»1 aro,,(C) ds •
an
a.re known as the mutual capacities of the boundary contours (some properties
of the numbers Plk are considered in problems 1101-1104).
268
PROBLEMS ON COMPLEX ANALYSIS
1495. Find the harmonic measures wk(z), and also the quantities
ek(C) and p 1k, defined in the remark to the preceding problem, for:
(1) The circular ring 1 < fzl <µ;
(2) An arbitrary doubly connected domain D, assuming that
the function which maps this domain onto the ring is known.
1496. Let D be the domain of problem 1494 and v(z) the bounded
potential of an electrostatic field which assumes the constant values
ak (k = 1, 2, ... , n) on the boundary contours (conductors) rk. Prove
the following assertions :
n
(1) v(z)
=2
IXkwk(z).
k=l
(2) If the domain D is bounded and I'n is the external
contour, then
1
v(z) = 1Xn- 2:7t
J----a;av(C)
log
1
IC-zl
ds;
r
if, however, D contains the point oo, then
1
v(z) = v(oo)- 2:7t
J----a;av(C)
log
1
IC-zl
d.s.
I'
Prove that the right sides of these formulae are equal to IXk (k
= 1, 2, ... , n) in the domains complementary to D bounded by the I'k.
(3) The values of the charges of the layer induced on the
I'1 are given by
2qi
=-
1
2:7t
J----anav(C)
n
ds
~
= .L.J Ptk IXk ,
r1
k=l
where
HINT.
See problem 1101, (1).
(4)
2~ JJ(gradv)
D
HINT.
See problem 1102.
n
2
dxdy= 2Pik1Xi1Xk.
i,k=l
APPLICATIONS TO MECHANICS AND PHYSICS
269
(5) If w(z) is the complex potential of the field the density
of the induced layer is
e(C>
= - _1 ov(C) = ± - I lw'(C)I
on
2n
2n
·
Let D be an arbitrary multiply connected domain with boundary I',
consisting of the Jordan contours I'1 , ••• , I'n· There exist conformal mappings
of the domain D onto each of the following canonical domains with the
given uniqueness conditions (a, b are arbitrary points of the domain D, A is
an arbitrary complex number):
(1) Onto the plane with parallel cues. The mapping function f (z) is
uniquely determined by giving its pole a and the coefficient A of the expansion
(a#: oo),
(a= oo).
(2) Onto the plane with radial cuta (this refers to the plane with
cuts along segments on rays originating at the coordinate origin, or with cuts
along concentric circular arcs with centre at the coordinate origin). The
function f(z) is determined by its zero a, the pole b and the coefficient A of
the expansion
I
z~b
f(z) =
Az+
+ei(z-b)+ ...
C-1
z
+ ...
(b
#: oo),
(b = oo).
(3) Onto a clialc with radial cuts or with cuts along concentric circular
area (centre at the coordinate origin). The function f(z) is determined by
the conditions
f(a) = 0, f'(a) = l
and the specification of a contour I't, which is transformed into a circle.
(4) Onto the annulus with radial cutB or with cues along concentric
circular arca (centre at the coordinate origin). The mapping is determined apart
from a magnification and a rotation by the specification of the contours which
are transformed into the interior and exterior boundary circles.
See, for example, G. M. GoLUZIN, The geometrical theory of functionB of a
complea: variable (Geometriche11caya teoriya funlctBii TcomplelcBnogo peremennogo),
Chapter V, Gostekhizdat, 1952.
270
PROBLEMS ON' COMPLEX ANALYSIS
1497. Assuming that the function which conforme.lly maps the
domain D onto the plane with parallel cuts, and the harmonic
(k = 1, 2, .. ., n)t, a.re
measures C.Ot(Z) of the boundary contours
known, find the potential of the electrostatic field produced in the
domain D by the dipole (a; p) when the boundary I' of the domain
is earthed.
1498. Assuming that the Green function of the domain D is known,
find the potential of the electrostatic field in this domain formed
by the point charge (a; 2q) (a e D) which has given potentials 0!1c
on the boundary contours I't (k = 1, 2, ... , n). In the case when
the domain D is simply connected obtain the formula of problem
1460.
1499. Determine the nature of the electrostatic field determined
in the multiply connected domain D by the complex potential
w = /(z), where /(z) is the function which maps this domain onto
the plane with cuts parallel to the real axis.
1500. Determine the nature of the electrostatic field defined in
rk
the domain D by the complex potential w
= 2qi log
- 1-
/(z)
if the
function /(z) maps the domain D onto:
(1) The plane with cuts a.long concentric circular arcs with
centre at the coordinate origin;
(2) A disk with cuts a.long concentric circular a.res with
centre at the coordinate origin;
(3) A circular annulus with cuts along concentric circular
arcs with centre at the coordinate origin.
In ea.ch case find the flux (of the field intensity) through the boundary contours.
1501. Sketch the distribution of the equipotential lines and
the lines of force of the electrostatic fields:
(1) Formed in an infinite doubly connected domain by
a dipole at oo;
(2) Formed in a bounded doubly connected domain D by
a point charge.
In ea.ch case the potentials on the boundary contours are constants.
1502. (1) Express the potential v(z) of the electrostatic field
formed in the multiply connected domain D by the charge distribut These a.re determined by means of Green's function; see, for example,
§I of the appendix by M. Schiffer to the book: R. CotraANT, Dirichlee'sPrinciple,
Oonformal Mappings and Minimal Surfaces, Interscience, New York, 1950.
271
APPLICATIONS TO MEOH.ANICS AND PHYSICS
tion 2qa: on I'a: (
±
q" =
k=l
o)
in terms of the harmonic measures
wa:(z) of its boundary contours. (The potential is constant on each
contour I'1;.)
HINT. Use the results of problem 1496.
n
(2) Find the expression for the potential v(z), if
= q =F 0 and there is a point charge (a; -2q).
HINT.
2
q"
k=l
The determination of the potential v(z)+2q g(z, a) reduces to part
(1).
1503. Find the potential v(z) in the circular ring r 1 < izl < r 1 ,
if the charge distributions on its contours are given as 2q1 , and 2q2,
where in the case q1 +q11 = q =F 0 there is also the point charge
(a; - 2q).
HINT. The Green function of the circular ring can be determined from
the solution of problem 1489 by choosing an appropriate circulation. From
the Green functions one can determine the induced charges on the boundary
contours.
§ 3. Applications to the plane problem of heat conduction
The plane problem of the steady distribution of temperatures in a tield is
characterised by the analytic function w (z) = u+w (u is the temperature),
known as the oompk:l! potential of a thermal field. The vector Q
= -k grad u = -kw'(z) (k is the coefficient of thermal conductivity, constant in what follows) is known as the wctor heat flow. The heat flow through
the contour a is equal to
J
Qncls
c
=
-k
J°: ds = - cJdv
k
c 0
(n is the outward normal to the contour 0, traversed in the positive direction). Since the function u is single valued, for a closed contour 0 the heat
flow is also given by ik w'(z) dz. If close to the point a we have
f
c
w(i) =
[ ...
C-1
+--+eo+ci(z-a)+ ...
111-a
]
q
1
+2n k log--,
111-a
1 -defines at the point a a sourcs (a; q) of strength q,
the term 2 qk log - :n;
z-a
and the term (c- 1)/(111-a) defines a doubkt at the point a.
272
PROBLEMS ON COMPLEX ANALYSIS
The following similarities exist with fluid flow and the electrostatic fileld:
I
Complex
potential
Vector field
Thermal field
Fluid flow
= u+i11
w{z) = u+i11
V=gradu
= w'(z)
iw(z) -11+iu
E = - gradu
= w'(z)
Potential
function
Equipotential
lines
Potential function
Equipotential
lines
-11 is the force
function
Lines of force
w(z)
Q = -Tcgradu
=
u
u
=
const
17
11
= const
-Tcw'{z)
Temperature
Isothermals
J
Electrostatic field_
=
Stream function
Stream function
Stream lines
Stream lines
Source (a; q)
Source
Doublet
A thermal field
with given sources,
doublets and isothermal boundary
contours
Doublet
Dipole
A fl.ow determined by the complex potential
iw(z), with given
vortices and doublets, streamUning
the boundary
contours
A field with given
charges, dipoles
and equipotential
boundary contours.
(a; - ! ) Point charge
(a; 2~)
1504. Formulate the symmetry principle for the continuation
of a heat source a.cross a rectilinear or circular segment of the boundary of a domain. Find the temperature distribution in an arbitrary
simply connected domain D, if it is known that within this domain
there is a source (a; q) and the temperature on the boundary has
the constant value 0.
In problems 1505-1508 find the temperature distribution in the
given domains from the given sources assuming that the temperature is constant on the boundary of the domain.
1505. In the upper half-plane Im z > O; the source (a; q).
1506. In the circle lzl < R; the source (a; q).
1507. In the half-strip lxl <a, y>O with the source (ih; q) (h >0).
1508. In the rectangle lxl <a, IYI < b with the source (O; q).
APl'LICATIONS TO MECHANICS AND PHYSICS
273
1509. (I) Give an interpretation of the Green function g(z, a)
of the domain D in terms of the theory of heat conduction.
(2) Assuming the Green function of the domain D to be
known find the temperature distribution in this domain ifit is known
that in D there is a source (a; q) and on the boundary contours
rk (le= I, 2, ... , n) the temperature has the constant values Uk.
Express the answer in terms of the harmonic measures ro1:(z) of the
boundary contours.
1510. Find the temperature distribution inside the circular
ring r 1 < !z! < r 2, if it is known that inside the ring there is the
source (a; q) and on the boundary circles the temperature has the
constant values: u 1 on the circle !z! = r1 and u 2 on the circle !z! = r 2•
HINT. See the similar problems 1489 and 1508.
ANSWERS AND SOLUTIONS
CHAPTER I
!
1. (1) -i; (2) -i; (3)
(1+3i); (4) -8.
2. (1) 3, n/2 (here and below only the values of a.rgz a.re given);
(2) 2, n; (3) y'2, n/4; (4) y'2, -3n/4; (5) y'29, tan-15/2;
(6) y29, -tan-1 5/2; (7) y'29, n-tan-15/2;
(8) y'29,tan-1 5/2-n; (9)
(10) y'(a•+b2 ), tan
b
ta.n-1- - n for a
a
<
-1 b
-
a
lbl,.:!..~
0 and b
~2 (l+i), ±-~2
(5) ±l±i, ±
(6) ±
~2
(8) y'2 [ cos
~2
2
b
for a> O, ta.n-1-
a
<
(1-i); (4) ±
(1-1-i), ±
'
+n
for a
<
0 and b ~ 0,
0.
1
iy'3
y'3
i
3. (1) 1,-2±-2-; (2) ±2+2·
(3) ±
= .:!..sgnbt·
2 b
-i;
~2 (y(3)+i). ± ~2
(J-'(3)-i), ±J-'2i;
~2-(1-i);
(y'[y'(2)+1]-iJ"[y'(2)-l);
(7) ±(2+i);
(2k+: )n
(2k+{)n]
+isin
3
3
(k
=
0, 1, 2);
[
(2k+l)n-tan-1 3/4
. . (2k+l)n-tan-13/4]
(9) H
JI 5 cos
5 - - - - - +isin
5
(k = O, 1,2, 3, 4).
(k
=
O, 1, 2, ... , n-1).
t sgn b denotes Kronecker's symbol :
sgnb
sgn b
=
=
1
for b> 0,
-1
for b
274
<
O.
275
ANSWERS AND SOLUTIONS
12.
z8
2n . • 2n)
= z 1 + (Z1-Z1) ( COSn±iBID
n •
13. z, = z1 +z 3 -z 2 •
14. The ratio (z8 -z1 )/(z1 -z1 ) must be a real number (the condition is
necessary and sufficient).
15. The anha.rmonio ratio (z10 z1 , z1 , z4 ) =
z,-za:
Z1-Z4
Zz-Za
Z9-Z4
must be a real
number (the condition is necell8a.ry and sufficient).
16. SOLUTION. In the proof it can be assumed (without loss of generality)
that the straight line in question is the imaginary axis and that all the points
considered are on the right of it (if this is not the case it is necessary to mul·
tiply all the z1: by some number of the form cos a:+isin a:). It is then obvious
1
that Re z1: > 0 and Re - > 0 for any k.
Zk
19. The interior of the circle of radius R with centre at the point z = 21o;
the exterior of the same circle; the circumference of the same circle.
20. An ellipse with foci at the points z = ±2 and major semi-axis 5/2.
21. The interior of the left hand branch of the hyperbola with foci at the
points z = ±2 and real semi-axis 3/2.
22. The straight line perpendicular to the segment connecting the points
z1 and z1, and passing through the middle of this segment.
23. (1) The straight line Ill = a and the half-plane on the right of it; (2) the
half-plane below the straight line y =a.
24. The strip -1 < y < o.
25. The interior of the angle with vertex at the coordinate origin and sides
which form with the real axis angles equal respectively to a: and fJ; the
interior of the same angle with vertex at the point z0 •
26. The parabola y1 = 2a:+ 1.
27. The half-plane bounded by the straight line a:+y = 1 and containing
the coordinate origin.
28. The straight line passing through the points z1 and z1 (the point z1
being omitted) ; the circle with the segment connecting the points z1 and z1
as diameter (the point z8 being omitted).
29. (1) The interior of the domain bounded by the segment 0.;;;:;; a:.;;;; 2n
of the real axis and one turn of the Arohimidean spiral r = ~ ; (2) The set
of points determined in part (1) and the complement to the interval (0, 2n)
of the real axis.
80. (1) A family of circles touching the imaginary axis a1> the coordinate
origin, and the imaginary axis itself (the equation of the family is; O(z1 +y1 )
= 111); (2) The family of circles touching the real axis at the coordinate
origin and the real axis itself.
31. (1) The family of hyperbolas 1118-y• = O; (2) The family of hyperbolas
lllY = 0/2.
32. Each curve is a circle, being the geometrical locus of points the ratio
of distances from which to the points z1 and z1 is constant (the circle of Apollonius with respect to the points z1 and z1 ).
33. The family of circular arcs with ends at the points z1 and z1 (this family
also contains the two rectilinear segments with ends at the points z1 and z1 ;
one of these segments contains the point at infinity).
276
PROBLEMS ON COl'rlPLEX ANALYSIS
34. (1) Each curve is the geometrical locus of points, the product of the
distances of which from the points z =--L.ancLz.-= l is constant (lemniscates
with foci z = ± 1). For ..1. > 1 the curves of the family are simple closed
curves, for ..1. < 1 they separate into two simple closed curves which for ..1. -+ 0
shrink to the points ± 1. For ..1. = 1 we have the lemniscate of Bernoulli;
its equation in polar coordinates is rl = 2 cos 2•.
(2) Lemniscates with foci at the points z1 and z20 where z10 z1 are the
of the equation z1+az+b = 0. The lemniscates consist of a single curve if
..1. >
..1.
=
V(.
lza;zil),
v(
and of two curves if ..1.
V(
lza;zil).
For
lzi;Zil) we have a lemniscate of Bernoulli with the double point
(z1 +z1 )/2.
35. The spiral of Archimides r = •·
36. The logarithmic spiral r = e•.
37. (1) :ii; (2) 2:n/3; (3) 2:n; (4) :ii; (5)
a:
38.
<
y
o.
a:•+y•
.
~+i11
E = :z:1+y1+1 ' 11 = a:1+y1+1 , C = a:1+y1+1 ; z = a:+i11 = 1-C •
at.(! ,o, !)· (-! ,o, !). (o. !,!). (v; ,- ~2 , !).
All the four points are situated on the equator, their longitudes being respec·
tively equal to 0, :ii, ; , - : (the longitudes are calculated from the initial
meridian, situated in the ~. (:-plane).
40. The circle of radius tan CP/2+:ii/4) with centre at the point z = O.
The "South" pole corresponds to the coordinate origin, the "North" pole
to the point at infinity.
41. ( 1) Half meridians of longitude cx; (2) Parallels of latitude p = 2 tMC 1 r
-:n/2.
42. (1) Diametrically opposite points on the Sl\llle parallel; (2) Points
mutually symmetrical with respect to the initial meridian (that is, differing
in the sign of the longitudes); (3) Points mutually symmetrical with respect
to the equatorial plane (that is, with the same longitude and with latitudes
differing in sign only).
43,. Z1.Za = -1.
44. Points obtainable one from the other by a rotation of the sphere through
180° about a dil\llleter parallel to the real axis of the z-plane.
45. (1) The eastern hemisphere; (2) The western hemisphere; (3) The hemisphere - ~ < ex < ;
(cx is the longitude); (4) The hemisphere ; < lcxl <:ii;
(5) The southern hemisphere; (6) The northern hemisphere.
46. A family of circles touching at the "north" pole (the pole of projections); a straight line passing through the coordinate origin corresponds to
a great circle, and a straight line parallel to it at a distance d, from the coordinate origin corresponds to a circle situated in a plane inclined at an
angle tan- 1 d, to the meridional plane.
47. The straight line corresponds to a circle passing through the "North"
pole.
277
ANSWERS AND SOLUTIONS
50. k(z,a) =
lz-al
k(z,co) = l/y(l+lzl2).
y(l+lzll)y'(l+laJZ)
51. The circle of Apollonius
I I=
z-zi
z-z 2
;(l+lzil 9l , in particular the
(1 lz1 12 )
lz 21, and the circle lz-z 1 1 = y(l+Jz111),
+
straight line lz-z 1 1 = lz-z11,iflz11=
if Z1 =CO.
54. (af+bf +of+df) (a~+b~+o~+d~) = (a1a 1 - b1b1 - o1 o1 - d 1 d 2 ) 1
(a1bs+a1b1 +c11ts-01di) 1+ (a1c1+a101 +dib1-dab1)1 + (a1da+a2d1 +b101
-b1C1)I.
55, SOLUTION, Let us multiply the arbitrary quaternion q and the quaternions
qi, qj, qk by another arbitrary quaternion q1 • The components of the four
quaternions obtained satisfy all the conditions of the problem except, perhaps,
the diagonal condition. This last condition is satisfied by the choice of the
components of the quaternions q and q1 • Here is an example of a matrix
constructed by Euler:
+
68
( -17
-29
31
41
79
28
-23
8
-11 -77
59
-37)
32
61
49
with the sum of the four squares equal to 8515.
s.. ,
_!!.,
!!.1
_!!.,
~,
57. 1, e 111, e 9 , e 2 , y2 e 4 , y2 e 4 , y2 e 4 ,
y2
58. ±i; (-l)k
59. e 2, .1; el, -3; e8, 4-2:n;; e-s, 2:n;-4; a, 4J-:n;, if 4>
f 4> :< O; 1, -4>, if 14>1 < :n;, and :n;, if 4> = :n;.
•
Slll
4
•
0, and 4>+n,
(2) - - - - - - - -
60. (1) - - - - - - -
sinx/2
sinx/2
sin (n+ l)x cos nx
2
2
n is an odd
, if"
cosx/ 2
• 2
sin2nx
2sinx
_s..,
. (n+l)x . nx
sm--2--sm2
(n+l)x cos nx
2
2
(3) - - - ;
>
e
(4 ) s~ nx;
smx
(5)
(n+l)x . nx
cos--2--sm 2
number;
61. (1)
- - - - - , - - - - , if n is an even number.
cosx/2
sin~{J
sin{J/2
n{J}
(
cos oi:+-2- ;
(2)
sin~{J
sin{J/2
n{J}
. (
sm oi:+T .
65. (1) sinz = sin(x+iy) = sinxcoshy+icosxsinhy,
lsinzl = y(sinh1y+sin1x);
lcoszl = y(cosh111-sin1x);
(2) cosz = cosxcoshy-isinxsinhy,
I _ y(sin22x+sinhl 2y) .
2y
(3) tan z -_ -l sin 2x+isinh
,
.
, ltan z .
2(cos2x+sinhly)
2 cos1x+sinh2y
lsinhzl = y(sinh1x+sin1y);
(4) sinhz = sinhxcosy+icoshxsiny,
278
PROBLEMS ON COMPLEX ANALYSIS
(5) coshz
== cosha:cosy+isinh:uiny,
jtanhzi= y(~h•!m+sin•2y).
2(sinhla:+cosly)
(6) tanh z = sinh!m+isin2y,
cosh2:ii+cos2y
(2) isinh2;
88. (1) cos2coshl-isin2sinhl;
(4 > 8+ 15i •
17 '
lcoshzl = y(cosh•:11- sin•y);
sin4-isinh2
<3l 2(cos1 2+sinh1 1) 1
sinh 8-i sin 4
(5) 2(cosh2 4-cos1 2)'•
(6) '·
67. Imes= 0, if lmz =Ten;
Reel:= 0, if lmz = (2k+l) n/21
Im cosz = O, if Rez =kn or lmz = O;
Re cosz = 0, if Rez = (2k+l) n/2;
Imeinz = 0, if Rez = (2k+l); or lmz = O;
Re sinz
=
0, if Rez = kn;
=
Imtanz
kn
Retanz = 0, if Re z = T;
O, if lmz = O;
Im coshz = 0, if lmz =Ten, or Rez = O;
Recoshz = 0, if Imz = (2k+l);;
Imcothz
k: ;
= 0, if Imz =
Recothz = 0, if Rez = 0.
In each case k is an integer (k = 0, ±1, ±2, ... ).
68. (1) log4+2km, (2k+1)ni,
(4)
m;
(2) ( 2k+
~) m, ~i;
(3)(2k± !)m;
~ log1s+(2kn-tan-1 :)i, ~ log1s+(c2k+l)n-tan-1 :Ji.
69. The set of values of 2Logz comprises only part of the set of values
of Log (z1 ) (see [2, Chapter III, sec. 19]).
70. (1) 4m;
(2) -2m;
(3) o;
(4) bi.
71. (1) cos (2ky2.n)+ i sin (2knJl'2);
(2) 2r'8 [cos (2k+l)ny2+isin (2k+l) ny2;
(3) elk•(coslog2+sinlog2);
(4) ellklr;
(2k-.!)n
(5) e
II
;
(6)
l~i e(2Hf)n I
(7) 5etau-sf+lk•[cos(log5-tan-1: )+isin (1og5-tan-1 :)];
(8) -5etan-•t+(2k+l)•[cos(log5-tan-l: )+isin {tog5-tan-1 : )].
Everywhere k is an integer (k = 0, ± 1, ±2, ... ).
78. The sets of values of al« and (a«)1 are identical, but generally speaking
are not identical with the set of values of (a1 )«; the general case when the
279
ANSWERS AND SOLUTIONS
index 2 is replaced by the arbitrary complex number p, is considered in [2,
Chapter III, sec. 20].
77. (1) Imcos-1 z = Imsin-1 z = 0, if z is a real number and lzl =::;;;; l;
Imtan-lz = O, if z is a real number;
(2) Resinh-1 z = 0, if z is a purely imaginary number and lzl =::;;;; 1.
n
5n
'i8. (1) 5+2kn, 5+2kn;
(3) 2kn±ilog (2+y3);
n
(2) 2kn±3;
(4) 2Tcn-ilog (y2-l),
(21c+l)n-Hog (y2+1);
(5) 21 [ tan-1 2l +(2Tc+l)n] + ~• log5;
(7)
!
log 5+ [
~ tan-
1
2+{ le+
~) n] i.
(6) log(y5±2)+(21c±
!)ni;
Everywhere Tc is an integer
(Tc = 0, ± 1, ±2, ... ).
79. The straight line segment: a: = 1, - 2 =::;;;; y =::;;;; 0.
80. The parabola y =a:•.
81. The right-hand half of the parabola y = a:1, traversed twice.
82. A left semicircle of radius a with centre at the point z = O.
83. The branch of the hyperbola y = l/a:, situated in the third quadrant.
8'. (1) The upper semicircle of radius 1 with centre at the point z = O;
(2) The quarter circumference of radius 1 with centre at the point z = O,
situated in the first quadrant.
815. (1) The cycloid: a:= a(t-sint), y = a(l-cost); (2) The first (counting from the coordinate origin) arc of the curtate (a < b), prolate (a> b)
or ordinary (a= b) cycloid: a:= at-bsint, y =a-boost.
86. (1) The images of the straight lines a: = 0 are for 0 ¢ 0 the parabolas
u=
01 -~,
401
=
for 0
0 the semi-axis
t1
= O, u =::;;;; O; the images of the
straight lines '!/ = 0 are for 0 ¢ 0 the parabolas u =
v•
40i" -01 ,
for
0 = 0
the semi-axis " = 0, u ~ 0; the image of the straight line '!/ = a: is the semi-axis
u = 0, "~ O; the images of the circles lzl = R are the circles lwl = R 1 ; the
images of the rays arg z =a: are the rays arg w = 2a:; the straight lines a:= 0,
'!/ = 0 for 0 :F 0 and the rays arg z = a: are mapped one-one; (2) The originals
of the straight lines u = 0 are the hyperbolas a:l-yl = 0 (for 0 = 0, a pair
of straight lines), the originals of the straight lines" = 0 are the hyperbolas
ey =
~
(for 0 = 0, a pair of straight lines).
87. (1) The images of the straight lines a:= 0 are the circles u 1 +v•- ~ = 0,
for 0 = 0, the axis u = O; the images of the straight lines '!/ = 0 are the
circles u 1 +v8+ ~
=
lzl =Rare the circles
O, for 0 = O the axis "= O; the images of the circles
lwl
= l/R; the images of the rays arg z =a: are the rays
arg z =-a:; the image of the circle lz-11=1 is the straight )ine u =
~;
280
PROBLEMS ON COMPLEX ANALYSIS
(2) The originals of the straight lines U=O are the circles w+y•- ~ = O,
for 0
=
O, the a.xis a:= O; the originals of the straight lines
circles a:1 +y1 + ~
=
0, for 0
=
O, the a.xis y
=
t1
=0
are the
O.
= z+ _!._
maps the circles lzl = R + I onto the ellipses
z
1 ) 8 = I, and the circle lzl = l onto the segment t1 = O,
1 )8 + (
(
R+R
R-R
88. The function w
u8
t1 8
- 2 :::;;; u :::;;; 2; the function w
I
ellipses (
u 1 )·1 + (
R-R
= z - _!._
maps the circles lzl = R + l
z
onto the
8
ti
1
8
= 1, and the circle lzl =
I onto the segment
R+R)
u = 0, -2:::;;; ti:::;;; 2.
89. The original of the family u = 0 is the family a:(a:l+y•+I) = O(w+y1);
the original of the family t1 = 0 is the family y(w+y1 -I) = O(a:•+y•) (Fig. 58).
Fm. 58
281
ANSWERS AND SOLUTIONS
90. On the ray going along the negative part of the real axi11 from the
.
.
point
w = - 41 to t h e pomt
w = oo.
91. (1) The circles €/ = eC, the rays (J = 0, the logarithmic spiral €! = e•;
(2) The curves y = e""+2kn.
92. (1) The family of straight lines a: = 0 is transformed into the family
v1 =-4a1 (u-a•+
!)
(parabolaswithfociatthepointw= -!,a=O+
to which also belongs the ray going from the point w = -
1
4
!),
along the negative
part of the real axis (a = 0); the family y = 0 is transformed into the family
of confocal parabolas
118
from the point w = -
!),
= 401 ( u+o•+
~
to which also belongs the ray going
along the real axis in the positive direction;
(2) The family a: = 0 is transformed into a family of circles of Apollonius
with respect to the points w = - 1 and w = 1 (including also the imaginary
axis); the equation of the family of circles of Apollonius is: (u - a)1 +v• = a•
-1; !al > l (a = coth 20); the family y = 0 is transformed into a family
of circular arcs with ends at the points w = -1 and w = 1, including also
the corresponding parts of the real axis; the equation of the pencil of circles
is ul+(v+b)• = l+bl (b =cot 20);
,.
•Co.
(3) The family a: = 0 is transformed into the family of spirals(/ = eClthe axis a: = 0 corresponding to the segment (J = 0, 0 < (! ~ 1 ; the family
~-Cl
y = 0 is transformed into the family of spirals (/ = e'°
, the axis y = 0
corresponding to the ray (J = 0, 1 ~ (! < oo.
98. To the straight lines y = 0 correspond the curves u = a:+e"" cos 0,
I}= o+e"" sin 0, to the rectilinear segments a:= 0 there correspond arcs
of the curves u = o+ec cosy, v = y+ec sin y (Fig. 59).
94. (1) To the family lwl = R there correspond the circles r = (cos ~)/log B
(B of:. 1) and the imaginary axis (B = l); to each ray arg w =a. there corresponds a family of circles r =
2:~a.
;
for a.= 0 the real axis belongs to this
family (for k = O); (2) The hyperbolas a:8 - y 2 =log B and 2zy = a.+2kn.
H. (1) z = 0 and z = 2; (2) z = 0, z = l/m, z = i/n (m and n are any
integers); (3) All the points of the plane; (4) All the points of the disk lzl ~ 1.
97. (2) When lim Zn equals 0 or oo.
n-+00
zRez
100. Only /(z) = -lzl- (/(0) = 0).
101. (1) and (2). Continuous but not uniformly.
102. (2) No; (3) Yes.
101.
au
av
au
av
'"ar = ai. ai"= -ra;:·
282
PBOBLEMS ON COMPLEX ANALYSIS
I
......~~~~-ci--~-+--i----~~
~.._~~~~
I
I
I
:r
0
y=~
1/
..... ...... .....
......
c
......
.....
:r=f·4
.....
' ',:r=f
'\
\
'
I
\
,y=1
\
l:r=-4
I
\
I
\v•O
FIG. 59
283
ANSWERS AND SOLUTIONS
109. o.
118. (1) No, if u #: const; (2) /(u) = au+b.
119. lf(z)I is not a harmonic function, arg/(z) and log IJ(z)I are harmonic.
88u
1 8u
l 8 2u
120. Liu = a;a
+-;:a at1i• ; u = 01 log r+o••
+-;-a;-
UU. P1 =a:, qi = y; P1 = a: 2 -y•, qi = 2a:y; Pa = a:8-&::y1, qa = 3a:ly
-'If'; P• = a:t-6a:ly1 +y•; q, = 4a;ly-4a:y8; 'Pn = ,.n cos 114', qn = ,.n sin r&t/I.
122. tJ (a:, y) = 2a:!f+y+0.
y
128. v(a:,y) = - a:•+y• +o.
124. (a) v(a:,y) = argz+O; (b) v(a:,y) = argz+2mn+O (the expression
for arg z in terms of a: and y is given in problem 2, (10)).
(b) v(a:, y) = argz125. (a) v(a:,y) = argz-arg(z-1)+2mn+O;
arg (z-1)+0;
(c) v(a:, y) = arg z-arg (z-1)+2mn+0.
n
126. (a) v(a:, y) =
n
2 cx1c arg (z-z1c)+2n 2 mu1+0;
k-1
k=l
n
(b) v(a:, y)
2
=
n
2
cx1c arg (z-z1:)+2nm
k-1
k-1
n
cx1:+0 {if
2
cx1:
=0
the function
k-1
v(a:, y) is single valued in the given domain!).
127. (1) It exists; (2) It exists; (8) It does not exist.
128. /(z)
= z9 + (5-i) z- j_z +Oi.
129. /(z) = ze• +2icosz+z8-iz+Oi.
180. /(z) =
2~
+iz 1 +3i+O.
181. /(z) = 2ilogz-(2-i)z+0. Everywhere 0 is an arbitrary real constant.
182. u = 01:i:+o1 •
188. u = 0 1caa:+by)+o•.
134. u
= 0 1tan-1.!.
+01 •
a:
135. u = 0 1a:y+02 •
186. u = 0 1 log (z8 +y1 )+01 •
01a:
187. u = - 1- 2 +oz.
a: +Y
188. u = 0 1 y[a:+ J"(za+y•)] +01 •
189. It does not exist.
140. /(z) = el«:z: 2 e•.
141. f (z) = el«e•'.
..
142. f(z) = Aes.
148. f(z) = Aze" (ex is an arbitrary real constant, A is an arbitrary positive
constant).
143. az+.il, m+.il, .ilea•, .ilelair,
284
l'ROBLEMS ON COMPLEX ANALYSIS
146. iaz+A, az+A, Aela, Ae0 •
147. ailog z+A, a log z+A, Ae"II011:, Aea1011:.
148. a log z+A, ailog z+A, Aeaiou, Aeallou.
a
ai
.!.
.!!..
149. -+A, -+A, Ae 11 , Ae 11 (a is an arbitrary real constant, A is
Iii
z
an arbitrary complex constant).
160. For w = z1 : (1) 0 = O, Tc= 2; (2) 0 = :ri, Tc=
4
(4) (J = :ri - tan-1 8
,
I
2 ;
(3) (J =:ri/4, Tc= 2y2;
Tc = IO.
0, Tc= 3; (2) (J =- 0, Tc= 3/16; (3) (J = :ri/2, Tc=- 6;
4
(4) 9 = - 2 ta.n-13' Tc = 75.
For w = zl: (I)
(J =
UH. (1) A compreBBion for lzl
I
I
< 2' a stretch for lzl > 2;
(2) A compression for lz+ll
I
< 2'
a stretch for
I
lz +II> 2;
> I, a stretch for lzl < I;
(4) A compreBBion for Re z < 0, a stretch for Re z > O;
(5) A compression for lz - II > 1, a stretch for lz - II < I.
(3) A compression for lzl
152. S
= Jf1f'(z)l 8 da:dy,
G
ua. y2(e8"- I).
154. 2e 2 (e1 - I).
L
=f
1/'(z)lds.
I
155. The domain D is the ring e < lwl < e•. It is not possible to apply
the formula of problem 152 since the mapping is not one-one.
CHAPTER II
1156. w = (l+i)(l-z).
157. w = (2+i) z+l-3i.
158. (1) z0 = -1+3i, 8 = 0, k = 2, w+l-3i = 2(z+l-3i);
(2) z0 = 2+2i, 8 = n/2, k = 1, w-2-2i = i(z-2-2i);
(3) there is no finite fixed point;
(4) if a = 1 there is no finite fixed point; if a+. 1, then
(
w1 -az1
w1 -az1
w1 -az1 )
Zo = - - - , 8 = arga,k = lal,
=a z - -- ,
1-a
l-a
l-a
w----
(5) Ifa = 1 there is no finite fixed point; if a .p 1, then z0 = _b_,
l-a
8=arga,k=lal,w- 1 ~a =a(z- 1 ~a)·
U9. (1) w = az+b; (2) w = -az+b; (3) w = -i(az+b); (4) w == az+bi.
Everywhere a and b are real numbers and a > 0.
160. (1) w = z+bi or w = -z+l+bi;
(2) w = z+b or w = -z-i+b;
(3) w = z+b(l+i) or w = -z+l+b(l+i).
Everywhere b is a real number. It is possible for correspondences to exist
between points lying either on a straight line parallel to the boundaries of the
strip, or on parallel straight lines symmetrical with respect to the centre line
of the strip. The mapping is not uniquely determined if the points lie on the
centre line of the strip.
161. (1) w = (z-a)/h;
(3) w =
(4) w =
(2) w = -z+,,,a+h +i;
y(l+k1 ) -i(-j-+tan-ik)
b
e
z,
y(l+k1 ) -1(.!!.+tan-1k)
.
e
2
(z-ib1 ).
b1-b1
162. w = el 11 Rz+w0 •
163. (1) The family of straight lines u = l/a parallel to the imaginary
axis (not including the imaginary axis itself);
(2) The family of straight lines 11 = - l /b, parallel to the real axis
(not including the real a.xis itself);
(3) The family of circles b(u•+v•)+u+v = 0, touching the straight
line tJ = -u at the origin (also including this straight line);
(4) The pencil of straight lines v = -ku;
(5) The family of circles passing through the coordinate origin and
through the point Wo = l/Zo (the straight line which passes through the points
w = 0 and w = w0 also belongs to this pencil);
(6) The cissoid u 1 = -vl/(v+ 1).
285
286
PROBLEMS ON COMPLEX ANALYSIS
164. (1) Into a family of circles touching at the point w = h the straight
lines, respectively parallel to the imaginary and real axes (including also
these straight lines); the equations of these families a.re;
+ (v-hg)8]-(u-h1) = O;
(0-y0 )[(u-h1 ) 1 + (v-h8 ) 1l+ (v-h1 ) = 0,
(O-:i:o)[(u-h1 ) 1
where z0 = :i:o+iyo, h
= h1+ih8 ;
(2) Into the family of circles with centre at the point w = h ( lw-hl
=
!) andafamilyofraysissuing from the point w ...
h (arg (w-h)
=-cc).
165. (1) The equation of a family of circles of Apollonius with respect
I I=
to the points z1 and z1: z-zi
z-z1
.A. The ends .A. and B of the diameter lying
on the straight line passing through the points z1 and z1 (Fig. 60), divide the
J..<I
I
I
'i
I
I
I
i--'-1
I
1i.
FIG. 60
segment a:i z1 internally and externally in the ratio .A. If use is ma.de of the no·
tation indicated in the diagram (0 is the centre of the circle with diameter
AB), and we put lz1 -z1 1= d. then for A< 1 we have the relations:
r1
R
.Ad.
_Aid.
d.
A= cos cc=-=-, B =--.-, r1 =--.-,
"1 = - -1;
R
"•
1-,..•
l-A1
l-.A
.
.
.
z-z1
(J
(2) Circular a.res passmg through the pomts z1 and z1 : arg-- = .
Z-Zz
The arcs corresponding to the values (J = cc and (J = n-cc are complements
and form a complete circle;
(3) To the polar net there corresponds (Fig. 61) a net consisting of the
287
.ANSWERS AND SOLUTIONS
Iz-z. I==
z-z
1
circles of Apollonius -
z-z1
R and the arcs arg-z-z.
= 8 orthogonal to
them (if 8 > 0, then the area a.re situated on the right of the direction z~~
if8 < 0, they are on the left of it);
B
FIG. 61
(4) to the upper semicircle there corresponds the right angle indicated
in the figure.
166. The semicircle lwl < 1, Im w < O.
167. The domain containing the point w = 0 and bounded by arcs of the
lw+ !i I= 3/4.
removal of the points of the disk Iw -
circles
lwl
=I
and
168. The domain obtained from the lower half-plane (Im w
! + ~ I<
<
0) by the
~2 which are in this half.
plane.
169. (1) The domain bounded by the straight line Rew= 1 and the
. 1
circe
Iw- 211 =2'l tangential to 1t;.
I ~I
(2) The domain bounded by the tangential circles w -
and
lw- ! I= ! .
=
~
288
PROBLEMS ON COMPLEX ANALYSIS
170. The doubly connected domain the boundary of which coneists of the
=
straight line Re 10
!
and the circle / 10 -
171. (1)
10
=- :
(2)
10
= ~(~-1)+hi
da-~
z
(3 )
10
=
172.
(1)
=_
(3)
10
175.
10
(2 )
2i(z+l) •
4z-l-5i '
(l+i)z+l+3i
(l+ilz+ 3 +i ;
173. (1) w =
10
= :
or
/
= : •
+hi;
10
= ~(~-1)+1+hi;
~-da
z
d1(z-da) .
z(d1+d2 )
10
174. (l)
+l+hi or w
:
(2)
= (-1+3i)z+l-i.
(l+i)z-l+i
=
'
(1+2i)z+6-3i
5(z-i)
•
10
10 =
(2) 10
iz+2+i
z+l ;
1-i
(3) w = - 2 -(z+l).
(l-4i)-2(1-i)
= z2z(l-i)-(4-i)
.
;
z (3-i)- (1 +i) •
(l+i)(l-z)
=
= IZ-1
.z-i ;
the upper half.plane maps onto the unit disk.
176. (1) 10 = (az+b)/(oz+d), where a,b,o, dare real numbers and ad-bo> O;
(2) 10 = (az+b)/(cz+d), where a,b,o, d are real numbers and
ad-bo < 0;
(3) 10 = i(az+b)/(cz+d), where a,b,o, d are real numbers and
ad-bo < O.
2
2z+l)
(2)10= -2 ( - •
177. (1) 10 = 2 _ 21 ;
z-2
178. w = (B-z)f(B+z); the image of the upper semicircle is the quadrant
u> O,v < O.
179. (1) 2 ~i ;
180. (1)
lzl =
(2) :
+i.
2; (2) the straight line a:
=
! ; Iz - ~ I= ~ ;
(3)
.!. ;
(5) lz-21ol = y(lz0 18 -l) (that is, this circle is
2
symmetrical to itself with respect to the unit circle);
(6) (a:Z+y•)•- (a:2-y•) - O (a lemniscate); (7) a curvilinear triangle
(4) a:a:o+YYo =
'h vertices
'
'
l
wit
at t h e pomts
=-,
Z1
l
l
=-,
=-,
Zs
and wi'th
Bl'des
.
which
are arCB o f
Zs
circles passing through the point z = 0 (one of the arcs may be a segment of
a straight line).
182. (1) O(a:)
=
cx+2 arg (a:-/3); (2) 10'({1) =
e(•-f) 1 / 2b
289
ANSWERS AND SOLUTIONS
(3) If b ;;;;;;,. 2, then the whole half-plane is compressed, if b < 2, then
the domain within the circle lz-pj < y(2b) is stretched. (The circle lz-pj
= y(2b) is termed iaometrio.)
z-2i
t(.!.+e)z-(a+bi)
188. w = (z-i)/(z+i); (2) w = i (
b" •
2 .; (3) w = e 2
z+
= Bi--.
+w0 •
z+1
w = -(z-2i)/(z+2i)
184.. w
185.
z- a- i)
i
z-i
zi+2
186. w = - 4 21 _ 2 _ 41 •
187. w =Tee
-~++ars~) z-zi,
where k is a real number the sign of which is
z-z.
identical with the sign of arg z1 /z1• Rays issuing from the point w = 0 in the
half-plane Re w > 0 correspond to circular arcs, situated within the disk
lzl < l and passing through the points z1 and z1 in the z-plane; the semicircles
with centre at the point w = 0 in the half·plane Re w > 0 correspond to
arcs of the circles of Apollonius with respect to the points z1 and z1 situated
within the disk lzl < l.
w-b
z-a
= el11.-=
188. ---w-b
z-a·
z-a
z-a
w-ii
189. - - =i--_-.
w-a
190. (1) O (</>)
=
oi:-<J>+2arg(el<f>-a)
=
oi:-<1>+2tan-1
sin<1>-Asin8
A
, where
COS</>- COS 8
a= Aelll.
=
(2) w'(O)
(3)
(l-lal 1 )el•,
w'(a)
=
el•
l-ja!•"
Ifa#= O, then the domain lying inside the disk lz -
!I< V(,:i.
-1}
is stretched, and the domain outside this disk is compressed. (The circle
lz-
~
1-v(
1; 11 -1) is isometric.) If
(4) max
191. (1) w =
(4)
IdZI=
dw
2z-l
2_
w-a
1-i.iw
21
=
;
= 2iz+_!_ ·
el•~.
1-'iiz
w-b
.ffl-bw
z-a
JU. R1----=el«R1--- .
193. w =- (l-z)/(z+2l
R~-ih
o,
Iaz I=
. dw
mm
l+lal
1-lal'
(2) w
a=
2+iz '
then lw'(z)I
1-lal
l+lal.
(3) w
=
-iz·,
== 1.
290
PROBLEMS ON COMPLEX ANALYSIS
19'. w
=
2 .
z-2+i
2·
u+ 2 -
i
z-a
R 8e•11 - - · (2)
R•-az'
195 (1) w =
.
z-a
11 -·
RB-az'
w-b
.
- - = e'
R•-bio
z-a
(3) w = R 1 - -
RB-az'
where a is a real number and Jal < I.
(l+k)z-(za+kzi) a, where k =-. f(l-lzsl 1)elar1(1-i.z,),
(1-k)z- (z1 -kz1)
l-Jz1J1
196. w =
JI
Jal=
lz1-z1l · - - - Jl-z1z1J+y[(l-Jz1J2)(1-lz111)] ·
az-l+Jl'(l-a2)
l-y'(l-a2)
197 • w = ± [l-y(l-a1 )]z-a' !I= 2
a
200. (1) w = z; (2) w
=
z+2-y3
.
l+( 2 -y 3 )z; (3) w =
t
z-2+)1'3
l-( 2 -y'3 )z"
toa. Circles passing through the point Zo and having at this point a tan·
gent determined by the vector h.
(R-lci)z-R1
204. w = z-(R+lci) , where k is a real number, k
O (fork= oo it
+
is necessary to put w = 1).
209. If Jal <sin ; , then the transformation is elliptic. If we use the
.a
notation: Jal = sin 2 sin(J, 11 = e
Ao
tan T = tan
w-zi
W-11
=
elA.
.a
2
II
II
fJ ,
tan 2
z1 = e
1(11+.!_.!!.)
fJ •
I
2 oot
2
cos (J, then the transformation can be written in the form
I-Ii •
1-1.
If Jal = sin 2
.a
1
is of the form - -
W-Zo
If Ja I
1(11+~-.!.)
1
,
then the transformation is parabolic and
= - - +iki0
Z-Zo
where z0 = e
I(11+ .!_.!.)
11
I ,
A
2
h = tan - ·
> sin ; , then the transformation is hyperbolic. If we use the notation:
• J.
I I . fJ 1 =e
mn2=asm,z
1(11+f-P) • zI = e 1(11+-i-,.+.8) ,
then the transformation is written in the
W-Zi
form - - -
w-z1
= KZ-Z1
--z-z ·
1
291
ANSWERS AND SOLUTIONS
210.
I'=
r
= 2 tan-1:;
~ + 0 [(:
1'-
r
r = ~ +o [(
n sin,,
:n
for sma.11
~;
for small : .
z..);
Zo
= 2 tan-1 (I +a:. tan
l-a:0 cosy
l-a:0
2
I' < y, if Zo < 0, and I'> y, if Zo > 0.
215. (l)w= -20/z; (2)w= -(2iz+I+2i).
211.
= Y+2 tan-1
216. w
==:~~~::==~el«,
l! =
~ + v[(~r-ll
217 h = 5/4· w = 2el« 4z- 3 or w
4z+3
·
'
= el« 4z+ 3
4z-3 ·
z+24
2
2z
218. w =---el« orw = - - - els (! = - 3 .
z+24
3z
219. w
z-z'
= .il--!
or
z-z
z-~
w = .il--, where .il is an arbitrary complex number,
z-z1
1
I
Z1 = z1+
u1=
~~-~
d
I
• Zs= z1+
~~-~
d
,
d
= lz1-z1I•
:d ('1+d1-r:- y ([d -(r1+r )1) [dl-(r1-r1) J),
1
1
1
I
~= 2ffM+d1-f1-f-y([d2-(r1+r1)I] [dl-(r1-r2)1]);
I)
µ ( orµ
=
I
(d+r1-u1) (r1-u1)
(d+r8 -u1 ) (r1-u1)
--------
I
I"
.HO. (1) µ = 2; (2) µ = 5+2 y6.
223. The group will be finite if a/n is rational.
225. The fundamental domains (some of their possible forms) are shown
shaded in Fig. 62. Equivalent boundary sides are connected by arrows.
Points with figures are fixed points of the rotations leading to the groups (the
figures indicate the number of rotations). For the last five examples the parallelogram of the doubly periodic semigroup is indicated; in example (7)
this is a square, in examples (8) and (9) it is a rhombus with angles 120°
and 60°.
RElllARX. It can be shown that apart from a linear transformation groups
(3)-(9) exhaust the groups of linear transformations with one limit point
(known as the limit point of a set of points equiva.lent to one another)t.
w-l
z-1
w-i
z-i
226. (1) w = elotz; (2) - - = el«--; (3) - - = el«--.
w+l
z+l
w+i
z+i '
t See L. R. FoRD, (1929), Amomorphic junctionB, §§ 59,60, McGrawHill (reprint Chelsea, 1951).
292
PROBLEMS ON OOMl'LEX ANALYSIS
w-a =e"'~
4
( ) l+aw
l+az ·
232. (1) and (2). The construction is obvious; (3) The equidista.nts of the
"straight line" a.P (a. and p a.re the "infinitely distant" points of this straight
line) are circular arcs with ends at a., 8 (known as hypercycles); (4) The li·
miting lines of a pencil of "parallel straight lines" with common point "at
infinity" a. are circles touching (internally) the unit circle at the point a. (they
are known as oricyclea).
288. (1) For the construction of a "rectilinear" triangle with angles tf> 1,
r1> 1 , r1> 8 , construct a circular sector OAB with central angle LI = n- (r1> 1 +r1> 1
+r1> 8 ) draw the "straight line" AB, and through the points A and B "straight
lines" at angles t/> 1 and t/> 3 to AB, intersecting at the point O. The triangle
ABO is the one required.
234. (1) w
= i(z1 -a1 );
235. (1) w
= z1/a1 ;
(2)w
=
y(z- ~ }-iy'(p/2).
(2) w = y(z/a); (3) w
I
I)
I
2
2
4)
J)
l
7)
)'•
= 2A (z+l
21 _ 2
8)
FIG. 62
ANSWERS AND SOLUTIONS
293
286. The domain is bounded by Pascal's lim~on u = B(ooslfi+moos 21/1),
"= R(sinlfi+m sin 21/1). If the coordinate origin in the w-plane is
transferred to the point w =-Rm, then we obtain the equation of the ~n
in the usual form (in polar coordinates): (! = B(l+2m cos 0). For m = 0
Pa.sea.l's lim~on becomes a circle, for m == 1/2, a cardioid with cusp w
= -B/2. The images of the circles lzl = r < 1 are also lim~ns of Pa.seal
the polar equations of which are easily obtained by transferring the coordinate origin to the point u• = -Rmr2 : (! = Br{l+2mr cos 0). The images
of the radii of the circle arg z = a: are parabolas passing through the coordinate origin: m(u sin 2a:-tl cos 2a:) 2 +R sin a:(u sina:-t1 cos a:) = 0. To the radii
a= 0 and a = 1r correspond the segments of the real axis O~ u~ B(l +m)
and R(m-1) ~ u ~ O.
237. The domain bounded by the parabola u = -t18 and the curve
Q = 2 cos (0/3), 101 < Sn/4 (Fig. 63).
Fm. 63
238. (1) The domain bounded by the epicycloid u =
"= R
z=
B( ooslfi+ oo:nl/I ) .
(sin lfi+ sinnnl/I ) , having (n-1) cusps which are the images of the points
n-V(-1); (2) the exterior of the hypooyoloid: u = B (cos lfi+ oos7~1/1 ) ,
V
.
)
•+l
t1=B(sinl/l-smnnl/I , having(n+l)ousps(imagesofthepoints z=
1).
289. (1) lml ~ l/n. The domain is bounded by an "elongated" epicycloid
(epitroohoid), that is, the trajectory of a point at a distance mB from the
centre of a circle of radius B/n, rolling outside a circle of radius B(n-1)/n.
(2) lml ~ l/n. In the first case the exterior of the unit circle, and
in the second case its interior is mapped onto the exterior of a "shortened"
hypooyoloid (hypotroohoid).
294
PROBLEMS ON COMPLEX ANALYSIS
a
~J
(V
l-J-_
241. (1) w = ( _
z )' ;
1-z
242. (1) w
=
ici '
2 (y(4)+l)ea za
(2) w = ---""---:-""""'---1 +3 j/4
c4)-2) e
l
240. (1) w =z•;
(2) w
=-
z•+ l ;
(3) w
2z
2z1 -J-3iz-J-2
<2> w = 2z1 -3iz+2 •
z1 -J-2iz-J-l
iz•+2z +i;
=
-2zl+3z-2 .
2zl-J-3z+2
243. w = [(z-1)/(z-J-1)]8.
l
244. (1) w =
1
z• +R•
.!.
( .!.
):
(2) w
=(
z•-R•
z•+R•
a
245 • w = -(2z-J-y(3)-i)I.
2z-y(3)-i
246. w = - ( 2z-J-y(3)-i
2z-y(3)-i
(
250.
)a.
151.
152.
153.
254.
2z+ y(3)-i)•
247 " w
= 2z-y(3)-i •
.. ~ 8
=
249. w
= [ z-y(2)(l-J-i)
Q• • w
z~ -R~).
w=e~'(z+l
)f.
z-1
w = y[(z+ 1)/(1-z)].
w = y[(z-J-i)/(i-z)].
w = y[(z-z 1 )/(z1 -z)].
w = y[(z-J-R)/(z-R)].
tel
a
255. w = e - ... y(z-i) •
256. w = y(z1 -J-h1 ).
·(2z-J-y(3)-i)I.
i 2z-y(3)-i
z-y(2)(1-i) ]'
158. w = (
~:::~:
r.
157. "' =
•
y(zl-J-hl)
-----"-'Z
For one choice of the branch of yz this function
gives the solution of problem 158, and for the other the solution of problem
159.
159. See the answer to problem 158.
260. w = z/(1-z)•
261. To the circles
4u•
---,1
+
""
(R+ ~) (R- ~)
lzl = R (Fig.
1
=
64) there correspond the confocal ellipses
l (to the circle lzl = 1 corresponds the segment
11 = 0, -1 ~ u ~ l); to the rays arg a = a: there correspond branches of the
confocal hyperbolas u 1 /cos•a:-'11 /sin1 a:= l (to the ray arg z = 0 corresponds the
ray"= 0, u;;;;;.. 1, to the ray arg z = n, the ray "= O, u~-1; to the rays
argz =±.;!..corresponds the axis u
2
=
0).
ANSWERS AND SOLUTIONS
v
FIG. 64
162. (1) and (2). The exterior of the ellipse
4t1•
1
+
4111
1
(R+ !) (R- ! )
=
1
(Fig. 65, (1), (2)) ;.
(3) and (4). The whole plane with a cut along the segment [-1, l]
(Fig. 65, (3), (4));
(5) and (6). The wholepla.newithcutsalong the rays (- oo, -1] and
[1, oo], on the real axis;
(7) The lower half-plane;
(8) The upper half-plane;
(9) The upper half·plane;
(10) The upper half of the interior of the ellipse
4'11
4111
(R+Rl).+( B-R1) =l;
1
(11) The lower half of the interior of the ellipse
4u1
4v1
(R+ ! ).+(R- ! )1=1;
(12) The right half of the interior of the ellipse
4u1
4v1
(R+
!
r
+ (R- ! )' =
with a cut along the segment [ 1,
1
!(R+ ! )];
296
PROBLEMS ON OOMPLEX ANALYSIS
(13) the domain between the branches of the hyperbola
u 8 /sin1 r1.-v 8 /cos8 r1.=l
FIG. 65
4ul
263. (1) To the circles lzl =R correspond the ellipses
(R-
~
4tJ!
+ ----,-,-
r
(R+
~r
= l (to the circle lzl = l corresponds the segment u = 0, -1 =::;;;; v =::;;;1), to
the rays arg z = r1. correspond the branches of the hyperbola v1 /sin1 r1.-u,lfcos•r1.
= l (to the rays • = 0 and • = :n; corresponds the axis v = 0, to the ray
• = :n;/2 corresponds the ray u = 0, v ~ 1, to the ray • = -:n;/2 corresponds
the ray u = 0, v=::;;;; -1);
4ul
4tJI
(2) Tothecircleslzl=Rcorrespondtheellipses (
1
a•) +( as)•
R+R
R-R
= l (to the circle lzl = a corresponds the segment v = O, -a=::;;;; u =::;;;;a), to the
rays arg z = r1. correspond the branches of the hyperbola
a 1 cos1 r1.
= l (to the ray arg z = 0 corresponds the ray v - 0, u ~ a,
to the ray arg z = :n; corresponds the ray v = 0, u =::;;;; -a,
to the rays arg z =
±;
corresponds the axis u
=
0);
(3) The families of confocal ellipses and hyperbolas obtained from
the corresponding families of problem 261 by a rotation through an angle y
and a magnification with coefficient lol (the centre of the magnification is
the coordinate origin).
.297
ANSWERS AND SOLUTIONS
264. (1) w
(2) w
ef•
= -(z+
y(z1 -c1));
c
= _2_b (z+ y[z1 a+
(a1 -b1 )]); in both cases for one choice of the
branch of the root we obtain a mapping onto the exterior of the unit circle,
and for the other onto the interior;
(3) w =
265. w
=
az-by[z1 -(a1-b•)]
a•-b•
.
A(z+ y[zl- (a1 -b1 )]), where A is an arbitrary comple:ir num•
a-b
ber; µ = y(al+b1 )-y(bl+kl)
! (a+ !)] a>O;
rays [-oo, ~ (a+ !}
] [-1,+oo)
266. The whole plane with a cut along the segment [-1,
the whole plane with cuts along the
if
if
and
a< O.
270. w
-v( ! [(~1 +
271. [1+
272.
;. )+ (z•+ :.)]).
4(l~h)](w+ ~) =
1
w+;; =
4a
(l+a)•
(
1)
z+z-
(e;• e:•)- 2(1h~h).
+
(1-a)1
-2 (l+a)•;
,
{l+a)1
w (0) = ~;
the length of the arc corresponding to the cut equals 2 cos-1
6a-l-a1
(l+a)I ; it is
equal to n for a= 3-yS.
1
l
a+-+b+2(z+-H-(a+-l-)+(b+-H
a
b
1
; w'(O) = - - 4- - ;
273. w+ w = 2
(
l ) (
l)
a+a + b+b
298
PROBLEMS ON COMPLEX ANALYSIS
the lengths of the arcs corresponding to the cuts are equal respectively to:
274. (1) The image of the circle 0 is a circular arc with ends at the points
± 1, inclined at the point l at an angle 2ct to the real axis; the exterior
of the given circle is mapped onto the whole plane with a cut along the given
arc;
!I
u
Fm. 66
(2) The image of the circle 0 is (see Fig. 66) a closed curve (Zhu·
kOtJBkii profile) with cusp at w = 1, the tangent at this point ma.king an
angle 2ct with the real axis; the circular arc with ends± l discussed in
part (1), is contained in the domain bounded by the Zhukovekii profile; the
exterior of the circle 0 is mapped onto the exterior of the Zhukovekii profile.
276. (1) The image of the circle 0 is a closed curve consisting of two circular arcs with common ends at the points ± 1, the tangents to these arcs
at the point l ms.king with the real axis angles respectively equal to 2ct-M
and 2ct+ (n-cx)c5; the exterior of the circle is mapped onto the exterior of the
domain bounded by the given arcs; the image of the circle 0' (see Fig. 67)
is a closed curve with an angular point at w = 1, the tangents at this point
being inclined to the real axis at angles respectively equal to 2ct-M and
2oi:+ (n-cx)c5; the image of the circle 0 is contained in the domain bounded
by the image of the circle O'; the exterior of the circle O' is mapped onto the
exterior of the image of this circle.
(2) The interior of the circle 0 is mapped onto the exterior of the
domain bounded by a.res of circles passing through the points -1,1, the tangents to these circles s.t the point l forming with the real a.xis angles respectively equal to: (a) 2oi:+ (n-cx)c5, 2oi:+ (2n-oi:)c5, if the function w(z) is
defined in the z-pl.a.ne with a cut a.long an arc of the circle 0 lying in the
299
ANSWERS AND SOLUTIONS
lower half-plane, (b) 2oc-d, 2oc- (n+oc)d, if the cut defining the function w(z) is made sloug an arc of the circle 0, which lies in the upper hslf.
plane.
lJ
FIG. 67
w-l
276. w+l
=
(
zelY+ie'~
)
_,1.. ,wherey= oc, if P> O,andy-oc+nif P<O.
1
zel7+ie •
277. The whole plane with cuts along the rays y = 0,
y = 0, a:~ 1/2.
278. The hslf-plane a:> 1/2 with a cut along the
1/2~ a:~ I.
279. The whole plane with cuts slong the rays y = O,
y = 0, -oo < a:~ -1.
280. The angle -n/n < arg z < n/n with a cut a.long
a: ~ -1 /2 and
segment y
l
~a:<
0,
oo and
the ray y = O,
ii!~ a:< oo.
281. (1) The whole plane with cuts along the rays
=
lwl ~
V
4,
I
arg w = 2kn (le= O, 1, ... , n-1); (2) w= (l+z•)11 .
f/C4)z
n
282. (1) w = y[z•+v<z'-1)]
(2)
W=
=-
1 -[y'z1 +1>+vcz•-1)];
y'2
f[l+y'(l-z')] = -1-[y(l+z1 )l+Jl'(-z1 )].
z
zy2
1
281. (1) w
= (oc•+oc-•)-n f/{zn +r•+ y[(zn+z-•)1 -
Solution. The function C=
! (z• + 21~)
(ocl'+oc-•) 8]}.
maps the sector onto the lower hslf.
300
PROBLEMS ON OOMPLEX ANALYSIS
plane, the points oi: a.ndoi:e :'passing into the points ±2-(oi:n+_.!._), Next,
ix"
2
it is necessary to compress the half-plane 1'
=
1
2
the unit semicircle
required.
1'+ y(pl-1).
184. w = ..
and map it onto
(ix"+ ix-")
The function w
=
y
T
is the one
.!.( zl +z-I + V![(zR +z-S)a- (oi:I +oi:-8)']).!n
n
n )= ( oi:I +oi:-s
n
(2) w
n
11
..
n
n
n
11
I[ y(b
Ji'(z'+c')+ y(a•+o•)] and
+o
y(z +c')
JI
w =-
T =
C
1
1) -
1
I
p {y(z•+o•)+oi:+y[y(z1 +c'+oi:)•-tJ8]}.
I
I
where oi: = 2" [t'(a1 +o')-y(b'+cl)], tJ = 2" [y(a1 +o1)+ y(b1 +o1)].
185. w
= y[y(z1 +c')+ y(a1 +o1)].
188. W=-*(v[(:::r+1J+v[(:::r-1J)·
187. w =
y'[ y~oi::{:;•::.;~~(:•-I) l
. . M<I.•-i/[y(~~)-~(:~f)l
for b = 1,
... » •.
1D
=
v[v(~+cot1 ;) +cot;];
·-V[v(!~~i+~t::fil ~- :~:.
l+b
bi= l-b.
301
ANSWERS .AND SOLUTIONS
'.Che function ~ = zl maps the upper half-plane with cuts along the
segments [O, I+i], [0,-I+i] onto the domain of problem 285 which we also
map onto the upper half-plane. By the symmetry principle these functions
map the domain given in the conditions onto the plsne with a cut along some
segment. It remains to map the exterior of this segment onto the ex'terior
of the unit circle.
SOLUTION,
•
n
n
!
; 2 [(z+y(z•-l))"i" +(z+y(z1 -I))-"i" + 2]
291. w =
i
= y2
~
..!!..
[(z+y(z1 -1))2« + (z+ y(z 1 -1)- •«] •
.!.
SOLUTION.
By means of the function T =
inverse function of z =
~ (z +
!),
domain Ii-I> I, Im T> O,which
~
~«,
where
C=
z+y(z•- I) is the
the upper half-plane is mapped onto the
(T+
~ ), mapsontotheupperba.lf-plane.
Applying the symmetry principle we obtain the mapping of the interior of
the right hand branch of the hyperbola onto the whole plane with a cut along
the ray ( - oo, -1]; this latter domain is easily mapped onto the upper half.
plane.
I
REMARK.
The factor y 2 does not play any pa.rt, since the transformation
w' = kw (k > 0) maps the half-plane onto itself.
rs
""''tr
292. w = [e::~<z+J"(zl-l))]IP-[e-l«(z+y(z 1 -l)]-'ii, where
z+y(z8
-c8) ]P
:193. w ... [ e-1«
c
, where c =
p
v'<a'+b•),
fJ = n-ix.
b
a:= tan-1 0
,
= n/[2 tan-l(a/b)].
294. (1) The domain is constructed as follows: the ring rf < lwl < rf
is cut a.long the segment rf :s;;;; u :s;;;; rl of the real axis and to the lower edge of
the cut there is attached pa.rt of the same ring: rl < lwl <
0 :s;;;; arg w <
< 2a:; if a: =- n then the second ring is complete and its free edge must be attached to the free edge of the first ring (in this case we obtain a two-sheeted
ring ti < lwl < rll;
(2) If a: :s;;;; 1, the inequality lzl-1 I < a determines two domains (see
problem 34), each of which is mapped onto the single-sheeted circle lw-11<
<a; if however a> I, the inequality lz1 -ll <a determines a single domain
which is mapped onto the two-sheeted circle lw-11 <a (in order to construct
this two-sheeted circle it is sufficient to cut two identical specimens of the
circle lw-11 <a along any radius and attach the lower edge of the cut on
the first specimen to the upper edge of the cut on the second specimen and
the upper edge of the cut on the first specimen to the lower edge of the cut
on the second specimen).
295. (1) The domain is constructed as follows: to thew-plane cut along
the segment [ -1,l], there is attached the interior of the ellipse
r:,
302
(
4u•
1 )1
R+B
PROBLEMS ON COMPLEX ANALYSIS
+(
4t11
1 )' = 1, also cut along the segment [ -1,1], the lower edge
R-R
of the cut plane being attached to the upper edge of the cut ellipse, and the
upper edge of the cut plane to the lower edge of the cut ellipse-;
I:+ ~ I
(2) The two-sheeted domain
< R' (the
defines the interior of a circle if R < 1; a half-plane
terior of a circle in the case when R > 1). The cut
attachments go along a curve connecting the point w
I:+~ I
I:+ ~ I
inequality
< B'
for R = 1 and the ex·
and the corresponding
= l to some boundary
point of the domain
< R1 •
296. (1) and (2). The surface consists of two sheets of the z-plane cut
along the segment [-1,l], the lower edge of the cut on the first sheet being
attached to the upper edge of the cut on the second, and the upper edge of
the cut on the first sheet to the lower edge of the cut on the second.
19'7. (1) and (2). The surface consists of two sheets cut along rays going
respectively from the points -i, O, i to infinity, for example, pa.rallel to the
real axis in the positive direction. The lower edges of the cuts of the first sheet
a.re atta.ohed to the upper edges of the corresponding cuts of the second sheet,
and conversely.
298. The surface consists of three sheets of the z-plane cut along the
rays (- oo, -1] and [l, oo) • .Along the ray (-oo, -1] the attachment proceeds
as follows: the upper edge of the cut on the first sheet is attached to the
lower edge of the cut on the second sheet, the upper edge of the cut on the
second sheet to the lower edge of the cut of the third sheet and the upper
edge of the cut on the third sheet to the lower edge of the cut of the first
sheet. Along the ray [l, oo) the lower edge of the cut on the first sheet must
be attached to the upper edge of the cut on the second sheet, the lower edge
of the cut on the second sheet to the upper edge of the cut on the third sheet
and the lower edge of the cut on the third sheet to the upper edge of the cut
on the first sheet.
299. (1) The polar net e = const, tJ = const;
,_,,
(2) The logarithmic spiral e = e-A:- (for Ii= 0 the ray tJ = b);
(3) The angle oc < tJ < p (for IX = 0 and p = 2n the plane with a cut
along the positive part of the real axis);
(4) The whole plane with a cut along the logarithmic spiral Q = e•;
(5) The sector e < 1, O < tJ < IX (for IX = 2n onto the unit disk with
a cut along the radius " = 0, 0 ~ u ~ l);
(6) The domain e > 1, 0 < tJ < IX (for IX = 2n onto the exterior
of the unit circle with a cut along the ray"= o. 1 ~ u < oo);
(7) The domain e« < (! < efl, y < tJ < d (for d-y = 2n this domain
is a concentric ring with a cut along the segment tJ = y, e« ~ (! ~ efl).
100. The angle 0 < arg (z+n) < n/n; the strip 0 < '!/ < n.
301. (1) The rectangular Cartesian net u = 0,
O; (2) A straight line;
(3) The strip 0 < " < IX; (4) The half-strip u < O, 0 <" < 1X;
(5) The rectangle log r 1 < u <log r 1, 0 <" < 2n.
"=
I
~.
I
I I
'1o ·
802. (5) b ==a tanhT , l =a ta.n2
303
.ANSWERS .AND SOLUTIONS
303. (1) The family a:= 0 is transformed into the family of confocal hyperbolas with foci at the points±
1, (
c:::
si::
0 -
0 =
into the family of confocal ellipses with the same foci
l);
the family y = 0
(cos~• 0 +~: 0 = l) ;
(2) The upper half-plane; (3) The fourth quadrant;
(4) The right half-plane with a cut along the segment [O, l];
(5) The whole plane with cuts along the real axis along the rays
(- oo,- l] and [1, oo);
u•
t1•
(6) the interior of the ellipse cosh• h
+Binh• h = 1 with
cuts along
the segments [ - cosh h,-1] and [1, cosh h].
804. (1) The half-strip - ;
(2) The strip - ;
.
<u < ~ , v>
0;
<u< ; ;
(3) The half-stnp 0
<
(4) The strip-;< u
u
n
<2, v >
O;
< o.
805. (1) The family a:= 0 is transformed into the family of confocal
ellipses with foci at the points ±
l (cO::• 0 + Sin~i 0 = l} ;the family y = 0
into the family of confocal hyperbolas with the same foci (c::: 0 - s;: 0
=
l) ;
(2) The whole plane with cuts along the real axis along the rays ( - oo,
- ll aod [l, oo);
(3) The upper half-plane.
806. (1) The strip-n/2 < v < n/2; (2) the half-strip 0< v < :r/2, u > O.
807. (1) The family a:= 0 is transformed into a family of circular arcs
with ends at the points w = ±1, including also the corresponding parts of the
imaginary axis; the equation of the pencil of circles: (u - a) 1 +v• = l+a•
(a= cot 20); the family y = 0 is transformed into the family of circles of Apollonius with respect to the points w = ± i (including also the real axis); the
equation of a family of circles of Apollonius: u•+ (v - b)1 = b1 - 1, !bl> 1
(b = coth 20);
(2) The upper half-plane with a cut along the segment 0 E;; t1 E;; 1
of the imaginary axis;
(3) The whole plane with a cut along the segment - l E;; v ~ 1 of
the imaginary axis;
(4) The semicircle lwl < 1, Re w > 0;
(5) The unit circle.
308. (1) The whole plane with two cuts on the real axis: along the rays
(- oo, - 1) and [l, oo );
(2) The right-hand half.plane with a cut along the ray [l, oo) of the
real axis.
304
:PROBLEMS ON OOMPLEX ANALYSIS
.. (l-1).r
309. w =
"' Cz+2)
e--,1--
312, w
(z-l) n
310. w = - cosh -,,,--.
318.
= e 3 (z-2)
fD =cos~.
2nl.r
.r-2
111. w = e
815 (l)
•
w
,
n (z+2)
y(2;-1 (y(.r)- l)
81'. w
•
= _ 1 +iy'S
2
=e
.r-1
"' !..±!
m
(z+Si),
tanh 4 (z - i) ,
(2) w = - e
+2 -
•
••
"'.!...±!.1
e
z- +2+i
'
ni.r+I
2e
(3) w
l+e
316. (l)w
817 ,
u•
,., __
z-I
=---.-+...,I •
.r-I
= y(elnl.r+e-snil);
=, /
JI
[cos m - cos
1+cosm
nh] .
318 • w =,/[cos nz- cos nhi].
cos m + cos nha
319, w = y(cos 2z + cosh 211.).
JI
f( coscos2z+I
2z+cosh 211. ) •
, f( cos 2z+cosh 211.t ) .
JI cos 2z+cosh 211.
320 , w = ,
JI
321 • w =
w=y[;::!:=::].
(2)
8240 w = i / ( c o s f - c o s f ) .
cos--oos-
z
325 •
vI(
=-.
323. w=
V(
3•..s. w = i cosh
326, w
sin7n),
l+sin-z
127.
l+oos 4;
4n
cos-21-
nyz
2a-.
-
10
cos
= i / ( c o s ! - cos!) '
cos--cos-
z
1
82!. w
10
W=
a
l/(e-T -e~)
=f
e2n/a - e2nlfz •
i/(cosh !!..-cos!!..)
h
z •
1-cos~
z
)
a4n
a
·
nyz
= t anh• 4;-
(see the remark to the answer to problem 291).
<
• 1 sin z
. Binz maps the ha'"
. y > 0, - n/2
329. w = em-osh . The fiunction
u-Btrip
c
a:
a; < n/2 onto the upper half-plane: the points ±n/2+ai then become the
ANSWERS AND SOLUTIONS
305
points ± oosh a. Hence it is easily obtained that the function w = sin-1 sin z
oosha
will map the given half-strip onto itself so that the rays a: = ± n/2, a.::;;; y < oo
will correspond to the rays u = ± n/2, 0.::;;; " < oo. On applying the symmetry
principle an infinite number of times we see that the function found is
the one required.
b sin-l i sinh z
330. w
=
cosha '
sin-1 (l/oosh a)
. 1 1
331. w = sm- "'CciSiil.I
.. / [( .
+V
1
sm-
1
)1
"'CciSiilJ -
(
1
• 1 sin z ) ]
sm- "'CciSiil.I
. 1 sinz
sm- oosha·
V!['"'-:~ -~-:~].
332. w = ..
sm- cosh a - sm- cosh a
333. w == sin-1 e21s . Solution. The function C= ells maps the strip 0 < a:
< n/2 onto the upper half-plane and the function w = sin-1 Cmaps the upper
half-plane onto the half-strip - n/2 < u <n/2, "> 0. The function w = sin-1ells
which maps the strip onto the half-strip transforms the rays a: = 0, - oo < y
< O; a:= n/2, - oo < y < 0 into the rays a:= n/2, 0 < y < oo; x = n/2,
0 < y < oo, respectively. Applying the symmetry principle an infinite number
of times we see that this function effects the required mapping.
134. (1) If b < 2n the curvilinear rectangle: 1 <!I< e", 0 < (J < b; if
b = 2n the ring 1 <!I < e" with a out along the segment [l, e"]; if b = 2Tcn
(Tc= 2, 3, ... ) a many-sheeted domain consisting of k rings 1 <!I < e", out
along the segment [l, e"] and attached in such a way that the lower edge
of the out of the first ring is attached to the upper edge of the out of the
second ring, the lower edge of the out of the second ring to the upper
edge of the out of the third ring and so on; if b = 2Tcn+P (Tc= 2, 3, ... ,
0 < p < 2:ir), then to the free lower edge of the last ring of the surface being
constructed it is necessary to attach along the segment [l, e"] the curvilinear
rectangle 1 < !I < ea, 0 < 0 < /J;
(2) The domain with an infinite number of sheets consisting of rings
1 < !I < e", out along the segment [l, e"] and attached by the method
indicated above;
(3) The infinitely many-sheeted domain consisting of the rings
1 < !I < e" out along the segment [l, e"], enumerated by means of the integers
( ... , - 2, - 1, 0, 1, 2, ... ) and attached so that tile lower edge of the out
of each ring is attached to the upper edge of the out of the ring the number
of which is one unit higher. This domain is the part of the Riemann surface
of the function Log w which lies above the ring 1 <!I< e".
381i. (1) The two-sheeted domain obtained by joining together two right
half-planes, each of which is out along the ray " = 0, 1 .::;;; u < oo; the edges
of the outs are attached criss cross, that is, so that the lower edge of the out
of the first sheet is attached to the upper edge of the out of the second sheet,
and conversely;
306
PROBLEMS ON OOMPLEX .ANALYSIS
(2) The two-sheeted domain consistlng of two planes with cuts along
the real axis along the rays -= < u =s;;;; -1 and 1 ~ u < CXl and attached
criss cross along the cuts -= < u =s;;;; -1. The edges of the cuts 1 =s;;;; u < oo
remain free.
336. A two-sheeted domain consisting of two planes cut along the imaginary
axis along the segment -1 =s;;;; " =s;;;; 1 and attached so that the left-hand edge
of the cut of the first sheet is attached to the right-hand edge of the cut of the
second sheet. The remaining edges are free.
887. The Riemann surface is infinitely many-llheeted and has two logarithmic branch points above the points w = 0 and w = =· Domains of single·
valuedness in the 111-plane corresponding to sheets of the w-plane with cuts
along the positive real axis are bounded by the circles 2Tcn(x'+y•)+y = O
(Tc= 0, ± 1, ± 2, ... ).
888. r 1 = <Xl, r 1 = r 8 = 1; the image of the disk 11111 < r 1 is the whole
plane with the point w = -1 deleted; the image of the disk lzl < r 1 (and of the
disc 11111
<
r 8 ) is the half-plane Re w
> -
!.
1
1
889. (1) r1 ="j"; (2) r1= 2 lal; (3) r1=l.
1
848. (1) r 1 = 1/4; (2) r 1 = 4 lal; (3)
= 2 - y3.
144. r 1 = n, r1 = 1.
1
345. (1) r8 = 1/2; (2) r1 = 2 lal; (3) ra = I.
r,
CHAPTER III
847. (1) For example, the whole plane with a segment removed;
(2) For example, the domain bounded by two circles touching internally.
354. BoL'O'TION. The .Jordan curve y can be considered as the topological
image of the segment LI (or, if the curve is closed of the circle LI). Let us assume
that y contains the circle K : lz - z0 1< (I. On LI the circle K corresponds to
some continuum, consequently, the segment [a, b]. The arc Yab CK joining
the images of the points a and b on y, also corresponds on LI to a continuum, moreover it belongs to [a, b] and contains the points a, b, consequently
it coincides with [a, b]. But this is impossible. However, continuous curves
exist which contain interior pointst.
858. BoL'O'TION. (1) We divide up y into sufficiently small parts, draw chords,
map them linearly onto the corresponding parts of the segment 0 :s;;;; t :s;;;; 1
and obtain the polygonal path n; then map the segments connecting points on
y and n which correspond to identical values oft, continuously onto the segment
OE;;
A:s;;;;
l;
(2) and (8). We first replace y (in the circle or in the ring) by a homotopic polygonal path.
862. SoL'O'TION. Let 0 be the centre of Q, y an arbitrary closed path in
Q not passing through 0, p an auxiliary circle with centre at the point 0 such
that the path y lies in the ring between p and Q, and lJ = ""fl' (0'), where
p', 0' are the images of p, O. In the circle Q the path y is homotopic to an
.,,,,. (0) times traversed circle p (see, 858, (3)) consequently, y' the image of
y is homotopic in Q' to an .,,,,.(0) times traversed path P', hence .,,,,.,(0')
= ni.(O) lJ. In particular, choosing as y the original of some circle y' around
0', we have ni.·(0') = 1, whence we conclude that lJ = ± 1. Now let a be an
arbitrary point of Q. As above we conclude that fly(a') = .,,,,.(a)lJ4 , where
da has a meaning similar to lJ for 0. We show that lJa = lJ. For this we draw
in Q the circle p, enclosing 0 and a. Then: f6fJ(a) = f6fJ(O) = 1, f6JJ• (a') = .,,,,. (0),
""fl' (a') = iJ0 , ""fl' (0') = lJ, whence it follows that iJ4 = lJ. In the case
of a continuously differentiable mapping with a Jacobian different from
zero the infinitesimal behaviour is equivalent to an aftine transformation
(see problem 875).
363. SOL'O'TION. Divide up the domain G by Jordan arcs into the domains
Gt, (i1 = 1, 2, ... , tii) with diameters less than 1. If /(z) of:. a, on these arcs
then denoting by y11 the boundary of G1,1 , we have
n,
d arg [f(z)-a] =
d [arg /(z)-a] of:. 0,
f
2 J
,,
l,=111,
hence a domain
<P.1
(s1 is the value of i 1 ) bas been found, such that
Jd [arg /(z)-a] of:. O.
1••
t Bee, for example, A.B. P.um:HOMENKO, What is a curve! (Ohtotakoelmiya),
Chapter I, § 2, Gostekhizdat, 1954.
307
308
PROBLEMS ON COMPLEX ANALYSIS
Divide up Gl, by .Jordan arcs into domains Gi11, (i1 = I, 2, ... , n 2 )
with diameters less than 1/2, then, repeating the process, construct the domains GR,s,ia with diameters less than 1/3, and so on. Then either on one
of the arcs a point will be found where /(z) = a, or we obtain a sequence of
domains G},, G.i,s,, G~18,s,, ... , enclosed one within the other which shrink
to some point z0 e G, consequently the images of these domains in the w·
plane shrink to the point /(z0 ) and as all the time d arg [/(z)-a] ,p 0 along
the boundary of these domains, it follows that /(z0 ) = a. Since by supposition
f (z) i= a on y, it follows that z0 lies within G.
375. a 1 = b2, b1 = - a 2 or a 1 = - b2, b1 = a 1 •
384. It is possible to construct the required quasi-conformal transforma·
J
R+lal
tion with the characteristic P ~ R _ jaj •
385. u
=
x/(cos 1%), v
= y-x
tan 1%, p
=
(l+sin 1%)/ cos 1%•
FIG. 68
386. SOLUTION. By means of the function C= log [(z-a)/(z-b)] (the
values z = a and z = b correspond to the points A and B) we conformally
map the given domain onto a strip of width n+Po· We compress this strip
into a strip of width n (a quasi-conformal mapping with characteristic p
=
I+ Po) and, using the inverse function z((;), we map the latter strip onto
-
:i;
the half-plane. Then the arc AM (Fig. 68) of length a will occupy the position
of the segment AM' of length x; the point M moves along the circle of Apollonius with respect to the points A and B (shown dotted), we have :;
cos
2
Po
2
Po
J!..
2
= - - - ~ cos2- (prove
cos 2
this!). In the resulting half-plane we widen the
-
2
vertical half-strip on the segment AB as base to the length of the arc AB,
and moreover in such a way as in the result to preserve lengths everywhere
on the arc AB. The resulting quasi-conformal mapping has the charac_... t'ic p""='
te1'1B
(i +-n/30)
sec
22·
/Jo
309
.ANSWERS AND SOLUTION'S
!
388. A =
[(a - d)+i (c+b)], B-=
!
[(a+d)+i (c-b)], F =
389. b =Ba. In particular, it is possible to take w
aii=w-Bw.
= w+Bw,
!
{/+ig).
and also
391. The highest terms of the equation of the problem determine a qua.siconforma.l mapping with two pairs of characteristics (P, Q), (Pi. Q1 ), depend·
ing on q1 (z) and q1 (z). The transformation C(z) is a quasi-conformal mapping
.
.
. .
•
p-l Biii
with one pair of cha.ractenstics pf} and q1 = - - -1 e
•
p 1-I
The coefticient fill= Pi+I e
11111,
P+
(see problem 379).
(see problem 378).
CHAPTER IV
197. (I) 1 1
1. =
=
~a+i, 1 1
i
in
= 1+2; (2) 1 1 =2' 1 1
n
= -2;
(3) 11
,.
id',
-nB1 •
198. (1)
vs(
1-
~ ); (2)
2; (3) 2i;
899. m.
400. 4/3.
Bn+1
401. (1) n+l [(- l)n+l_ I], if n
(4)
o.
+ -1;
m, if n
=
-1;
+
(2) and (3). 0, if n
-1; 2ni, if n - -1.
402. (1) -2(1 - i); (2) 2(1 - i); (3) -2(I+i); (4) -4; (5) 4i.
408. (1) 2ni; (2) -2n; (3) 2di; (4) 2di.
+ -1;
404. (1) 2ni, if n
n+I
-2n1 if n = -1.
elGuri-I
405. I+et , if
IX+
-2n', if n
-1;
=
2ni, if
-1; (2) (-l)n+l 2ni if n
n+l
ct=
+ -1;
-1,
407. (1) letl::s;;;;n/2; (2) sinpet;;;::.O.
421. (1) n/3; (2) -n/3; (3) o.
422. If the contour 0 contains the point 0 as an interior point and does
not contain I and -1 then I= -2ni; if it contains only one of the points
-1 or I and does not contain the point 0 then I = ni. It is clear from this
that the integral may assume five different values (-2m; -m; 0; ni; 2ni).
423. 2n - I, if n > I; 2, if n = I.
424. ni/2.
429, (1) 2/3; (2) 1 - 2i/3.
425. (sin a)/a.
445. Convergent.
446. Convergent.
a}
(
426. e" 1 + 2 ·
447. Divergent.
427. (1) I; (2) -le; (3) I-le.
448. Non-absolutely convergent,
449. Non-absolutely convergent for ~ :;(; 2lm (k = O, ±1, ±2, ... ),
divergent for ~ = 2kn.
450, Converges absolutely.
459. e.
451. Diverges.
460. I.
452. Absolutely convergent.
461. I.
462. I.
458. Divergent.
454. Absolutely convergent.
468. 1/4.
455. B =I.
464. l/e.
456. 00,
465. I, if !al ::s;;;; I; Iflal, if !al> 1. 457. o.
466. I.
458. 2.
467. (1) B; (2) B/2; (3) oo; (4) 0; (5) B"; (6) B, if !Zol ::s;;;; I, and B/llol•
if lz.I > I.
.
310
311
ANSWERS AND SOLUTIONS
<
r1/r2•
468. (1) R ;;;:i:: min (r1, r 1); (2) R ;;;:i:: r 1r 1; (3) R
1
1
469. (1) z/(l-z) ; (2) -log (1 - z); (3) 2 log [(l+z)/(1-z)]; (4)log (l+z).
470.
471.
472.
473.
Divergent at every point.
Convergent (non-absolutely) at every point except z =l.
Abeolutely convergent.
Convergent (non-absolutely) at every point except z
=-
I.
lkirl
474. Converges (non-absolutely) at every point except the points z = eP
(k = 0, 1, ... ,p -1).
475. Converges (non-absolutely) at every point except the points z
= (1 ± iy3)/2 and z = -I.
476. Converges absolutely.
477. For example, Cn = (-l)n.
479. SoLtrTION. Let us first investigate the convergence at the point z = 1.
~ (-l)b'n)
with denominators k 1, k•+ 1, ... ,
The terms of the series ,,::,,
n
n-1
(k+l) 2 -1 have the sign (-l)k; let us denote the sum of these terms
by (-l)kO'k and prove that O'k-+ 0 monotonically. We have
2k+l
(k+l)1 -k1
=--;ca--+Oask-+oo,
k•
O<ak<
On the other hand
, t-0'1<+i
[~. - (k~l)lJ + Lk.~l) - (k+:>·+1 J+
=
... +
[
]
1
(k+2)1 -3 -
1
(k+l) 2 - l -
> (2k+l) [
1
(k+l)•-1 -
1
(k+2)1 -2 -
]
1
(k+2)1 -3 -
···
1
(k+2) 2 - l
1
(k+2)2 - l
1
(k+2)1 -2 -
>
O.
'l'his proves that the given series converges for z = I. If lzl = 1, but z #: 1,
then use must be made of the convergence test of problem 440, putting
a,,= (-l)[vnlzn, bn
= .!..
n
We have
Sn= -z-z1 -zl+z'+ ... +(-1)
... ±
where p
=
zP'.
[yn)
1-z'
l-z
1-zl
...
-z--+z'---z•--+
1-z
1-z
1-z
5
zn =
1-zSP+l
1-z
± (z(p+t>'+
••• +z")
[yn] - I.
< ll~zl +2p+3, consequently for every z there
< kp < kyn and this completes the proof.
00
00
"'1 212n +1
"'1 212n
Hence lsnl
that lsnl
482. ~ (2n)!, R = oo.
n~o
483. ~ (2n+l)!, R
n=O
exists a k such
=
oo.
312
PROBLEMS ON COMPLEX .ANALYSIS
00
00
"1
21111-1z••
484. ,L.J (-1)11+ 1 ( 2n)! , R =co.
n=O
00
486. a«
2( : )(:
n-o
(n
=
n=O
where (
~) = 1,
(:) = ct(ct-l)
[1+_!_~+
2
i
00
2
n=l
f,
anzn R =
488. ,L.J (-1)" bn+i,
(-l)n-1 1.3 ... (2n-3)(~)1, R
2. 4 ... 2n
1
Ia.I
b
n=O
[";1]
wherec,.= "1
,L.J
n
(-1)'11(
2m+l
m=O
}2•-tm-131111, R=yl3.
00
490.
,J; (-1)" (n-l)z",
R =I.
n-2
00
"1 z•n+1
491. 2 ,L.J 2n+l ,
R
n-o
00
492.
2
n=O
=
zsn+1
(-l)•-,
2n+l
1.
R =I.
00
493
"1(-l)"I. 3.s ... (2n-l)z1n+1
•,L.J
2•.n!(2n+l)
'
R=I.
n=O
00
494. log 2-
2( l+~n-):,
R = 1.
n=I
00
"1 zsn+1
495. ,L.J n!(2n+l), R
= oo.
n=O
00
"1
z•n+1
496. ,L.J (-1)" (2n+l)!(2n+l), R = oo.
n=O
·~!(ct-n+l)
R =!al.
1, 2, ... );
487. l+i
y2
r
"1 2••-lz'rt
485. 1/2+ ,L.J (2n)! , R =co
=
1.
313
ANSWERS AND SOLUTIONS
497.
!
00
+2 2(-1)11+1
<;:-;t,
B = 3.
11=1
498
2-(i+ z-1 + (z-1)1 + (z-1)1 + (z-1)'
21
21
2'
2'
+ •..
• 4
00
I ~
] (
(z-I)lll-1 (z-I)llc
211c
+
2llc
+ ... = T ~
... +
(z-1)11
)
11+1
• R ""'
n-o 211+¥<-1> +1]
l
s.
00
499
.!..+ ~ (-1)11 (n- 3) (z-l)ll
B = 2
• 4 L,;
211+•
•
•
n=l
00
Ht. -I;iy3
2(1~3) (z-1)11,
B =I.
n-0
00
501. 2<-1>11+1 <z-:1>11.
n-1
511.
B =I.
oosin(1+tm)
n! 2 (z - 1)111, B = oo,
,2;
n=O
z'
508. I+zI + 3
+ ...
z•
z'
504. 1-4-96+ ...
I
5
3
505. I+z•- 2 z1+ 6 z'-4z1 + ...
5
5
13
506. 1+z+z1+ 6 z11- 8 z1+ 30 z11+ ...
z•
507. z+ 21 +
2zl
9z5
31 + 51 +
...
00
z
zl
z'
508. log2+-+----+
2
8
192
... = log2+
I
Co= 2'
509. 2 [
00
2°"
--z11+1,.
n+I
n-o
+nc1+n(n-I)o1 + ... +nlo,.- 1+2nlo11 =I, n;;;;i. I.
~ + (I+ ! )~ + (I+ ! +
!)~
+ ... + (I+
! + ! + ...
I
) z11
]
··· + n-1 -;;-+ ··· '
314
110.
PROBLEMS ON COMPLEX ANALYSIS
-'n1+2[-2niz+(l-2m); +{1+ ~ -2m)z; + ...
... +{1+ ~ + ... + n~l -2ni}: + .. .].
t See the literature mentioned on page 82.
315
ANSWERS AND SOLUTIONS
•
_ n(n+l)-1.2 z3
[n(n+l)-1.2][n(n+l)-3.4] za+
5,.. 8 . z
3!
+
5!
...
[n(n+l)-1.2][n(n+l)-3.4] ... [n(n+l)-(2k-112k] 21:+i+
·I I 1
(2k+l)!
z
•.. ' z < .
529. z/(1-z1 ).
... +
m1
m1 (mB-21 )
m 1 (m1 -21 ) ••• [m1 -(2n-2)8]
530. 1-2!z1 +z'- ... -H-l)m
(2n)!
X
41
xza•+ ...
581. w
= F(a, b, c, z) =
ab
a(a+l)b(b+l) 1
1+ T.Cz +
z + ···
21 c(c+l)
... (b+n-1) 11 I I
... + a(a+l) ...n!(a+n-l)b(b+l)
z; z < 1 .
c(c+l) ... (c+n-1)
583. SOLUTION.
After differentiating the hypergeometric
obtain the equation satisfied by the function
d1C
dC
C=
!
equation we
F(a, b, o, z):
z(l-z) dzl +[(c+l)-(a+b+3)z]dZ-(a+l)(b+l)C =- O.
Since the function
!
(1)
F(a, b, c, z), being the derivative of the function
F(a, b, c, z) which is analytic at the point s = 0 is also analytic at the point z = 0,
and every solution of equation (1) analytic a.t the point z = 0 must be of the
form kF(a+l, b+l, o+l, z) (see problem 581), where k is a constant, it
follows that
d
dZ F(a, b, o, z) =
kF(a+ 1, b+ 1, o+ 1, z). Putting z = O, we
find that k = ab/o.
538. (2) 2n!M.
545. (1) 4; (2) 15; (3) 3.
546. (1) A zero of order k+Z; (2) A zero the order of which is not lower
than min (k, Z); (3) A zero of order k - l, if k > l; a regular point not being
a. zero if k = l, and a. singularity if k < l.
547. The points z = ± 3i a.re zeros of the first order.
548. The points z = ±3i a.re zeros of the first order; the point at infinity
is a. zero of the second order.
549. z = 0 is a zero of the second order; z =kn (k = ± 1, ± 2, .•• ) are
zeros of the first order.
550. z = ± 2 a.re zeros of the third order; z = 2kni (k = 0, ±1, ±2, ... )
are zeros of the first order.
551. z = 2kn (k = 0, ±1, ±2•... ) are zeros of the second order.
651. z = ± n are zeros of the third order; all the remaining points of the
form z =kn (k = 0, ±2, ±3, ... ) are zeros of the first order.
553. z = n/4+kn (k = 0, ± 1, ±2, ... ) are zeros of the first order.
564. It has no zeros.
555. z =kn (k = 0, ± 1, ± 2, ... ) are zeros of the third order.
556. z = 0 is a zero of the second order; z = kn (k = ± 1, ± 2, ... ) are
zeros of the third order.
316
PROBLEMS ON COMPLEX .ANALYSIS
557. z
(k
= 0 is a zero of the third order; z = ; (kn) and z =
!J
(kn) (1 ± iy'3)
=- ±1, ±2, ... ) are zeros of the first order.
558. z - (2k+ 1) ;
159. z =-
(k = 0,
jf[ (2k+l);]
± 1, ± 2, ... )
and z =
!
are zeros of the third order.
-i/[<2k+l) ;] (1
± ly'3)
(k"" O,
±1, ± 2, ... ) are zeros of the third order.
160. z = 4 is a zero of the third order for one of the branches.
161. Here two functions are given: one of them has zeros of the second
order at the points z = 2kn ± n/6, the other has zeros of the second order
at the points z = (2k+ 1) n ± n/6 (k = 0, ± 1, ± 2, ... )
168. (1), (2) and (3). It does not exist; (4) It exists {/(z) =
164.
161.
domain
166.
z~ 1 ) •
(1) It exists (J(z) = z1 ); (2) it does not exist.
There is no contradiction, as the point z = 1 does not belong to the
of analyticity of the function.
The limit point can only be the point at infinity.
00
167.
SOLlJ'TION.
From the expansion /(z) =
2 c,,(z -
z0)• it follows that
n-o
-
u(z, 71) = co+co
2
Putting here z
00
+2
{(c.[(z-:t:oHi(1/-11o)]n +c;;[(z-zo)-i(1/-yo>l"} •
n=l
= 1110 + C~ Zo,
y =-Yo+
C~ Zo, where Cis suftl.ciently close
to z, we obtain (establish this!)
u(
1110
+ C~Zo ,
1/o
+ C;.,zo) =
! COo"+/CClJ
and, after replacing Cby 111, we arrive at the equality already proved.
169. z1 +2+0i.
170. ze:r - i/z+Oi.
171. (l+i)z - 3i+O.
172. sin z - cosh z+O (0 is an arbitrary real constant).
CHAPTER V
!27 (; r
co
573. -
for lzl < 2 ;
n=O
for lzl > la!.
co
575.
! +2
co
z~l +
zn for lzl < 1; -
n-o
2
(-1)• (z-1)•
n=O
00
for 0 < lz-11< 1; -
~~
.LJ
zn
for lzl > 1 .
n-:a
for 0 < lz-al < lb-al ;
for !al < lzl < !bl ·
co
577. - 1-+i
z-2
~ (-1)• <2 +i)•+l-( 2 -i)•+i (z-2)• for 0 < lz-21<Jl'li;
.LJ
5n+1
n-o
co
00
22
(-1)• z!n -
+
1
for 1<lzl<2.
n=O
n=l
578. - _ i _ - __l __
4(z-i)
4(z-i)B
2 ;:
co
~ (n+S)i• (z-i)•
.LJ
for 0
< lz-il < 2;
2n+c
n=O
co
2
n=l
317
(-1)• z•=+a for lzl
>
I.
318
579.
PROBLEMS ON COMPLEX ANALYSIS
± [z-
00
!
:~n]
(a+bl+ ,};
!z! > !bl,
for
where
n-1
00
00
580. ,}; e;n"
+ ,}; en21n,
n=O
n==l
where e-n =
(-lri [(-
Y
n
~)+ f
m=l
(- ! )(- !) 2-•].
m
n+m
Cn=2-"c-a(n=l,2,. .. ), C0 = : 2 [1+,k:(-})2-•].
00
1
for 0
581. ..!_ +z+z•+ ,};
n=l (n+2)!zn
2
< !zl < oo.
00
582. ,}; (-1)"
n=O
1
n!(z-1)11
0
for
< lz-11 <
oo;
00
1- :
+,}; C-n=-"
for jzj
>
1,
n=2
n-1
where C-n
=
'\_1 (-l)k+l
-1+ ~ (k+l)!
( n-1)
(n
k
=
2, 3, ... ) .
lc-1
00
00
583. ,}; Cnz"+,}; C-nz-n, where
n=O
Cn == e_n
n==l
00
= ,}; k!(n~k)!
(n = 0, 1, 2, ... ) .
lc=O
00
584. ,}; Canz 2n
n=O
00
+,}; C-2nz-111
where C1n
= C-an
n=O
00
=
(-1)"
'\.1
l
~ (2k+l)I (2n+2k+lil
k=O
(n
=
0, 1, 2, ... }.
319
ANSWERS AND SOLUTIONS
oo sin{1+ n2n)
585. - ,J;ni-(z-l)"
for 0< lz-11
<
oo;
n=O
cosl , sinl-2 cosl
.
21 zl
·-sml- --ZT
+
6 sinl-5 cos2
3! zs ~+
£
or lzl
...
>
I.
00
(see problem 615), for 0
<
<
lzl
,J; nan
3
--+2n; - +2
zan 1
z
n=l
+
+-2-]zan-1
~[(-1)" 22nR1n
n•n
(2n) !
.L.J
for n
<
lzl
<
2n .
n=l
00
687. \-, b" - all for fzf
.L.J nzn
>
max (la!, !bl) .
n=l
,2: ;.."
00
588.
0
where
o_,,. =
21k-a
(
- i 221:- 1- - 3-
22k-s
+-5-
- ...
n=2
... +(-l)1'+l
C-11i+il
=
20-11c (k = 1, 2, ••. ), for fzf
>
2; i tan-1
!·
00
n=O
where o- 1 = 2i tan-1 1/2, 0-1Tc
=
21lci ( tan-1 1/2-
00
,J; ;: + ,J;
k-1
,2:
m=O
2k~l ).
0
;_,." ,
n=l
( 2m1;l~):m+t),
= 20- 11: (k = 1, 2, ..• ), for 1 < lzl < 2.
689. (1) Yes; (2) Yes; (3) No (the point z = 1 is not an isolated singularity);
(4) No; (S) No; (6) No; (7) No (in any ring round the point z = 0, the function
is not continuous); (8) No; (9) Yes; (10) Yes, if ex is an integer or zero; no
in all the remaining cases.
590. (1) No; (2) Yes, both branches can be expanded; (3) No; (4) Yes, all
three branches have expansions; (5) No; (6) Two branches of the four determined
by the conditions (l+yl) = ± y'2 have expansions; (7) No; (8) No; (9) No;
(10) Yes, all six branches have expansions; (11) No; (12), (13) and (14) Yes,
all the branches have expansions; (15) No; (16) No; (17) All i;ne branches have
expansions except for two determined by the values sin-1 y'2/2 = n/4.
•'-c1t+11
y
594. Onto the whole plane with a cut along the ray e-loit, :
~t <
oo.
+ ... , then denoting by w 0 the value not
595. SOLUTION. If f(z) =
assumed by the function f(z) in the unit circle, we consider the funct.ion
z+c 2z2
320
PROBLEMS ON COMPLEX ANALYSIS
f(z) / ( 1- f :: } • It is single valued in the circle lzl
expansion
1
~~
= z + ( o1 +
~o } z1+ .. .
<l
(why?) and has the
By the theorem of problem
Wo
1598:
lc1 1~
2 and
lo•+ ~o I~ 2.
Hence
lw0 ! ~
!.
597. z = 0, z = ± 1 are simple poles, z = oo is a regular point (a zero of
the third order).
1598. z = (1 ± i)/y'2,z = (-1 ± i)/y'2,aresimplepoles; z - oo isa.regula.r
point.
099. z = 1 is a pole of the second order; z = oo is a pole of the third order.
600. z = 0 is a pole of the first order; z = ± 2i are poles of the second
order; z = oo is a regular point (a zero of the fifth order).
801. z = ± i are simple poles; z = co is an essential singularity.
801, z = oo is an essential singularity.
803. z = oo is an essential singularity.
80,, z = 2km (k = ± 1, ± 2, ... ) are simple; z = oo is a limit point of
poles.
·605. z = 0 is a pole of the second order; z = 2kni (k = ± 1, ±2, ... )
are simple poles; z = oo is a limit point of poles.
808. z = (2k+ l)m (k = O, ± 1, ± 2, ... ) are simple poles; z = oo is a limit
point of poles.
807. z = kni (k - O, ±1, ±2, ... ) are simple poles; z = oo is a limit point
of poles.
608. z = 0 is an essential singularity; z = oo is a regular point.
609. z = 0 is an essential singularity; z = oo is a simple pole •.
810. z = 1 is an essential singularity; z - oo is a regular point.
811. z = 0 is an essential singularity; z = oo is an essential singularity.
812. z = 1 is an essential singularity; z = 2km (le= 0, ± 1, ± 2, ... )
are poles of the first order; z = oo is a limit point of poles.
813. z =kn (k = 0, ± 1, ±2, ... ) are simple poles; z = oo is a limit
point of poles.
81'. z = 0 is a pole of the second order; z = oo is an essential singularity.
815. s
=
(2k+l) ; (k
=
0, ±1, ±2, ... ) are simple poles; z - oo is a limit
=
O, ±1, ±2, ... ) a.re poles of the second order;
point of poles.
818. z
=
n
(2k+l) 2' (k
s = oo is a limit point of poles.
817. z = 0 is a pole of the third order; z =kn (k = ± 1, ± 2, ... ) are
simple poles; z = oo is a limit point of poles.
818. z = kn (k = ± 1, ±2, ... ) are simple poles; z = oo is a limit point
of poles.
819. z =kn (k = O, ± 1, ± 2, ... } are simple poles; z = oo is a limit point
of poles.
820. If a #: mn+n/2 (m = 0, ± 1, ± 2, ... ), then z = 2kn+a and
z - (2k+ 1) n-a (k = 0, ± 1, ± 2, ... ) are simple poles; if a = mn+n/2
321
ANSWERS AND SOLUTIONS
then form even, z
=
n
2kn+ 2 and form odd, z
=
n
(2k+ l)n+ 2 are poles of the
second order; z = oo is in each case the limit point of the poles.
621. If a#- mn(m = 0, ±I,± 2, ... ) then z = (2k+l)n±a (k = 0, ±I,
± 2, ... ) are simple poles; if a = mn then for m odd z = 2kn, and for
m even z = (2k+l)n are poles of the second order, in each case z = oo is the
limit point of the poles.
622. z = I is an essential singularity, z = oo is a regular point (a zero
of the first order).
623 and 624.
z= :n
(k
=
±I,± 2, ... ) are simple poles;
z= 0
is the
limit point of the poles; z = oo is a simple pole.
625. z = 0 is an essential singularity; z = oo is a regular point (a zero of
the first order).
626. z = 0 is an essential singularity; z = oo is an essential singularity.
627.
z= :n
(k =±I, ± 2, ... ) are essential singularities;
z= 0 is the
limit point of the essential singularities; z = oo is an essential singularity.
628. z = 2/(2k+ l)n (k = 0, ±I, ± 2, ... ) are essential singularities;
z = 0 is the limit point of the essential singularities; z = oo is a regular point.
629. z = I/kn (k = ±I, ±2, ... ) are essential singularities; z = 0 is
the limit point of the essential singularities; z = oo is an essential singularity.
630. z = 2/(2k+ l)n (k = 0, ±I, ± 2, ... ) are essential singularities; z = 0
is the limit point of the essential singularities; z = oo is a regular point.
631. (1) The point z = oo is a pole of order k =max (n, m), if n #- m;
if, however, n = m, then z = oo is either a pole of order k ~ n, or a regular
point; (2) A pole of order n-m, if n > m and a regular point if n ~ m;
if n < m, then z = oo is a zero of order m-n; (3) A pole of order n+m.
I
633. Examples& (I) z1 ; (2) --;- +z; (3) l/(1n - I).
z
834. (I)
a
Z-CI:
(a #- 0) or az+b (a#- O);
of the numbers am different from zero) or
ao+a1z+ ••• +anzn
(Z-C1:1) (z-ix1 )
...
(Z-IXn-1)
(an #- 0, IXk #- a1 for k #- l).
636. (1) z0 is a removable singularity; (2) z0 is a pole of order n, if cf>(z)
is single-valued in the neighbourhood of the point z0 , and a pole of order nm,
if cf>(z) is m-valued in the neighbourhood of this point; (3) z0 is an essential
singularity.
637. (1) The point z: is a pole of order n if y' is a rectilinear segment, and
a regular point of multiplicity n, if y' is a circular arc, that is, J(z) - f(zt)
322
PROBJ,EMS ON COMPLEX ANALYSIS
= (z-zcT)" t/>(z) where t/>(z) is analytic in the neighbourhood of the point zf and
t/>(zf) ¥< o. If z: = oo, then this condition is written in the form /(z) - /(oo)
= z-nt/>(z), where t/>(z) is analytic at infinity and t/>(oo) .fi O;
(2) Z: is an essential singularity.
689. (1) -1; (2) O; (3) O; (4) O.
Ha. (1) An essential singularity at z = oo; the exceptional value is 0 (and
ro!); ez-+ 0 if a:-+ - oo (art:-+ oo, if a:-++ oo);
(2) An essential singularity at z = O; the exceptional value 0 (and oo);
-1
-1
e 11 -+ 0, if z -+ 0, for example, along the path y = 0, a: < o (e 11 -+ oo as
z -+ 0, along the path y = 0, a: > 0);
(3) An essential singularity at z = 0; there are no exceptional values
(not counting oo); cos .!._ -+ oo for a:
z
== 0,
y -+ 0;
(4) An essential singularity at z == oo; the exceptional values are i and
-i; tan z-+ i, if y-+ + oo, and tan z-+ -i, if y-+ - oo;
(5) An essential singularity at z == oo; the exceptional value is -1;
tan• z-+ - 1, if '!J-+ ± oo.
1
6'8. For example, /(z) = (l+z)I = 1-2z+3zl- ... ; 11111 = -n, 'an+i
= n+l.
= o.
650.
61H.
652.
658.
654.
(!
(! =
(! =
(!
(!
660.
(!
655. (! =
656. (! =
657. (! =
658. (! =
659. (! =
n, a= a.
1, '1 = 3.
= 1, '1 = 3.
= 3, '1 = 2.
1
= 2m, '1 =
SOLUTION.
Similarly
oo
ztn
I
2
--=n=O (2 .n)I
2
(cos yz +
8
where ot
"\"'1
zn
nL::.o
(2•.n)!
= 21
J; (2~n
=
1
.n).
n=O
V-1,
1/2,
'1 =
1.
l .
00
whence
== Jl'5.
= 1.
= 1.
1, '1 = y2.
2, '1
1, '1
1, '1
=
(cos
COB
iyz),
'Vz)+cos i j/z).
1/4 (cos otVz+cos ot1 Yz+cosot3 V'z+cos ot•Vz),
zn
2~
= -4l~
,,c.; cos afc 'l /z and
k=l
00
whence
n=O (.....-.n).
so on.
661. (! = 00.
662. SOLUTION. It is easily seen that it is sufficient to consider values of
z> O.
I
I
f e.z:t'dt
Then _o_ _
art:
<I·
'
on the other hand, if 0
<
ot
<
1, then
Jart:''dt
°
e"%
323
ANSWERS AND SOLUTIONS
1
=
1
J
ez(t1 -1Z)dt;;;;i:
J
eZ(t 1 -1Z)dt-+oo ifz-+OO,
v«
o
Consequently, fl= 1, a= I.
66'. fl* = max (fl1• fl•>·
665. (1) fl*= fl• a• .::;;;a1+a1; (2) fl*= fl, a•= max (a1, <11)·
666. (1) fl*.::;;; fl· a•.::;;; 2a; (2) e•.::;;; fl· a•.::;;; a.
667. SOLUTION. Let M(r), fl• a be the maximum of lf(z)I on the circle lzl = r,
the order and type of the function /(z), and let M 1 (r),(11 and a1 be the correspond·
If
ing characteristics for the function f'(z). From the equality /(z)
= Jf'(t)
di
0
+J(O) it follows that M(r).::;;;rM1 (r)+IJ(O)i and, consequently,
J' (z) == ~
2m
f 'c;c>d~
.,-z)
r
radius r+~ (~ >
where as
r
e.::;;;e1 •
Since
can be taken the circle with centre at z and
0 is arbitrary), it follows that M 1 (r).::;;;
M(2r+:: (r+~l
,
that is fll.::;;; fl• and hence, fl1 = Cl• Hence also from the inequalities given
above we conclude that a 1 = a. Another possible method of proof is based
on the theorem given on page 99.
669. fl = 1, a = l/e.
675. fl = 1, a = 2.
670. fl - a, a"" oo.
676. h(</I) = cos"'·
671. fl= a, a= 0.
677. h(</I) = !sin"''·
672. fl = o.
678. hC•> =
678. fl = o.
679. h(</I) = !cos .,.
674. e == 1, a..,. 1.
680. h(</I) = cos n</I.
681. h(<fl) = cos </I, if cos <fl;;;;:, O; h(<fl) = 0, if cos <fl < 0.
682. h*(<fl) = h(</I), if h(</I) > 0; h*(<fl) = 0, if h(</I) < 0; h*(<fl).::;;; 0,
if h(</I) = o.
!sin"''·
CHAPTER VI
The annulus 1/2 < lzl < 1.
The exterior of the unit circle (lzl > 1).
lzl < 1.
The half.plane Re z < - 1.
The real axis.
The whole plane except for the points z = 0, ± 1, ± 2, ...
lzl > 1.
lzl < 1.
The whole plane except for the unit circle (lzl :fo 1) •
..!... Clk+i)m
692. The whole plane except for the points z = 4 n e
n
(k, n = 1, 2, ... ).
683.
684.
685.
686.
687.
688.
689.
690.
691.
00
698.
SOLUTION.
If the series
J; an
converges, then for
lzl <
1 the conver·
n-1
00
gence of the series
2 .2;
00
=-
00
an-
n=l
n=l
If the series
gence R
00
~ 1Gnzn is obvious and from the identity ~
~
n-1
an(_!_
zl
r
n it follows that the series also converges for
1-h-)
co
co
n=l
n-1
J; an diverges then the series J; anzn
:s;;; I.
For
anzll
~ 1-zn
n=l
-zn
lzl >
lzl > 1.
has the radius of conver·
co
1 the divergence of the series
~ anzn follows from
~ 1-zn
n=l
00
the fact that if this were not so the series ~ - 1 Gn
would converge, and
~
-zn
n-1
co
consequently also the series
2( ~
1
zn -
co
;.:_z:n } =
2
an. If, however
lzl < 1,
n-1
n=l
2
00
then the modulus q of the ratio of the general terms of the series
co
L ~z~
anzn and
n=l
is contained within the limits 1 -
n-1
both series converge or diverge together.
324
lzl :s;;; q :s;;; 2
and consequently
ANSWERS AND SOLUTIONS
325
co
2 bnzn, where bn = 2a,,,
694. (1)
the summation extending to those
n=l
indices p, which are divisors of the number n, including 1 and n. The radius
of convergence R =min (r, 1), where r is the radius of convergence of the
co
series
2 anzn .
n-1
co
co
(-l)D ~ (logk)D
~~
kl
k=l
~
695. ~ aa(z-2)n, where an =
n=O
R = l.
696. 1/2, if lzl < 1, and - 1/2, if lzl
697.
(l~z)B,
if lzl
<
1, and
(l~z)•,
>
I.
if lzl
>
I.
698. z, if lzl < 1, and 1, if lzl > 1.
699, z/(z - 1), if lzl < 1, and l/(z - 1), if lzl > I.
701. (1) and (2). Uniformly convergent in every circle lzl ~ r < 1 and in
every domain lzl ;;;:.: R > 1;
(3) Converges uniformly on the whole real axis; diverges elsewhere.
708. Uniformly convergent on the circle lzl = 1; diverges everywhere
else.
704. Uniformly convergent in any half-plane Re z;;;:.: "· where tJ > 0.
O.
705. Uniformly convergent in any half-plane Re z;;;:.: l+", where
706. Uniformly convergent on the real axis; diverges everywhere else.
707. Uniformly convergent on every segment [2k:n:+s, 2(k+ 1) :n; - s)
of the real axis (Tc = 0, 1, 2, ... ).
718. flJc = 0, flJa = +oo.
708. No.
715. flJc = flJa = -oo,
719. flJc = 0, flJa = I.
716. flJc = -oo, flJa = I.
720. flJc = flJa = -1.
717, flJc = -oo, flJa = +oo.
721. flJc = flJa = +oo.
724. flJc = 0; it diverges at all the points of the boundary.
725. flJc = flJa = -2; it diverges at all the points of the boundary.
726. flJc = flJa == O; it converges (non-absolutely) at the points z = (2k+ l)m
(Tc= 0, ±1, ±2, ... ) and diverges at the remaining points of the bound-
">
ary.
727.
728.
ary.
flJc
a:c
= a:a = O; absolutely convergent at all the points of the boundary.
= a:a = 0; converges non-absolutely at all the points of the bound-
736. The integral converges uniformly in every strip 0 < a.~ Re z ~ A <
oo.
787. The integral converges uniformly in every half-plane Re z;;;:.: a. > O.
738. The integral converges uniformly in every strip a. ~ Re z ~ 2 - a.,
where a.> O.
739. The integral converges uniformly in every strip a. ~ Re z ~ l - a.,
where a.> O.
740 and 741. The integral converges uniformly in any closed interval
of the real axis not containing the coordinate origin.
742. The integral converges uniformly in the half-plane Im z ~ 0 with
the semicircle lzl < r deleted, where r is any arbitrarily small positive number.
<
326
PROBLEMS ON COMPLEX .ANALYSIS
°'
'748. The integral converges uniformly in the intervals 0 < :s;;; z :s;;; {J < 1
and 1 < y :s;;; z < oo.
'744 • .An example is:/(t) =- e11 for n < t < n+e-n' (n =I, 2, ••. ) and/(t) = 0
for all other values of t.
'74li. :I:c = a:a = 0.
'749. flJc = - oo; a:a =-+ oo.
'746. :l:c = :I:a = - 00.
'750. :l:c = - l; a:a =+ oo.
'751. :l:c = 0; flJa.,. I.
'74'7. :I:c = a:a = + oo.
'748. a:c = - oo; a:a =I.
'752. Divergent at a.11 the points of the boundary.
'753. Absolutely convergent.
'714. Convergent at the point z = 0, convergent (non.absolutely) at the
remaining points of the boundary.
'755. Non-absolutely convergent at all the points of the boundary.
'765. SOLUTION. On evaluating the integrals present in the obvious inequalities
:!..
~
n
I
f
sm•n+l:i;da;
<
0
I
f
0
sin•na;da;
<
I
f
sin•n-la;da;,
0
. [ (2n)!! ] 1 l
n [ (2n)ll ] 1 1
we obta.m (2n-l)ll 2n+l < 2 < (2n-l)ll .2n. In order to prove
Wallis's formula it remains to establish that the difference between the extreme
members of this set of inequalities tends to zero as n .,.. oo.
'76'7. (1) No; (2) Yes.
'769. (1) Divergent; (2) Diverges (to zero); (3) Convergent; (4) Convergent.
'7'70. Converges non-absolutely.
'7'71. Divergent.
'7'72. Convergent if p
>
!,
the convergence being absolute if p
> l;
divergent if p .s;;; 1/2.
'7'78. Absolutely convergent if p
'7'74. Absolutely convergent.
'7'76. lzl < I.
'7'7'7. lzl < 2.
778. lzl < oc.
'7'79. lzl > 1.
> 1; divergent if p .s;;; 1.
'780. lzl < l/e.
'781, lzl < oo.
'781. lzl < oo.
'781. 1•1 < oo.
'784. lzl < oo.
'791. SOLUTION. (1) From the series C(a) =- I+ ~ + ~ + ... by subtract·
ing the series for 2-•C(a),weobtain (1-2-•)C(a) = l+
the terms
!,
~s ,
:. + :. + •.. ;
for which n is divisible by 2 a.re absent from the right-hand
side of this equation. Similarly (I-2-•)x (I-3-•) C(a)
the terms
~+
=
l+ :.
+:.
+ •.. ,
for which n is divisible by 2 or by 3 being absent from the
right-hand side. Generally
(I-Pi"') (1-p;"') ... (1-p;;;')C(a)
=
l+
J; ! ,
(1)
327
ANSWERS AND SOLUTIONS
the summation on the right of (1) extending to those indices n (greater than
unity), which are not divisible by any of the numbers p 1 , p 8 , .. ., Pm· It is easy
to prove that for Re a ;;ii. 1 + c5 (c5 > 0) the sum of the series on the right hand
00
side of (1) tends to zero as m-+ oo, and, consequently, C(a)
fl (1-p;•) = 1.
m-1
(2) Since it follows from the test of absolute convergence (see problem '768)
n
00
that the product
(1-p;;;•) converges for Re a;;;;i. l+c5, the function C(a)
m=l
has no zeros for Re a> 1.
'792. SOiiO'TION. It follows from the proof of the preceding problem that
n
00
for any c5 > 0 we have
(1-p;Ci+"»
= C(l~c5)
• From this it is easily
n-1
00
concluded that lim
fl (1-p;<i+">) = O.
Hince
(l-p;;-1)
<
(1-p;<i+">),
1-oO n=l
00
it is obvious that the product
fl (I-p;;1)
n=l
00
quently, the series
I
n=l
p;;-1 also diverges.
diverges (to zero), and conse·
CHAPTER VII
798. res [/(z)]:r= ± 1
79'. res [/(z)]:r=I
=
795. res [/(z)]:r= -
1
=
-1/2; res [/(z)]:r=o = l; res [/(zll:r=co
i
= 4;
-1/4; res [/(z)]:r= -I
=
res [/(z)]z=co
O.
= O.
(2n)I
(-l)•+i (n-l)I (n+l)I; res [/(z)l:r=co
=
(2n)!
= (-l)• (n-l)!(n+l)I ·
796. res [/(z)]:r=o = l; res [/(z)]:r=±l = -1/2; res [/(z)]:r=co = 0.
797. res [/(z)]:r=o = O; res [/(z)]z=l = l; res [/(z}]z=co = -1.
798. res [/(z)]:r= - 1 = 2 sin 2; res [/(z)]:r=co = -2 sin 2.
1
5 4 (sin 3-i cos 3);
799. res [/(z)]:r=o - 1/9; res [/(z)]:r=sl = -
res [/(z)]:r=-al = - 514 (sin 3+i cos 3); res [/(z]:r=co = 217 (sin 3~3).
800. res [/(z)]:r= d:+i,.
=
-1 (le= 0,
±
1,
±
2, ... ).
I
801.
802.
808.
88'.
res [/(z)]:r=lcn = (- l)k (le = 0, ± 1, ± 2, •.. ).
res [/(z)]:r=kn = 0 (le = 0, ± 1, ± 2, ... ).
res [/(z)]z•lcn = -1 (le = 0, ± 1, ± 2, ... ).
(1) res [/(z)]z=1 =res [/(z)]:r=co = O;
(2) res [/(z)]:r=1 = -res [/(z)]z=CO = -143/24.
co
805. res [/(z)]z•o
==
-res [/(z}]z=co
=
2
nl (nl+ l)I •
n-o
806. res [/(z)]z=o = res [/(z)]:r=co = O.
807. res [/(z)]z= - 1 = -res [/(z)]:r=co = -cos l
808. res [/(z)]:r= -a = -res [/(z)]:r=co
=-sin 2
41n
[2co (2n-l)!
(2n)!
n-1
809. res [/(z)]:r=o = 1/2; res [/(z)]
810. res [/(z)]z=o
..!!.
=
(-1)• .
(n+l)I' if n
=
0, if n
= 0 or
n
l
d:ni = -2le
z--.-
< 0, and also
.
if n
co
")1
41n+1
]
+ -.J (2n)I (2n+l)I •
n-0
• (le=
:n:i
1,
±
2, ... ).
> 0 is odd; res [/(z)]z=o
> 0 is even; res [/(z)]:r=co =
328
±
-res [/(z)]:r=o·
329
ANSWERS AND SOLUTIONS
811. res [J(z)]
1
z-bi"
=
l
(-l)k+1-k
1 1 (k
:rr;
=
± 1, ±2, ... );
00
2 ~ (-l)k
1
:rr;• ,L.J ---,CS- = - 6.
k=l
812. res [f(z)]z=fcl"' = (-l)k2k8:rr;1 (k = 1, 2, ... ).
.
.
2tk(211k-l)
(2k)!
B 1t,
813. res [f(z)],.=o = 0, if n is odd, res [J(z)],.=o = (-l)k+l
if n = 2k (k = 0, ± 1, ± 2, ... ), where B1 t are the Bernoulli numbers (see
1
problem IH5); res [f(z)]
(k+ 1) = (k = 0, ± 1, ± 2 ... ).
res [J{z)]:r=oo
=
I"
z=
( k+2
l )":rr;n
814.
815.
816.
by the
l; - 1.
O; 2.
-2e2knai, if yl = 1 and Logl = 2km; 0 for the branch defined
value yl = -1.
(a-b)2
s17. ±-8- ·
818. (1) a.-{J (for all the branches); (2) e'% - efl (for all the branches).
819. (1) 2b' + 2 .131 - - /51 + ... , if Log l = 2km;
l
(2) -21+
820. res [f(z)],.=o
if Tan-1 oo =
1
sTI·-
l
5 • 61 + ... (for all the branches).
.
= k:rr;, if
Tan-1 0 = k:rr;;
res [J(z)]s=oo = -
(2k+l):rr;
•
2
( 2 k+l):rr;.
2
821. res [f(z)],.=o = 0, if n;;;.. O; res [J(z)]z=o =Log
P'
if n = -1, and
res [f(z)]z=o = - 1 - (a."+i-f1n+1), if n~ -2; res [J(z)],.=oo = _l_ (a,11+1
n+l
n+l
-[Jn+ 1), if n;;;.. O; res[f(z)],.=oo = -2k:rr:iifn = -1 and Logl = 2km,and
res [J(z)],.=oo = 0, if n ~ -2.
822. -2 0001·
o_11f>' {a)
O-t<f>(k-i)(a)
828. (1) .Alf>(a);
(2) 0_11f>(al+-1- 1- + ...
1) ! - ·
+-vc=
82'. (1) n; (2)-n.
825. (1) n<1>(a); -nlf>(a).
826.
~.
<1>'(a)
827 • .AB.
k
_
828. ~(-1)" no-n(/JR*.~_:i)rt+i.
,L.J
n-1
-·
m
829. - y 2 •
880.
831.
832.
883.
SH.
885.
-2:rri.
-m/121.
ni.
-2m/9.
l.
o.
330
PROBLEMS ON COMPLEX ANALYSIS
2n+1
•
836. - - - • i f n;;;:i: -1,
(n+I)I
and 0 if n < -1.
838. 0, if,. < I; ± I if,. > l (the sign depends on the choice of the branch
of the integrand).
I
841. entl/(l+e").
839. 7v(I+y'2).
842. 2n/v(a1 -I).
843. 2na/(a1 -b1 ) 111 •
840. n/(l+a).
844. (2a+b)n/[a(a+b)]111 •
845. 231/(1-a1 ) if lal < l; 231/(a8 -l) if !al> l; 0 (the principal value)
if !al = 1, a
± l (for a = ± I the principal value does not exist).
.
n(a1 -f-l)
.
n l-a11
846. n(a'+l)/(1-a1 ), if lal <I; a'(a•-I), if !al> I; 2 a'(a•-I)
(the principal value), if !al = 1, a
±I (for a = ± I the principal value
does not exist).
847. 2n/nl, if n ;;;:i: O; 0, if n < 0.
848. nisigna (for a = 0 the principal value of the integral equals 0).
849. -2ni signima.
851. -n/27.
852. n/4a.
8..,. 3 . 1.3.5 ... (2n-3) n , if n> 1 ,. n /2 , if n -_I .
2.4.6 ... (2n-2) 2
855. n/y2.
n
•
854.
857. ( - l)k-1/(2ih-z)k.
ab(a+b)
+
+
858. (1) ~(cos 1-3 sin I);
3e'
ne-lallJ
860.
•
2 lbl
(2) ~ (3 cos I+sin 1),
3e1
861.
..:!.. e-Jalllsign a.
2
859. ~ (2cos2+sin2) •
2e'
863. ni, if t> O; 0, if t = O; -ni, if t < 0.
864. n (2sin2 - 3sin3).
865 • ..:!.. (cos l - l/e8).
5
866 • ..:!.. [sina+e-Cav'a)/I (sin (a/2)+y3 cos (a/2))].
3
867. n(e-labl-1/2) sign a.
869. ~ [2- (2+ labl)e-fallf] sign a.
labl
868. ~ (1-e- ) sign a.
4b'
2b8
870. n (lbl-lal)·
871. n/2.
872. 3n/8.
I'(p) cos
873. (1)
aP
n:
n:
I'(p) sin
; (2) - - - - ; i n order to verify the correctness
aP
331
ANSWERS AND SOLUTIONS
the answer for -1 < p < 0, it is sufficient to note that for these values
of p the integral converges and the function in the answer is analytic.
87'. - 1
p
n
r (1)
-p cos -2p
876. _l_ r
p-l
875. -1
p
•
(2-) cos n/2p (for p = l the integral is equal to n/2).
p
878. n/sin p:;i.
881.
882.
n •
r (1).
-p ~m -2p
880. n/2 cos np •
2
n(l-p)
- - - (for p = 1 the integral equals 1/2).
4cos np
2
~ ~pl
smpn sm l
:::: :
, if l #:-
o, and~
~~: :::
smpn
888.
if l =
o.
-.-:;g[211 ( 1smpn
.!.)-1]
.
2
n
( . pn
pn
)
889. - . - - sm-+cos--1 ,
smpn
2
2
886. np(l-p) •
28-11sinpn
pn
) ,
887. - .n- - ( 2L2 COS--1
smpn
4
~/4/y3,
890. 7'J'
891. If a does not belong to the interval (-1,1), then 1 = -- y(a:- l),
where y(a1 - l) > 0 for a> 1 (in· the plane with a cut along the seg·
ment [ - 1, l] the quantity v(a 8 -1) is single valued) for a = ± el•,
1
n
= ± y( 2 sinor;)
1(7-.!.)
e c
1
ni
;
for a= iy, I= y'(l+ y•) sign y; for a< -1,
I= y(a~-l); for -1< a< 1, I= 0 (the principal value).
892. If b does not belong to the interval (0,1), then I = ,..:!-bll- 1 (b- l)-ll •
smp:;g
where (b-1)-11 > 0, bll- 1 > O for b > l; if 0 < b < 1, then I = -nb11-1
X(l-b)-czcotp:;g (the principal value).
898. Io= 2 ; 2 , Iu = ; ( ; 2 -Sit). Iat+1 = - ; ( : 2 -Sat+1) where
. h esumof "terms of t h e series
. 1- l
1.3
( 1) 1.3 ... (2S1-l)
8 v 1Bt
2
2 . 4 - ... + - v 2 .4 ... 2,, •
n
n 1.4 ... (3n-2)
8 4 I
9 . o = y3, In= y3
3.6 ... 3n
(n = 1,2, ... ).
+-
SH. n/(n sin : ) •
898. 2 !aJ•;( 2!al) ( : log !a!-1- 3: } •
896 ~l I I
• 2!al og a•
899 ' -:n. .
n
900.
897. 8nlal (n8+4 log• !al).
2
1og 2.
901. :;il/4.
332
PROBLEMS ON COMPLEX ANALYSIS
-+-1 -1 -(for
903. (1) - 1
I-a
:ri;
a= 1, I= 1/2); (2)
2a(log2a+~2)
oga
- _I_.
I+a2
T' ,r -,:..J.
2n
.... ..:i_, {:: L<-'~ log2 a+
k=O
k+ -
:ri;2
2
2n-l
:ri; ~
--+-
l
905. - l {
2n l + a 2
907.
n
2
2a
:ri;a
tanh 2
}.
2k+I
l )2
(
log2 a+ k+ 2 :ri;2
(-I)t+1
k=O
.
1/(2cos ; ) .
909.
:ri;2e""'
908. (e""'+l)B
:ri;
910. -log (l+a),
a
I+a .
a
:ri;
a
if 0 <a< l; - l o g - - if a> 1.
if t> I; 0, if t< I ;
lognt
913. (1) t"/nl; (2) ----;;!•
·r
i
I
h
t= I, ten
=0
for n > land I= 1/2 (the principal value) for n = 1.
916. cost.
914. eattn /n!.
917. t-sin t.
915. sin t.
ect
ebt
eat
918 · (b-a) (c-a)
(a-c) (b-c)
(a-b) (c-b)
+
+
919.
_!._ ( 1- ..!...)",
t
n!
if t
~ 1;
0, if t::;;; 1.
924. e-terfyt.
e-lt
e-t
922. (1) y'(nt):
(2) y(:ri;t) •
925. sin t, if t
< :ri;;
0, if t ~ :ri;.
u
923. erf yt, where erf u
= }
y 1'
926. 1, if 0
< t <a;
J
e-X'dx.
0
0, if t =a; -1, if a< t
< 2a; - ~ ,
if t
0, ift > 2a.
927. n+l, if na
<t<
(n+l)a; n+
~,
if t
=
na; (n
928. 1-erf-x- (see the answer to problem 923).
2yt
= 0, 1, 2, ... ).
= 2a;
333
ANSWERS AND SOLUTIONS
2
--
oo
1
929. -+2
a
ntn•t sin
( -l)ne
a•
n-1
~
- -a- •
nnr
J:
931. 2sinh t •
t
t
930. -Ei(-t), where Ei(t)
=
932. 2'" •
du.
a
-oo
933. 0, if b
<
O; cosh-1 ~. if b
a
1
936. y(z•+I)[z+y(z•+I]n
> 0.
(y(z1 +1)
>
0 for z > 0).
937. (1) y(aLb•) if a> b; 0, if a <b; (2) O, if a> b;
y(b•~a•)
, if
"< b.
938. ; J 0 (i)laly(t1 -b1 ).
131. If the real part of even one pole is positive, then lim /(t) = oo; if
t-+00
the real parts of all the poles are negative, then lim /(t) = 0; if some poles
t-+00
are situated on the imaginary axis and all the remainder have negative real
parts, then /(t) oscillates as t -+ oo, the amplitude of the oscillation increasing
without limit if there is even one pole on the imaginary a.xis of higher order
than the first, and remains bounded if all the poles on the imaginary a.xis
are simple. If there is only one pole on the imaginary a.xis at the coordinate
origin, then /(t)-+ oo, if the pole is multiple, and /(t) -+res [ellt<[l(z)]:r-O• if
the pole is simple.
HI. /(t) ,._, (at-3)/a'.
142. J(t) ,._,
iw
elmt-t/('2aiw-,,,..,.
y(2aiw-w1 )
J -e-x-t
-dt
x
1'3.
SOLUTION.
Let us represent the integral in the form
t
-x
00
_,._,
+ J _e- - dt. Since on asymptotic expansion in negative powers of a;, e-x....., 0,
x
t
then by the solution of problem 144 it follows that the second integral is
asymptotically equal to zero. By the definition of the principal value of an
integral,
. [-J• -dt+
e-t
Jx e-t
--dt=e-Zlim
_cftJ
Jxe-x-t
t
t
t
-x
a-+O
-x
•
x
= e-x lim
B-+0
J
•
e-t
- -et
- dt
t
J
x
= e-x
0
e-t et
--=--dt
t
334
PROBLEMS ON COMPLEX ANALYSIS
(in the last integral the integrand is continuous). Moreover,
1
JC
J
e-t;et dt
=
0
J
JC
JC
e-t;et dt+ Je~t dt+
0
1
JC
J ~t
dt ...... 0(1)+
1
J~
dt.
1
Integrating by parts, we obtain
JC
J ~ dt
JC
o+aX( ! + ~
=
(n;.l)!) +nl Jtn~l dt,
+ ... +
1
1
where 0""' O(n) is a constant. It remains to establish, that
JC
lim a:• e-JCJ ....!.._ dt = 0,
»-+CO
1
tn+l
and this is easily proved by l'H6pital's rule.
948.
J es"-ra dt,
Let us consider the integral
BOLtTTION.
contour 0 is represented in Fig. 69 (for Re
g
*>
c
where t.h1>
0). This integral is
-
z
'I
FIG. 69
equal to zero and hence (z
•
= .11+'1/)
co
00
Jas"-"dt = Jas"-"dt- Jes1-cr+11> dt.
1
0
0
JC
yn as";
The first integral on the right-hand side equals
.2
let us write the
l800Jld in the form
Jt+ig
co
I
-2
JC
J
00
I
1
· - - das•-cr+11>1 = 2z
_ -12
JC
as•-(t+l1)•
dt.
(t+ig)B
Repeating the integration by parts we obtain the required expansion. For
the remainder we have the estimate
1
1.3 ... (.2n-l) Jco as•-(t+l1)• dtl
2"
(t+ig)••
JC
I
<
1.3 ... (2n-l)Jco eX"-tl dt.
2"
It+'1/l18
JC
335
ANSWERS AND SOLUTIONS
Again integrating by parts, we obtain
eir•-t•
1
J
de < - -, whence
x ICt+ly!••
21zl11t:i:
00
it follows that the expansion is asymptotic. The case Re z
similarly. If however, Re z = 0, then
IJoo
es•-(t+l7)•
I< Joo y••
e-t•
_ y (n)
-de - - -
----dt
0
9'9. f(t),.,.,
!
sin
(t+iy)llt
0
2y11t •
(roe+ =)- Cl.I~ >~
x
(et~) - :~~) + ... +c-1>• r!:::+f,) + .. .];
(2<.ot)1
for small t, /(t) - 2y(t/n) [ 1- 1 . 3 . 5
2
n
1 co
951. /(t),.,., 1+(-1)"
8
e1
r(:) -
.5.7.9 + 1. 3(2<.ot)'
...
]
111:1
,
2y(t1n).
r(3n+ ! )
0
for small t,f(t) =
< O is treated
;
t3n+-}
I
t•
I'(4)
e1
+I' (~l)
I
- ...
4t8
111:1
3yn.
952. 1.
957. 0.
953. o.
958. 1.
959, n.
95'. 4.
956. l; 3.
961. SOLUTION. Since the sequence offunctions /n(z) converges to the function ell• everywhere except at the point z = 0, then for any small circle K
with centre at a point 111 ¢ 0 and not containing the coordinate origin either
within it or on its boundary, for a given e > 0 it is possible to find an N
such that the inequality f/,.(111)-el/lll < e is satisfied for all the points of the
small circle K for n > N. It remains to choose B < min!ell•f, where 0 is the
EC
circumference of the circle K, and apply Roucbe's theorem.
REMARK. The assertion of the problem follows directly from Hurwitz's
theorem (see, for example, (2, Chapter VIII, § 2].
965. O.
966. 2; 1.
967. One root in each quadrant,
968. Two roots in ea.ch of the second and third quadrants.
970. In the domain fl> O, or:> +vfl (the domain I in Fig. 70) m = O;
in the domain fl> O, or:< + y{J (the domain II) m = 2; in the domain
fJ < 0 (the domain III) m =- I.
336
FROBLEMS ON COMPLEX ANALYSIS
971. In the domain ex
> O, fJ > _!._
(the domain I in Fig. 71) m
ex
the domain where either ex ~ 0, or ex
>
0,
fl
< _!._ (the domain II), m
ex
= 2.
:Fm. 72.
FIG. 71
972. In the domain ex
= O; in
>
! + -V{ + ! )
{J2
~
(the domain I in Fig. 72)
+-i/({J +!) (the domain II) m=2;
in the domain where either ex<!
(P2+ !). or ! -y({J +4) ~ex< O,
m,=0; in the domain O<a:<
-v
2
1
337
ANSWERS AND SOLUTIONS
{J > O (the domain III), m = 1; in the domain
{J < 0 (the domain IV) m = 3.
! - y(fJI+ ~ ) <
cc
<
0,
97,. The domain containing the positive semi-a.xis a and bounded by the
t COB Tt
lines
a+b =
0 and {a= -
~in Tt-'
0::;;;;
t::;;;; ; •
b~ sinTt'
975. The dome.in situated in the first quadrant and bounded by the lines
b =
o
and {
a=tanTt,
:n;
t•
o==:;;t==:;;TT.
b= - - - ,
COB Tt
:n;
976. The finite domain bounded by the segment b = 0, 0::;;;; a ==:;; 2T
and the a.re
{
a=tsinTt,
b = t• cos Tt,
:n;
0 ::;;;; t ::;;;; 2T .
w
w2 d
wn dn- 1
986. z=a+{[J(a)]2 }+ ... -:- 1 - d n-l {[J(a)]H}-t....
21. -d
1 ,J(a)+.
a
n. a
00
987 z
•
= ..!...w-)- ~(-l}n-1
L.J
2
n=2
n(n+lL::_. (2n-!Lwn.
22n- 1 .n!
'
oc
z = - 2 _..!...w+ ~ (-l)n_n(n·l-1) ... (2n-2) wn.
L.J
2
2•n- 1 .n1
n=2
w l
w1 1 d
988. z = a+ 112 (a2 - l)-j-2!2i" da [(a2 -1) 8 J+
wn
... +n!
00
1
...
Tn
dn- 1
dlJn-1 [(a•-1)"]+ ...
00
990. (1) z = ~ nn-:~n-l w";
n=l
(2) ebz = b
..2,1 (an-~~)n-lwn.
n=O
00
~wft
dR-1
991.
z = a+ L.J n! dan-1 (sin"a).
992.
SOLUTION.
n=l
lz- ;
The function w = (z -:n;/2)/(sin z) is analytic in the disk
I< r, if r < :n;/2 and in this disk there are no zeros other than z
= :n;/2.
On the circumference of this circle Jwl ~ 2r/(e• +e-•), hence the disk Jwl <fl
of radius (} < 2r/(e• + e-•) is mapped one-one onto a corresponding neighbourhood of the point z = :n;/2 and the expansion of z(w) converges in this
circle. The function 2r / (er + e-•) has a maximum at r = r*, where e•"*
338
FROBLEMS ON COMFLEX ANALYSIS
= r:+ 1
,._ 1 (r* = 1°19 ...
< n/2), and this maximum is equal to
• 2"* •
~+~
= y'(r*2 -l) = 0·6627 ...• Thus, the required radius of convergence is not
less than 0·6627 ... At the same time the point w, at which dw/dz = 0, that is,
tan z = z -n/2, cannot be inside the circle of convergence of the expansion
= t we transform the equation dw/dz = 0 into the
form cot t = -t or e11r = (it+ l)/(it-1). Consequently, it = r* is a root of
the equation dw/dz = O. The corresponding value is given by w = t/cos t
2r*
= -2ir*/(e'•+e-r•), whenJe lwl = e'• +e-r
• = y'(r*2 -l) = 0-6627 ... Thus,
of z(w). Putting z- ;
this point is located on the circumference of the circle of convergence a.nd
the radius of convergence is consequently equal to y'(r*1 - l) = 0-6627 ..•
1018. n 1 /(sin2 na).
2~2
1014 • n 8 .cot na
smna
1015. n 1 /8.
1018. n 8 /32.
1
1016. 2a2 (l+na coth na).
1019. n2
1017.
(
l+
m:
~ ab •
smna
na).
CHAPTER VIII
==
==
==
1024. (1) .F+(z) = (z-a)11, .F-(z)
O; (2) .F+(z)
0, .F-(z) = -1/(z-a)ll;
(3) .F+(z) = l/(z-a)11, .F-(z)
0. In all three cases an integral of
Cauchy type is transformed into a Cauchy integral.
1025. (1) .F+(z)
(2) .F+(z)
= •(z)
1 -),
"\"1 Uk (z-ak
L.J
-
1 -) +
= L.J
"\"1 Uk (z-ak
-
1 -);
"\"1 Uk (z-ak
L.J
u(z),
1 - ) is the principal part
Uk { z-ak
of the expansion of the function • (z) in the neighbourhood of the pole ak;
.F-(z)
=-
•(z) +
1 -)
"\"1 9k (z-ak
L.J
.F-(z) = -
+ u(z).
g(z) is the principal part of the expansion of •(z) in the neighbourhood of
the point at infinity, in which the constant term is included.
z-2
z+logz-3
i
, .F-(z) = - z•+l •
1026 • .F+(z) =
z•- 4
1027 • .F+(z)
2z ,
= cotz-..!..-z
z1 -:ii8
3
3
- 2:n • .F+(-:n) = 2:n; .F-(z)
1028 • .F+(z) = -2 111 ")
(z+i
in particular, .F+(O)
2z
1
= - -;- z1 -:nl
·
.F-(z) = - __z_.
2(z-i)
,
00
1029 . .F+(z)
=
1
2
"\"1
L.J
= 0, .F+(:n) =
00
(a -ib ) z11
n R1111
' .F-(z)
= -
1
2
11=0
"\"1 R• (~+ib11)
L.J
zn
11=1
00
the limiting values are: .F+(Relll)
=
! 2 (~-ib,.)
el..O
11-0
=
ao
4
°"
+ ..!..2 •(Ref9) + .i
"\"1 (-b cosn8+~ sin n8);
2L.i
11
11=1
00
.F-(Re"'>
=-
!2
(~+ib11) ff.111J
00
ao
1 (RellJ >+2
i .L.J
"\"1 (-b,.cos7W+a11
--"
. 8).
=-4-2•
smn
11=1
lOao. (1) If the point z lies in the circle Qk:
339
lz-Tc:nl < ~, then l1(z)
2
340
PROBLEMS ON COMPLEX .ANALYSIS
=f(z-lcn), 1 1 (z) = (-l)k/(z-kn). In particular, if lzl < n/2, then 1 1 (z)
1 1 (z) = f(z). If the point z lies outside ell the closed Qt, then J 1 (z)
1 1 (z) = O.
(2) Let Qk be the disk lz-knl < n. Then 11 (z) = J(z-Tcn)
=
=
+J[z-(k-l)n],
1 1 (z) = (-l)k[J(z-kn)-f(z-(k-l)n)], if zeQk-iQk;
11 (z) =f(z-kn)+f[z-(k+l)n],
1,(z) = (-l)k[/(z-kn)-/(z-(k+l)n)], if zeQkQk+ 1 ;
1 1 (z) =/(z-kn), 1 2 (z) = (-1)"/(z-kn), if z lies within the domain
Q1<-Q1;_ 1Qk-Q1<Q1;+1 ; 11(z) = 0, 1 1 (z) = 0, if z lies within the complement
to all the Qk •
__l_
z-lt,
::1:
_ _1_
1-C
_!._
1031. F(z) - 2ni log z+l
, F (C) - 2m log l+C ± 2 ,
1-C
1
F(C) = 2ni log l+C (-1
F(O)
=
.
1
1
< C< l); F(±i) = ± 4", F::!:(O) = ± 2',
O.
1032. F(z)
= ~Log
z+BB
... m
z-
-
a single-valued branch in the z-plane
with a out along 0 is defined by the value Log 1
Log : : :
= log
I:::a. I
=
O.
+iLlc arg CC-z), where Lie arg (C-z) is the increment
of arg <C-z> along
~1ogJC+B1
+.!.,
2ni
C-B
4
F+(C) =
~logJC+BJ
_ _!_,
2m
~-B
4
F-(C) =
1
1C+B1
1
1
F(C)= 2mlog
C-B +7;
F(0)=2·
F+(iB) = 3/4, F-(iB) = -1/4, F(iB)
=
1/4; F'(O) = -i/nB.
1033. F(z) =~Log z+BB (a single-valued branch in the z-plane with
...ni
:za ou.t along 0 is defined by the value Log 1 = O; for lzl > B it is identical
with the similar branch of problem 1032)1
F+(C) =
2~ log1~=:1+1/4,
F(O) = -
!"' , F'(O) =
_
F-(C)
=
2~ log1~:;1-3/4;
_i_,
nB
t At first glance the solution is obviously:
~
2ni
1
J~=~[log
C-z
-1
2m
(1-z)- log (-1-z)]
=~log
2ni
z-l •
z+l
However, it is necessary to verify that the final transformation really
leads to the given branch of the logarithm, since the equation log z1 - log z1
= log z1 /z1, is not generally true. This remark must be home in mind in what
follows also.
341
ANSWERS AND SOLUTIONS
1034. (1) 0, if lzl < r or izl > R, and l/z", if r < lzl < R;
(2) l/zn, if Im z > n, and 0, if Im z < n; (3) 0, if !Im zi < n, and -1/zn,
2~zn
> n; (4)
ifllmzl
[Log
!:=: - k~l ( 2k~;)-~•k-i] •
if m = [;] •
2
Log It+z =log/:!l_+z/+iLfc{arg (C-z)-argC} is the single-valued branch
R-z
R-z
in the z-plane with a cut along 0, defined by the value Log 1 = 0. The
limiting value of this branch on 0 has imaginary part n/2 on the left and
(- 3; ) on the right. This determines the limiting values F±(C) on 0;
F(O) =
~ l+(-l)n-l ; (5) the function F(z) is the same as in the prenRn
2ni
ceding section, only Log RR+z is the single-valued branch in the z-plane
-z
with a cut along the semicircle 0, with the same value Log 1 = 0.
The limiting value of this branch on 0 has imaginary part 3n/2 on the
left and ( - ; ) on the right. This defines the limiting values of F± (C) on 0;
F (O)
=
_1_ 1+ (- l)n-1.
nRn
2ni
identical in
R-z
the disk izl < R, generally they are different; thus, for example, at oo in
the first case the branch has the value -ni and in the second +ni. However,
In parts (4) and (5) the branches of Log R+z are
REMARK.
F (oo) = O.
1035. Log b-z = log / b-z1 + iLlc arg (C- z) is the single-valued branch
a-z
a-z
in the z-plane with a cut along 0, defined by the value Log I = 0 (the
branches are defined in the same way in problems 1036-1040).
b-z
1036. b-a+z Log a-z.
1037.
n
'\1
L.J Ankzk-1 +zn
b-z
Log
a=z ,
Ank =
bn-k+l-an-k+l
n _ k -i- 1
k=l
1038.
b-z
oo ~n
cnAnkzk-1 + q, (z) Log - - .
~
a-z
n=lk=l
b-z..)
b-z
- .
- - Log 1039. -l - ( Log a-z0
a-z
z-z.,
n
1040.
b-z + ~ Ak (z-z0 ) kb-z
- - Log--"
n [ Log a-z.,
a-z
~-Zo)
l
k=2
l
At= k-1
[
]
l
l
(b-z0 )k-1 - (a-z 0 )k-f '
F (zo)
= -An+i·
1]
,
342
PROBLEMS ON OOMPLEX .ANALYSIS
1041. (1) F+ (z) =log (z+R), F- (z) =log( l+
. ,,,
,,,
!) ;
,,,
,,,
F+ (C) =log !R~+•-p F- (CJ =log 2 oos2-l2;
(2) F+ (z) =log (R-z)+ni"t, F-(z) =log
F+
(1- :);
m- log 2R lsin : I+' n~t/I , F- (CJ= log 2 lsin ~I+ i n~t/I
of the angle 4> =
11org
C corresponds to the conditions of the problem).
1042, (1) F+ (z) == O, F- (z) = log ( 1F- (C) =log (1-
n;
(2) F+ (z) =log
!);
F+ (C) = O,
z~l'
F- (z)
=
1
(the sense
~
~
R+z
1 [
= O;
F+ (C)
c
logC-I' F-(C) =
zlll:-1
1043, F (z) = 2ni ,.:::- ";;i' Log R-z - 2 ,.:::- (2k- l) Rd-l
n=l
]
O.
where m
k=l
=[;], and Log [(R+z)/(R-z)] has the same value as in problem 1034,
(4). This determines the limiting values .ll':i:(
00
If
~
C) on O.
If lzl
<
R, then
C C-+1
~ .
1l' (z) -= ,.:CJ onzn, where On = 2 ni log C- l
n-o
c
1044 F+ (z) = -
:i [
2+y'z log
~~~: ] ,
:~:: ] - rz.
F- (z) .... - :, [ 2+y'z log
1045. F+ (z)seO, F-(z) =Log z-b (the branch of the logarithm is de-
z-a
termined by the conditions of the problem).
1, F- (z) == 1- y'[(z-a)/(z-b)].
1046. Jl'+(z)
1047. F+(z) = z-la- (1-A)b, .ll'-(z) == z-Aa- (l-A)b-(z-a)A(z-b)l-.1.,
==
1048. F+ (z) =Log z-Cbo, F- (z) =Log z-Co,
z-
z-a
1050. -:i:l+z-3·5.
1011. (1) .!.. (tog - 21 - )
2
z-1
1
;
(2) .!.. (tog _-r_ )
2
1-T
1
-
n• •
2
t If Log z and log z are the branches indicated in the conditions of the
problem, then Log z =log (-z)+m.
343
ANSWERS AND SOLUTIONS
1058. If
l•I> 1 and
~
1
11
R+z
zeO, then F' (•) =2-, log--Log-R +F'1
m
z-1
-z
where F'1 (z) is an analytic function for
(•),
l•I > 1, and the branch of Log BB+z
-z
is chosen just as in problem 1084, (4). From this the behaviour of F(z) at the
points ± B at the ends of the arc 0 is obvious.
BF
Ba: =
1059. F'(z) = - 211 log log 1/r+2• log log l/B,
~~·
- 2 log log l/r
~··
- 2 1
+m.oglogl/R, u
.ll.. = -2'loglogl/r-2 1
+ 2'loglogl/R,
ogr
•
ogr
BF'
elll<f>
BF'
1
1
1
;;z= of>(Z) =logl/r' az=-2loglog-;--logr+2 loglog1f,
1068. D (/)is the area of the domain (}', onto which/ (•) maps the domain (Jo
I
I
Rl-C• ; (2) log
1069. (1) log B(z-C)
J "(C)
1·-c1
•-.C ;
,e.. -e"'I
(3) log e•• _ e•C.•
2K
1
1976" "(•) = 2n
" (oo) =
rl-B•
~ (0- 4>)+rl dO'
0
2n
2~
.Bl-2Br
J "m d8.
0
1078.
(1) /(•) - 4>(•)+'1'
('!'), /
1 (•)
= -4> (':)- 'P(z),
where
v(O) - Im/ (0) = Im [4> (O)+'P (oo)];
(2)
t1
/(z)
=
-iof>(z)+if/l(~B),
/1(11)
""'iof>(~I) +i'P (z),
(0) =Im/ (0) ... Im [-iof> (O)+i'P (oo)].
1079. (1)/(z) = z•, / 1 (z) = -B••/•" (here and in the answers to problems
1080-1088 the value of v(O) is taken as equal to zero).
1080. /(•)
z"
= B••,
/ 1 (z) =
-l/z11 •
1081. /(•) = - log ( 1- ;. ) (log 1 =- 0), / 1 (z) =log (1-1/z).
1082. /(•)
= y(R~-•) (/(0) =
! ), /1(11) = -
y[z/(z-1)].
1083. /(z) ===log B, / 1 (z) === - log R.
J u(t)
1085. u(z) = n
00
1
-oo
J u(t)c:U + iO.
00
yc:U
1
(t-a:)•+y• , /(•) = ni
~
-oo
(for the existence of the first integral the piecewise continuity and boundedness of u(t) in the whole interval (- oo, oo) is sufficient, for the second, in
344
:PROBLEMS ON COMPLEX ANALYSIS
addition, for example, the function u(t) must be of order ·It~« (a.> 0) at infi.
nity).
1086. /(z)
== u(z)-f-it1(z)
J
J
~
=
1
2f
~
n(t-z)
u(t) coth - 2-
1
dt -2f
-~
n(t-z)
U1(t) te.nh-2dt,
-~
where u 1(t) =- u(t+i) (for the existence of the integrals it is sufficient, for
example that u(t) and u 1(t) should decrease at infinity like
lti~+« , a.>
0) •
1088. Circles inside the disk lzl < 1, tangential to the circle lzl == 1 at
the point el9.
1090. Arcs of circles, connecting the points el9,, el91 inside the circle lzl < 1.
z-b
1
1
1091. <»(z; a, b) = -arg - - , <»(z; -oo, b) = - arg (z-b); <»(z; a, oo)
n
= 1- 2-arg
(z-a).
n
z-a
n
The geometrical meaning of these harmonic measures
is the angle divided by n subtended by the segment or ray at the point z.
1
y
1092. 1- - arg z for the ray arg z
1
= 0 andarg z for the ray arg z = 1"
y
_ 2
z-R
2
_1 2Ry
1093. <»(z, LI) - -;:; arg z+R -1 - 1--;:; tan R•- izl• , <»(z, I')
= 1-<»(z, LI). The level lines are circular arcs connecting the points± R,
from the points of which the diameter LI subtends an angle; (l+<») in the
case of <»(z,LI) and n{1-;) in the case of ro(z,I').
z-R
arg - B •
z+
2
yz-yB
-;:; arg yz+yR •
2
1094. <»(z, I')
= -n
1095. m(z, I')
=
loglzi-logr
loglzi-logR
for lzl = R and
for izl = r.
log R-log r
log r-log R
n
ll09. -cos nl/>.
1099. u(z) ==
Cy<»y(z).
lllO.
a sin 4>
Y-1
l-2a COS 4>+a1
nos. sin,,,,,..
1096.
2
llll. arg (1-ae14')
=
-sin-1 y(l
m-1
Ill!.
!_ .J: otm
•
n1a.
1
21
sm
2
:asin 4>4>+a
sin 2(m-k) 4>.
k=O
nl/> • (n+ 1)4>
sm 2
-
COB
1) •
345
ANSWERS AND SOLUTIONS
lll5. -2 log 2 cos "2 .
"'
ll22. h(I/>)
= sin 21/>.
Ill&. -2 log 2 sin : •
ll23. h(I/>)
= sin 31/>.
k
h11; .Approx.
k
h11; Approx.
0
1
2
3
4
0·00000
0·58779
0·95106
0·95106
0·58779
0
1
2
3
4
5
0·00000
0°80903
0·95107
0·30902
-0·58779
-1·00001
ll24. h(I/>) = sin 41[>.
ll25. h(I/>)
=-
cos"·
k
h11; Approx.
k
h11; .Approx.
0
1
2
3
4
5
0·00000
0·95104
0·58777
-0·58778
-0·95104
0.00000
0
1
2
3
4
5
-1·00000
-0·95107
-0·80902
-0·58779
-0-30902
0-00000
ll26. h(I/>) = - cos 34>.
k
h11; .Approx.
0
1
2
3
-1·00002
-0·58780
0°30902
0°95108
0°80903
0.00000
4
5
346
PROBLEMS ON COMPLEX ANALYSIS
k
hk
0
1
2
3
4
5
-0·84925
-0·78768
-0·64004
-0·44958
-0·23107
-0·00000
hk
Approx.
Actual
-0·85247
-0·78556
-0·64039
-0·44940
-0·23120
-0·00000
4 (
cos3tf>
cos51/>
)
1128. h(tf>) = --;;- COSl/>--3-1 -+-5-1--··· •
k
hk
0
1
2
3
4
5
-1·1868
-1·1240
-1·0611
-0·8430
-0·6248
0·0000
hk
Approx.
Actual
-1·1662
-1.1345
-1·0815
-0·8585
-0·5694
0·0000
1131. (2) T[tJ>(x)] = -i[tJ>(x)-21/>a(x)-tf>(oo)],
T[!p(X)] = i[tp(x)-2tpb(x)-tp(oo)],
where lf>a(z) and '/lb(z) are the principal parts of the expansions of tf>(z) and
'Jl(Z) in the neighbourhoods of the points a and b.
No.of
problem
h(x)
1132
sin h
f(z)
f1(z)
eiM:
-e-IA:
-ie-IAll
1
- z-bi
1133
-cos.Ax
-ieiAz
1134
b
- z1+b2
1135
x(l+x2)
(y2)(l+x')
1
z+bi
y2-iz
(y2)(1-z 2-i(y2)z)
y2+iz
(y2)(1-z1+i (y2)z)
1136
1
Im (x-a+i'b )"
1
-(z-a+ib)n
1
(z-a-ib)"
1137
1
Im (x-a+i'b) n
1
(z-a+ib)"
1
(z-a-ib)"
347
ANSWERS AND SOLUTIONS
U.'1. If h (:i:)
== 1
in the interval (-1, 1), then h (:i:)
for l:i:I > l; if h(:i:) = 1-l:i:I,
if h (:i:)
=
h (:i;) =
then
= - -1
n
:i:-1
log-:i:+l
.!.. [:i: logw-l -log:i:-l] ·
r
n
1
:i;I
sign :i:, then h (:i:) = -; log w-1 •
:i:+l '
1141. By the rectangle formula: h(l) = 0·60716; h(2) = 0·34003; h(3)
= 0·20116; h(4) = 0· 14595; h(5) = 0· 11524. By the asymptotic expansion:
h(2) = 0·34; h(3) = 0·2015; h(4) = 0·14595; h(5) = 0·11524.
1143. By the rectangle formula: h(l) = 0·49985; h(5) = 0·19257. Exact
values: h(l) = 0 · 5; h(5) = 5/26 ~ 0·19231.
1146. (2) T[l/I (t)]
T ['I' (i)]
= - i [• (t)-21/1,.(t)- l/loo~l/l-cc],
= i ['JI (i}-2'1'b ('ij- '1'oo+2'1'-oo],
n
1/111 (z) =
z-a) ,
'1 (- l)k- 1 At
L.J
dA:- 1 {
(k-l) 1 dzA:- 1 coth2-
k=l
m
'Jib (z}
'1 (-l)k-1 B1: ·dk-1
= L.J
(k- l} 1
z-b) ,
(
dzk-l coth - 2-
k=l
n
where
m
,J; (z~:)A: ,
2
k=l
k=l
(z~~)A:
are the principal values of the expansions
of the functions If> (z) and 'I' (z) near the points a and b.
1147. (The notation is the same as in problem 1146)
(1) l/l(z)-i Im lf>oo+2lf>-oo;
(2) 'I' (i)+i Im '1'oo+2'1'-oo;
(3) l/l(z)-1/111 (z)-i Im l/lcc~l/l-oo +4>11 (i);
(4} 'l'(z)-'l'b(i)+i Im 'l'oo~'l'-oo +'l'b(z);
(5)
1
I
z::a2
z-a
z-o
I
coth2-+ 2 coth2- ;
(6) coth (z-a) -
1
2
z-a
coth2- +
I
2
z-a
coth2- .
1148. -:;ill"'.
1150. -n•i-n.
1149. -n1/i-•.
1101. nl/-r:•.
1151. =F n 1i-P+9 (the sign - if p and p+q have the same sign, the sign+
if they have different signs; 0 is counted as the sign +).
348
PROBLEMS 0111' COMPLEX ANALYSIS
:ii•
ll5S. - T+bi •
I
llH. -:n1 oosAToosµT,ifA > µ;:ii 1 sinATsinµT, if A<µ; - ~ cos 2AT,
if A=µ¢ O.
ll55. =F i:n1
:~;
(minus, if b > O, plus, if b < 0).
ll56. log l-T + ..!. [(1og-T-) 1 -:ii•].
T
2
1-T
ns7.
~+Tlog l:T + ;•[(1og 1 ~T}9-:n1].
nss.
~[(1og1~Tf-:n•]+ ~n~k[~1og1-;T +(~ + k~1 + ...
k=l
... + ~; 1 )]
ll59.
-1!
T•-p-l
p=O
Lpl
! [1og :~: -:ii•],
1
T [
R+T
]
R+T
1180, "j" log8 R-T-:n1 -2R log R-T·
ll61.
j[tog•:~;-:n•],
ll62. 2R [m- logR+T] + .!.[1og•R+T -:ii•].
T
R-T
2
R-T
ll63.
R+T
]
1
21 [ log•R-:n-:n
.
R+T
+2:n.Rt+TlogR-T'
llH. __!._log•R+T - 2m.
21'8
R-T
T
1)1 -
i1 ~ (~~1~:.J.
k-0
CHAPTER IX
00
1175. /(z) =
~
.L.i
(z-a)"
n-o (1-a)"+i
;
this expansion continues /(z) analytically
inside the circle lz-al < 11-al which does not lie entirely within the circle
lzl < 1, if a does not belong to the interval (0,1).
1176. /(z)
00 (~)"(z+~)"
= log -32 + .L.i
~
; the
n
. . Iz+ 211 < 23..
sanes
circle of convergence of this
n-1
18
1191.
SOLUTION.
The substitution e' =a: reduces the integral to the form
00
/(a)
=
J
sin; da;
(11:8
=
eB 101 X),
Integrating by parts, we obtain /(a)
1
00
= cos 1-
J
1
Re
cos a: da:. The latter integral converges in the half-plane
::z:11+1
a> -1.
1194. 0 < Re z < 1, -1 < Re z < I.
1196. The point z = 1 is a simple pole with residue one.
00
1197.
SoLUTION.
Let us use the notation / 1 (z) =
J
e-tl/l(zt)dt. The ana-
o
lyticity of the function / 1 (z) in the circle lzl < 1 follows from the result of
problem 535 and the genera.I properties of Laplace integrals (see page 110).
Integrating by parts (n+l) times and using the inequalities of problem 535,
we obtain for lzl < 1
n
/i(Z)
oo
= - 2,1 zk [e-tlfi(lc)(zt)r +zn+ 1 f
k=O
O
e-tlfi(n+ll(zt)dt
O
n
=
oo
J; anz"+zn+i J e-tlfi(n+l(zt)dt.
k-o
o
From the estimate for llfi(n+l)I it follows that the second term on the right
hand side of the last equality tends to zero as n-+ oo (lzl < r). In order to
prove the second assertion let us take any point z e G. Then, as is not difficult to prove, inside and on the boundary of the circle on Oz as diameter
there a.re no singularities of the function /{z). Hence for a sufficiently small
d > 0 within and on the boundary 0 of a circle of radius (lzlf2)+d, concentric with the one already constructed, the function /(z) is also analytic. Thus
3-19
350
PROBLEMS ON COMPLEX ANALYSIS
the equality o,, =
2~ j ~:~ dC holds for the ooemcients On of the expansion
00
/(z)
=
2
o,,zn and consequently,
n-o
co
~ znt",!(C) converges uniformly on 0, it follows that lfl(d)
£..I n!1o•+i
n-o
Since the series
=~
J!(C)entcdf. The maximum of the quantity Re (z/C) on Oie equal to
me
..
jzi'~,,,.- q <
1 (for the proof of this assertion it is sumoient to consider the
case when z is real and positive as a rotation round the coordinate origin
does not change the quantity Re (z/C)) and consequently, ll/l(d)I < .Aett (..4.
00
is a constant). It follows from this that the integral
J e-tl/l(zt)dt
converges.
0
1201. If Log<1 l z-za is the branch of the function Log z-zi, which isana·
z-z1
z-z1
lytio in the Z·pla.ne with a out along the arc y 1 and which tends to 0 at
111 ...
oo, then .F-(z) - (a-b) Log<1 lz-za, F+(z)
111-1111
=
(a-b) Log<ll z-zi +2nbi;
z-z1
within G+: F-(z) = (a-b) Log<1l z-z. -2m (a-b) for analytic continuation
z-z1
through y 1 and
F-(111)
= (a-b) Log<1l z-za for analytic continuation through
z-z1
Ya·
1208. (1) z = ± 3; (2) z = ± 2i; (3) z = ±2i. In all three oases the va.luee
of w' (z) are different.
1205. Above z = 1 there are two elements: z = I+t,
wi
1-(1-t)i
1
----t-= 2 +se+ ... ,
ltl
<
1,
and
Wz=
I+(I-t)i
t
= .!_ _ _!_ _ _!_t+
t
2
8
....
0
<
ltl
above z = 2 there is one algebraic element: z = 2+t8, w =
-t•+ ... , ltl <1.
<
1;
I~it = I-it
351
ANSWERS AND SOLUTIONS
1206. Above z = 1 there a.re two algebraic elements: z = l+t•,
to=± iv2(1-
~Y= ±ii12(1-: + .•. ).
algebraic element: z =
ular elements: z =
o+t•, to=
o+t,
!ti< 2; above z = 5there is one
~ (1- ~ t'+ ...).
to=± 2i(1+ ;2 + •.. ).
there is one algebraic element: z =
t-',
ltl
ltl
<
2, and two reg·
< 4;
to= +-t+
above
... , 0
< ltl
z=
00
< ~(1/5),
1207. Above z == 1 there are two algebraic elements: z = l+t•,
!t- ...
w-± (l+e>i- ±(1+
),!ti< 1.
1208. Above z = oo there are two algebraic
= ± +c1-at)i (1-bt)i = ± +(1- a~b t+ ... ).
0<
ltl
<mine~,. 1~1).
1209. z = 0 is an essential singularity, two-valued: z =
1
l
"""l+7+ 21t•+ •..• 0 < ,,, < oo.
UUO. Above z-= 0 there is an algebraic element:
1
1
1
= ,.-Ft+51t- ...•
0
< ltl <
1212. Above z
to= 1(1-
t•,
1
to= er
sint
z = t 8, w = ~
oo.
l!ll. Above z ... 0 there is the algebraic element: z
1
t
= 7 - 3 + ... , 0
t-•, to
elements: z =
= t•,
to= cot t
< ltl < n.
=- 0 there
~~ + ...).
is the algebraic element: z =
t•,
ltl < yn.
1218. to= 1/4, w = oo are algebraic branch points of the first order.
1214. w = ± 2 a.re algebraic branch points of the first order; w - oo
is an algebraic branch point of the second order.
1215. w
1216. w
= 1/4, w = oo a.re algebraic branch points of the first order.
= O, w = oo are algebraic branch points of the first order.
1217. to=
!
e:l:ltz (a = cos a:) are algebraic branch points of the first order.
1218. w = oo is an algebraic branch point of the (n - l)th order. To
every zero of the derivative w'(z) of order k there corresponds an algebraic
branch point of the function z(w) of the same order.
1219. To the zeros of the derivative w'(z) there correspond algebraic branch
points just as in the preceding case. To the poles of the function w(z) of order
greater than one there correspond algebraic branch points the order of which
352
l'ROBLEMS ON COMl'LEX .ANALYSIS
is one less than the order of the pole. If at co the function w(z)
= w0 +
;;k + ...
(la> 1), then z = oo corresponds to an algebraic branch point of order (la - 1)
abovew = w0 •
II
1220. The surface for z(w) is the same as for yw; its branch points are
situated above w = O, and w = oo and correspond to z = -n, z = oo. To
sheets of the w-plane with the cuts 0 < u < oo, ti = 0 there correspond angles
2n/n with vertices at the point z = -n. As n .,.. oo these angles are transformed into horizontal strips of width 2n, the function w(z) into eS, and the
surface for z(w) into the surface for Logw.
II
1221. The surface for z(w) is the same as for yw. To the sheets of the
w-plane with the cuts - oo < u < 0, ti = 0 there correspond figures formed
from two circular segments with angles 2n/n at the points z = a, z = b
(Fig. 73, where n = 3)t.
1222. The surface for z(w) is obtained by joining together two sheets of
the w-plane with the cuts lul < 1, ti =- 0, corresponding to the domains lzl < 1
and lzl > 1. The branch points are situated above w = ± 1 and correspond
to z = ± I. To the polar net lzl = r, arg z -= 4' there correspond the ellipses
and hyperbolas with foci ± 1 :
:ii•
y•
-----=l
cos• 4'
sin1 4'
•
1223. w(z) = z/(l - z)1 = z+2z"+3z8 + ... is a well-known extremal
function in the theory of single-valued conformal mappings (see[4, Chapter
XIII, § l]). It maps the unit disk lzl < 1 onto the w-plane with the cut
- oo < u < -1/4, ti - 0. The surface z(w) is obtained by joining together
two such sheets.
1224. The surface for z(w) is obtained by the sequential attachment of
2n sheets of the w-plane with the cuts 1 < lul < oo, t1 = 0. It has 2n branch
points of the first order above w = ± 1 and 2 branch points of the (n - l)th
order above w = oo. The basic mapping is shown in Fig. 74. The function
w(z) is automorphic (invariant) with respect to groups of linear transformasirl
tions, generated by the transformations T = wz (w = e 11 ), 8 = l/z. These
transformations correspond to transformations of the surface for z(w) into
itself during which the sheets are cyclically permuted and the projections·
of points onto the w-plane are preserved.
1225. The surface for z(w) is 2n-sheeted with a branch point of the
(2n - l)th order above w = 0, corresponding to z = oo, and with 2n branch
points of the first order of which n are situated above w = oo and correspond
"'
2d
to the points Zk =en wit (w =en, la= 0, 1, 2, .... n-1), and n above the
t In the figures to the problems of this chapter corresponding points
are denoted by identical letters. As a rule, the symbols D and !J' are used
for the points at infinity.
ANSWERS AND SOLUTIONS
FIG. 74
353
354
points
PROBLEMS ON COMPLEX ANALYSIS
•
Wk=
1
(2n-1)
•
2n Zkcorrespondingtothe
points
•
Zk
= ny(l/(2n-ll) rut. The
basic mapping is shown in Fig. 75. The rest is obtained by continuation by
the symmetry principle.
FIG. 75
1226. The surface for z(w) has n sheets with a branch point of the (n-1)-th
order above w = oo, corresponding to z = oo, and n - 1 branch points of the
first order, situated above
Wk
=
Zk ( 1-
!)
and
corresponding to
Zk
=
wit
1nl
(w = en- 1 , k = 0, I, ... , n-2). In order to construct the surface it is necessary
to attach to its zero sheet (the w-plane with n - I radial cuts originating
from the WJt) along each cut criss cross one sheet (the w-plane with one radial
cut). '!'he basic mapping is shown in Fig. 76. To the disk lzl < I there
corresponds the interior of an epicycloid (for n = 2, a ce.rdiod).
FIG. 76
1227. The surface for z(w) has (n+l) sheets with a branch point of the
= oo, corresponding to z = oo, and n+ I branch
(n - l)th order above w
355
ANSWERS AND SOLUTIONS
points of the first order situated above w1c = i1c ( 1 +
~) and corresponding
•••
to z1c = ~ (ro = e n+i, Tc = O, I, ... , n). The basic mapping is shown in Fig. 77
The domain lzl
a segment).
<
1 corresponds to the exterior of a hypocycloid (for n = I
©
(I)
Zo (2) !J
FIG. 77
1228. The surface for z(w) has n-sheets with a branch point of the (n - l)th
order above w = oo corresponding to z = oo, and n - I branch points of the
(-I)k
Ten
first order situated above Wk = ~ and corresponding to Zk = COS
n
(Tc= 1, 2, ... , n - 1). To construct the surface it is necessary to attach to
1 , t1 = 0) sequen·
< u::;;;; - -2n-1
its zero sheet (the w-plane with the cut - oo
L
tfall.y n-2 sheets with the two cuts 2
1 ::;;;;
lul
< oo, t1 = 0,
ot them to attach another sheet having the cut - oo
then to the last
< u::;;;; - 2.~ 1 , t1 = O,
1 ::;;;; u < oo, t1 = 0 if n is odd. In the
if n is an even number, and the cut -2n-t
mapping w(z) ellipses and hyperbolas with foci ± 1 pass into ellipses and
hyperbolas with foci±
2.~ 1 • The change of the semi-axes is obtained from the
relations z = const, w = const. Fig. 78 shows the dissection of the z-plane into
domains corresponding to the half-planes t1 > 0 and t1 < 0 (the former are
shaded) for n = 5.
356
PROBLEMS ON COMPLEX .ANALYSIS
FIG. 78
1229. w = 0, w = oo are logarithmic branch points, w = 1 is a simple
pole for one of the branches of the function z(w).
1230. The function z(w) has one algebraic branch point of the :first order
above each of the points w = e:l:ll and two logarithmic branch points above
each of the points w = 0 and w = oo. Its Riemann surface is obtained by
joining together two specimens of the Riemann surface of the function
Log w along cuts made on them above the arc connecting on the W·plane the
points w = e:l:1tl.
1281. The function z(w) has above w = ± y.2 an infinite set of algebraic
branch points of the first order and above w = oo two logarithmic branch
points.
1282. The function z(w) has one algebraic branch point of the first order
above each of the points w1c
= sin z1c
(the ZJ: are the roots of the equation
Zt
tan z = z; they are all real), two logarithmic branch points above w = oo and
an indirectly critical singularity above w = 0, the limit of the given algebraic
branch points (see R. NEVANLINNA., Ei.ndeuffge analytiache Funktionen, page
238, Berlin, 1936).
1
1233. z = oos-1 w =-. Log[w-j-y(iol-1)). The surface for z(w) has an
i
infinite number of sheets with two logarithmic branch points above w = oo
and algebraic branch points of the first order above w = ± 1 corresponding
to z = kn (k = 0, ± 1, ± 2, ... ) ; obtained by joining together an infinite number
of W·planes with the outs 1 :::;;;; lul < oo, " = O, which correspond to the vertical
strips kn< a:< (k+l)n (Fig. 79).
1234. z = sin-1 w = n/2-cos-1w. The smi"ace for z(w) is the same as
for cos-1 w (Fig. 80).
ANSWERS AND SOLUTIONS
U'
IJ'
FIG. 80
357
358
PROBLEMS ON COMPLEX ANALYSIS
1235. z = ta.n-110 = 21. Log ~+-w· The surface for z(w) has an infinite
'
•
1D
number of sheets and two logarithmic branch points above w = ±i; it is
obtained by joining together an infinite number of 10-pla.nes with the cut
u = 0, lt1I.;;;;; 1, corresponding to the vertica.Istrips kn< a: < (k+l)n (Fig. 81).
©
R'
D'
FIG. 81
1238. z = cot-lw = n/2 - tan-110. The mrfa.ce for z(w) is the same aa
for Ta.n-1w (Fig. 82).
©
D
D'
D'
FIG. 82
359
ANSWERS AND SOLUTIONS
+
' coehz =
f
1237 z - cosh-lw = L og (w ·l(w8-l))·
z(w) is the ;rune as that for cortw (Fig. 83).
COHZ.
·
The surface for
FIG. 83
• The surface
· u.
.l(w+l))· sinh z = - t• sm
z = sinh-1 w =Log (w+rding one 'by a rotation through n/2 about
z(w) is obtained ~rom t?e prece
for 1238.
the coordinate origm (Fig. 84).
Flo. 84
360I
l'ROBLEMS ON COMPLEX AN
239. z
=
tanh-1
1
ALYSIS
z(w) is obtain
w = 2 Log I+w.
the coordinate
edori
from
_,,
tanhbyz =
gm
(Fig.
85)
r tan-lw
. the
. su.-.ace
rol-w'
0
,
-
.
i"tan iz.
uotatioo
e surface
fo
n /2 about
""°Thugh
'
Jti
. 1240 • 21 = coth-1 w == I L
w+I
ooth z
(F;g. ; 86),
1s the sa.me as that for tanh-1
2 og0 w-1
©
Fxo. 86
= i cot iz'
Th'......,, fm(0
)
361
ANSWERS AND SOLUTIONS
1241. The surface for z(w) is constructed as follows: we make on thew-plane
the horizontal cuts -'.Xl<u.;;;;;1, v=(2k+l)n, (k=0,±1, ±2, ... )
md along each of these we attach one sample of the w-plane with the same
(single) cut. The surface constructed for z(w) is based on the fact that w(z)
maps ea.ch strip 2kn < y < (2k+l)n onto the strip 2kn < v < (2k+2)n
carrying the W·plane joined along the cut -1.;;;;; u < oo, tJ = (2k+ l)n (see
Fig. 87; the sign + denotes that the domains have to be joined together).
©
JV
(8) ·A{-/+!!/)
~~~+~~V=O
©-------
FIG. 87
12,2. (1) Let F' be the Riemann surface onto which the function w = R(C)
maps the C-plane. In order to construct the Riemann surface of the function
z(w) it is necessary to join together infinitely many specimens of the surface
F' with a cut along the arc connecting on F' the points w(O) and w(oo) (similar
to the construction of the Riemann surface of the function Log w). The resulting Riemann surface has two logarithmic branch points at the ends of
the arc of attachment and an infinite number of algebraic branch points belonging to the surfaces F'.
(2) To construct the Riemann surface of the function z(w) we join together
an infinite number of specimens of the surface F' alternately along the cuts
going from the points w(±l) to the point w(oo) (similar to the construction
of the Riemann surface of the function sin-1 w). The resulting Riemann surface
has two logarithmic branch points above w(oo) and in addition to the algebraic
branch points of part (1) it has also an infinite number of algebraic branch
pointsofthefirstorderatthepoints w(±l) (ifw(+l)orw(-l)isan algebraic
branch point of order k of the surface F, then for z(w) it will be an algebraic
branch point of order 2k+ 1). The investigation of the function z(w) can also
be reduced to the preceding case by the substitution z1 = iz.
1243. All the Riemann surfaces of w(z) have two sheets and have algebraic
branch points of the first order above the points: (1) z =a, z = b; (2) z =a,
z = b, z = c, z = oo; (3) z = ak and z = oo, if n is odd. For the construction
of the surfaces we take two sheets of the z-plane with cuts going from the
points indicated above to oo, and join them along identical cuts.
362
PROBLEMS ON COMPLEX ANALYSIS
1244. All the Riemann surfaces for w(z) have three sheets and have algebraic
branch points of the second order above the points:
(1) z =a, z = oo; (2) z =a, z = b, z = oo; (3) z =a, z = b, z = e; (4) z = ak
and z = oo, if n is not a multiple of 3. For the construction of the surfaces
we take three sheets of the z-plane with cuts going from the given points
anl
to oo, on which we define the three single-valued branches w, (J)'W, w'w (w = e 8 ).
On going round the branch points w(z) acquires a factor won account of one
of the factors of the root, hence the order of the attachment of the cuts is
the same, cyclic, along all the cuts (see the sketch in Fig. 88).
FIG. 88
1245. The Riemann surface for w(z) has n sheets with algebraic branch
points of the (n - l)th order above z =a, z = b, z = e, z = oo. The surface
is obtained by attaching n sheets of the z-plane with cuts going from the
points z = a, z = b, z = e to oo. The attachment is cyclic, simultaneous on
all three cuts. The sheets correspond to the single-valued branches of the
function wkw (w = e2 711/n, k = 0, 1, ... , n-1).
1246. The Riemann surface of w(z) has six sheets with an algebraic branch
point of the fifth order above z = oo, two algebraic branch points of the second
order above each of the points z = a, z = b and three algebraic branch points
of the first order above z = e. The surface is obtained by joining together
six sheets of the z-plane with a cut along a curve going from a to b, from b to
c and from e to oo. These sheets correspond to the single-valued branches:
W1 +w1, (J)'Wl +w2, ro9w1 +w1 , W1 -w2, WW1 -w2 , ro9w1 -w2, where w = etni/a
3
and w1, w1 are the single-valued branches y[(z-a)/(z-b)2] and y(z-c).
The detour round a cyclically connects sheets 1, 2, 3 and 4, 5, 6, that round
b sheets 1, 3, 2 and 4, 6, 5 (connected twice in half-sheets), that round o the
half-sheets 1, .2, 4, 5 (shown in Fig. 89); 2, 3, 5, 6; 3, 1, 6, 4; that round oo cyclically the sheets 1, 6, .2; 4, 3, 5 (see the diagrams in Fig. 89).
1247. The Riemann surface for w(z) has she sheets with two algebraic
branch points of the second order above z = 0, one algebraic branch point
of the first order above z = 1 and an algebraic branch point of the fifth order
above z = oo. In order to construct it two samples of the surface of the function
8
yz, have to be joined together, each of them having on one of the sheets a out
along the ray y = 0, 1 =:;;;; :c < oo.
1248. The Riemann surface of w(z) has two sheets with algebraic branch
points of the first order above the points z = kn (k = 0, ± 1, ± 2, ... ) ; above
z = oo the surface has a transcendental singularity the limit of algebraic branch
points. In order to construct the surface take two sheets of the z-plane with
cuts going from the algebraic branch points to oo (for example, along rays
parallel to the imaginary axis) and join the sheets together along identical cuts.
363
ANSWERS AND SOLUTIONS
1)
c
a
2) c
u
9)
6)
ww1+wz.
(J)W1+Wz
tiJzw,+wz
:
b
:
a
w1+wz
w,-w,
a
<U!U1-Wz
c
t.JW1-Wz
a
(J)ZWt -W3
(J) 3W1-w1
a
1111-w8
w1+w1
6J1W1+11J3
@\
c
(J)W1+W3
@
w,-w,
(llzW,+Wz
<Jzw,+Wz
b (J)W1+W3
:
w,-w,
b <.tJZw1-w3
:
tiJW,-Wz
b W,-Wz
c
FIG. 89
(J)W1-Wz
@
w,-w,
c
(l)Zw1 +w1
w
cuw,-w,
w,+Wz
tiJZw,-w1
(J)Z W1-Wz
b (,)IJJ1-Wz
w1+w1
(J)Zw1-w1
(J}l1J1 +Wz
b w1 +w1
(,)zw,+Wz
4) c
5)
w1+Wz
c
<UW1 .,.Wa
364
PROBLEl\IS ON OOMPLEX ANALYSIS
1249. t F':z and F'w are each obtained by joining together two planes with
cuts [ -1, l] (Fig. 90).
FIG. 90
1250. F's is obtained by joining together two z-planes with the cuts (-oc
< z :s;;;; O, y
= 0), F'w two w-planes with the outs [ -
! , !]
(Fig. 91).
©~
~
0
I
1
FIG. 91
1251. F'• and F'w are each obtained by joining together three planes having
a
a
a
in cyclic order outs along two of the segments [O, y4], [O, roy4], [O, ro8y'4J,
where <.o = e11tl/1 (Fig. 92).
t In the answers to problems 1249-125' by F'w are denoted the surfaces
for z(w), that is, surfaces above thew-plane and by F., the surfaces of the function
w(z) above the z-plane.
365
ANSWERS AND SOLUTIONS
FIG. 92
12H. F: is obtained by joining together two z-planes with the cuts
[o, !]. Fw by joining to thew-plane with the cuts y:[i 3 , 3 ,+ioo],
1
3
] two oth er W·PIanes wit
"hh
.
1'1= [ - t·.3-Y
t e cuts y1 and y1 respectively
2- , -•·oo
(Fig. 93).
FIG. 93
l!li3. Fz is obtained by joining together two Z·planes with cuts [-1, I],
Fw by joining together three w·planes having respectively one cut
[-ib, ib], two cuts [-a, a], [O, ib] and two cuts [-a, a], [O, -ib], where
y'5-l
J1'5+1
a= - 2-y'(y'5-2), b = - 2-y(Jl'5+2). The mapping and the three
W·planes with cuts are shown in Fig. 94 (the :figures are not to scale).
366
PROBLEMS ON COMl'LEX ANALYSIS
i
FIG. 94
1254. (a) n odd. F 11 has n·sheets and has n alirebraic branch points of the
(n-l)th order above the points 11 = wk(w = e"Ttn, k = 1, 2, ... , n), Fw has
Jin sheets and n 1 algebraic branch points of the first order, n over each of
•
the points w = r{'y(l/4) (17 = ea•lfn, v = 0, 1, 2, ... , n-1), corresponding to
the points of F11 above z = 17"', and n algebraic branch points of the first order
above w = oo, corresponding to the algebraic branch points of F 11 above the
1 = wt. The lD6pping is shown in Fig. 95 (for the investigation use is made
©
A(cfl)
0
B(I)
0~,;,.;:,.-c~===:~
c
FIG. 95
or the substitutions
c= zn,
w = w").
The function w(z) maps the circle 1111
n
1 onto thew-plane with n radial cuts from the points 17"'y (1/4); this is a well·
known extremal function from the theory of single-valued conformal map·
pingst.
<
t Bee G. M. GoLtTZIN, The geometrical theory of juncliona of a compl«I; t1tll'iable,
(Geometrichealcaya teoriya ftmktaii komplekmogo peremennogo), page 184,
Gostekhizdat 1952.
367
ANSWERS AND SOLUTIONS
(b) n even. If n .. 2m, then w(z) factorises into two
functions:~.
z•
= ± - - - . In order to construct F., take n "half-sheets" fzf < 1 and n "half·
l+zlnl
sheets" lzl > 1. Let us denote them by Hl and H~ respectively, and the boundary arcs defined by the points wt, by YI:· Let us attach to Ht along YI: the
half-sheet Hf, along Yt+i the half-sheet B:. a.long Yl:H the half.'sheet H: and
so on cyclically. Th.is also determines the order of attachment Fw of the 2n
sheets representing the w-plane with n radial sections indicated above.
1255, w = zl/•, z = 0 and z = oo are algebraic branch points of the second order.
1256. w
= zn1/•1
~) ,
( "1 is an irreducible fraction, equal to
m1
m
z ""' O
and z = oo a.re algebraic branch points of the (m1 -l)th order.
1257. z = 0 and z = oo a.re algebraic branch points of the first order; z = l
is an essential singularity of one of the two branches of the function.
z
z1
1258. w = 1- Si+ 5i - ... is an integral function; z = oo is an essential
singularity.
1259, z =- 0 and z = oo a.re algebraic branch points of the first order;
a ... 1 is an essential singularity of one of the two branches of the function.
the limit of its simple poles at the pointszrc =
u~::r. a1; =
; +nk (k -
o.
± 1, ... ).By problem 638, the domain of indeterminateness at the point z = I
is identical with the whole plane.
1260. If n = 0, then z = 0 and z = oo are removable singularities and
w
1; if n < 0, then z = 0 and z = oo are logarithmic branch points where
lim w = lim w = 1 and z = 1 is an essential singularity for one of the branches
,, .... o z-oo
of the function; if n = 1, then w = z; ifn > 1, then z = 0 and z = oo a.re
logarithmic branch points, the domain of indeterminateness of w (z) at these
points coinciding with the whole extended w-plane.
1261 and 1262. z = 0 and z = oo are logarithmic branch points with the
domain of indeterminateness for w(z) coinciding with the extended w-plane.
1263. z = 1 and z = oo are logarithmic branch points where lim w
==
=Z-+00
lim w =
1264. z
,, .... 1
oo and z ... 0 is a simple pole for all the branches of w(z) except one.
=
0 and z = oo are logarithmic branch points and lim w
z....O
... umw = oo.
Z-+00
1265. The branch points a.re the same as for sin-1z (that is, an infinite set
of algebraic branch points of the first order above z = ± 1 and two logarithmic branch points above z = oo, lim w = O; z = O is a pole of the first
Z-+00
order for all the branches of the function except one).
1266. z = ± i are logarithmic branch points, lim w = oo; z = 0 is a pole
z .... :I: I
of the second order for all the branches of the function.
1267. The surfaces w(z) and z(w) are the same as for the logarithmic function (logarithmic branch points above the points 0 and oo ). The mapping
is easily obtained by means of the parametric representation z = eC. w = ~.
368
PROBLEMS ON COIVIPLEX ANALYSIS
1268. The Riemann surface for w(z) has an infinite number of sheets with
one logarithmic branch point above each of the points z = a, z = b and two
logarithmic bmnch points above z = oo. The surface is obtained by joining
together an infinite number of sheets of the z.plane with cuts going from the
points z = a, z = b to co. These sheets correspond to the single-valued branches
of the function w+2:iin(n = 0, ± 1, ± 2, ... ). On going round the points
z = a and z = b these branches pass successively into one another and this
determines the character of the attachment of the sheets.
1269. The Riemann surface for w(z) has an infinite number of sheets with
one logarithmic branch point above each of the points z = a, z = b, z = c
and three logarithmic branch points above z = oo. The construction is the
same as the preceding.
1270. The Riemann surface for w(z) has an infinite number of sheets with
one logarithmic branch point above each of the points z = kn (k = 0, ± I,
± 2, ... ). As sheets, z-planes with cuts going from the points z = k"l to oo
(for example, along vertical straight lines) can be taken. Two sheets are joined
together simultaneously along all the cuts, on one side, as in the construction
of the surface of the logarithmic function. At oo there is a transcendental
singularity, the limit point of the logarithmic branch points.
1271. (1) In every connected part of the Riemann surface of the function
!; = <t>(z) above the z-plane to which corresponds a connected part of the
Riemann surface of the inverse function z = 4>-1 (/;) above Gt, w(z) represents a single analytic function; (2) In every connected part of the Riemann
surface of the function !; = lf>(z), located above Gz: (this is the domain Gt,
transferred to the z-plane), w(z) represents a single analytic function; (3)
The same as in part (2). The particular case of w(z) indicated in the conditions of the problem always represents a single analytic function.
1272. w(z) consists of the two analytic functions ±z.
1273. w(z) consists of the two analytic functions ±zt/s.
1274. w(z) consists of the p analytic functions wkzm1/111,
(w
=
e21ti/P,
m n1
p = H.O.F of (m,n); k = 0, 1, ... ,p-1; m 1 = p'
n )
= -p·
1111
1275. w(z) consists of the n integral functions wkez:/11 (w = e " , k = 0,
l, ... , n-1).
1276. w(z) is a single n-valued function with algebraic branch points of
the (n-l)th order above z = kn (k = 0, ± 1, ... ). At oo it has a non-isolated
singularity, the limit of the algebraic branch points.
1277. w = nLogz+2nik(k = 0, l, ... ,n-1) is n distincti analytic func·
tions.
1278. w(z) consists of the functions z+2nik (k = 0, ± l, ... ).
1279. A single infinitely many valued function with logarithmic branch
points above z = 0, ± l, oo.
1280. w(z) is a single infinitely many valued function with one logarithmic
branch point above each of Zk = 2nik (k = 0, ± 1, ... ). At co it has a nonisolated singularity, the limit of the logarithmic branch points. The Riemann
surface of the function w(z) is simply connected and is obtained by joining
together an infinite number of sheets of the w-plane with cuts (no two having
369
ANSWERS AND SOLUTIONS
points in common) going from the zk to oo
simultaneously along all the cuts, but only
1!81. The Riemann surface is the same
logarithmic branch points above z = nk (k
1282. The Riemann surface is the same
logarithmic branch points above
z= n:
(two sheets are joined together
along a definite side of them).
as in problem 12110, only with
= 0, ± 1, ... ).
as in problem 1280, only witb
(k
=
0,
± 1, ..• ).
1283. w(z) consists of the functions ±z+2kn (k = 0, ±1, ••. ).
1284. w(z) consists of the functions z+kn (k = 0, ± 1, ••. ).
1285. (1) Let r 1 = mi , rz = ms, r = r 1r 8 = m be irreducible fractions
n1
n8
n
and p = (m10 n 1 ) (the greatest common divisor of m1 and n 1 ), q = ('m2 , n 1).
Then (z'1)'2 consists of p distinct n-valued analytic functions, equal to
p
p
II
yl.z' = y'l.(y'z)m
=
p
q
II
yl.y'(zm), and (z•2)•1 consists of the q functions yl.z'.
One of them, namely zr, always belongs to both cases. In particular, (z2/ 8 ) 8 / 1
3
= ± z, (z3/B)B/8 = yl.z.
(2) Let r 1 = m 1/n1; r1 = m 1/n8, r = r 1+r2 = m/n be irreducible
,,
fractions and p = (n1, n 1 ). Then z•1z•2 consists of p distinct n-valued analytic
functions equal to yl.z'.
n.
' n1 •
n~ = ...!. and n 2 = 'P
'P
'P
Then w(z) consists of p distinct analytic N-valued functions with the para. 1et N = ni.n.
. t h e preceding notation
(3) U smg
--· ,
"
metric representation: z = tN, w = y'l (tm1~+tm211~). The Riemann surfaces of
N
all these functions above the z-plane are the same as that of the function yz.
n,
1286. LetN=L.C.M.ofm and
= .!!...
,, • Then w(z) is a single
p= (m,n),q
=(m+n,
mn)and~+.!.
'P
p
m n
N-valued analytic function having N /n algebraic
branch points of the (n-l)th order above z = 0, N/m algebraic branch
points of the (m-l)th order above z = l and N/q algebraic branch pointl'
of the (11-l)th order above z = oo.
1287. One nm-valued analytic function having one algebraic branch point
of the (n-l)th order above z = 1, n algebraic branch points of the (m-l)th
order above z = 0 and one algebraic branch point of the (nm-l)th order
above z = oo.
1288. Two distinct four-valued functions differing in sign with the same
•
Riemann surface as the function yz. These functions have each one branch
for which the point z = 1 is a pole of the first order.
1289. An infinitely many valued function with one algebraic branch point
of the (n-l)th order above z = 1 and n logarithmic branch points above
each of the points z = 0, z = oo. In order to construct the surface it is necessary to join n surfaces for Log z with the cut [l, oo) on one of the sheets
of each. The curves ensinn8 = kn(k = 0, ± 1, ± 2, ••. ) divide the w-plane
into domains corresponding to the half-planes y ;;e: 0 (Fig. 96, for n = 2.
C= w8 is an auxiliary plane).
370
PROBLEMS ON COMPLEX ANALYSIS
FIG. 96
1290. An infinitely many valued function with one logarithmic branch
point above z = 1 and an infinite number of only logarithmic branch points
above z = 0 and z = oo. The Riemann surface is obt,ained by joining an infinite number of surfaces for Log z with the cut [1,oo) on one of the sheets.
Above z = 0, z = oo the surface has only logarithmic branch points though
they are infinitely many, and above z = 1 ordinary points and one logarithmic
branch point. The curves eu sin ti= (21c+l):n and t1 = 2Tc:n (Tc= 0, ±1, ±2,
... ) divide the w-plane into domains, each corresponding to the z-plane with
the cut -co <a:~ 0, y = 0 and the complementary cut 1 ~a:< oo, y = 0
for the domains whose boundary contains one of the straight lines t1 = 2b
(Fig. 97; C= ew is the auxiliary plane).
@
Q
Q
~
(2)~g:
-~
(1)
(4aew) :
~I
Q'
D'
FIG. 97
1291. An infinitely many valued function with the same Riemann surface
as Log Log z.
1292. The Riemann surface has infinitely many sheets and has an a.lgebraic
branch point of the (n-l)th order above z = o. only algebraic branch points
of the first order above z = ± 1 and 2n logarithmic branch points above z = oo.
In order to construct the surface it is necessary to join n surfaces for sin-1 111
with the cut [0,1] on one of the sheets of each of them. The curves n8
= Tc:iifn and
cos n8 = Ten (Tc = 0, ± 1, ... ) divide the w-plane into domains corresponding to the half-planes y ~ 0 (Fig. 98, where n = 2, C= w'
is the auxiliary plane).
en
371
ANSWERS AND SOLUTIONS
FIG.
98
1298. The Riemann surface is obtained by joining an infinite number of
the surfaces for Log z with the cuts [O, l/e],[e, oo) on one of the sheets (the
surfaces for Log z are joined together in pairs, first along one and then along
the other of these cuts). The surface has above z = 0, z = oo an infinite set
of logarithmic branch points and above z = l /e, z = e ordinary points and
infinitely many algebraic branch points of the first order. The curves
Im sin w
cosu sinh
(2k+I)n and u = n/2+kn (k = 0, ± 1, ± 2, ... )
divide the w-plane into domains each corresponding to the z-plane with the
=
"=
< :x: < 0, y = 0 and with the two complementary cuts: 0 .-:;;;; :x: .-:;;;; ..!._
e
= 0 and e .-:;;;; :x: < oo, y = 0 for the domains whose boundary contains one
out - oo
fl
of the straight lines u
=;
+kn (Fig. 99; C=sin w is the auxiliary plane).
(C-sln w)
1
I
ltll
FIG.
99
129'. An infinitely many valued function. Its Riemann surface is obtained
by joining together infinitely many specimens of the Riemann surface of
n
tbe function J"(z)-1, provided with a cut along the ray l .-:;;;; :x:
<
oo, y = O
372
PROBLEMS ON COMPLEX ANALYSIS
on one of the sheets. The attachment proceeds as in the construction of the
Riemann surface of the logarithmic function.
1295. An infinitely many valued function with one only algebraic branch
point of the first order above z = 0 and z = oo and two logarithmic branch
points above z = I.
1296. An infinitely many valued function with two logarithmic branch
points above z = -1 and an infinite number of only algebraic branch points
of the first order above z = 0 and z = oo.
m
2:n;i
1297. If 111:=m/n, then w=- Log z+-k(lc=O,l, ... ,m-l) are
n
n
m distinct analytic functions; if 111: is irrational then w = 111: Log z+2nik
(k = 0, ± 1, ... ) are infinitely many distinct functions.
1!98. Two distinct infinitely many-valued functions with the same Riemann surface as Log z.
1299. Two distinct analytic functions equal respectively to 2 Log z and
2 Log (-z).
:n;
:n;
1800. w = 2 +2nk and w = 2 cos-1z- 2 +2nle (le = 0, ± 1, ... ).
1801. w
:n;
= T+:n;le (le=
0,
±
1, ... ).
1302. w = i (:n;/2+nk) (k = O, ± 1, ... ).
1303. w is an infinitely many valued function with the same Riemann
surface as Log Log z (see, problem 1290). If C=
= Log z = log r+ilfi
(I/I = Arg z), then w = el Log' = ei(1oce+lll+s"fk). For fo(z) = e-ll+iloce (0 =s;;;;
(J =s;;;; 2:n;) given in the problem the sets of limiting values represent respectively: (1) and (2) the circle lwl = e-";
(3) the circles lwl = e-11/1 as 1/1-+ +ooand lwl = e-•11/1as 1/1-+ -oo;
(4) the ring e-•"I• =s;;;; lwl =s;;;; e_,.I~.
For other groups of branches the factor e-11111 (k = ± 1, ± 2, ... )is obtained.
1304. At all the points w (z) represents a single analytic function if lal < l;
if !al ;;;;i: 1, then in parts (1) and (3) the function separates into n, and in parts
(2) and (4) into an infinite number of distinct analytic functions.
1805. If !al < 1, then in both cases w (z) repi-esents a single analytic function; if !al ;;;;i: 1, then in part (1) there will be n and in part (2) an infinite
number of distinct analytic functions.
1306. (1) Let z = reie and C= eelll = Log z. Then w (z) consists of many·
valued analytic functions, equal respectively to X (z) eil0 1 e-B+s11k(lzl < l;
g = y(log2r+lfi 2 ), (J = a.rg log r+ilfi), where - oo < r < oo and ; < (J < 32:n;;
(le= 0, ± 1, ... ).They each have one logarithmic branch point above z = 0,
in the neighbourhood of which the set of indeterminateness is a ring (in
particular, for k = 0 the ri.TJ.g is X (O)e-1"/I :s;;;; lwl :s;;;; x(O) e-7</I.
(2) Let z-1 = rei<f> and (; = (!6iB =Log (z-1). Then w (z) consists
of single-valued analytic functions equal respectively to z(z)eiloge-B+sxk
ea"'
(lzl < l;
<
3:n;
</>
e=
< 2'
0
y(log2r+ (<t>+2nn) 2 ),
<
0
< 2:n;;
k, n = 0,
different analytic functions.
(J = arg [log r+i(<t>+2nn)],
±
where ;
1, ... ). For different k, n these are
If x (1) = lim x(z) exists, then the set of
z--.1
.ANSWERS AND SOLUTIONS
373
indeterminateness as z-+ 1 is a circle (in particular, for k = n = 0 it is the circle
lwl = X (1) e-").
1307. (1) Ifj(z) has no zeros of odd order, theny{/(z)) factorises into two
integral functions. If a 1 , a 1, ... are the zeros of odd order of f(z), then the
Riemann surface for y (/(z)) has two sheets with algebraic branch points above
ai. a1, ... and above oo, if /(z) has a pole of odd order there. If f(z) is a transcendental function, then above z = oo there are two essential singularities
of single-valued character, if f(z) has an even number of zeros of odd order,
and one of the essential singularities is of two-valued character if the number
of such zeros is odd or infinitely great.
(2) If /(z) has the zeros a 1, a1 , .. • then the Riemann surface for Log
f (z) has one logarithmic branch point above each of a10 a2, .. • and no other
points. If f{z) has no zeros, then Log f(z) factorises into an infinite number
of integral fu.nctions cliffering from one another by terms of the form 2nki
(k = 0, ± 1, ± 2, ... ).
(3) If f(z) has zeros then the Riemann surface for [/(z)] 11 is the same
as that for Log f(z) (see part (2)). If /(z) does not have zeros, then [/(z)]11 factorises into an infinite namber of integral functions cliffering from one another by factors of the form e•111c«i (k = O, ± 1, ± 2, ... ).
1808. The two-sheeted circle lzl < I with algebraic branch points at the
00
zeros of the function
.J:znl,
n-1
in particular at the point z = 0.
1309. (1) The two-sheeted circle lzl < 1 with a single branch point at
z = 0 (part of the Riemann surface of the function yz. situated above the
circle lzl < I);
(2) The part of the Riemann surface of the function Log z situated
above the circle lzl < I ;
(3) The part of the Riemann surface of the function Log z situated
above the ring
!<
lzl
<
2.
CHAPTER X
1324. (1) At the vertex ..4. 1, 111:
=
O; (2) At the vertices ..4., and ..4. 1, 111:1
= lits
= O; (3) At the vertex ..4.8 , 111:8 = O; (4) At the vertices ..4. 1 and ..4.4, 111:1 = «4
= -1; (5) At the vertex ..4. 1, 111:1 = -2, at the vertex ..4.4 , «4 = O; (6) Atthe
vertices ..4. 1, A8 and ..4. 4 , 111:1 = 111:8 =«4 = O; (7) At the vertex A 1, 111:1 = -2,
at the vertices ..4. 4 and ..4. 8 , «4 = 111:8 = O; (8) At the vertex A 1, 111:a = -2, at
the vertex
«4 ... 111:-2.
A,,
1826. It is necessary and sufficient that a1: =
or infinity) and
= na ... na =
i:
2..
(n1: is a natural number
n1:
(1-2..) = 2, which is possible only for n
k-1
== 4 with ni
"'"
n 4 = 2 (that is for a rectangle) and for n = 3;
"i
na
na
1
2
2
2
3
00
00
00
2
3
6
4
3
4
3
p
a strip
a half-strip
a right-angled triangle
an isosceles right-angled triangle
an equilateral triangle
=~·log z+a, z =- e•(w-11)/ll (a is a real parameter), s(w) is ape-
181'7. (1) w
n
riodic function with period co "" 2m; the group G is generated by the transformation T (w) = w+co; its fundamental domain B consists of a doubled strip
and one of its boundary sides;
(2) w = Afnlog [(l+•)/(1-s)]+a, z = tanh [n(w-a)/2h] (a is areal
parameter); z (w) is a periodic function with period co = 21K; the group G and
its fundamental domain Bare the same a.sin part 1.
1828. w = sin-lz, z = sin w; 1 (w) is a periodic function with period cu
... 2n; the group G is generated by the transformations T(w) = w+cu,
S(w) ... -w; its fundamental domain B consists of the strip 0 < u < n and
the boundary half-lines u = 0, u = n, ":=:;;; O.
z
1819t. (1) to = 0
z-•I• (l-1)-1/1dz,
where
0 = co/[B(l/6, 1/2)]
f
1
=
(B(p,q)
0
= JzP-1 (1-z)f-ldz
is Euler's integral of the first kind); z(w)
0
is a doubly periodic function with periods 2co and 2weBirlfe; the group G is
generated by the transformations: T(w) = w+2co, S(w) = we•"lf•;
t Diagrams of the fundamental domains are given in Fig. 62, page 292.
374
375
.ANSWERS AND SOLUTIONS
ita fundamental domain B conaiata of the doubled triangle and two different
boundary aides;
z
(2) w = 0
Jz-f(l-z)-l dz, where 0 = ro/[B(l/4, 1/2)]; z(w) iaa doubly
0
periodic function with periods 2w and 2wi; the group G ia generated by the
transformations: !Z'(w) = w+2w, S(w) == iw; ita fundamental domain B con·
aiata of a square with aide w and two boundary aides of one of the triangles
forming the square;
z
(3) w = 0
J:1rl(l-z)-fdz where 0 = ro/[B(l/3, 1/3)); z(w) is a doubly
0
periodic function with periods 2hl and 2he•d/e, where h = wy'3/2; the group
Gia generated by the transformations: !l'(w) = w+2hi, S(w) = fQ8l'Cf/1; its
fundament.al domain B conaiata of the doubled triangle and two dift'erent
boundary sides.
1330. (1) The "triangle" with two vertices at the points w = 0, w =cl
= I'(rr.)I'({J)/I'(rr.+/J) and angles nrr., nfJ at these vertices. If rr.+fJ < l then
the three vertices are finite; if rr.+fJ ';?!: l, the three vertices lie at infinity; if
rr.+fJ = l, then cl = n/ain rr.n, and the "triangle" haa the shape of a half-strip,
oblique, if rr.:f./J; in the case rr.+/J = 2 the aides of the "triangle" which issue
from the vertices of the base are parallel, directed in opposite directions and
cl=
8=~(:~.\); if rr. = fJ = : , then the "triangle" conatitutea the exterior
of a straight half-strip (Fig. 100).
(2) A "triangle" with one finite vertex at the point w == 0, with the
angle nrr. and with two vertices at oo. Two sides of the "triangle" form rays,
iaauing from the origin, the third aide ia a straight line standing at a distance
h ==
s~nfJ r~:~:; l)
from the origin. In the case rr.
-
1 a strip of width
n
is obtained; in the case rr.+/J = l two aides are parallel and h = n; in the
case rr. == 2 a half-plane is obtained with a cut along the real positive semi-axis
ainnfJ
. particu
• lar, ,.i. = 4, if/J = and ,.i. == /1(/J+l),
m
l
"2'
andi.,. -= n, if/J
"" -1 (see Fig. 100).
1831. (l) See Fig. 101, (l); (2) see Fig. 101, (2).
1332. (I) w =
(2) w
(3 )
= .!. [ain-1 yz-y(z-zl)];
n
w =
(4) w
.!.
(ain-1 y'z-(l-2z) y(z-zl)];
n
nh [ 1og
= -2h
n
l+Jl'z
1-y'z
2 y'z]
u
u
2h
=n
(tanh-1 y'z-y'z);
v
[tan-1 t' a+tanh-1 t' z-2 z];
)
(5) w = ia ( -yz · -z-3
2 --1 •
377
ANSWERS AND SOLUTIONS
z
1333. (1) w =
q-1
- -naJ zi- 11 (dz
z-1 )II
• If 8
l
1
Xlog(1-,: )•where
(2) w
h
= -,
q
then w
j )'dz.
L)] , t
X
v-o
llYftl
t= (z:l ) ' andtv = eT (v
= ,';,
= -na~l
-p
tv
=
0,
1, ... ,q-1).
If6 = p/q, then w = : [ - fJ(;_l)
(z:l
l
q-1
+ ~ t~ log ( 1v-o
138'.
w
=
1335. w-
~'
where
c= ! (l~t• +log~~:). t = v( z:l ).
~ {vcz•-1)+sin- ~)·
1
B
1337. (l)w=
,..
a= l+ HI;
) Je«-1 (l-t•)-11 dt.
2d
(IX
1331. w ..,
ol=
and tv have the same values as in part (1).
where
2,
z
1-oi:
o
! {Htan-1 y(a~z)+htanh-1 ! v(a~z)}'
- _!
(")
"' w - n
A•
{H tan
-1
whmi
z
y(a1 -z8)
where
l+ HI;
z
ytdt
(3) w = 0 J (t-l)(t-a)(t+b), whmi 0, a, b are determined from
0
the equations:
On
OnJfa
OnJfb
(a-l)(a+b) =hi, (a-l)(a+b) =ha, (b+l)(a+b) =A,.
1338. (1) w =
"'
n
n
J
.!.
l
(2) w
Z
I
--
( 1
2 ) J (l-tn) ndt;
B -, 1-- o
=0
z
(l-tn) n
- 11dt+ 1, C-1 = 0
2-1/"nsin n/n
l
l )•
B -;-+2•2
= (l
378
PROBLEMS ON COMPLEX ANALYSIS
1389. (1) w
Z
.
BUo•!)o
_J
w 1
(2)
f (1-t1)-&/1 (l+t1)1/1dt;
5 21/I
=
0
(1-t1)&/1(l+t')-l/1
2-l/II'(7/10)
_
t•
dt+l, C-1 - 0 = I'(g/lO)I'(4 / 5) •
z
1840, Onto a many pointed star with angles
:n;-
2"
n
-i.n and :n:+i.:n: alter.
ilately, with centre at the coordinate origin and one of the vertices of the
first type of angle at the point
2-1/n
w=--
n
z
1341. w
= 000
f
{z
J
=
2m, and
n(
m
{z
,
O=
(z•-tan1 :n-1/•dz, if n
k-1
z
001
n
m-1
,
"' =
r(l-A _..!..)r(l+i.) sin n(l+A)
2
n
2
2
,
sin :n:/n
z1- tan1 :k)}-1/n dz, if n - 2m+ 1, where
k-1 \
----"--,
B(!,1-!)
00
=
-2i(2n)-lfn e-lafn and 0 1 = -2i.2-lf•,
1342. The para.meters are determined by means of equation (3) for the
bt, from the equalities lf(bk>I = It (k = 1, 2, ... , n) and the direction of one
of the sides of the star. One value of «k ie chosen arbitrarily.
1348. w
=
O(z-l)ll«!z+l)ll«-1'
0
= :
111:-1•(1-111:)••-1,
1345. The parameters a!.'e determined from the values of lf(bi)I, lf(ota)I,
known from P and one of the directions of the sides of P being given. Three
of the parameters (ak, b,, "J• d8 ) are chosen arbitrarily. If one of the parameters «t or "J is equal to infinity, then (4) and (5) apply if the factor and term
with this parameter are neglected. If one of the parameters b1 or d8 is equal
to oo then (4) and (5) remain in force without alteration.
1847. w
= O(z+l)•1(z-1)•1,
1348. w
=
0
ay'{;~:~.). o =
= h(b+l)-•1(1-b)--.,
b
= «1-«1 ,
«1+«1
J"(hH), a= J"(h/H).
1849. w = oz1-1•cz•-1)•, 0 = hb••-1c1-b•)-•, b = y(l-2111:).
1350. w
=
J"(hH) (
:t:t( :=~r,
where a and bare determined from
379
ANSWERS AND SOLUTIONS
the system of equations
a•ztb«•
= i/(h/H),
IX1 (
1851. (1) w = [Tn(z)]l/n, where Tn<z>
are Chebyshev polynomials;
T~(z)~z8 - l)
(2 ) w = [
~ -a) =
CX1 (
~ -b).
1
= "2 [(z+Ji(z1 -l)")+ (z-Ji(z1 -l)")]
]1'"
1852, The parameters lll'e determined from the valuesofRef(bl) and Re/(d8 )
known from P and the position of one of the sides of P being given,
Three of the parameters (a,., b;, oJ• d8 ) are chosen arbitrarily. If one of the
parameters
or OJ equals oo, then (6) and (7) still apply if the corresponding terms are neglected. If one of the parameters b; or d 8 equals oo, then
(6) and (7) apply without alteration (see the answer to problem 1845).
1853. The parameters are determined from the values of Ref(b;) and
Re f(d8 ). Three of the parameters a,., bi, OJ• d 8, are chosen arbitrarily. In formula (10) two of the parameters ak, b;, "J• d8 are chosen arbitrarily.
1854. The parameters are determined from the values of Ref(b1) and the
position of one of the sides of P. Two of the parameters are chosen arbitrarily.
1855, The parameters are determined from the values of Re f (bi) and
Rej(d8 ) and the position of one of the sides of P. Three parameters are chosen
arbitrarily. In form1ua (11) two parameters corresponding to the vertices
are chosen arbitrarily.
a,.
h
h
:Tr
:Tr
1856. (1) w =....!log (z+l)+ ....!. Iog (z-1)
0 = -
hi log (b+l)- halog (1-b), b = ht-ha;
:Tr
(2) w = ....! log
:Tr
=
hi+~
:Tr
h
b
+ 0,
h1
(z1 -I)+:Tr
h1
hg
log z+O, 0 = - - log (l-b2 ) - - log i,
:Tr
:Tr
v(2hih:hs):
z+a1 +h11
(3) w = -h11og--- og -z-aa
- - . The parameters a 1, a 1 are
n
l+a1z
n
1-a1z
determined from the
h1(!1 -a1)
system of equations: a/11a.'•2 = e-,.,
=h1(~1 -a.);
!
+~)].where f(z) is the mapping of part
(4) w
=f [
(5) w
=~log Tn(z),
:Tr
(z
where Tn(z) are Chebyshev polynomials (see
the answer to problem 1351);
h
(6) n =-[log(z-l)+z];
:Tr
(3);
380
PROBLEMS ON' COMPLEX ANALYSIS
z+I
2
.
.
(7) w =log I-z +Az,A- b'-I' where b is determined from the
.
b+I
2b
equation log b-l + b'-I =cl;
(8) w =log (z+I)-Azl-z+const, where A=~ and a is determi·
ned from the equation log a+
~(a-~) +cl= O. In particular, if cl= 0,
then
a- I;
(9) w =
h/n log (z-a)+O/(z-a)+Az+const, where A = h/2na,
0 = 27:m (l-a 8) and a is determined from the equations
log I-a+.!. = ncl • In particular, if h = 0 then a = 0, A = 0 = d/4.
I+a a
h
1357. The upper half-plane Im z > 0 is mapped onto the rectangle with
± K, ± K+iK',
vertices
corresponding to the points
±
I,
± ~.
In the
•·plane the same half-plane corresponds to the vertical half-strip with the
base (-n/2, n/2), the points given above corresponding to the points
±
n/2,
±
n/2+i log
l~k'
. The whole z-plane with the cuts (- oo,-1],
[I, oo) is mapped onto the rectangle with vertices K ± .X',-K ± iK',
with the corresponding pairs of points I/k and -1/k at the sides of the
outs. In the •-plane a vertical strip is obtained passing through the
points ± n/2 (Fig. 102).
@
iK'
Ill
D~C
---JV
A~1 8
B'
C'
I
D
C'
I
c"
'
B'
-f
/(
-if('
FIG. 102
z
1158.
10
= .ii
J
0
dz
y[(I-z1 )(1-klzl)]'
z = sn ( ': , k) , where .ii and k are
"'
determined from the relations K'/K = b/a, a= .ilK.
381
ANSWERS AND SOLUTIONS
1859. See Fig. 103. The inverse function is z = en (w, k).
JJ" D'
@)
21<
c
-ti<'
If C'
D
©
D'
-1
])"'
Ill
D
I
A'
D"
A'
])
·k'
Ill
IY
D"
IV
FIG. 103
1860. See Fig. 104. The inverse function is z
2il<'~----
IV
= dn
(w, k).
@)
-tl<''----'----1
-2tK'1---...i..---1
FIG. 104
1362. The mapping of the z1 -plane onto the Ui-plane is shown in Fig. 105
(the w 1 -plane on this same diagram refers to the answer to problem 1868).
1368. The relation between the z1 -plane and the w1 -plane is shown in Fig.
105. The expressions for l and h are indicated in Table 5 on page 383.
1864. SOLUTION. After applying formulae (12), (13) oo the expression w(z)
given in the hint oo the problem we obtain:
w (
!
)-w(l)
= 0"" O(k'blK'-E'),
w
==
whence b
0{E(u)-(1-;,) u}
= 1/ky(E'/K'). Then
= 0 {z(u)+ 2 ~,} •
(The correspondence between the w, z and u-planes is shown in Fig. 106).
;,...•
'[email protected] ~
f
0
I
0
Arf O
lkK'
I(
i•~ @i)
It I
a~r
l
FIG.
-:Jr
-b -1
105
0
FIG. 106
I
b
if'
383
.ANSWERS AND SOLUTIONS
TABLE
D
---~ ____l _____________
I -~
k
E
1
7C
~
(E-k' 1K)
k
k'
E'
~'
1
7c'
h
K'-E'
I
~
(E'-k 1K')
k
K-E
!,
(E'-k 8K')
(E-k' 1K)
v
k' E
!,
ik'
.!_E'
_!_ (K-E)
ik
1
T
k
(K'-E')
k
For the determination of k and 0 we use the equalities
0{E-(1- !: )K} == :~,
h ... ; [w(b)-w(l)] = o{/ y(1 ;~;:e•) lit-;: /}lcc 1
-e•t;1-k'•t•)l}'
a= w(l)
=
0
where t 0
0
,/D't
1
= Vy"(l-klb•) = If
K' •
1365. See Fig. 107. The position of the points B, B' on the u-plane
is determined from the condition dw/du = 0, which leads to the equation
h
FIG. 107
t For details of the solution and graphs for the determination of the
parameters see: A. BETZ, KonfONne Abbildu11g, Berlin, 1948.
384
PROBLEMS ON COMPLEX ANALYSIS
cn•(1l,k)
E' = 0 . After t hi s smce
.
.
k) +-Ku = fl ±i"K' h as b een tiound we d etern:une
sn2 (u,
the ,·nlue of h: h = Z({J, k). If the parameter k is replaced by k', then the
mapping reduces to the mapping of problem 1864 for 0 = l and a = n/2K
with the substitution u 1 = K' +iu, w 1 = iw.
1866. Cases (1), (2), (3) and (4) are represented in Fig. 108, 109 respective·
ly. In all cases for comparison the mapping has been given onto the u·plane
by means of the normal elliptic integral of the first kind (1). The continuation
of the mapping of the first quadrant I of the Z·plane by the symmetry principle
leads in the w-plane to a strip with a rectangular cavity (see the domains
I + II in Fig. 108, 1, 2), to a strip with a rectangular projection (see the
domains 1+11 in Fig. 109, 1, 2) and also to other domains (some of them are
shown in the diagrams). For the fundamental dimensions H, l, h which then
arise see Table 6. Note that the second case reduces to the first and the fourth
to the third by replacing z by t: klz1 +k'lt 8 = 1. Then instead of 'I', k in cases
(2) and (4) there occur the values 'I''= - -1 'k k' 8, k' and the corresponding
+•
w-figures are obtained from the w-figures for the cases (1) and (3)
by means of entire linear transformations with coefficients of magnification
(length) 2
k1 +'1' (length),
kl+•
.
. .
(length)i = - ""kl• (length), = ~ (the subscripts md1cate the cases).
The relations between the u and w-planes for the cases 'I' = -1 and 'I' = -kl
are given in Fig. 110. Using Table 6 on page 887 and the hint to problem
1865, we obtain for 'I' = -1:
1 [k'I u-E(u)+ sn cnu
u dn u ]
w = k'•
1-(E-k'1 K)
= -k'I
'
l
and for
11
_ ...!._ [
h
' E ( t"k'u,
= -y
1 )'
k'
1-(E'-klK')
= -k'I
= -k1 :
w - k'• E(u)
-k1 sn
u u] -_...!._k' E ( k , u, ikk' ) ' z -_ ...!._
K' -E' H _ _!_JC
k'• (
),
- k'• •
en
dn u
1867. SOLUTION. It follows from the condition .1 > 0 that e1, e1, Sa are
real and different and g8 > 0. Let us consider that e1 > e1 > e1• The upper
half-plane Im z > 0 is mapped onto the rectangle with vertices 0, co,
co- co', - w' (it is assumed that Im co' > 0), corresponding to the points
co
oo, e1 , e1 , Sa· The median line of the rectangle corresponds to two semicircles
(Fig. 111) : the first with centre at the point ea, with respect to which e1 and
e1 are symmetrical (consequently, its radius is y[(e1 - e1) (e1 - e1)], and
the second with centre at the point e1 , with respect to which e8, e1 are symmetrical (its radius is y[(e1 - e8) (e1 - e8)]. Continuing the mapping of p (w)
by the symmetry principle we arrive at the half·periods co, w' of this function:
00
w=J~~~~~-dx~~~~~­
y'[(x-e1) (x-e2) (x-ea)]
ANSWERS AND SOLUTIONS
@)
©
•
385
IY
®
JI
II
f
Ill
9
IY
/Y
m
II
2)
Fm. 108
18
JV
t /JI
z)
FIG. 109
lJ
'11~-1;
A
0
c
· ·
@
8
I<
FIG. 110
!vi
jvj
n -,/
-
!
v = -k 2
The quantity v'
REMARK.
H
1
i
I
I
I
I
i
I
I
I
(v, k)
I
!
'
!
i
'
i
'
I
I
(v, k)
-k'' E
1
II1
k' II ( I k'
k'+" 1 v. )
II1
I k'i:•+ v nc'v. k '>
1-
I
l
6
(v, k)
I
(
k'•
F1
(K'
-
E1 ·
)
E-k K)
,.
k' II (I I
k•+v i v • k)
II1
k• II ( I I k'
k'+v i v • )
- n 1 (v,k)
l
i
I
I
i
i
'
i
I
I
k'+v
v -k'•.
= -- __
h
;,, (E' -k 2 K')
H=H•
H-H•
H•-H
H•-H
In the column "l" the integrals are underst.ood in the sense of the principal value.
-
I
fv+ll lv+k'j
-1
V=
2
jv+Il lv+k'I
I
2
lv+ll lv+k'J
v>O
I
n -,/
< v<
0
!vi
n -,/
< v <-k2 I
- k2
--1
2
Iv!
lv+Il lv+k'I
n-,/
2
I
V<
-1
H•
I
1'
TABLE
~
00
~
Cll
z
1-3
....
0
q
Cll
0
t"
t::I
~
Cll
ttl
t;J
~
~
Cll
388
PROBLEMS ON COMPLEX ANALYSIS
From the consideration of Fig. 111 we arrive at the relations:
p(w) =
ea +
e,. - 6a
sn8 (
If g1 > 0, then ea < e1 < 0
K < K'); if Ua < 0, then
K
row,
k
)
, k• = ~ .
e1 - 6a
< e1 and co< lw'I (for k < k' and consequently
ea < 0 < e. < e,., and consequently O> > IW'I•
FIG. 111
If g1 = 0, then 6 1 = 0, e1 = -e1 and co= lw'I· In this case the mapping is
also symmetrical with respect to the vertical axis. To the whole z·pla.ne
with the cuts ( - oo, ea], [e1 , oo ), [O, ioo) there corresponds the triangle
(0, 2co, 2co'), which forms half of the period parallelogram (this is now
a square) (Fig. 112).
Let us also note that in the case of arbitrary e,., 6 8, ea (61 +sa+6a = 0) half
the period parallelogram corresponds to the z-plane with cuts, in general
curvilinear, originating at e,., e8, ea and going off to oo (see the diagram in
Fig. 113).
389
ANSWERS AND SOLUTIONS
2(J)
·-·-·-·-,
IV
I
! @)
©
I
i
i
FIG.
112
//"
F10. 113
390
PROBLEMS ON OOMl'LEX ANALYSIS
1368. SOLUTION. The basic mapping is shown in Fig. 114. It is obtained
by means of the symmetry principle from the mapping of the semicircle II.
Noting that dw
=-
y'[(z-11 )(z-e
dz ) (z-ea )] and t.a.kingarg (-1)""'
1
1
±n
A
Az
"FIG. 114
we have arg dw
= ± n+ arg
1
dz -
"2
2a arg (z-111;). whence it follows that
k-l
arg dw has the values -
; , 0, ; , - n respectively, on the sides of the
''qnadrilateral" PBMO, which leads to the mapping shown in :Fig. 114.
-
391
ANSWERS AND SOLUTIONS
Thus, for example, on the arc PB we have (Fig. 115)
arg dw = -
( n)
1
n
n+ aa+2 -2(-111:1+ix1+«a)
= -2
+
and so on. For the determination of the complex conjugates of the half.
c
e,
FIG. 115
periods w, w' of the f1mction p(w) we have:
co
w+w' =
w-w' =
J
••
y[(a:-81)
(a:~81) ·c;ii°-88)]
'
2' Je" _____dai_ _ _ _ _=-co
y'[(ei-a:) (e1-a:) (e.-:r)] •
If g1 = O, then 8 1, 8 1, e, are the vertices of an equilateral triangle and the
period parallelogram has the form of a rhomb with an angle of 60° at zero;
if g8 < 0 (then 8 1 < 0), and an angle of 120° at zero if g8 > 0 (then 8 1 > 0).
In both oases half the period parallelogr~ corresponds to the 111-plane with
symmetrical radial out.a going from the points e1, e1, e1 (see Fig. 116, 117).
1369. (1) w = -
where h
["~;)]'.
the half-periods are w, iw;
(2) w = p'(z)/2181 18/1,
= ay3/2;
the half-periods are co ... ke-111/6, m' = kel•/6,
(3) w ... p'1 (z)/4el, the half-periods are the same as in part (2).
392
PROBLEMS ON OOMPLEX ANALYSIS
A
A,
FIG. 116
A,
FIG. 117
393
ANSWERS AND SOLUTIONS
1170. (1) z = sn (u, la), w
K'
= fJffu/K', logµ= 2n K;
!Reul< K, llmul < K';
(2) the mapping function is the same as in part (1) except that
0 < Reu < K, llmul < K', logµ= nK'/K;
(3) reduces to part (1) by means of a linear transformation; then we
have la= ~.il~~~!=~~ where .il = (a, b, c, d) is the cross-ratio of the given
point.a;
(4) reduces to part (3) by means of the mapping I = y'(z1 +h1 );
nK'
(5) z = sn(u,la), w = i,elnu/2K, logµ="'! K ; - 3K <Re u < K,
0
< Imu < K';
(6) reduces to part (5) by means of the mapping I= y'(l+z•); here
008 Ot;
(7) the mapping function is the same as in part (5) except that in
la=
nK'
onecaseO < lmu < !K',andintheother !K' <Imu < K'; logµ=-; K;
(8) z = la sn• (u, la), w
<
= e-lnu/K,
log µ
= n K'
K ;
!Reul < K,
<
K';
(9) the mapping function is the same as in part (8) except that in
one case 0 < Im u < i K' and in the other case i K' < Im u < K';
0
Imu
nK'
logµ=]" K;
(10) is reduced to part (8) by means of the transformation
t = (z+l)/y'(l+e); here la= l/y'(I+e);
(11) isreducedtopart (7)bymeansofthemapping I= (1-z)/(l+z);
here la = sin ~ ;
2
(12) is reduced to part (2) by means of the mapping e = - i sin z;
here la= sinhH1 /sinhH1;
(13) is reduced to part (7) by the mapping t = sin z; here
la = cos tanh H;
(14) is reduced to part (5) by means of the mapping t =sin z and
y'i.-y'(.il-1)
a subsequent linear transformation; here la= vi.+y'(.t-1). where
.il _ (l+sin Pl (1 - sin at).
2 (sin P- sin at) '
(15) z = 0 [ Z(u)
+ dnucnu]
,w=
snu
e- 1"" /K , log µ = 2n -K' ,
K
where la is determined from the equations 2~ : Z (p) = tan or:, dn1
(see problems 1864 and 1365).
1871. e = (1 - en u)/sn u. The vertices K± iK', - K
the points e±f«, - e±f«, where cos or:= la.
1372. This reduces to problem 1871 by the mapping u
p=
E /K
± iK' pass into
= 2nK
sin-1 z. Here
394
PROBLEMS ON OOMPLEX ANALYSIS
-ir
lo isdeterminedfromtheequationsq ... e
K'
K =
1
(a+b)I' whereb = y'(a'-1).
I - dnu
2K .
k
, where u = sm-J z, IRe u) < K, llm ul < K'.
snu
:n;
The parameter lo has the same value as in the answer to problem 1371,
The foci z == ± l pass into the points ±k/(I+k').
dnucnu]
I+cnu
.
137,. (1) z = 0 [ Z(u) +
, t = - - - ; k IS determined as
snu
snu
in problem 1370, (Iii). See also problems 1385 and 1371;
1373. t =
(2) z
=
o[
Z(u) +
2~,], t =
(l+on u)/sn u; for the determination
of k see the footnote to the answer to problem 136&;
(3) z =
l:
E (1cu,
~) = ; [E(u, k) -
dn (u, k)] t 1 + on u
h
lo •
+on(u,snk)(u,k)
• = sn u • w ere
IS
i(l.11' - k•K') - k'lu
determined f
J!J' - k 1K'
b
J!J _ Tt'•K =
(see problem 1383 and Table 6).
a
rom
th
t'
e aqua ion
CHAPTER XI
1876. Steady motion with velocity V = a: - i{J. At oo there is a doublet
with moment p = 2:n:c. The stream lines are {J~+a.y = O; the equipotential
lines are ~ - fJy = O.
1377. At the point z = 0 there is a critical point (a stagnation point).
at oo there is a multiplet of order 2n (also a stagnation point); r" cos M = 0
are equipotential lines, r 11 sin M = 0 are stream lines (z = rei<f>); V = nzis-1.
1878. At the points z = 0 and z = oo there are vortex sources: (0; Q. I').
(oo; -Q, -I'); log r = -
~ <1>+0 are equipotential
lines; log r =
~ "+ 0
are stream lines. Both families of lines are logarithmic spirals; in the cases
I'= O or Q = 0 one of these families are circles r = o. the other are rays
II'+iQI 1(<1>-tan-1 ..!l+~)
• = O. The velocity is V =
e
r
a (z = rel<f>).
2:n;r
1879. At the points a, b there are vortex sources: (a; Q. I'), (b; -Q; I');
the lines of the field are logarithmic spirals about the points a and b (Fig.
'\\
\
•
\
\\
FIG. 118
118); log IJ = -
z-a
lines ( --b- =
~ 8+0 are equipotential
lines; loge=
r-•Q
eel• ) .
~ 8+0
;_;;
are stream
The velocity is V = - 2- . _
•
- :n:i (z- a) (z-b)
1380. At the point z = 0 there is a doublet with moment p = 2n;
r = 0 cos If> are equipotential lines; r = 0 sin </> are stream lines. V = -elf</> /r1•
Vco = 0. At the points 2 ± i the velocity equals - (3 ± 4i) /25.
1881. (1) and (2). At the points 0 and oo there are doublets with moments
±2:n:R8 and 2:n: (the upper sign refers to part (1). the lower to part (2)). The
field lines are cubic curves: ~ ± R1~/(a:&+y1 ) = 0 are equipotential line11.
z-
395
v~
N COMFLEX ANALYSIS
PROBLllMS O
1 The -
96
Jl' ,,.
- ·
.
3 =F Ry/(zl+y•) = O are stream
rm.._
'.. Pomts 1- F>g.
_ ± Ri
forvpart1 'F7•
(2 ) are cnt1cal
Y
z = ± R for part (1) and z 119 and lJO).
I
i
i
l
I
i
i
;
i
!
i
I
i
.
\
397
ANSWERS AND SOLUTIONS
1383. At the points ± a there are sources of strength 2n, at oo there is
a source of strength -2n. The equipotential lines are the lemniscates lz9 - a 1 1
= O;
the stream lines are the hyperbolas x1 -
~xy-y1 =
a1 with centre
at the coordinate origin which pass through the points ±a. V
V ±la =
the coordinate origin is a
± ..£;
a
o
=
2:
ZS-a' ,
critical point (Fig. 122).
a
FIG. 122
FIG. 123
1384. At the points ±a,
± ai
there are sources: (±a; 2n),
(± ai;
-2n);
re+oatr2 cos 2e+a' = O (le!> 2) are the equipotential lines (the straight
lines y = ± x also belong to them); re+oa2r2sin 2<1>-a' = 0 are the stream
4.az 2
lines (the coordinate axes also belong to them). V = ~· V00 = 0, the
..--a
point z = 0 is a critical point (Fig. 123).
1385, There are sources at the points ±1, 0 and at oo; (±
l; Q), (C,;-Q)
X ANALYSIS
• tential lines (for
PB
the eqwpo
398
(
1 - o+2cos 21/> (0 > 0) are. clear= ·'0
and r = l/y'O);
I'
®· -Q); ..+7....... to ...
' alues or o the curves
· elude
lines (they m
OBLEMS ON COMPLE
=
vl( o+tan<J>) ... ... .......
JI 0-tan</>
Q [
""° ..............
Jarge..
r
=-
d the circle
.
r = 1). V = 2n
The ......
z1+1 ]. Voo
z(zl-1)
== o.
=-~~~'~1~%,~
'w$%~--.
·~.~~.'l.h,
0,
±l
399
ANSWERS AND SOLUTIONS
1887. At the points ± (1± i)/y2 there are sources of strength 2n. at the
point z = 0 a source of strength - 4n, at oo a source of strength - 4n; r'+
~
= a - 2 cos~ (0 > 0) are equipotential lines (for a < 4 the curves sepa·
rate into four components, for 0 > 4 into two; for large values of a they
t
a
are "almost" the circles r = yO and r = l/yO); tan24> = (Or'-1)/(r'+l)
are stream lines (they also include the coordinate axes, the bisectors of the
coordinate angles and the circle r = 1). V = 2 [
z~~~ll)],
Voo
= O.
The
points ± 1, ±i are critical points (Fig. 126).
FIG, 126
FIG. 127
1888. At the point z = 0 there is the source (O; Q), at the point z
blet and the source (oo; -Q); y 1
are the equipotential lines;
~(C-11.1r)
=ea
-a;•
(
= oo a dou-
Q
cos-1 4> + 2n log r =
a)
r= { 0 - ! 4') /a sin 4' are the stream lines; the
. al asymptotes:
stream rmes have thehonzont
a
YJt:-+OO-. -;•
o-!L
2
YJt:-+-oo-.--a-'
V =a+ Qel4>
V:
Q '
" al
'
(F'ig. 127) •
2nr, oo =a; z = - 2na lB a critic pomt
1889. At the point z = 0 there is the vortex (0; I'), at oo the doublet
and vortex (oo; -I'); r == (O-I'4>/2n)/acos4' are equipotential lines;
400
PROBLEMS ON COMPLEX ANALYSIS
~(ay-C)
~=er
1( ")
Xe <fl+ I , V00
(
I'logr
)
-y• sin-l<fl- - - - · = 0 are stream lines; V
2n
= a; the critical point is z = 2I'"
,;, (Fig. 128).
r
=a+ --X
2nr
FIG. 128
1390, The fiuid stream lines the circle of radius B; Voe
=a,
the circulation
!a
is I'; the critical points are determined by the equation Zkr = 4
(11
1
± y'(16nla Bl-I'9)). If I'< 4-naR, then lztrl = B, that is, both stagnation
points lie on the circle lzl = B; if I' = 4naR, then the stagnation points merge
into one; if I'> 4nq.R, then IZtrl > B (the second stagnation point is in this
case inside the circle lzl = B). See, for example, [3, Chapter III, § 49].
1391. w(z) = Ve-l11z+
i: I't:~Qk
k=l
log (z-ak)· At infinity there is a doublet
m
n
with moment 2nVe-f11 and a vortex source of strength Qoo =
and
k=l
n
intensity I'oo = -
2 Qk
2 rk.
k=l
1
r
189!. (1) No; (2) Yes; (3) Yes (for example, the floww =-+..--=-log z
z
...m
has a stream line originating from the coordinate origin).
1893. In a schlicht conformal mapping a vortex source is transformed
into a vortex source of the same strength and intensity. A multiplet is transform!i!d into a set of multiplets of the same order inclusively. A doublet is
401
ANSWERS AND SOLUTIONS
transformed into a doublet with the following relations between the moments:
(1) (a; p)-+ (a:; pc1 ), (2) (oo; p)-+ (a:; pc_ 1 ), (3) (a; p) -+{oo:
(4) (oo; p)-+ { oo;
L)
C-1
:J,
1394. In a conformal mapping onto n sheets a vortex source is transformed
into a vortex source with the strength and intensity decreased by a factor
equal ton.
1895. The law of the change in the vortex source is (a* is the point symmetrical to the point a): (a; Q, I')-+ (a*; Q, - I') in the case of a stream line
and (a; Q, I')-+ (a*; -Q, I') in the case of an equipotential line. The law
giving the change in a doublet is more complicated. In the case of a rectilinear
stream line: (a; p)-+ (a*; p'), where the vectors p, p', are drawn through
a and a* respectively, symmetrical with respect to the stream line. In the
case o! a circular stream line lzl
=
R: (a; p) -+ {a*; -
~ p ).
if a of: 0 and
(a; p) -+ ( oo; :. ). if a= O. In the case of rectilinear and circular equipotential
lines using the same notation we have respectively: (a; p)-+ (a*; -p');
(a; p)-+ (a*; ;: p ). (O; p)-+ (oo; -
p/R1).
1896. (1) In all cases the singularities of the flow must be symmetrically
situated with respect to the circle lzl = R (see problem 1895). In particular the axes of the doublets must be situated on this circle, they
must be tangential to it. If vortices occur on lzl = R they must also
have doublets. Moreover, the total strength and intensity must be equal to
.
~
1 ~ •
zero, for which:
~ Qa:+ 2"~ Q1 = 0, where Qk are the strengths of the
sources within lzl = R and Qj are the strengths of the sources on lzl = R;
= 0 where the
are the intensities of the vortices on 1z1 = R.
(2) The singularities of the flow must be symmetrically situated with
respect to the circle lzl = R (see problem 1875). In particular, the axes of
doublets situated on lzl = R must be orthogonal to it. If there are sources
on lzl = R they must also have doublets. Moreover the sums of the strengths
Iri
rj
and of the intensities must be equal to zero, for which:
2 I't+ ! Ir';
= O, where the I't are the intensities of the vortices inside lzl = R and
the rj are the intensities of the vortices ~n lzl = R;
Qj = 0, where the Qj
are the strengths of the sources on lzl = R.
2
1897. (1) w
(3) w
= Vz+c
= ~
(c is a constant); (2) w
log [(z-a) (z-ii)]+c (at oo there is a source of
strength -2Q);
p
(4) w= 2n
z-a
= -2r:n:i.log--_-+c;
z-a
1
p
1
z::a+ 2n z-a+o;
402
PROBLEMS ON COMPLEX .ANALYSIS
p
1
p 1
+--+--+liz+c
2:n z-a 2:n z-i
7
n
.J; Q1c);
(at oo there is a source of strength P
+ -r+•Q
-.-logz+ c,
k=l
(6) w = -2where Imp= O. A flow with
2:ii•
:nz
a vortex without a doublet is impossible.
r
z-a
1398. (1) w = - 2 • log R• _ +c;
:it'
-az
pl
p*l
.
(2) w = - - - + - - - + c , if
2:n z-a
2~z +
and w =
2:n z-a*
a+ 0
: : z+c, if a = O { p* = :. ) ,
n
1899, (1) w
= .}; ~! log [(z-a1c) (R1 -a1cz)]+c,
k-1
n
(2) w
n
if .}; Q1c
=
O;
k=l
=.}; ~!log [(z-a1c) (R -a1cz)]
1
k-1
m
+.}; ~~ log (z-a;)+c,
1-1
1400. (1)
r
z-a
w = -2 • log R• _ +c;
:iii
-az
(2) w =
L -1-+
p* - 1-+c ( a*
1
= Ra , p* = - iR•1 p ) ;
2:n z-a*
R 1 Velac
(3) w = Ve-lacz+ - - +c;
z
R•Veloc
I'
(4) w = ve-lacz+--+-. log z+c.
z
2m
1401
' w
140Z. w
2:n z-a
= _.!.__ 1
2:iii
= 2~
(z-ia) (az+i) ,
og (z+ia) (az-i) '
log (z 8 +a 8 )+c.
0 •
1408. w=
2~ log(z -l)+o.
1
403
ANSWERS AND SOLUTIONS
1404.
Q
w= 2n
z -l
log --+c.
z1 +I
1
The flow is possible if
tX
1405, w
= ± ~ {if I
2
= 2~
log
(I+!) +c.
n
I't iF 0, then at oo there is
1'=1
n
,2; I't) .
a vortex of intensity -2
k-1
n
1407. w
~ [r"+'Q"
= £.J
2ni
r"-'Q"
- ]
log (z-ak)+ ~-log (Rl-atz)
k ... 1
1'
I
p•
I
RI Vel«
+-.
--+-.
--+
ve-l«z- -z- +a
2n z-a 2n z-a•
( a•
RI
= ii'
n
The flow is possible if
I't = 0 •
k-1
1408. Let t =- /(z) ma.p D conformally onto the unit disk
to ""' fl)[J(z)], where
1
p•
= Ba• p ) •
2
!ti < I. Then
n
2
Q" = 0 •
k-1
1409. Let I = /(z) map D conformally onto the domain
11: = /(at), with the indispensable condition
ltl > I with the
normalisation /(oo) = oo, /'(oo) > 0. Then with the condition
we have w =- !J>[J(z)], where
n
2
k-1
Q" = 0
404
PROBLEMS ON COMPLEX ANALYSIS
1410. In the notation of problem 1409, w = !li[J(z)], where
Ve-lex
Veiac
I'
!li(t) = /'(oo) t+ f'(oo).t + 2mlogt+c.
For I'
=
O, w(z) maps the exterior of 0 onto the exterior of the segment
[- !'~:), /'~:)]
w(oo)
=
of the real axis of the w-plane with the normalisation
oo, w'(oo) = Ve-lex.
1411. (I) w(z) =
_!_b [(az-by(z1 -c1)cosa:+i(bz-a y(z1 -c1))sina:]+const
a-
(Voo = Velex, c = y(a1 -b1));
(2) w(z) =
...!'.'.._ [(az-b y(z1 -c1)) oos a:+i(bz-a y'(z1 -c1)) sin a:]
a-b
rm log (z+ y(z -c )}+ const.
+ 2
1
1
1412. (I) w(z) = V(z cos GC-iy(z1 -c1 ) sin GC)+const (Voo = Vefex);
(2) w(z = V(zcosa:-iy(z1 -c1 ) sin GC)+
{m log (z+v (z -c ))+ const,
1
1
where I'= -2ncV sin GC (c is a point of departure).
1418. Let theZhukovskiiproflle be obtained by the mapping z = (C+C-1 )/2
of the circle IC-Col= I I-Col= R > I, Co= I-Be-IP (0 ~ fJ < n/2). Then
for circulation I' and V oo = Veicx ;
w(z) = VR ( z-Co+ y(z1 -I)
2
B
Relex
)
e-lex+ z-Co+ y(z•-I)
r
+ 2 mlog [z-Co+Y (z1 -I)J+c,
where I'= - 2n BV sin (or:+ {J)(I' is determined from the condition w'(I)
= 0 corresponding to the Zhukovskii-Chaplygin postulate).
1414. w(z) = y[z - (p/2)]+o is the streamlining of the parabola from
outside; w(z) = i cosh y(2z-p)/2yp is the streamlining of the parabola from
inside.
I
1!. _ mcx
1!. mcx
1415. w(z) = y.I [(z+ y(z1 -c1 ))1P e IP -(z-y(z•-08))1/1 e I/I ]+const
is the streamlining of the right hand branch of the hyperbola from outside
b
(tan GC = - , fJ = n-a:, c = y(a1 -f-b1 ) and
a
i
..!!...
1!..
w(z) = y 2 [(z+y(z1 -c1 )).., + (z- y(z1 -c1 )).,. l+const
is the streamlining of the right hand branch of the hyperbola from inside.
1416. w(z) is determined from the equation z = enwto+nw/v (the values
of the stream function on the streamlined half-lines are taken as equal to± v).
405
.ANSWERS AND SOLUTIONS
1417. w = cosh-1z =Log [z+Jl'(z1 -l)] (the values of the stream func·
tion on the streamlined half-lines are taken as equal to 0 and n).
1418. (1) A fl.ow with period n; at the points kn (k an integer) there
are sources of strength Q; the points ; +kn are stagnation points. The velocity at infinity in a strip of periods is V00 = V(:i: ± i oo) = =F
~ i.
See
Fig. 129 for the stream lines and equipotential lines.
(2) The same, only instead of sources at the points kn there are
vortices of strength I' and V(:i: ± i oo) = =F I'/2n. In order to construct the
field the stream lines and equipotential lines in Fig. 129 must be interchanged.
FIG. 129
1419. A fl.ow with period n; at the points Tm there are doublets with
moment p; the velocity is V(:i: ± i oo)""' O. For the stream lines see Fig.
130.
1420. If V1 = iV, then the solution is possible and unique for V1
=i{v- 2~):
·(v--;;-Q) +
_ I'+iQ 1
"' -
. n(z+ti) +
• n(z-a) + -I'+iQ 1
ogsm
2w
2ni
ogsm
2w
'
2ni
1421. w
=
p
n(z-a)
2n cot 2w
P
2n
1
0•
cot n(z+a) +iVz+o.
2w
1422. Let t = /(e) map S conformally onto the rectilinear strip St so that
D 1, D 1 pass into points at infinity of St. If there exist non-zero derivatives
/'(D1 ),J'(D8), the velocities V1, V1 touch the boundary of S at infinity and
one of them is given arbitrarily, then for St the problem reduces to problems
1420, 1421; the solution exists and is unique.
406
PBOBLEMS ON COMPLEX il.ALYSIS
f>·O
arg p·f
FIG.
130
407
ANSWERS AND SOLUTIONS
c
A. •
~8 .8A
~B
Cc:e,
]J
A
~B
Cce8
A
~
e1 cCc:e1
A
z)
1)
]) c =
A
3)
FIG. 181
A
40S
PROBLEMS ON OOMPLEX ANALYSIS
1413. (1) It is necessary and sufficient that the numbers M and 0 should
be real. In this case the curves Re u = ± <» are equipotential lines. In Fig.
131 the mapping t = J(u) = C(u)+Ou is shown for various real a. Figure
131 corresponds to the case <» > I<»'[. By the solution of problem 1367, we
1
then have ea<O<e1 <e1 and k8>2·
(2)
o'(~)
2<»
JM= :
<»
t
=JM=
71
o~( "2:,or:)
+c. For or:= 0 and M
1 °~(2:)
CM--u = <»
2<» 01 ( 2: )
FIG.
1427.
J(u)
=
':!
C(u-or:)+
!':
==
2n the mapping
is shown i::1. Fig. 132
(note
that
132
C(u-P)+Ou+c. For the function /(u) to
be elliptic it must be of the form /M =
':!
[C(u-or:)-C(u-p)]+c. Iflmu
==±Im <»' is a stream line, then M must be real, and if they are equipotential lines, M must be purely imaginary. For or: and p only the values
0, <»1; (k = 1, 2, 3) a.re possible. For or: = 0, p = cv1;o M = 2n
!M
l
p'(u)
a,(u)aJM
= C(u)-C(u-<»1;) = 711;-2 p(u)-e1; =fJt+ a(u)a1;(u)
(i. ;, lo are permutations of I, 2, 3). The points u
= ~"
(mod <», <»') a.re
ANSWERS AND SOLUTIONS
400
critical, that is. f'(u) = 0 there. For the basic mappings see Fig. 133. The
rectangles given there are mapped onto the half-plane bounded by the
horizontal straight line (k = 1), the half-plane bounded by the vertical
FIG. 133
straight line (k = 2), and onto a two-sheeted quadrant joined along the
horizontal half-line corresponding to the dotted curve on the rectangle
(k = 3). These mappings are continued by the symmetry principle.
410
PROBLEMS ON COMPLEX ANALYSIS
! r,
1428. The periods of the B.ow are 4K and 2iK', doublets 2mK+ (2n+l)iK'
with
moments
2n( -
critical points (2m+ l)K+niK' (m,n integers).
For the mapping see Fig. 134.
FIG. 134
1429. The periods of the flow are 4K and 2K+2iK', the doublets are the
(-l)n+m
, critical points 2mK
ik
same as in problem 1428 with moments 2n
+2niK' and (2m+l)K+(2n+l)iK' (Fig. 135).
i
i
i.
i
FIG. 135
1430. The periods of the flow are 2K and 4iK', the doublets are the same
as in problem 1428 with the moments 2n(-I)n+mi. the critical points are
mK+2niK', The basic mappings are shown in Fig. 136.
411
ANSWERS AND SOLUTIONS
@ii])
I !
FIG. 136
I'+iQ
...m
1431. /(u) =--.,.-log
co+w',
and {J =
w'
a(u-oc)
.
(
/J) +ou+o. In particular, for
au-
ix=
neglecting the additive constant and the factor
0, w,
r:;:.Q,
varying 0 and transforming to the a-functions, we obtain respectively,
log a(u) +Ou, log a 1 (u) +Ou, log aa(u)· +Ou. If /(u+2w)
~M
r
~M
"Q
f (u) = _t.;_ log
2m
~M
81(u;a)
..w
= /(u)
then
+ e.
81( u2:!)
1432. (1), (2), (3) doubly periodic flows with sources of strength 2n, and
-2n at the zeros and poles of the functions sn u, en u, dn u (Fig. 137).'
1433. A doubly periodic flow with quadruplets at the zeros of p(u) (Fig.
138).
143'. Periodic flows with the period 2w (period of the velocity!) with
sources of strength 2n at the zeros of 81<M (Fig. 139, (1) for 81 and for 81 on
displacing to the right by
to the right by
!).
!;
Fig. 139, (2) for 61 and for 68 on displacing
r
r
1435. (1) /(z) = 2ni log z+o; (2) /(z) = 2ni log t(z)+o, where t(z) maps
the domain D onto the circular ring with preservation of the directions o(
traversal of the boundary contours;
(3) j(z)
I'
Z-Z1
= ~log--+o,
...m
z-z
where z1, z1 are mutually symmetrical
1
with respect to each of the circles (that is, they are the points of intersection
FIG. 138
ANSWERS AND SOLUTIONS
413
of circles orthogonal to the two given circles with the straight line joining
their centres), where the point z1 lies inside the circle with circulation I';
(4) /(z)
= 2~
log t(z)+o, where the function t(z) maps the domain
D onto the ring with preservation of the direction of traversal of the contour
with circulation
r.
~)
I)
FIG. 139
1436.
/(z) = !Ii { :
log
z),
where
The function /(z) maps B onto the exterior of two parallel segments, at a dfs.
tance
l~I
from one another (Fig. 140). The ends of the segments are
determined from the condition !Zi'(u) = O.
414
PROBLEMS ON COMFLEX ANALYSIS
©
@
"''
D
C
:~
8
A
C
D
FIG. I40
1437. /(z) = !li[u(z)], where
!li(u)=_!.._[kel«
2w
and u(z)
=a+~+
...
z
8~(T,J-)
81{ u2:a)
+ie-lai
8~(~)]+c,
81( u2:a)
is the function which maps the domain D onto the
rectangle.
=!Ii(: log z), where
1488. /(z)
t
A = -
0
-•.w
,B=
:n;
•
~
(c_ 8 -c_ 1 ). The problem is possible if
:n;
C- 1
is a real
number and the difference c--a-c..1 is purely imaginary. If A of: 0, then the
function /(z) maps R onto the exterior of a horizontal ray and a segment
parallel to it and distant
1°-•; 0 - 11 from
it. The ends of the segment and
the beginning of the ray are determined ftom the condition !li'(u) = 0 (Fig.
HI, (I)), (the case B = 0 is shown in Fig. HI, (2)). If, however, A = 0,
that is, there is only a doublet then R is mapped onto the halt".plane
bounded by the horizontal straight line and having a cut along a horizontal
segment at a distance '0; 1 1 from the straight line (Fig. HI, (8)).
415
ANSWERS AND SOLUTIONS
b •
q
c
©
a
FIG 141
b
416
PROBLEMS ON OOMPLEX ANALYSIS
1439. (1) A solution is possible if I'1 -I'1 =I'; with this condition
f(z) = 4> (
:i
log z ), where
4>(u} =
~log
2m
(2
r.
cu
+2cu u+c,
u+a:)
01 (2cu
f~-a:)
01 -
cu
a = - . log a
m
[O { u-a: )]
1
. necessary to b ear m
. mm
. d t'--t
.
t of 1og - \ 2W
(i•t 18
,.... the mcremen
---
[ 01 ( u~a: )]
on varying u from 0 to 2cu equals 2m, and on varying u from 2w+iw' to
iw' it equals 0). The stagnation points of the flow are determined from the
equation
where A = I'1 /I', and are situated on the sides of the rectangle with vertices (0, w, w+w', w') and the rectangles symmetrical to it. In the case I'1
= 0 the function 3(z} maps R onto a circle with a cut along an arc of the
circumference (Fig. 142; I'> 0). In the case I'1
=
-I'1
=- ~
the function
FIG. 142
3(z) maps R onto the two-sheeted domain formed by joining the exteriors
of the circles
131 > 1 and 131 > ~
along the segments from - oo to
a
&ml'o
3o ""' - e--Y- where 'l'o is the value of 'I' at the stagnation point. The function
11(3) maps this two-sheeted domain onto the exterior of two lemniscates, and
417
ANSWERS AND SOLUTIONS
~I'.
/(z) = -
..:n;i
log [s(z)-s0 )(s(z)+s0 )] (Fig. 143; I'> O,
a•< e;
it is neceesa.ry
to join the half-strips in the e-plsne along the common cut). In the general
case 'I'= 0 on the lower base of the rectangle in the u-plsne and • varies
FIG. 143
r
r
from - 2 to 2 +r1, snd on the upper base
and • is varied from 0 to
(I'> Q).
r.
1
r." loge
"'= 2nr log a+
2
i
418
PROBLEMS ON COMPLEX .ANALYSIS
(2) /(z) = !l>[u(z)], where
u-a)
u-a ""
1--
(J (
I'1-I'1 l
2W r. o
+~
og-(--::.-) + 2 ,.. u+ ,
i
(J
2w
and u(z) - a+!:_+ ... maps the domain D onto a circular ring.
z
14150. The trajectories of the vortex are closed curves within the rectangle
with centre of symmetry at the point a+bi. For a vortex close to the centre
of symmetry these trajectories are nearly ellipses. If the vortex is at the point
a+bi itself these trajectories are stationary.
Solution. Let Zo = m0 +iy0 be the position of the vortex in the rectangle considered. Continuing the flow by the symmetry principle we obtain a doubly
periodic flow with periods 2w == 4a, 2w' = 4bi, having in the rectangle of
periods two vortices of intensity r at the points ± Zo and two vortices of
intensity -I' at the points± i 0 • Hence
w(z)
(J (z-zo)(J (z+zo)
r
= - - . log
2ni
1'° 1'°
91 (z~0 }(J1 (z!zi0 )
•
consequently•
w'(z)
==
f'ni
{[ccz-z0 ) -
ia (z-z.>]+(ccz+z.i>- i, (z+-zo>]
- [ C(z-i0 ) - 2: (z-z0 )[ -
[
C(z+zo)- 2: (z+io)]}.
• to t h e lim'it as z -+ z.,, we
·
tirom t his -2r . -1- and passmg
Sb
u tractmg
m z-zo
obtain:
dmo
. dyo
I'
,.
2a;
•
,.
2i
'"(Om
I'
p'(2mo)-p'(2iYo)
""dt-'dT = 2ni [,( o+ 2•Yo)-,( Yo)-, -o>l = 4m p(2lllo)-p(2iyo) '
whence
419
ANSWERS AND SOLUTIONS
(since p(2iy0 ) is a real number, p'(2iy0 ) is a purely imaginary number). (From
(•) it follows that p' (2m0 )d3:0 = ip' (2iy0 ) dy0 • Butfor the analytic function p(z)
the derivative p'(z) =
apa
a:
=.;.
i
apa
y
and it follows from the preceding that
dp(2111o) = dp(2iy0 ), consequently,
p(2111o)-p(2iyo) =
a.
(••)
Such are the trajectories of the vortex within the rectangle. In order to investigate them, let us note, that when a:0 increases from 0 to a, the value of
p(2m0 ) decreases from +oo to 8 1 = p(2a), then increases to +oo, and when
y0 increases from 0 to b, the value of p(2iy0 ) increases from - oo to 8 1
= p(2bi), then decreases to -oo. We conclude from this that for C > 8 1 - 88
the trajectories (••) represent closed curves symmetrical about the point a+bi;
for 0 = 8 1 -e1 the curve shrinks to a point - the stationary case. In particular, if 1110 = a+,, Yo = b+17 and E, 17 are small, then p(2a + 2E)
AS
e8 +
;!
(2E}1 p''(2a), as p'(2a) = O; p(2bi+2Ei) ""'e8 +
;!
(217i)1 p"(2bi),
as
p'(2bi) = 0 and the equation (.. ) assumes the form p"(2a),1 +p"(2bi)17'
""' !
[0-(e1 -e8 )] that is the trajectories of the vortex are close to ellipses.
1451. u = ru:-{Jy, "= {Ja:+rx,y, E = -iii; a dipole (oo; -iC).
1452. u = 2q<t>, "= 2q log.!.; E =
r
(co; -2q).
~
el<f>;
r
point charges (a; 2q) and
lz -a1
z
-b
1458. u=2qargz-a• t1=2qlog z-b; E=
2q(b-ii)
_
_ _
(z-'ii)(z-b)
point charges (b; 2q) and (a; -2q).
1454. u = -2q arg (zl-a8), " = 2q log lz1 -a1 1; E = -4qi/(r-a8);
point charges (a;-2q), (-a; -2q) and (oo; 4q) (see Fig. 122).
1455, u =
J!1
sin (<t>-cx)," = E1 cos (<fl-ex); E =
r
r
2 1PI el(•4>-«);
~
a dipole (O; p) (Fig. 144).
1456. u = ( r
dipoles (O; =F
± ~1 )
iB8)
cos .,,, " = ( r
± ~a )
sin 4>; E = - i ( 1 =i=
~8
e•l4>) ;
and (-o; -i) (see Fig. 119, 120).
1457, u= -py+2q4>, t1=pa:+2qlog_!._; E =
r
-p-f-~e14';
r
point charges (O; 2q) and (oo;-2q); a dipole (oo; p) (see Fig. 127).
1458. u = -py+
1:" 2q1;4>1;,
k=l
"=pa:+
"
l
2
2q1; log-;
k=l
r1;
E = -p
+ 2"
2q1; el4'1; '
k-1 r1;
where z-a1; = r1; e1"'"; point sources (ak; 2q1;); a dipole (oo; p) (Fig. 145).
420
PROBLEMS ON COMPLEX ANALYSIS
i
i
/
FIG. 144
Fm. 145
421
ANSWERS AND SOLUTIONS
1469. (1) The value of the point charge is preserved; the law of variation of the moment of the dipole is the same as in problem 1893;
(2) The sign of the charge is reversed; the law of variation of the
moment of the dipole is the same as in problem 1896 on continuation across
the equipotential line.
1460. " ""' 2q g(z,a).
1461. w
= 2qi log z-Ro +c.
z-zo
.
RI-Roz
1462. (1) and (2) w = 2qi log R(z-Zg)
1
1463. w
= 2qi log /(z)
1464. w
= 2qi log f;z) +c,
1466. w
= 2qi log/(~) ,
1
..
M
2d
--
B
f
(.!..2 • .!.)
t
4
+c •
z-y~-~
a-b
+ const, where /(z) =
,
where f(z) = [z- y(z1 -R1 )]/R.
where t
= /(z)
is determined from the equation
y(l-1') ...
.:I•+ _d (see problem 1838 for
ei
2
n
=
4 and
1374 for k .... l/y2).
1-cn(!z,k)
1
1466. w = 2qi log /(z) +c, where /(z) =
and k is desn (
! z, 1c)
termined from the equation K' /K = b/a (see problem 1871).
1467. w
=
1
2qi log/-() +c, where f(z) =
z
6 ( Z-Zo)
1
4a
(J ( Z-Z9 )
1
4a
61 ( z;az1) 61 ( z;aza)
and w = 2a, w' = 2ib, z1 = (4a-a:0 )+iy0 ,
z1 = a:0 +i(4b-y0 ) (see problem 1460).
z1 = (4a-a:0 )+i(4b-Yo),
.RI
~) ,
1468. w = -pi
- + - -~
. + c ( a+ 0, a*= -,p*
=-::a
z-a
w =
~
z-a
q
a
_ii.z!. z+c (a =0), c is a real number. Compare with problem 1868, (2).
RB)
pi
p*i
(
R•
1469. w = - - + - - . + c a+ oo, a*=-, p* =-=a p ,
z-a
z-a
q
pi
pi
w = z--Ri"z+c (a= oo). See problem 1400, (2).
1470. w
= e(z cos a.+i sin a. y(z8-R1))+const.
q
422
PROBLEMS ON COMPLEX ANALYSIS
1471. w =
where o1
~b [(ae-by'(z1 -o1)) cos ix-i(bz-ay'(z1 -o1)) sin ix]+ const
a-
== a 1 -b1•
2Ke
K
K'
b
1472. w = - - (cos ix+isin tXCnu), where u = - z and-= - (see
asnu
a
K
a
problem 1466).
1473. (1) If pi= eela., then
/(z) = et'(a>{[,;z)+t(z)]oosix+i[ ,;z) -t{z)]sinix}+o;
R•pi
(2)/(z)=pil(z)- t(z) +o=e
{[t{z)+t(z)
RI] oosix+i [ t(z)-t(z)
R•] sinix}+o,
where pi= <!8'°'· The functions in the square brackets effect the normalised
conformal mapping of D onto the exterior of the horizontal and vertical segmenta
respeotivelyt.
n
a)~(:.
= ,J;
2q1ci log /(z,1a1c) +if'(a,
a)-J/(z, a)] +o,
k-1
where /(z, a1c) and /(z, a) conformally map D onto the unit disk with the
normalisation /(a1c, a1c) =/{a, a) = 0, /'(a, a)> 0 and o is a real number.
1474. w
I-
, if
Imz~ 0,
- log-1- 1- 1 , if
z-a
Imz~O;
log
1476. v(z, a) =
lz~'iil
{J
nI (a:-ix)•+(JI
e(a:, a) = -
(a= ix+i{J).
1477. (1) Inside the cirole
v(z, a)
RI-_
azl
R
= log I
=
1
I
.Rl +log l;I] if a¢ O,
,-[log/
z- i.i
logR,
ifa = 0,
Outside the circle v(z, a) = - log-1- 1- 1 • The density is
z-a
I
e(Rel", a)=
-"'iiR
R 1 -lal 1
.Rl-2Rlal cos (8-ix)+lal1
In particular for a = 0 it has the constant value -
(a= falel«).
2~
and is identical
t See the appendix by M. Schiffer to the book: R. Cc•URANT, Diriohlet'a
Principle, ConJormal Mappings and Minimal Surfaoea, § 1, sec. 2, especially
pages 242-3, Interscience, New York, 1950.
423
.ANSWEBS AND SOLUTION'S
with the potential of a layer having the constant value log R inside the circle
and the value log lzl outside the circle.
I
log
(2) v(z, a)=
IR1 -iiil
R
, a of:. oo, lzl ;;;;;. R,
- log-1- 1- 1 , a of:. oo, lzl
z-a
v(z, oo) = {
e(Re19, a)= -
~ R;
log lzl, if lzl ;;;;;. R,
.
log R, if lzl ~ R;
l
!all-RI
2nR Rl-2Rlal cos (8-0t)+lal1 (a"" lalef« and a of:. oo).
U a = oo the same potential is induced all in the preceding case for a = O.
lz+y(z1-R1 )1
1
1478. v(z, oo) =log
, e(a:, oo) = - 23tf(R•-a;I) (la:I< R).
2
lz+y(z1-ol)l
, outside the ellipse, "(z, oo)
1479. v(z, oo) =log
2
.... - log 2(0t-/1) inside the ellipse. The density is e(z, oo) = -
CC
2ny(l~-c•I)
is on the ellipse, e' = 0t•-f18).
1480. e(C) = - 12n
1
1481. e(C) = 2d
og(C, oo)
em=
1485, l/2R.
•
1
1482. e(a:) = 2ny(R8-a:I) (la:I
1483.
1484. R.
on
<
R).
1
2ny(IC8-0•1> (el= a 1 -b1 ).
1486. l/2(a-b).
1487. a.
1490. If c.o(z; LI) is the harmonic measure of the interval LI of the reel
axis at the point z with respect to the upper bslf·plane:
c.o(z; LI) =
-1 - 1 dt (see problems 1091 and 1489), then (omitting
0n. log1t-z
n
°
.!.. J.d
a real additive constant): (1) w = ±..1og (z-a)," = </lc.o(z; LI), LI= (- oo, a);
3t
(2)
z-b
1 [
w=-log--,t1=</lc.o(z;LI),
LI= (a, b); (3) w=</11 log(z-a1)
<fl
n
z-a
n
z-a1
z-an ]
.n
+<fl1 log--+ ... +<fin log
, ti= 2 </IJ;C.O (z; Ll1;), £11; =- (a1;-1, a1;),
z-a1
z-an-1
k=l
a_1 = - oo; (4) w and " are obtained from the expressions indicated in
the answer to part (3) by replacing <flt by </11;-</10 , by the addition of
i</1 0 1io w and of </lo to v. The potential can also be represented in the form
n
2
k-o
</11;ro(z; £11;), Lio= (!Jn, oo).
424
PROBLEMS ON COMPLEX ANALYSIS
id
1491. (1) w(z) = - 1- - logz+const,
ogµ
r1
µ=-;
rl
d
t1(Z) =- - - log r+const,
logµ
id
where
d
It (z)I + const,
ogµ
ogµ
where µ is the modulus of the domain D (see page 30) and t(z) conformally
maps D onto the circular ring.
1493. w(z) is determined by the formula indicated in the text of problem
(2) w(z) = - 1- - log t(z)+const, t1(z) = - 1- - log
t(z) = yµ
1492, where: (1)
(2) and (S)t (z)
Z-:!11
= y(µ)--,
z-:111
mined from the equation :iii+
,
µ =
rl-R1 -a1
a
µ = ;
)I
z = sn (u, la), µ =
nK+,u
(6) t =- e K , z
y[( ; r
-1] i
:i:+B1 = 0 (:111 < :111);
K+u
ny
(5) t = e
,
(see problem 1370, (l));
+
( B-:111
-B-- , :111, :111 a.re deter·
-:111
a+b
(z+y'(z1-c1 )), µ = - - , c
= -1c
(4) t(z)
:~ ~~::=!:~
= y(a 1 -b1);
c
K
•1r.•
e
(!Re ul < K. llm ul ~ K'
2Kh [
u en
=- Z(u)+ dn snu
n
u]
an~
, µ = e IC
where la is determined from the equations: KZ(µ)
nh
= 2ii'"'
JC
dn1 P = K
(see problem 1870, (16)).
log lzl
log lzl
1495. (1) w1 (z) = 1- - 1- - , w,(z) = - 1 - ;
ogµ
ogµ
ei(C) = -e1(C) =
I
1
on lzl = 1,
2nµ logµ
-
1
Pu= PH= -Pi·=
2iiµlogµ
(2)
;,
W 1 (z)
;,
el( ,.) = -ea (") =
=
==
10glt(z)I
1ogµ
onI'to
lt'(C),_l
logµ
t(C)
t'(C)
-
l
=logµ;
on lzl = µ;
l-w 8 (z), w8 (z)
II I
-P11
1
j
1
Pu= Pza = -P11 =-Pu= logµ'
t(C) logµ onI'1;
where t(z) conformally maps D onto the ring 1
corresponding to the circle !ti = 1.
< !ti < n, the contour I'1
425
ANSWERS AND SOLUTIONS
n
J; oc1cw1c(z), where f(z) conformally
1497. t1(z) =Im f(z)-
maps D onto
k-1
the plane with horizontal outs, and
f (z)
1498. t1(z) = 2gg(z, a)+
=
I
pi
Z-iJ + ... •
if
piz +
if a
... ,
n
~
L.J IXk°'lc(z),
a #: oo,
w1c(z) =
=
oo.
1
-2k
J---a,;-da·
og(C, z)
r1c
k=l
1499. If /(a) = oo, then the field is formed by the dipole (a; p) where
I' is determined from the expansion off (z) close to the point a;
+ ... ,
'( ) 1--1!!.__
z-a
z =
piz+ ••. ,
if a #: oo,
if a
= oo.
1500. (1) If /(a) == 0, f(b) = oo, then the field is formed by the point
charges (a; 2q), (b; -2q) the flux of the vector intensity through each boundary
contour being equal to irero;
(2) If /(a) = 0, then the field is formed by the point charge (a, 2q),
the flux of the vector intensity through a boundary contour, corresponding to
a circle, in the direction of the outward normal to the domain D being equal
to 4:.irq, end through every other contour being equal to zero;
(3) The field is regular everywhere. The flux of the vector intensity
through boundary contours, transformed into circles, in the direction of the
outward normal to Dis equal to ±4:.irq (+for the contour, transformed into
the outer circle), end through every other contour it is equal to zero.
1501. See problems 1487 and 1489.
n-1
1502. (1) t1(Z) =
n-1
J;
the system
J; or:lc°'1c(z)+c, where the or:1c a.re uniquely determined from
k=l
Plkott
=
2qi (i
=
1, 2, ... , n-1) (see problem 1104) e.nd c is
k=l
an arbitrary real number. The problem is equivalent to the construction of
the :Bow in D, streamlining the boundary contours I'1c with circulations 4:.irqrc
(k = 1, 2, ... , n), if oo e D, and by the circulations 4:.irq1c (k = 1, 2, •.. , n - 1),
-4nqn, if oo D (I'n is the external contour).
(2) t1(z) = t10 (z)-2qg(z, a), where t10 (z) is determined as in part (1),
form the charges of the layer 2q1c+2qt, where
e
!lt = _ _!_
2:.ir
uoa.
"(z)
=
2q1 log
~
+c, if !l
f
r1c
=
og(C,a) dB.
on
0, and
r1
t1(z) = 2(q1-.Aq)log"jZj'"-2qg(z,a)+c,
426
PROBLEMS ON COMPLEX ANALYSIS
8 (log z+log a)
1
2:.iri
og a g(z, a) = lzlA log - - - - - - 1
loge'
8 (logz-loga)
l
2:.iri
l
where A -
for.,,.= (loge)/:.iri, where e
<
a
<
1, if q:;tO (in the notation of problem 1439
the Green's function g(z, a) = Im 4i ( :
log 111) for I'= 2:.ir and I'1 = -2:.ir;
the latter from the condition 'P = 0 on the boundary of the ring).
15H. The source (a; q) is transformed into the source (a*; -q), where
a• is the point symmetrfoal to a. The function u = -2q log -11 1
+c, where
:ii
(111, all
f (111, a)
conformally maps the domain D onto the unit disk (here and in what
follows the coefticient of heat conductivity Tc is assumed to be equal to 1).
I
1505. u - -2q log 1 z-ii +c.
:ii
111-a
1506. u =
2~ log1;;~:)1 +c.
q
1507. u = 2 _ log
••
1508.
u=
:Tc log
sin~+isinh :.irk
2a
2a
+c.
. m
'sinh nh
s1nra-•
2a
I:~!~~: j+c,
. d ..-om
.....
.
termme
t h e relat1on
KK
e=
sn [
!
(111 + ib),
u],
where Tc is de-
a.
= 2b
1509. (1) The Green's function g(z, a) of the domain D can be considered
as the temperature created in D by the heat source (a; 2:.ir) when the temperature on the boundary is equal to zero;
n
(2) u(z) = -2q g(z, a)+
:ii
2 u1;w1:(111),
k-1
where w1(111) is the harmonic
measure of I'1;.
q
U1-U1
lzl
1510, u = -2 g(z, a)+ 1 ( I )log- +uv where g(z, a) is the Green
:ii
og r 1 ,.,
f't
function (see the answer to problem 1508).
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