# 2016febComplex-Analysis-Problems ```COMPLEX ANALYSIS
PROBLEMS
CLAUDIA TIMOFTE
COMPLEX ANALYSIS
PROBLEMS
To my students
Preface
The present book is a collection of problems in ordinary differential equations.
The book is based on some lectures I delivered for a number of years at the Faculty
of Physics of the University of Bucharest and covers the curriculum on ordinary
differential equations for the students of the first year of this faculty.
The material follows the textbook . Each chapter contains a brief review of
the corresponding theoretical results, worked out examples and proposed problems.
Since the ”learning-by-doing” method is a successful one, the student is encouraged
to solve as many exercises as possible. The basic prerequisites for studying ordinary
differential equations using this book are undergraduate courses in linear algebra
and one-variable calculus.
It is my hope that this book will serve as an useful outlook for the students of
the first year of the Faculty of Physics of the University of Bucharest.
I would like to thank to my students for their continuous questions, comments
and suggestions, which helped me to improve the content of these notes.
Claudia Timofte
7
8
.
Contents
1 Complex Numbers
13
1.1
The Complex Number Field . . . . . . . . . . . . . . . . . . . . . . .
13
1.2
Sequences and Series of Complex Numbers . . . . . . . . . . . . . . .
30
2 Complex Functions
39
2.1
Functions of a Complex Variable . . . . . . . . . . . . . . . . . . . .
39
2.2
Limits of Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . .
40
2.3
Continuous Functions . . . . . . . . . . . . . . . . . . . . . . . . . .
42
3 Differentiable Functions
45
3.1
Holomorphic Functions . . . . . . . . . . . . . . . . . . . . . . . . . .
45
3.2
Harmonic Functions . . . . . . . . . . . . . . . . . . . . . . . . . . .
59
4 Complex Integration
63
4.1
The Complex Integral . . . . . . . . . . . . . . . . . . . . . . . . . .
63
4.2
Cauchy’s Theorem. Cauchy’s Integral Formula. Applications . . . .
73
5 Taylor and Laurent Series
83
5.1
Sequences and Series of Functions. Power Series . . . . . . . . . . . .
83
5.2
Taylor Series . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
93
5.3
Laurent Series . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
98
5.4
Residues . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 109
9
10
CONTENTS
6 The Residue Theorem. Applications
115
6.1
The Residue Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . 115
6.2
Evaluation of Real Integrals . . . . . . . . . . . . . . . . . . . . . . . 119
Bibliography . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 133
Chapter 1
Complex Numbers
1.1
The Complex Number Field
If R is the set of real numbers, let us consider the Cartesian product C = R &times; R, i.e.
the set C = {(x, y) | x, y ∈ R}. We introduce a special algebraic structure on this
set, by defining the sum and the product of two such ordered pairs of real numbers
as follows:
(x1 , y1 ) + (x2 , y2 ) = (x1 + x2 , y1 + y2 ),
(x1 , y1 ) &middot; (x2 , y2 ) = (x1 x2 − y1 y2 , x1 y2 + x2 y1 ).
Then, (C, +, &middot;) becomes a commutative field, called the complex number field (the
proof is straightforward and it is left to the reader). An element of (C, +, &middot;) is called
a complex number.
Remark 1.1 It is important to mention that it is not possible to define an order
relation on C that is compatible with its ring structure. So, the complex numbers do
not form an ordered field.
If we consider the subfield of C consisting of all the ordered pairs with the
second element zero, i.e. S = {(x, 0) | x ∈ R}, then the map ϕ : R → S, defined
by ϕ(x) = (x, 0), is an isomorphism of R onto S. So, R and S can be identified. In
other words, the field R can be considered as being a subfield of the field C.
13
14
COMPLEX ANALYSIS
Now, if we identify, through the above mentioned isomorphism, the pair (x, 0)
with the real number x and the pair (y, 0) with y, it is not difficult to see that we
have
(x, y) = (x, 0) + (0, 1) &middot; (y, 0).
(1.1)
Let us denote
z = (x, y).
We consider a special pair, denoted by i:
i = (0, 1).
(1.2)
This pair is called the unit imaginary number. In this way, (1.1) can be written as
z = x + iy.
(1.3)
This is the algebraic form of the complex number z (sometimes, a complex number
written in the form (1.3) is said to be in rectangular form). So, the set of complex
numbers can be represented as being:
C = {z = x + iy | x, y ∈ R}.
Definition 1.2 For a complex number z = x + iy, the number Re z = x is called
the real part of z and the number Im z = y is said to be the its imaginary part. If
x = 0, z is said to be a purely imaginary number.
Definition 1.3 Let z = x + iy ∈ C. The complex number
z = x − iy
is called the complex conjugate of z and
|z| =
p
x2 + y 2
is said to be the absolute value or the modulus of the complex number z.
15
COMPLEX NUMBERS
Remark 1.4 Let us notice that, since (0, 1) &middot; (0, 1) = (−1, 0), we have
i2 = −1,
(1.4)
which clearly shows that C cannot be an ordered field.
It is often convenient to use a graphical representation of complex numbers. As
we shall see in this section, complex numbers can be geometrically represented either
as points in the Euclidean plane or as two-dimensional vectors on a diagram, called
the Argand diagram or the complex plane. More precisely, in a given rectangular
coordinate system in the Euclidean plane R2 , the complex number z = x + iy can
be identified with the point having the coordinates (x, y). The axes Ox and Oy
are called the real axis and, respectively, the imaginary axis. The number z = 0
corresponds to the origin of the plane and i corresponds to the point (0, 1). The plane
is referred to as the complex plane. This plane is also known as the Argand plane or
the Gauss plane. In this way, we establish a one-to-one correspondence between the
set of all the complex numbers and the set of all the points in the complex plane.
Geometrically, the conjugate z of a complex number z is the reflection of z in the
horizontal axis. The symmetric point of z with respect to the imaginary axis is −z.
It is easy to see that the points z, −z, −z and z are the vertices of a rectangle which
is symmetric with respect to the axes Ox and Oy. A complex number z = (x, y)
can be also represented as a two-dimensional vector pointing from the origin O(0, 0)
of the Argand plane to the point P (x, y). The point P (x, y) is the image-point
in the Argand plane of the complex number z and z is said to be the affix of the
point P (x, y). So, for each point in the complex plane, we associate the directed
−
line segment from the origin to the point, i.e. →
v = (x, y). Operations such as the
addition of complex numbers can be easily visualized by using this representation.
Definition 1.5 Let C∗ = C \{(0, 0)}. For any complex number z ∈ C∗ , any solution
θ of the equation
cos θ + i sin θ =
is called an argument of the complex number z.
z
z
(1.5)
16
COMPLEX ANALYSIS
Therefore, there is no argument for z = 0 and the argument (also called the phase) of
any complex number z 6= 0 is not uniquely determined, being defined only up to the
addition of integer multiples of 2π, i.e. the argument is a so-called multiple-valued
function.
Any two values of the complex argument differ by an exact integer multiple of
2π. To remove this ambiguity and to get a unique representation, a conventional
choice is to limit θ to an interval of length 2π.
We shall denote by arg z such a value of θ and we shall call it the principal or
the reduced argument of the complex number z. Also, we shall denote by
Arg z = {arg z + 2kπ | k ∈ Z}
the class of all the arguments of a given complex number z ∈ C∗ . Alternatively, we
shall use the following notation:
Arg z = arg z (mod 2π).
Definition 1.6 Any complex number z ∈ C∗ can be represented in the following
trigonometric form:
z = |z| (cos θ + i sin θ) ,
with θ ∈ Arg z.
Here, the argument must be understood as being a multi-valued function, i.e.
Arg : C∗ → P(R).
There is no standard choice for such an interval of length 2π for the reduced argument, the appropriate one depending on the problem we address. Still, it is
customary to use one of the following two typical conventions for specifying the
principal argument: either we can consider that −π &lt; arg z ≤ π or 0 ≤ arg z &lt; 2π.
17
COMPLEX NUMBERS
If we adopt the first convention, the principal argument arg z is defined by

y

arctan , x &gt; 0, y ∈ R,


x






y


arctan + π, x &lt; 0, y &gt; 0,


x






 arctan y − π, x &lt; 0, y &lt; 0,
x
arg z =





π/2, x = 0, y &gt; 0,








−π/2, x = 0, y &lt; 0,







π, x &lt; 0, y = 0.
In what follows, unless otherwise mentioned, we shall adopt the first convention.
So, for any complex number z ∈ C∗ , its principal argument will be the unique real
number θ ∈ (−π, π] given by (1.5). We shall write
z = |z| (cos θ + i sin θ).
Remark 1.7 Let us notice that arg is not a continuous function: it has a discontinuity along the negative real axis.
Remark 1.8 Using Euler’s formula
ei θ = cos θ + i sin θ,
(1.6)
we can write any complex number z ∈ C∗ in the so-called exponential or polar form:
z = r ei θ ,
(1.7)
where r = |z| and θ ∈ Arg z.
From Euler’s formula, we obtain immediately the following identities:
cos θ =
eiθ + e−iθ
,
2
sin θ =
eiθ − e−iθ
.
2i
18
COMPLEX ANALYSIS
If we take θ = π in Euler’s formula (1.6), we obtain a famous and amazing identity
eiπ + 1 = 0,
connecting several important constants in mathematics.
Remark 1.9 The modulus of a complex number is a multiplicative function, i.e.
|z1 z2 | = |z1 | |z2 |,
arg (z1 z2 ) = arg z1 + arg z2 ,
up to integer multiples of 2π, i.e. we can write
Arg (z1 z2 ) = Arg z1 + Arg z2 .
Also, the modulus is invariant under conjugation, i.e.
|z| = |z|
and the argument changes sign:
arg z = − arg z,
up to integer multiples of 2π.
Let us recall now two formulas for computing powers and roots of complex numbers. We start with the so-called de Moivre’s formula:
z n = r n [cos (nθ) + i sin (nθ)] ,
for n ∈ Z.
Also, let us turn our attention to the root extraction rule. Let a ∈ C∗ and n ∈ N,
with n ≥ 2. We consider the equation
z n = a,
(1.8)
19
COMPLEX NUMBERS
with a = ρ (cos θ + i sin θ) and z = r (cos φ + i sin φ). The equation (1.8) has n
complex roots:
zk = ρ1/n
cos
θ + 2kπ
θ + 2kπ
+ i sin
n
n
,
k = 0, n − 1.
So, there are n roots of the n-th order of any complex number z ∈ C∗ . They have
the same modulus and their arguments are equally spaced. From a geometric point
of view, they represent the vertices of a regular polygon with n sides.
Remark 1.10 In many problems, it is necessary to extend the complex number
system C by introducing a symbol ”∞” to represent infinity. So, adding an improper
number denoted by ”∞”, we define C∞ = C ∪ {∞},
∞ ∈
/ C. The convention
|∞| = +∞ extends the notion of absolute value from C to C∞ . Also, we define:
a + ∞ = ∞ + a = ∞,
for a ∈ C,
a &middot; ∞ = ∞ &middot; a = ∞,
for a ∈ C∞ \ {0},
a
a
= ∞, for a ∈ C∞ \ {0},
= 0, for a ∈ C.
0
∞
∞ 0
We shall not define:
, , 0 &middot; ∞, ∞ + ∞.
∞ 0
The extended complex numbers do not form a field.
Definition 1.11 A sequence (zn )n is said to converge to infinity (we shall denote
lim zn = ∞) if
n→∞
lim |zn | = ∞,
n→∞
i.e. for any R &gt; 0, there exists NR ∈ N such that |zn | &gt; R, for any n ≥ NR .
Definition 1.12 If C is identified with the Euclidean plane, then C∞ is called the
extended complex plane. We call z = ∞ the point at infinity.
Remark 1.13 The set C ∪{∞} can be visualized as a sphere (the Riemann sphere or
the extended complex plane). We consider the Euclidean space R3 , with coordinates
(X, Y, Z) and the XY -plane identified with C. Let S be the sphere centered at the
20
COMPLEX ANALYSIS
point (0, 0, 1/2) and with radius 1/2. The sphere S has the diameter equal to one and
touches the complex plane at the point O = (0, 0, 0). We denote by N = (0, 0, 1) the
north pole of the sphere, i.e. the antipode of O. For each point z ∈ C, let M be the
point obtained by the intersection with the sphere of the segment joining z and M . To
a sequence (zn )n converging to the point at infinity it corresponds a sequence of points
on S, converging to N . Therefore, the ”image” of z = ∞ is N . This correspondence
between the points of the extended plane and those on the sphere S is one-to-one
and is called the stereographic projection. S is said to be Riemann’s sphere. The
stereographic projection is a conformal mapping. Therefore, if we identify, via the
stereographic projection, the points in the complex plane with the points on S \ {N }
and, further, ∞ with N , we get a bijection between the extended complex plane C∞
and S. Such a construction maps the unbounded set C into the compact set S by
adding one point, the point at infinity. The Riemann sphere is also called the onepoint compactification of C. The Riemann sphere has many useful applications in
various fields, such as physics, cartography, crystallography, quantum mechanics or
geology.
Remark 1.14 The usual topology we shall consider on C will be the one induced by
the metric
d : C &times; C → R+ ,
defined, for any z1 , z2 ∈ C, by
d (z1 , z2 ) = |z1 − z2 | .
This topology coincides with the one induced by the Euclidean metric on R2 . Here,
| &middot; | : C → R+ is the complex modulus, which is taken to be the standard norm on the
real vector space C. If we consider the Euclidean space (R2 , | &middot; |), endowed with the
classical Euclidean norm, then there exists an isomorphism of vector normed spaces
which allows us to identify the normed spaces (R2 , k &middot; k) and (C, | &middot; |).
Definition 1.15 Let z0 ∈ C and r ∈ R, with r &gt; 0. The set
B(z0 , r) = {z ∈ C | |z − z0 | &lt; r}
21
COMPLEX NUMBERS
is called the open ball or the open disk centered at the point z0 and having the radius
r.
Definition 1.16 Let z0 ∈ C and r ∈ R, with r ≥ 0. The set
B(z0 , r) = {z ∈ C | |z − z0 | ≤ r}
is called the closed disk centered at the point z0 and having the radius r. If in the
above definition we take r = 0, then B(z0 , 0) = {z0 }.
Remark 1.17 We shall sometimes denote the open disk B(z0 , r) by U (z0 ; r). Also,
we shall denote by U̇ (z0 ; r) the set U (z0 ; r) \ {z0 } and we shall call it the punctured
disk centered at z0 and with radius r:
U̇ (z0 ; r) = {z ∈ C | 0 &lt;| z − z0 |&lt; r}.
Definition 1.18 For z0 ∈ C and r, R ∈ R, with 0 ≤ r &lt; R ≤ +∞, the set
U (z0 ; r, R) = {z ∈ C | r &lt; |z − z0 | &lt; R}
is called the open annulus centered at z0 and with radii r and, respectively, R.
Let us notice that
U (z0 ; 0, R) = U̇(z0 ; R),
U (z0 ; r, ∞) = C \ U (z0 ; r),
U (z0 ; 0, ∞) = C \ {z0 }.
Definition 1.19 Let z0 ∈ C and r &gt; 0. The set
C(z0 , r) = {z ∈ C | |z − z0 | = r}
is called the circle centered at the point z0 and having the radius r.
Example 1.20 As a concrete example, we can consider the circle with center at the
point z0 = (1, 1) and radius two, which is the locus of all the points z = (x, y) ∈ C
such that
(x − 1)2 + (y − 1)2 = 22
or, alternatively, the set of all the points z ∈ C with
|z − z0 | = 2.
22
COMPLEX ANALYSIS
Definition 1.21 If z1 , z2 ∈ C, the line segment joining z1 and z2 is defined by
[z1 , z2 ] = {z ∈ C | z = (1 − t)z1 + tz2 , 0 ≤ t ≤ 1}.
A polygonal line is a piecewise-smooth curve which consists of finitely many straight
line segments. We shall say that a set A ⊆ C is polygonally connected if for any
two given points a, b ∈ A there exists a polygonal path lying in A and having the
endpoints a and, respectively, b.
Remark 1.22 A nonempty open set A ⊆ C is connected if and only if it is polygonallyconnected.
Definition 1.23 A set A ⊆ C is called convex if for any pair of points a, b ∈ A, the
whole segment [a, b] is contained in A.
Definition 1.24 An open connected set in C is called a domain or a region.
Definition 1.25 A domain A ⊆ C is said to be star-shaped or a star-like domain
if there exists a point a ∈ A such that the line segment [a, z] ⊆ A, for all z ∈ A.
Such a point a is called a star-center.
Remark 1.26 Every convex set in C is connected. Also, any convex domain is a
star-like one. The converse is false, in general.
Definition 1.27 A domain D ⊆ C is said to be simply connected if every closed
path lying in D can be continuously deformed into a point without leaving D.
Exercise 1.28 Prove that the set of all matrices of the form
a b
,
−b a
with a, b ∈ R, endowed with the matrix addition and multiplication, is isomorphic
to the field of complex numbers.
Exercise 1.29 Let z1 , z2 ∈ C. Prove that
z1 = z2 if and only if Re z1 = Re z2 and Im z1 = Im z2 .
23
COMPLEX NUMBERS
Exercise 1.30 Show that Re (iz) = − Im z and Im (iz) = Re z.
Exercise 1.31 Prove that z = z if and only if z is a real number.
Exercise 1.32 Let z ∈ C. Prove that (z) = z.
1
1
Exercise 1.33 Show that
= , for any z ∈ C, with z 6= 0.
z
z
Exercise 1.34 Prove that z1 + z2 = z1 + z2 and z1 z2 = z1 z2 , for any z1 , z2 ∈ C.
Exercise 1.35 Show that Re z =
z+z
z−z
and Im z =
, for any z ∈ C.
2
2i
Exercise 1.36 Prove that |z1 z2 | = |z1 | |z2 |, for any z1 , z2 ∈ C.
Exercise 1.37 Let z1 , z2 ∈ C, with z2 6= 0. Show that |z1 /z2 | = |z1 |/|z2 |.
Exercise 1.38 Prove that |z| = |z| and |z|2 = z z, for any z ∈ C.
Exercise 1.39 Let z1 , z2 ∈ C. Show that ||z1 | − |z2 || ≤ |z1 − z2 |.
Exercise 1.40 Prove that |Re z| ≤ |z| ≤ |Re z| + |Im z| ≤
Exercise 1.41 Show that |Im z| ≤ |z| ≤ |Re z| + |Im z| ≤
√
2 |z|, for any z ∈ C.
√
2 |z|, for all z ∈ C.
Exercise 1.42 Let z1 , z2 ∈ C. Prove that
|z1 + z2 |2 + |z1 − z2 |2 = 2 (|z1 |2 + |z2 |2 ).
Exercise 1.43 Find the real and the imaginary parts of the complex number (1 + i)100 .
Solution. We notice that (1 + i)2 = 2 i. Thus, (1 + i)100 = 250 i50 = −250 .
Exercise 1.44 Compute the absolute value of the complex number z = 1 + 3 i.
Solution. It is easy to see that |z| =
√
10.
24
COMPLEX ANALYSIS
Exercise 1.45 Find the absolute value and the conjugate of the following complex
numbers:
a) z =
3+i
;
2 + 3i
b) z = (1 + i)4 ;
c) z = (2 + i) (7 − i) .
Exercise 1.46 Express the complex number z = (1 + i) /i in algebraic form.
Solution. It is easy to see that z = 1 − i.
Exercise 1.47 Write the complex number z = 1 − i in trigonometric form.
Solution. If we adopt the convention that the principal argument of a complex
number z ∈ C∗ is the unique argument that lies on the interval (−π, π], it is not
difficult to see that the complex number z = 1 − i can be written as:
π
π i
√ h
1 − i = 2 cos −
+ i sin −
.
4
4
Exercise 1.48 Write the number
1+i
z=√
3+i
in trigonometric form.
Solution. It is easy to see that the trigonometric forms of 1 + i and
√ h
π
πi
1 + i = 2 cos + i sin
4
4
and, respectively,
It follows that
√
h
π
πi
3 + i = 2 cos + i sin
.
6
6
√ h
2
π
πi
z=
cos
+ i sin
.
2
12
12
Exercise 1.49 Let a ∈ R and n ∈ N. Prove that
z = (1 + ai)n + (1 − ai)n
is real.
√
3 + i are
25
COMPLEX NUMBERS
Exercise 1.50 Compute the absolute values of
z1 = i(2 + i)(1 − 4i),
z2 =
Exercise 1.51 Compute (1 + i)16 and (1 +
√
(3 + 2i)(1 − i)
.
(2 + i)(3 − 4i)
3i)5 .
Exercise 1.52 Show that if the sum and the product of two given complex numbers
are both real, then either the numbers are real or they are complex conjugated.
Solution. Let
z1 = a + i b,
z2 = c + i, d.
Then, we get b = −d and ad + bc = 0, which implies that either b = d = 0 or a = c
and b = −d.
Exercise 1.53 Find all the roots of the equation z 4 + 1 = 0.
Exercise 1.54 Let z = 3 + 4i. Compute z 2 + 3z + 7.
Exercise 1.55 Find the cube roots of z = −27 and represent them on the Argand
diagram.
Exercise 1.56 Solve the equation
z 4 − 4z 2 + 4 − 2i = 0.
Exercise 1.57 Solve the equation
z 4 = −16.
Exercise 1.58 Sketch the following sets in the complex plane:
n
o
a) A = z ∈ C : |z − 2 + i| ≤ 3 ;
b) B = {z ∈ C : |z| = |z + 1|}.
Exercise 1.59 Prove that the sets D1 = {z ∈ C | |z| &lt; 1} and D2 = {z ∈ C | 1 &lt;
|z| &lt; 2} are open sets.
26
COMPLEX ANALYSIS
Exercise 1.60 Show that the upper half-plane, i.e. the set of all the complex
numbers with positive imaginary part,
D = {z ∈ C | Im z &gt; 0},
is an open set.
Exercise 1.61 Let A1 , A2 , . . . , An be open sets in C. Prove that
n
T
Ai is an open
i=1
set, too.
Exercise 1.62 Let {Ai }i∈I be an arbitrary family of open sets in C. Prove that
S
Ai is also an open set.
i∈I
Exercise 1.63 Show that any closed ball B(z0 , r) is a closed set.
Exercise 1.64 Let B1 , B2 , . . . , Bn be closed sets in C. Prove that
n
S
Bi is also a
i=1
closed set.
Exercise 1.65 Let {Bi }i∈I be an arbitrary family of closed sets in C. Prove that
T
Bi is also a closed set.
i∈I
Exercise 1.66 Any nonempty set X can be endowed with a metric. In particular,
on C, we can define the so-called discrete metric by putting, for any z, w ∈ C,
d0 (z, w) =
Show that d0 is indeed a metric on C.

 1,

0,
z 6= w,
z = w.
Exercise 1.67 Prove that the map d1 : C &times; C → R+ , given by
d1 (z1 , z2 ) =
is a distance on C.
|z1 − z2 |
,
1 + |z1 − z2 |
27
COMPLEX NUMBERS
Exercise 1.68 If z1 , . . . , zn and w1 , . . . , wn ∈ C, prove that
n
X
2
zi w i
i=1
≤
n
X
i=1
2
|zi |
!
n
X
i=1
2
|wi |
!
(Cauchy’s inequality).
Exercise 1.69 If z = (z1 , . . . , zn ), show that the following maps are norms on Cn :
a) kzk1 =
n
X
i=1
|zi |;
v
u n
uX
b) kzk2 = t
|zi |2 ;
i=1
c) kzk∞ = max |zi |.
1≤i≤n
The norm k &middot; k2 is usually denoted by k &middot; k.
Exercise 1.70 Let z, w ∈ C. Prove that
kz + wk2 + kz − wk2 = 2 kzk2 + kwk2
and
kz − wk2 + kz + w̄k2 = −4 (Re z) (Re w) .
Exercise 1.71 Prove that the map d : C &times; C → R+ , given by
d(z1 , z2 ) = p
2 |z1 − z2 |
p
,
1 + |z1 |2 1 + |z2 |2
(1.9)
is a distance. If z1 and z2 are finite points in C, the distance between their stereographic projection is given by (2.9). This distance is called the chordal or the
spherical distance between the points z1 and z2 . If z2 = ∞, the corresponding
distance is given by
d(z1 , ∞) = p
2
1 + |z1 |2
.
Exercise 1.72 Describe the sets of the points z ∈ C∗ satisfying the relation 1/z = z̄.
28
COMPLEX ANALYSIS
Solution. It is not difficult to see that we get the unit circle.
Exercise 1.73 Describe the sets of the points z ∈ C satisfying the relation |z| =
Re z + 1.
Solution. Since
parabola.
p
x2 + y 2 = x + 1, we are led to y 2 = 2 x + 1. Thus, we get a
Exercise 1.74 Draw the set of all the points z ∈ C satisfying the condition Re (z) &gt;
3.
Exercise 1.75 Draw the set of all the points z ∈ C satisfying the condition 1 &lt;
Im (z) &lt; 3.
Exercise 1.76 Draw the set of all the points z ∈ C satisfying the condition Re (z) &gt;
3.
Exercise 1.77 Draw the set of all the points z ∈ C satisfying the condition |z −2| ≤
|z + 2|.
Exercise 1.78 Solve the following equations:
a) z 3 = 1;
b) z 4 = 81 i.
Exercise 1.79 Describe the following sets:
a) A = {z ∈ C : Im(z) &gt; 2};
b) B =
ni
n
o
: n ∈ N∗ .
Exercise 1.80 Decide if the following sets are open, closed, bounded or compact:
a) A = {z ∈ C : 0 &lt; Re(z) ≤ 2, Im(z) = 0};
b) B = {z ∈ C : 0 ≤ Re(z) ≤ 2, Im(z) = 0}.
Exercise 1.81 Let A, B ⊆ C. Prove the following properties of usual operations
over sets:
1. Int A ⊆ A;
29
COMPLEX NUMBERS
2. if A ⊆ B, then Int A ⊆ Int B;
3. Int (A ∩ B) = ( Int A) ∩ ( Int B);
4. Int (A ∪ B) ⊇ ( Int A) ∪ ( Int B);
5. the interior of A is the largest open set contained in A, i.e.
Int A =
[
{ D | D ⊆ A, D open};
6. A is open if and only if A = Int A;
7. Int A is an open set;
′
8. A ⊆ A and A ⊆ A;
′
′
9. if A ⊂ B ⇒ A ⊆ B and A ⊆ B ;
′
′
′
′
′
′
10. A ∪ B = A ∪ B and (A ∪ B) = A ∪ B ;
11. A ∩ B ⊆ A ∩ B and (A ∩ B) ⊆ A ∩ B ;
12. A = A;
′
13. A = A ∪ A ;
14. the set A is the smallest closed set which contains A, i.e.
A=
\
{F | F closed, F ⊇ A};
15. A is a closed set;
16. A = A ∪ ∂A;
17. ∂A = A \ Int A;
18. Int A = A \ ∂A;
19. A is closed if and only if A = A;
20. ∂A ∩ Int A = ∅.
Exercise 1.82 Show that D1 = {z ∈ C | |z| &lt; 2} is a connected set.
Exercise 1.83 Prove that an open set is pathwise connected if and only if it is
connected.
30
COMPLEX ANALYSIS
Exercise 1.84 Let A ⊆ C be a connected set and B ⊆ C a set such that A ⊆ B ⊆ A.
Prove that B is connected, too.
Exercise 1.85 If A ⊆ C is a connected set, then A is also connected.
Exercise 1.86 Let (Ai )i∈I be a family of connected sets in C such that
T
i∈I
Prove that
A=
[
Ai 6= ∅.
Ai
i∈I
is a connected set.
Exercise 1.87 Show that the set of all the points z ∈ C with 2 &lt; |z| &lt; 3 is open
and connected, but not simply connected.
Exercise 1.88 Show that the punctured plane C \ {0} is not simply connected, but
the slit plane D = C \ {(−∞, 0]} is a simply connected domain.
Exercise 1.89 Prove that a star-shaped domain in C is simply connected.
Exercise 1.90 Let z0 ∈ C and 0 &lt; r &lt; R. Prove that the annulus U (z0 ; r, R) is
not a star-domain.
Exercise 1.91 Show that every path-connected set is connected. Find a connected
set which is not path-connected.
1.2
Sequences and Series of Complex Numbers
Definition 1.92 Let (zn )n ⊆ C. The sequence (zn )n is called convergent if there
exists z ∈ C such that, for any V ∈ V(z), there exists nV ∈ N such that zn ∈ V , for
any n ≥ nV . We shall use the standard notation z = lim zn .
n→∞
Theorem 1.93 A convergent sequence in C has only one limit.
Proposition 1.94 Let (zn )n be a convergent sequence in C and A ⊆ C. If there
exists n0 ∈ N such that zn ∈ A, for any n ≥ n0 , then
lim zn ∈ A.
n→∞
31
COMPLEX NUMBERS
Theorem 1.95 Let (zn )n be a sequence in C and z ∈ C. The following two assertions are equivalent:
(i) z = lim zn ;
n→∞
(ii) for any ε &gt; 0, there exists nε ∈ N such that |zn − z| &lt; ε, for any n ≥ nε .
Definition 1.96 A sequence (zn )n in C is called a fundamental or a Cauchy sequence if, for any ε &gt; 0, there exists nε ∈ N such that
|zn − zm | &lt; ε, for any n, m ≥ nε .
Proposition 1.97 A convergent sequence in C is also a Cauchy sequence.
Remark 1.98 The converse implication in Proposition 2.51 is, in a general metric
space, not true. However, in (C, |&middot;|), any Cauchy sequence is convergent. So, (C, |&middot;|)
is a complete space.
Remark 1.99 Any Cauchy sequence in C is bounded.
Remark 1.100 If a Cauchy sequence (zn )n possesses a subsequence (znk ) which is
convergent to z ∈ C, then (zn )n is convergent to z.
Definition 1.101 Let (zn )n≥0 ⊆ C. If
Sn =
n
X
zi ,
(1.10)
i=0
then the pair ((zn )n≥0 , (Sn )n≥0 ) is called a series (the series associated to the sequence (zn )n≥0 ).
Instead of the above pair notation, we shall use a simpler one:
X
zn .
(1.11)
n≥0
Sn is called the nth partial sum of the series (2.11), while zn is said to be its general
term.
32
COMPLEX ANALYSIS
Definition 1.102 If there exists S ∈ C such that
S = lim Sn ,
n→∞
then the series (2.11) is called convergent. In this case, S is said to be its sum.
If the series
X
zn converges to S, we shall write
n≥0
S=
∞
X
zn .
n=0
Definition 1.103 A series which is not convergent is called divergent.
Definition 1.104 A series
ries
X
n≥0
X
zn is called absolutely convergent if the number se-
n≥0
|zn | is convergent (in R).
Definition 1.105 A series which is convergent, but not absolutely convergent, is
called semi-convergent or conditionally convergent.
Proposition 1.106 If the series
X
zn is convergent, then the sequence (zn )n is
n≥0
convergent and
lim zn = 0.
n→∞
Proposition 1.107 (Cauchy’s Criterion) The series
X
zn is convergent in C if
n≥0
and only if, for any ε &gt; 0, there exists Nε ∈ N such that
|zn+1 + &middot; &middot; &middot; + zn+p | &lt; ε, for any n ≥ Nε and p ≥ 1.
(1.12)
Proposition 1.108 (The Geometric Series) For z ∈ C with |z| &lt; 1, the series
P n
z is convergent and
n≥0
∞
X
n=0
zn =
1
.
1−z
33
COMPLEX NUMBERS
Proposition 1.109 (Comparison Test) Let (zn )n ⊆ C and (wn )n ⊆ R+ . We assume that there exists n0 ∈ N such that
|zn | ≤ wn , for any n ≥ n0 .
If the series
X
wn is convergent, then the series
n≥0
X
zn is absolutely convergent.
n≥0
Proposition 1.110 (Abel-Dirichlet) Let (zn )n ⊆ C and (wn )n ⊆ R+ . We assume
that there exists M &gt; 0 such that
n
X
i=0
zi ≤ M, for any n ≥ 0.
If the sequence (wn )n is nonincreasing and
lim wn = 0,
n→∞
then the series
X
zn wn is convergent.
n≥0
Let us give now a very powerful convergence test, called Cauchy’s Test or the
Root Test.
Theorem 1.111 Let (zn )n≥1 ⊆ C and
L = lim sup
n→∞
Then, we have:
(i) if L &lt; 1, the series
X
p
n
|zn |.
zn is absolutely convergent;
n≥1
(ii) if L &gt; 1, the series
X
zn is divergent.
n≥1
Corollary 1.112 If there exists
L = lim
n→∞
p
n
|zn |,
34
COMPLEX ANALYSIS
then:
(i) if L &lt; 1, the series
X
zn is absolutely convergent;
n≥1
(ii) if L &gt; 1, the series
X
zn is divergent;
n≥1
(iii) if L = 1, we cannot decide upon the nature of the series
X
zn .
n≥1
We recall now another powerful test, called D’Alembert’s Test or the Ratio Test.
Theorem 1.113 Let (zn )n≥0 ⊆ C.
(i) If lim sup
n→∞
(ii) If lim inf
n→∞
X
zn+1
zn is absolutely convergent.
&lt; 1, then the series
zn
n≥0
X
zn+1
zn is divergent.
&gt; 1, then the series
zn
n≥0
Corollary 1.114 If there exists
L = lim
n→∞
zn+1
,
zn
then:
(i) if L &lt; 1, the series
X
zn is absolutely convergent;
n≥0
(ii) if L &gt; 1, the series
X
zn is divergent;
n≥0
(iii) if L = 1, we cannot decide upon the nature of the series
X
zn .
n≥0
Definition 1.115 Let
sider the series
X
X
zn and
n≥0
tn , with
X
wn be series of complex numbers. Let us con-
n≥0
n≥0
tn =
X
zi wj .
i+j=n
This series is called the Cauchy product series of the series
X
n≥0
zn and
X
n≥0
wn .
35
COMPLEX NUMBERS
X
Theorem 1.116 (Mertens) Let
zn and
n≥0
X
wn be convergent series of complex
n≥0
numbers. If at least one of the two series is absolutely convergent, then the Cauchy
X
product series
tn is convergent and
n≥0
∞
X
tn =
n=0
Theorem 1.117 (Cauchy) If
∞
X
zn
n=0
X
zn and
n≥0
!
X
∞
X
wn
n=0
!
.
wn are absolutely convergent series of
n≥0
complex numbers, then their Cauchy product series
X
tn is absolutely convergent,
n≥0
too.
Exercise 1.118 Let (zn )n ⊆ C. Prove that
Re zn → Im z,
zn → z if and only if
Im zn → Im z.
Exercise 1.119 Let (zn )n , (wn )n ⊆ C such that
zn → z, wn → w,
with z, w ∈ C. If wn , w 6= 0, show that
z
zn
→ .
wn
w
Exercise 1.120 If the sequence (zn )n converges to a limit z, prove that (zn )n converges to z.
Exercise 1.121 Find the limit of the sequence zn = in!+1 .
1+i n
Exercise 1.122 Prove that lim
= 0.
n→∞
2
Exercise 1.123 Let (zn )n ⊆ C, with zn = xn + iyn . Show that
X
n≥0
zn is convergent if and only if
X
n≥0
xn and
X
n≥0
yn are convergent.
36
COMPLEX ANALYSIS
Exercise 1.124 Show that the sequence zn = (1 + i)n is divergent.
Exercise 1.125 Find the limit of the sequence
zn =
√
n + 2 (n + 1) i
,
n
n ≥ 1.
Solution. Since zn = xn + i yn , with
xn =
√
n
,
n
yn =
2(n + 1)
,
n
we get lim zn = 2 i.
n→∞
Exercise 1.126 Prove that
(n + i)(1 + n i)
= i.
n→∞
n2
lim
Exercise 1.127 Prove that for any z ∈ C, the series
X
(−1)n
n≥0
z 2n
(2n)!
is absolutely convergent.
Exercise 1.128 Prove that for any z ∈ C, the series
X
(−1)n
n≥0
z 2n+1
(2n + 1)!
is absolutely convergent.
Exercise 1.129 Compute the sum of the series
X i n−1
n≥1
Exercise 1.130 Test the series
4
X in
for convergence.
3n
n≥0
.
37
COMPLEX NUMBERS
Solution. Using the ratio test, we get
lim
n→∞
zn+1
1
= .
zn
3
Hence, by the ratio test, the series converges absolutely.
Exercise 1.131 Test the following series for convergence:
a)
X
n≥1
X
1
;
2 + in
1
;
+ in
n≥1
X 3 + i n
√
c)
.
10
n≥1
b)
n2
Exercise 1.132 Show that the series
X 1 + i n (−1)n
n2
n≥1
is convergent.
X 1
X (−1)n
and
are convergent, it
2
n
n
Solution. If we recall that the real series
n≥1
n≥1
follows that the given complex series is convergent, as well.
Exercise 1.133 Show that the series
X 1 + in
n2
n≥1
Solution. If we recall that the real series
is divergent.
X1
is divergent, it follows that the given
n
n≥1
complex series is divergent, too.
Exercise 1.134 Show that the series
X
(1 + i)n is divergent.
n≥1
Solution. Notice that lim zn 6= 0.
n→∞
Exercise 1.135 Prove that the series
X (3 + 4 i)n
n≥1
5n n4
is convergent.
38
COMPLEX ANALYSIS
Solution. Since |zn | = 1/n4 , using the comparison test and the fact that the series
X 1
is convergent, it follows that the given series is convergent.
n4
n≥1
Exercise 1.136 Test the following series for convergence:
a)
b)
X 1 + i n
√
;
3
n≥1
X 1 n
n
.
i
n≥1
Chapter 2
Complex Functions
2.1
Functions of a Complex Variable
Definition 2.1 Let D be a nonempty set in C. A single-valued complex function
or, simply, a complex function f : D → C is a map that assigns to each complex
argument z = x + iy in D a unique complex number w = u + iv. We write
w = f (z).
The set D is called the domain of the function f and the set f (D) is the range or
the image of f .
So, a complex-valued function f of a complex variable z is a rule that assigns to
each complex number z in a set D one and only one complex number w. We call w
the image of z under f . If z = x + iy ∈ D, we shall write
f (z) = u(x, y) + iv(x, y)
or
f (z) = u(z) + iv(z).
The real functions u and v are called the real and, respectively, the imaginary part
of the complex function f . Therefore, we can describe a complex function with the
aid of two real functions depending on two real variables.
39
40
COMPLEX ANALYSIS
Example 2.2 The function f : C → C, defined by
f (z) = z 3 ,
can be written as f (z) = u(x, y) + iv(x, y), with u, v : R2 → R given by
u(x, y) = x3 − 3xy 2 ,
v(x, y) = 3x2 y − y 3 .
Example 2.3 For the function f : C → C, defined by
f (z) = ez ,
we have
u(x, y) = ex cos y,
v(x, y) = ex sin y,
for any (x, y) ∈ R2 .
Example 2.4 For the function
f (z) = |z|,
we have
u(x, y) =
p
x2 + y 2 ,
v(x, y) = 0,
which means that f is a real-valued function of a complex variable, its domain of
definition being C.
2.2
Limits of Functions
′
Definition 2.5 Let D ⊆ C, a ∈ D and f : D → C. A number l ∈ C is called a
limit of the function f at the point a if for any V ∈ V(l), there exists U ∈ V(a) such
that, for any z ∈ U ∩ D \ {a}, it follows that f (z) ∈ V . We shall use the notation
l = lim f (z).
z→a
Remark 2.6 If a complex function f : D → C possesses a limit l at a given point
a, then this limit is unique.
41
COMPLEX FUNCTIONS
′
Theorem 2.7 Let D ⊆ C, f : D → C and a ∈ D . The following statements are
equivalent:
(i) l = lim f (z);
z→a
(ii) for any ε &gt; 0, there exists δε &gt; 0 such that for any z ∈ D, z 6= a, with
|z − a| &lt; δε , it follows that |f (z) − l| &lt; ε;
(iii) for any (zn )n ⊆ D, zn 6= a, such that zn → a, it follows that f (zn ) → l.
Let us discuss now limits involving ∞. We notice that z → ∞ means |z| → ∞
and f (z) → ∞ means |f (z)| → ∞.
Definition 2.8 a) Let z0 ∈ C. We say that
lim f (z) = ∞
z→z0
if, for any M &gt; 0, there exists δ &gt; 0 such that, for any z with 0 &lt; |z − z0 | &lt; δ, we
have |f (z)| &gt; M .
b) We say that
lim f (z) = ∞
z→∞
if, for any M &gt; 0, there exists R &gt; 0 such that, for |z| &gt; R, we have |f (z)| &gt; M .
c) For a given l ∈ C,
lim f (z) = l
z→∞
means that for any ε &gt; 0, there exists R &gt; 0 such that, for any z with |z| &gt; R, we
have |f (z) − l| &lt; ε.
′
Exercise 2.9 Let f, g : D ⊆ C → C and a ∈ D . If lim f (z) and lim g(z) exist,
z→a
prove that
lim (f &plusmn; g)(z) = lim f (z) &plusmn; lim g(z);
z→a
z→a
z→a
lim (f g)(z) = lim f (z) lim g(z);
z→a
z→a
z→a
z→a
42
COMPLEX ANALYSIS
lim f (z)
f (z)
= z→a
,
z→a g(z)
lim g(z)
lim
z→a
provided that g(z) 6= 0 on D and lim g(z) 6= 0.
z→a
′
Exercise 2.10 Let f : D ⊆ C → C and a ∈ D . Show that

lim Re f (z) = Re l,

 z→a
lim f (z) = l if and only if
z→a

 lim Im f (z) = Im l.
z→a
Also, prove that
lim f (z) = l if and only if lim f (z) = l.
z→a
z→a
z
does not exist.
z→0 z
Exercise 2.11 Prove that lim
Solution. To prove that the above limit does not exist, we compute this limit as
z → 0 on the real and on the imaginary axis, respectively. In the first situation, i.e.
for z = x ∈ R, the value of the limit is 1. In the second situation, i.e. for z = i y,
with y ∈ R, the limit is −1. Thus, the limit depends on the direction from which we
approach 0, which implies that the limit does not exist.
Exercise 2.12 Compute the following limits:
a)
z 2 + 16
;
z→4i z − 4i
lim
z4 − 1
;
z→i z 2 + 1
b) lim
c)
lim (z 2 + 2).
z→1−i
Exercise 2.13 Find the value of the following limits:
z2 + 2
;
z→∞ z 2 + z − i
a) lim
2.3
z 3 + iz + 2
.
z→∞
z2 + i
b) lim
Continuous Functions
Definition 2.14 Let f : C → C and a ∈ C. The function f is said to be continuous
at the point a if for any V ∈ V(f (a)), there exists U ∈ V(a) such that f (U ) ⊆ V .
43
COMPLEX FUNCTIONS
Definition 2.15 The function f is said to be continuous on C if it is continuous at
each point a ∈ C.
Theorem 2.16 f : C → C and a ∈ C. The following statements are equivalent:
(i)
f is continuous at the point a;
(ii) for any ε &gt; 0, there exists δε &gt; 0 such that for any z ∈ C with |z − a| &lt; δε ,
it follows that |f (z) − f (a)| &lt; ε;
(iii) for any (zn )n ⊆ C such that zn → a, it follows that f (zn ) → f (a).
Definition 2.17 The function f : D ⊆ C → C is said to be uniformly continuous on
D if, for any ε &gt; 0, there exists δε &gt; 0 such that, for any z, w ∈ D, with |z −w| &lt; δε ,
it follows that |f (z) − f (w)| &lt; ε.
Exercise 2.18 (Heine-Cantor) Let K ⊆ C be a compact set and f : K → C be a
continuous map on K. Show that f is uniformly continuous on K.
Exercise 2.19 Let f, g : C → C. If f and g are continuous and α, β ∈ C, prove that
αf + βg, f g, f /g, for g 6= 0, are continuous, too.
′
Exercise 2.20 If f : D ⊆ C → C and a ∈ D , then f is continuous at the point a
if and only if the limit lim f (z) exists and equals f (a).
z→a
Exercise 2.21 If a is an isolated point for D, then f is continuous at a.
Exercise 2.22 Show that the function f : C → C, defined by f (z) = z 2 , is contin-
uous, but it is not uniformly continuous on C.
Exercise 2.23 Prove that the function f : C → C, defined by f (z) = 3z + 4, is
uniformly continuous on C.
Exercise 2.24 Let f : K ⊆ C → C be a continuous function. If K is compact,
show that f (K) is also compact. If K is connected, prove that f (K) is connected,
too.
44
COMPLEX ANALYSIS
Exercise 2.25 Compute the following limits:
z5 + 1
;
z→i z 2 + 1
a) lim
b) lim
z→0
1 − cos z
.
2 z2
Exercise 2.26 Show that:
z 2
2 z2
a) lim
doesn’t exist.
= 0; b) lim
z→0 z
z→0 z
Exercise 2.27 Test the following function for continuity
 2
z +9



, z 6= 3i,
z − 3i
f (z) =



4 + 6i, z = 3i.
Exercise 2.28 Test the following function for continuity


z3


, z 6= 0,
z Re z
f (z) =


 0, z = 0.
Chapter 3
Differentiable Functions
3.1
Holomorphic Functions
Let D ⊆ C be an open nonempty set, z = x + iy ∈ D and f : D → C be a complex-
valued function of a single complex variable z. In what follows, we shall use the
following notation:
f (z) = u(z) + iv(z)
or
f (z) = u(x, y) + iv(x, y),
where u and v are the real and, respectively, the imaginary part of the complex
function f . The function f can be considered either as a complex function depending
on a complex variable or as a complex function of two real variables.
Definition 3.1 Let D ⊆ C be an open nonempty set, f : D → C and z0 ∈ D. The
complex function f is said to be C-derivable at the point z0 (or, simply, derivable at
z0 ) if there exists
lim
z→z0
f (z) − f (z0 )
∈ C.
z − z0
(3.1)
′
This limit is called the derivative of f at the point z0 and is denoted by f (z0 ).
Definition 3.2 Let D ⊆ C be a nonempty open set, f : D → C and z0 ∈ D. The
function f is said to be R-differentiable at the point z0 = x0 + iy0 if there exist two
45
46
COMPLEX ANALYSIS
complex numbers A1 , A2 and a complex function f1 : D \ {z0 } → C such that
lim f1 (z) = 0
z→z0
and, for all z ∈ D \ {z0 }, we have
f (z) = f (z0 ) + A1 (x − x0 ) + A2 (y − y0 ) + f1 (z) |z − z0 |.
(3.2)
Remark 3.3 If f = u + iv is R-differentiable at z0 ∈ D, then it is not difficult to
see that f possesses first-order partial derivatives at z0 and we have

∂f
∂u
∂v


(z ) =
(z0 ) + i (z0 ) = A1 ,

 ∂x 0
∂x
∂x


∂f
∂u
∂v


(z0 ) =
(z0 ) + i (z0 ) = A2 .
∂y
∂y
∂y
Also, we can see that f is R-differentiable at z0 = x0 + iy0 if and only if u and v are
R-differentiable at (x0 , y0 ).
Remark 3.4 If we take

f (z) − f (z0 ) − A1 (x − x0 ) − A2 (y − y0 )


,

|z − z0 |
f1 (z) =



0, for z = z0 ,
for z ∈ D \ {z0 },
then we can obtain an equivalent definition of R-differentiability of a function f , but
with f1 continuous at z0 and f1 (z0 ) = 0.
Definition 3.5 Let D ⊆ C be a nonempty open set, f : D → C and z0 ∈ D. The
function f is said to be C-differentiable at the point z0 if there exists α ∈ C and a
complex function f1 : D\{z0 } → C such that lim f1 (z) = 0 and, for all z ∈ D\{z0 },
we have
z→z0
f (z) = f (z0 ) + α(z − z0 ) + f1 (z)(z − z0 ).
(3.3)
47
DIFFERENTIABLE FUNCTIONS
Remark 3.6 If we take

f (z) − f (z0 )


− α,

z − z0
f1 (z) =



0, for z = z0 ,
for z ∈ D \ {z0 },
then we can obtain an equivalent definition of C-differentiability of a function f , but
with f1 continuous at z0 and f1 (z0 ) = 0.
Remark 3.7 If f : D → C is derivable at z0 ∈ D, then f is continuous at z0 .
Proposition 3.8 A function f : D → C is derivable at z0 ∈ D if and only if it is
C-differentiable at z0 .
Theorem 3.9 (Cauchy-Riemann) Let D ⊆ C be a nonempty open set and let f :
D → C. The complex function f = u + iv is derivable at a point z0 ∈ D if and only
if f is R-differentiable at z0 and

∂u
∂v


(z0 ) =
(z0 ),

 ∂x
∂y
Under these conditions,



 ∂u (z0 ) = − ∂v (z0 ).
∂y
∂x
′
f (z0 ) =
∂f
(z0 ).
∂x
(3.4)
(3.5)
The relations (3.4) can be also written in the form:

∂v
∂u


(x , y ) =
(x0 , y0 ),

 ∂x 0 0
∂y



 ∂u (x0 , y0 ) = − ∂v (x0 , y0 ).
∂y
∂x
(3.6)
48
COMPLEX ANALYSIS
Remark 3.10 If f is derivable at z0 , using the Cauchy-Riemann equations, we get
′
the following expressions for f (z0 ):
′
f (z0 ) =
∂f
∂u
∂v
(z0 ) =
(z0 ) + i (z0 ) =
∂x
∂x
∂x
∂u
∂v
∂v
∂u
(z0 ) − i (z0 ) =
(z0 ) + i (z0 ) =
∂x
∂y
∂y
∂x
∂v
∂u
∂f
(z0 ) − i (z0 ) = −i (z0 ).
∂y
∂y
∂y
Also, we have:
′
f (z0 )
2
=
=
2 2 2 2
∂u
∂v
∂u
∂u
(z0 ) +
(z0 ) =
(z0 ) +
(z0 ) =
∂x
∂x
∂x
∂y
2 2 2 2
∂v
∂v
∂v
∂u
(z0 ) +
(z0 ) =
(z0 ) +
(z0 )
∂y
∂x
∂y
∂y
and
′
f (z0 )
2
=
∂u
∂v
∂u
∂v
(z0 ) (z0 ) −
(z0 ) (z0 ) = J(z0 ).
∂x
∂y
∂y
∂x
where J is the Jacobian, defined by
J(z0 ) =
∂(u, v)
(z0 ).
∂(x, y)
Remark 3.11 The Cauchy-Riemann equations can be written in other coordinate
systems. For instance, it is not difficult to see that in the system of coordinates given
by the polar representation z = r eiθ these equations take the following form:

∂u
1 ∂v



 ∂r = r ∂θ ,



 ∂v = − 1 ∂u .
∂r
r ∂θ
Alternatively, we can write these equations in the complex form
∂f
1 ∂f
=
.
∂r
ir ∂θ
(3.7)
DIFFERENTIABLE FUNCTIONS
49
Proposition 3.12 (Goursat) Let D be an open set in C, f : D → C, f = u + iv
and z0 ∈ D. If all the first order partial derivatives of u and v exist, are continuous
and satisfy the Cauchy-Riemann relations at z0 , then f is C-derivable at z0 .
Let us define now the following two differential operators (sometimes called the
formal derivatives of f at the point z or the Wirtinger derivatives):

∂f
1 ∂f
∂f


=
−i
,


 ∂z
2 ∂x
∂y


1 ∂f
∂f
∂f


=
+
i
.

∂z
2 ∂x
∂y
(3.8)
These expressions are only symbolic derivatives with respect to z and z.
Remark 3.13 Let D ⊆ C be an open set, f : D → C, f = u+ iv and z0 = (x0 , y0 ) ∈
D. Then, f is derivable at the point z0 if and only if u and v are R-differentiable
∂f
at (x0 , y0 ) and
(z0 ) = 0. In other words, the Cauchy-Riemann equations are
∂z
equivalent to
∂f
(z0 ) = 0.
∂z
Also, if f is complex differentiable at z0 , then, for any z0 ∈ C,
∂f
′
(z0 ) = f (z0 ).
∂z
The concept of C-differentiability at a given point is not enough for building an
interesting theory. Therefore, we shall be interested in dealing with functions which
are C-differentiable in a whole neighbourhood of a given point. A function f will be
called holomorphic at a point z0 ∈ C if it is C-differentiable in a neighbourhood of
z0 (this neighbourhood, usually, will remain unspecified). A function f will be said
to be holomorphic on a set D if it is C-differentiable in a neighbourhood of every
point in D. More precisely, we have the following definition.
Definition 3.14 Let D ⊆ C be an open nonempty set and f : D → C. The function
f is called holomorphic on D if it is C-differentiable at each point z0 ∈ D. The
50
COMPLEX ANALYSIS
function f is called holomorphic on an arbitrary set A ⊆ D if there exists an open set
D0 , with A ⊆ D0 ⊆ D, such that f is holomorphic on D0 . In particular, f : D → C
is holomorphic at a point z0 ∈ D if there exists r &gt; 0 such that B(z0 , r) ⊆ D and f
is holomorphic on B(z0 , r).
We shall use the following notation:
H(D) = {f : D → C | f is holomorphic on D}.
Definition 3.15 Let f : C → C. If f is holomorphic on C, then f is said to be an
entire (integer) function.
Example 3.16 The function f : C → C, defined by f (z) = z 3 , is holomorphic on
C. So, it is an entire function.
Example 3.17 The function f : C → C, defined by f (z) = z, is not C-differentiable
anywhere in C, but it is R-differentiable everywhere on C.
Example 3.18 The function f (z) = 1/z is holomorphic on any open set in C that
does not contain the origin.
Example 3.19 Any polynomial function P = P (z) with complex coefficients is an
entire function.
′
′
Remark 3.20 If f is derivable at the point z0 and f (z0 ) 6= 0, then arg f (z0 )
is the rotation angle of the tangent to a smooth path starting from z0 under the
′
transform f and |f (z0 )| is the linear deformation coefficient at this point (the linear
magnification ratio).
Remark 3.21 A function f is said to be holomorphic at infinity if the function
g(z) = f (1/z)
is holomorphic at z = 0. So, we can consider holomorphic functions in C∞ .
DIFFERENTIABLE FUNCTIONS
51
Remark 3.22 Holomorphic functions are sometimes called regular functions. Since,
as we shall see later, holomorphic functions are analytic, the terms analytic, holomorphic or regular are often used interchangeably.
We shall consider now multi-valued functions. As we shall see, a multi-valued
function will be a map that assigns a finite or an infinite nonempty subset of C for
each argument in its domain of definition.
Definition 3.23 Let D be an open nonempty subset of C. We denote by P(C) the
set of all the parts (subsets) of C. A complex map F : D → P(C) is said to be a
multi-valued function on D.
Definition 3.24 Any continuous map f : D0 ⊆ D → C, with f (z) ∈ F (z), for
z ∈ D0 , is called an uniform branch (or a determination) of the multi-valued function
F on D0 . A point z0 ∈ C is called a critical point for the multi-valued function
F : D → P(C) if there exists no r &gt; 0 such that on U̇r (z0 ) there exists a uniform
branch of the function F .
Example 3.25 We have already seen that the argument function
Arg : C∗ → P(C)
is multi-valued, i.e.
Arg z = {arg z + 2kπ | k ∈ Z} ,
where we choose the principal argument arg z in (−π, π]. In fact, Arg : C∗ → P(R).
The argument function has an infinite number of branches. Unfortunately, if z
travels along a closed path winding around the branch point z = 0, the argument
changes by an integer multiple of 2π and we move from one branch to another.
Intuitively, if we want a branch to be a continuous single-valued function, we must
avoid working on a domain in which there are closed paths going around the origin.
Therefore, we can define a branch cut in the Argand plane, i.e. a line or a path in the
complex plane that will play the role of a barrier not to be crossed and preventing z
52
COMPLEX ANALYSIS
from making a complete loop around any branch point. In such a way, the function
will remain single-valued. For instance, a standard choice is to take the non-positive
real axis as a branch cut for the argument function (the origin is a branch point).
If we set C0 = C \ {(−∞, 0]}, then arg : C0 → (−π, π) is a continuous map.
Moreover, if D is a domain in C0 , any uniform branch f : D → C for Arg z is of the
following form:
f (z) = arg z + 2kπ,
with k ∈ Z fixed.
Remark 3.26 Let us notice that the point 0 ∈ C is a critical point for the multivalued function Arg.
Remark 3.27 Our choice to consider the principal branch of the argument function
defined by the restriction −π &lt; arg z &lt; π is not unique. Still, this is the most
common convention made in the literature for the argument and for many other
related functions, such as the logarithm function or the power function. Instead of
working with this particular branch cut obtained by removing from the complex plane
the non-positive real axis, one could consider, for instance, any radial cut from the
origin to infinity. When dealing with other functions, we need to consider one or
even more branch cuts to ensure single-valued definitions for different branches, the
choice of such cuts being sometimes a very difficult task.
We shall define now the multi-valued logarithm function as being the ”inverse”
of the exponential function. The exponential function is periodic, with complex
period 2πi. The complex exponential is not a one-to-one map. We can divide the
complex plane into horizontal strips of height 2π in such a way that in each strip
the exponential function is one-to-one. More precisely, for z = x + iy ∈ C, we set
Sk = {z = x + iy ∈ C | x ∈ R, (2k − 1)π &lt; y ≤ (2k + 1)π},
k ∈ Z.
The sets Sk are called fundamental strips for the complex exponential function. It
follows that if z1 and z2 belong to the same strip Sk , then ez1 = ez2 implies that
z1 = z2 .
DIFFERENTIABLE FUNCTIONS
53
From the periodicity of the complex exponential or, equivalently, from the fact
that the argument function is multi-valued, we shall see that the complex logarithm
is a multi-valued function. Also, since there is no solution for the equation exp(z) =
0, we shall not define the complex logarithm for the point 0. We shall use the
notation ln for the real logarithm and, respectively, log, for the complex logarithm
(there is no confusion because we work only with logarithms to base e).
Definition 3.28 The complex logarithm function Log : C∗ → P(C) is a multi-valued
map defined by
Log z = {w ∈ C | ew = z} .
Then,
Log z = ln |z| + i Arg z.
(3.9)
The complex logarithm has an infinite number of branches. A standard choice is to
take the non-positive real axis as a branch cut for the logarithm function.
Let C0 = C \ {(−∞, 0]}. For z ∈ C0 and k ∈ Z, we set
logk (z) = ln |z| + i (arg z + 2kπ),
where −π &lt; arg z &lt; π. The map
logk : C0 → C
is a uniform branch of the logarithm function. For k = 0, we get the principal branch
of the logarithm
log z = ln |z| + i arg z.
(3.10)
Moreover, if, for any k ∈ Z, we consider the so-called fundamental regions for the
exponential function
Tk = {z = x + iy | x ∈ R, (2k − 1)π &lt; y &lt; (2k + 1)π},
then the continuous map logk : C0 → Tk defined by
logk (z) = ln |z| + i (arg z + 2kπ),
54
COMPLEX ANALYSIS
is a uniform holomorphic branch for the multi-valued logarithm function Log on C0
and for any z ∈ C0 and k ∈ Z, we have
′
(logk z) =
1
.
z
Example 3.29 For z = −5, we have
log (−5) = ln | − 5| + i arg (−5) = ln 5 + i π,
Log (−5) = ln 5 + i π(2k + 1),
k ∈ Z.
Example 3.30 For z = 1 + i, we have
√
π
log (1 + i) = ln 2 + i ,
4
π
√
Log (1 + i) = ln 2 + i
+ 2kπ , k ∈ Z.
4
Let us note that
log (z w) 6= log z + log w,
for any z, w ∈ C0 .
As a matter of fact, there are complex numbers z and w for which log (z w) is not
even defined (for example, we can take z = w = i). For this particular example, z w
moved into the domain of a different holomorphic branch. Still,
Log (z w) = Log z + Log w,
for any z, w 6= 0.
Remark 3.31 Let D be a simply connected domain with 1 ∈ D and 0 ∈
/ D. Then,
there exists in D a branch of the logarithm F (z), which is holomorphic in D and
for which eF (z) = z, for all z ∈ D. Each branch is an extension of the standard
logarithm defined for positive numbers.
Remark 3.32 Let us notice that we can make the logarithm function single-valued
in other regions of the complex plane by choosing a different branch for the argument
function. Typically, we can choose such a set as being the complement of a ray or
55
DIFFERENTIABLE FUNCTIONS
a path in the complex plane going from the origin (inclusive) to infinity in some
direction. In fact, such a line or a path which creates a domain of holomorphy for
a given function is called a branch cut and any point that lies on a branch cut is
called a branch point of that multi-valued function.
With the aid of the logarithm function, we can define complex powers of complex
numbers. Let α ∈ C. For any z 6= 0, we define the α-th power of z by
z α = eα Log z ,
i.e.
z α = eα [ln |z|+i (arg z+2kπ)] ,
k ∈ Z.
This is, in general, a multi-valued function. For each branch of the logarithm, we
obtain a branch of z α . Moreover, for any k ∈ Z, the map fk : C0 → C, defined by
fk (z) = eα logk (z) ,
is a holomorphic uniform branch of the map z α and
d
(z α ) = α z α−1 .
dz
Example 3.33 Let us compute 1i . Since Log 1 = ln 1 + 2kπi, i.e. Log 1 = 2kπi, we
get
1i = e−2kπ ,
k ∈ Z.
If the exponent α is an integer, we get the standard concept of a power function.
If α is rational, i.e. α = m/n, with m and n &gt; 0 integers with no common factors,
there are exactly n values of z α for each z ∈ C∗ . So, we have n branches of the
1
power function z α . In particular, for α = , with n ∈ N∗ , there are only n distinct
n
uniform branches of the multi-valued power function z 1/n . In this case, we write
√
n
z. If α is irrational or non-real, there are infinitely many branches of the complex
power function. If α &gt; 0, the power function is also defined for z = 0. Finally, if
α = 0, the power function is defined as being the constant 1.
56
COMPLEX ANALYSIS
Exercise 3.34 Show that the function f : C → C, defined by f (z) = z 2 , is C-
derivable at any point z0 ∈ C.
Solution. For z0 ∈ C, we have
lim
z→z0
z 2 − z02
= 2z0 .
z − z0
Therefore,
f ′ (z0 ) = 2z0 .
Exercise 3.35 Prove that the function f : C → C, defined by f (z) = z̄ is not
complex differentiable everywhere on C.
Exercise 3.36 Prove that the function f : C → C, given by f (z) = (z)2 is differentiable only at the point z = 0.
Exercise 3.37 Show that the exponential function exp : C → C defined by f (z) =
ex (cos y + i sin y), with z = x + i y, is analytic in C.
Exercise 3.38 Find the points at which the function f : C → C, given by f (z) =
x2 + i y 2 is complex differentiable.
Exercise 3.39 Find the points at which the function f : C → C, given by f (z) =
z Im z is complex differentiable.
Exercise 3.40 Find the points at which the function f : C → C, given by f (z) =
2xy − i (x + y)2 is complex differentiable.
Exercise 3.41 Prove that if f (z) and f (z) are both holomorphic in the region
D ⊆ C, then f is constant in D.
Exercise 3.42 Let D ⊆ C be an open nonempty set and z0 ∈ D. If f, g : D → C
are derivable at z0 , prove that:
′
′
′
1) f &plusmn; g is derivable at z0 and (f &plusmn; g) (z0 ) = f (z0 ) &plusmn; g (z0 );
57
DIFFERENTIABLE FUNCTIONS
′
′
′
2) f g is derivable at z0 and (f g) (z0 ) = f (z0 ) g(z0 ) + f (z0 ) g (z0 );
3) if g(z0 ) 6= 0, then f /g is derivable at z0 and
′
′
′
f (z0 ) g(z0 ) − f (z0 ) g (z0 )
f
(z0 ) =
.
g
g2 (z0 )
Exercise 3.43 Let D, E ⊆ C be open sets, z0 ∈ D and f : D → E, g : E → C. If f
is derivable at z0 and g is derivable at f (z0 ), show that g ◦ f is derivable at z0 and
we have
′
′
′
(g ◦ f ) (z0 ) = g (f (z0 )) f (z0 ).
Exercise 3.44 Show that the function f : C → C, defined by f (z) = z, does not
satisfy the Cauchy-Riemann equations.
Solution. Indeed, since
u(x, y) = x,
it follows that
v(x, y) = −y,
∂u
= 1,
∂x
while
∂v
= −1.
∂y
So, this function, despite the fact that it is continuous everywhere on C, it is Rdifferentiable on C, is nowhere C-derivable.
Exercise 3.45 Show that the function f (z) = ez satisfies the Cauchy-Riemann
equations.
Solution. Indeed, since
ez = ex (cos y + i sin y),
it follows that
u(x, y) = ex cos y,
v(x, y) = ex sin y
58
COMPLEX ANALYSIS
and
∂v
∂u
= ex cos y =
;
∂x
∂y
∂u
∂v
= −ex sin y = − .
∂y
∂x
Moreover, ez is complex derivable and, using (3.5), it follows immediately that its
complex derivative is ez .
Exercise 3.46 Prove that the function f : C → C, defined by
 5
z


z 6= 0,

|z|4
f (z) =



0, z = 0,
is continuous at z = 0, satisfies the Cauchy-Riemann equations at z = 0, but is not
C-derivable at z = 0.
Exercise 3.47 Prove that the function f : C → C, defined by f (z) =
p
|xy|, where
Exercise 3.48 Prove that the function f : C → C, defined by f (z) =
p
|xy|, where
z = x + i y, is not differentiable at the origin.
z = x + i y, satisfies the Cauchy-Riemann equations at the origin.
Exercise 3.49 Find the values of the complex numbers a, b and c for which the
function
f (z) = ax + by + i(cx + y)
is C-derivable on C.
Exercise 3.50 Let f be holomorphic on a domain D ⊆ C. Show that each of the
following conditions implies that f is constant on D:
(i) f ′ = 0 in D;
(ii) |f | is constant in D;
(iii) f is real-valued in D.
(iv) arg f is constant in D.
Are the above results still true if we replace D by any open subset of C?
59
DIFFERENTIABLE FUNCTIONS
2
Exercise 3.51 Let f (z) = z e−|z| . Find the points at which f is derivable and
compute its derivative at these points.
Exercise 3.52 Let f be an entire function such that its real and imaginary parts
u and v satisfy the condition u(z) v(z) = 2, for all z. Prove that f is constant.
Exercise 3.53 Find all the complex numbers for which the function f (z) = (z − 2)i
is holomorphic.
Exercise 3.54 Find the solutions to the following equations:
a) ez = π i;
3.2
b) sinh z = 0;
c) cos z = 0.
Harmonic Functions
Definition 3.55 A real function u : D ⊆ R2 → R of class C 2 (D) is said to be
harmonic on the domain D if it satisfies Laplace’s equation ∆ u = 0 in D, i.e., for
any (x, y) ∈ D,
∂2u ∂2u
+ 2 = 0.
∂x2
∂y
If f = u + iv is holomorphic on a domain D, then f is infinitely differentiable on
D, and, hence, so are u and v. In particular, u, v ∈ C 2 (D). However, here we only
assume this in order to state the following result.
Proposition 3.56 Let f (z) = u(x, y)+iv(x, y) be holomorphic on a domain D ⊆ C.
If u, v ∈ C 2 (D), then u and v are both harmonic functions on D.
Example 3.57 The function u : R2 → R, defined by u(x, y) = x2 + y 2 , is not
harmonic and, so, it cannot be the real part of a holomorphic function.
Definition 3.58 Two harmonic functions u and v defined on a domain D ⊆ C and
related through Cauchy-Riemann relations are called harmonic conjugated.
Remark 3.59 If f = u + iv is holomorphic, then u and v are harmonic conjugated.
60
COMPLEX ANALYSIS
Theorem 3.60 Let D ⊆ C be a simply connected domain and let u = u(x, y) be
a real-valued harmonic function on D. Then, there exists a real-valued function
v = v(x, y), defined up to an arbitrary constant, such that the complex function
f (z) = u(x, y) + iv(x, y)
is holomorphic on D. So, every harmonic function u(x, y) defined on a simply
connected domain is the real part of a complex-valued function f (z) = u(x, y) +
iv(x, y), defined for all z = x + iy ∈ D. In other words, for each function u =
u(x, y) that is harmonic in a simply connected domain D, there exists its harmonic
conjugated function v = v(x, y), defined up to an arbitrary constant term.
Exercise 3.61 Show that the function u : R2 → R, defined by u(x, y) = y 3 − 3x2 y,
is harmonic on R2 .
Solution. It is not difficult to see that
∂2u
= −6y;
∂x2
∂2u
= 6y.
∂y 2
So, ∆ u = 0.
Exercise 3.62 Prove that the function u : R2 → R, defined by u(x, y) = ex cos y, is
harmonic on R2 .
Solution. Since
it follows immediately that
 2
∂ u


= ex cos y,

 ∂x2

2


 ∂ u = −ex cos y,
∂y 2
∆ u = 0.
Exercise 3.63 Show that the function u : R2 → R, given by u(x, y) = e−x (x sin y − y cos y),
is harmonic on R2 .
61
DIFFERENTIABLE FUNCTIONS
Solution. It is not difficult to check that
 2
∂ u


= −e−x (2 sin y − x sin y + y cos y) ,

 ∂x2
Therefore,

2


 ∂ u = e−x (2 sin y − x sin y + y cos y) .
∂y 2
∆ u = 0.
Exercise 3.64 Find an entire function f : C → C for which
Re f (z) = x2 − y 2
and
f (0) = 3i.
Solution. We start by noticing that u : R2 → R, given by u(x, y) = x2 − y 2 , is
harmonic on R2 . Therefore, there exists v(x, y) such that f (z) = u(x, y) + iv(x, y)
is holomorphic on C. Let (x0 , y0 ) ∈ R2 . Using Cauchy-Riemann equations, we get
v(x, y) =
(x,y)
Z
−
∂u
∂u
dx +
dy + C1 ,
∂y
∂x
with C1 ∈ R,
(x0 ,y0 )
along any path joining the points (x0 , y0 ) and (x, y). Integrating along a particular
path consisting of two line segments parallel to the coordinate axes, we obtain
v(x, y) = 2xy + C, with C ∈ R.
So,
f (z) = x2 − y 2 + i (2xy + C), with C ∈ R,
i.e.
f (z) = z 2 + iC, with C ∈ R.
Imposing the condition f (0) = 3i, we get
f (z) = z 2 + 3i.
62
COMPLEX ANALYSIS
Exercise 3.65 Let u(x, y) = a x2 + b x y + c y 2 , where a, b, c ∈ R. Prove that u is
harmonic if and only if a = −c.
Exercise 3.66 Show that
4
∂ ∂
∂ ∂
=4
= ∆.
∂z ∂z
∂z ∂z
Exercise 3.67 Prove that the functions
a) u(x, y) = e−x sin y,
b) v(x, y) = sin x sinh y
are harmonic. Find the corresponding holomorphic function f = u + iv in each case.
Exercise 3.68 Prove that the functions
a) u(x, y) = x2 − y 2 + 2 x − 4 y,
b) v(x, y) = cosh y sin x
are harmonic. Find the corresponding holomorphic function f = u + iv in each case.
Exercise 3.69 Let v : R2 → R, defined by
v(x, y) = 3 + 4xy.
Find all the entire functions f : C → C such that Im f = v.
Exercise 3.70 Find the harmonic conjugates of the function v : R2 → R, defined
by
v(x, y) = x2 + 3xy − y 2 + x + 4y.
Exercise 3.71 Let u(x, y) = ln x2 + y 2 . Show that u is harmonic on C∗ .
Exercise 3.72 Prove that if f is holomorphic and nonzero in a region D ⊆ C, then
ln |f (x, y)| is harmonic in D.
Chapter 4
Complex Integration
4.1
The Complex Integral
Definition 4.1 A continuous function γ : [a, b] ⊆ R → C is called a path in C.
Remark 4.2 Sometimes, it is convenient to take γ : [0, 1] → C. Also, we can deal
with paths γ : [a, b] → C∞ .
The points γ(a) and γ(b) are said to be the endpoints of the path γ; γ(a) is called
the initial point (or the start point) of the path γ and γ(b) is said to be its terminal
point.
Definition 4.3 We say that a path γ : [a, b] → C lies in a set D ⊆ C if γ(t) ∈ D,
for all t ∈ [a, b].
We denote the image of the map γ by {γ} or by γ([a, b]), but we shall often prefer
to refer to either the function or the image of the path with the same symbol γ.
Definition 4.4 A path γ : [a, b] → C is called closed if γ(a) = γ(b).
′
Definition 4.5 A path γ : [a, b] → C is called smooth if γ is differentiable, γ is
′
continuous and γ (t) 6= 0. A path γ is called piecewise smooth if it is continuous
and splits into a finite number of smooth pieces with no common interior points.
63
64
COMPLEX ANALYSIS
Definition 4.6 A path without self-intersection points is said to be simple. A simple
closed path is called a Jordan path or, sometimes, a loop.
Theorem 4.7 (Jordan Curve Theorem) Let γ be the image of a simple closed path
in C. The set C \ γ has exactly two connected components. One of these, known as
the interior, is bounded and the other one, known as the exterior, is unbounded.
So, any loop in the complex plane separates the plane into two domains having the
loop as common boundary. One domain, the interior, is bounded, while the other
one, the exterior, is unbounded.
Example 4.8 The line segment joining the points a and b in C, with a 6= b, is
defined as being:
γ : [0, 1] → C,
with
γ(t) = (1 − t)a + tb
or
γ(t) = a + t(b − a).
Obviously, γ is smooth. Also, γ is a simple path.
Example 4.9 The circle of radius a, centered at the origin, is defined as being:
γ : [0, 2π] → C,
with
γ(t) = a eit ,
for t ∈ [0, 2π]. It is easy to see that γ is a smooth closed path.
Example 4.10 The path γ : [−π, π] → C, given by γ(t) = cos t, is not a simple
path.
65
COMPLEX INTEGRATION
Definition 4.11 The path γ −1 : [a, b] → C, given by
γ −1 (t) = γ(a + b − t),
is called the inverse or the opposite path of γ.
If we take γ : [0, 1] → C, the inverse path γ −1 : [0, 1] → C is given by
γ −1 (t) = γ(1 − t).
Example 4.12 A simple example of a smooth path is offered by the circle centered
at a point z0 ∈ C and having the radius r &gt; 0, i.e. the set
C(z0 , r) = {z ∈ C : |z − z0 | = r}.
Its positive orientation (counterclockwise) is given by the following parametrization:
γ(t) = z0 + reit ,
t ∈ [0, 2π],
while its negative orientation (clockwise) is given by
γ(t) = z0 + re−it ,
t ∈ [0, 2π].
Remark 4.13 In what follows, if not otherwise mentioned, we shall work only with
simple positively oriented paths.
Definition 4.14 Let γ1 , γ2 : [0, 1] → C be two paths such that γ1 (1) = γ2 (0). We
define their compound path (alternatively called the sum, the concatenation or the
composition of the paths γ1 and γ2 ) as being
γ : [0, 1] → C,
with
γ(t) = (γ1 ∨ γ2 )(t) =



 γ1 (2t),
0≤t≤


 γ (2t − 1),
2
1
,
2
1
≤ t ≤ 1.
2
66
COMPLEX ANALYSIS
In a similar manner, if γ1 : [a, b] → C and γ2 : [c, d] → C are two paths such that
γ1 (b) = γ2 (c), then their sum can be defined as being
γ : [a, b + (d − c)] → C,
with
γ(t) = (γ1 ∨ γ2 )(t) =

 γ1 (t),

t ∈ [a, b],
γ2 (t − b + c),
t ∈ [b, b + (d − c)].
Remark 4.15 Let us notice that γ ∨ γ −1 is a closed path.
Sometimes, a sequence of smooth paths {γ1 , γ2 , . . . , γn } such that the terminal point
of γk coincides with the initial point of γk+1 , for 1 ≤ k ≤ n − 1, is called a contour.
A contour γ is said to be closed if its endpoints coincide. A closed contour is also
called, in some textbooks, a loop. Let us mention that some authors use the term
contour for a closed piecewise smooth path (what we call here a closed contour).
Definition 4.16 Let γ : [a, b] → C and e
γ : [α, β] → C be two smooth paths. We shall
say that γ and γ
e are equivalent (we denote γ ∼ γ
e) if there exists ϕ : [a, b] → [α, β]
′
bijective, of class C 1 , with ϕ (t) 6= 0, such that
i.e., for any t ∈ [a, b],
γ=γ
e ◦ ϕ,
γ(t) = γ
e(ϕ(t)).
Remark 4.17 The above relation is an equivalence relation on the set of all the
paths of class C 1 . Each corresponding equivalence class is called a curve.
We can define also an equivalence relation for oriented paths. Indeed, by asking that
′
ϕ (t) &gt; 0, we get equivalent paths having the same orientation. Equivalent paths
have the same image and, moreover, equivalent oriented paths are traversed in the
same direction.
67
COMPLEX INTEGRATION
Definition 4.18 Let γ : [a, b] → C be a smooth path. The length of the path γ is
defined as being
l(γ) =
Zb
a
′
|γ (t)| dt.
This definition is independent of the parametrization of the path γ.
Remark 4.19 If γ is only a piecewise smooth path, then its length is the sum of
the lengths of its smooth parts.
Definition 4.20 Let D ⊆ C be a domain and γ1 , γ2 : [0, 1] → D. A map ϕ :
[0, 1] &times; [0, 1] → D is called a continuous deformation of γ1 into γ2 if ϕ is continuous
and, for any t ∈ [0, 1],

 ϕ(0, t) = γ1 (t),

ϕ(1, t) = γ2 (t).
Definition 4.21 Let D ⊆ C be a domain and γ1 , γ2 : [0, 1] → D. The paths γ1
and γ2 are called homotopic in D (we shall denote this fact by γ1 ≈ γ2 ) if we can
continuously deform γ1 into γ2 , without leaving D.
A closed path γ : [0, 1] → D is said to be homotopic to zero (we shall write γ ≈ 0)
in D if γ can be continuously deformed into a point, without leaving D.
The continuous map ϕ is also called a homotopy between the path γ1 and γ2 . It is
not difficult to see that homotopy defines an equivalence relation.
Definition 4.22 Let γ : [a, b] → C be a smooth path and f : γ([a, b]) → C a
continuous function. The complex integral of the function f along the path γ is
defined as follows:
Z
γ
f (z) dz =
Zb
′
f (γ(t)) γ (t) dt.
(4.1)
a
Remark 4.23 If γ is a piecewise smooth path, then the integral of f along γ is just
the sum of the integrals of f along the smooth parts of γ.
68
COMPLEX ANALYSIS
Proposition 4.24 1. (Linearity) If the functions f and g are continuous on a
piecewise smooth path γ, then, for any complex numbers α and β, we have
Z
(αf + βg)(z) dz = α
γ
Z
f (z) dz + β
γ
Z
g(z) dz.
γ
2. (Additivity) Let γ1 and γ2 be two piecewise smooth paths and f a continuous
function on their union γ1 ∨ γ2 . Then,
Z
f (z) dz =
γ1 ∨γ2
Z
f (z) dz +
γ1
Z
f (z) dz.
γ2
3. (Invariance) The complex integral is invariant under a re-parametrization of
the integration path.
4. (Orientation) Let γ be a piecewise smooth path. If f is continuous on γ, then
Z
f (z) dz = −
γ −1
Z
f (z) dz.
γ
5. The modulus of the integral of a complex function f along a path γ is bounded
by the integral of its modulus with respect to the arc length, i.e.
Z
γ
f (z) dz ≤
Z
γ
|f (z)| ds.
If, in addition, we suppose that there exists a constant M ≥ 0 such that |f (z)| ≤ M
on γ, then
Z
γ
f (z) dz ≤ M l(γ).
Definition 4.25 Let D be a nonempty open subset of C. A function F : D → C is
called a primitive or an anti-derivative of f : D → C if it is holomorphic on D and
′
F (z) = f (z), for any z ∈ D.
69
COMPLEX INTEGRATION
Remark 4.26 A primitive of a continuous function f on a domain D, if it exists,
is defined up to an arbitrary constant. So, any two primitives of f on D, if they
exist, differ by a constant.
Theorem 4.27 (The Fundamental Theorem of Calculus for Path Integrals) Let D
be a domain in C and f : D → C a continuous function which possesses an antiderivative F : D → C. If γ : [a, b] → D is a smooth path contained in D, then
Z
γ
f (z) dz = F (γ(b)) − F (γ(a)).
(4.2)
Corollary 4.28 If γ is a closed path in a domain D ⊆ C and f is continuous and
has an anti-derivative in D, then
Z
f (z) dz = 0.
γ
Remark 4.29 Using this result, we can prove that the function f (z) = 1/z does not
have a primitive in the open set C \ {0}. Indeed, if we take γ as being the positively
oriented unit circle |z| = 1, given by γ(t) = eit , with t ∈ [0, 2π], we get
Z
γ
f (z) dz =
Z2π
0
i eit
dt = 2πi 6= 0.
eit
Exercise 4.30 Find a parametrization for the circle C(1 + i, 1), oriented counterclockwise.
Exercise 4.31 Find a parametrization for the line segment from 1 − i to 3 i.
Exercise 4.32 Find a parametrization for the he rectangle with vertices at the
points 1 &plusmn; 3 i, oriented counter-clockwise.
Exercise 4.33 Show that any two positively oriented circles C(z0 , r) and C(w0 , R)
are homotopic in C.
70
COMPLEX ANALYSIS
Exercise 4.34 Show that the positively oriented circles C(0, 2) and C(2, 4) are not
homotopic in C \ {0}.
Exercise 4.35 Show that in a simply connected domain, any two paths with common endpoints are homotopic and any closed path is homotopic to zero.
Exercise 4.36 Compute the integral of the function f : C → C given by f (z) = z 2
over the line segment from 0 to 1 + i.
Solution. A parametrization of this path is γ : [0, 1] → C, γ(t) = t + i t. Therefore,
I=
Z1
0
(1 − i t)2 (1 + i) dt =
2
(1 − i) .
3
Exercise 4.37 Compute the integral
I=
Z
z dz,
γ
where γ is the line segment [−i, i].
Solution. The path γ can be parameterized as γ : [0, 1] → C, with γ(t) = (1 −
t)(−i) + ti, i.e.
γ(t) = (2t − 1)i, with t ∈ [0, 1].
Obviously, γ is a smooth path and f (z) = z is continuous along γ. Then, we have
I=
Z1
0
[−(2t − 1) i 2i] dt =
Z1
0
Exercise 4.38 Evaluate the integral
I=
Z
γ
where γ is the circle |z − 1| = 1.
z 2 dz,
(4t − 2) dt = 0.
71
COMPLEX INTEGRATION
Solution. The path γ can be represented as γ : [0, 2π] → C, with γ(t) = 1 + eit .
Obviously, γ is a smooth path and f (z) = z 2 is continuous along γ. Then, we get
I=
Z2π
−2it
1+e
−it
+ 2e
0
it
i e dt =
Z2π
0
i eit + ie−it + 2i dt = 4πi.
Exercise 4.39 Compute the integral
Z
In = (z − a)n dz,
γ
where n ∈ Z and γ is the circle |z − a| = r, with a ∈ C and r &gt; 0.
Solution. The path γ can be represented as γ : [0, 2π] → C, with γ(t) = a + reit. So,
γ is a smooth path and f (z) = (z − a)n is continuous along γ. We have
Z2π
I=
n int
r e
it
ri e dt =
0
Z2π
r n+1 i ei(n+1)t dt.
0
For n 6= −1, we obtain
In = 0,
while for n = −1, we get
I−1 = 2πi.
Exercise 4.40 Evaluate the following integral:
Z
I = z 2 dz,
γ
where γ is the quarter of the unit circle lying in the first quadrant (traversed counterclockwise).
h πi
Solution. We can parametrize the path γ as being γ(t) = eit , with t ∈ 0, . Using
2
this parametrization, we get:
Zπ/2
1
I=
e2it i eit dt = − (1 + i).
3
0
72
COMPLEX ANALYSIS
Exercise 4.41 Let |a| &lt; r &lt; |b|. Show that
Z
γ
1
2πi
dz =
,
(z − a)(z − b)
a−b
where γ is the positively oriented circle |z| = r.
Exercise 4.42 Compute the integral
I=
Z
(z 2 + 3 z) dz,
γ
where γ is the path connecting the origin to 1 + 2i along a straight line.
Exercise 4.43 Compute the integral
I=
Z
(z 3 + 1) dz,
γ
along the path γ : [0, π] → C, defined by
γ(t) = eit .
Solution. Since the function
F (z) =
z4
+z
4
is a primitive of the continuous function f (z) = z 3 + 1 in C and γ is a smooth path,
it follows that
I = F (γ(π)) − F (γ(0)),
i.e.
I = −2.
73
COMPLEX INTEGRATION
4.2
Cauchy’s Theorem. Cauchy’s Integral Formula. Applications
Theorem 4.44 (Cauchy) Let f be a holomorphic function in a simply connected
domain D ⊆ C and γ a closed path in D. Then,
Z
f (z) dz = 0.
(4.3)
γ
Theorem 4.45 (Path Independence Theorem) Let D ⊆ C be a simply connected domain. If f is holomorphic in D and γ1 and γ2 are piecewise regular paths connecting
two given points z1 and z2 in D, then
Z
Z
f (z) dz = f (z) dz.
γ1
γ2
Theorem 4.46 (Cauchy) Let D be an open nonempty set in C. If f is holomorphic
in D and γ is a piecewise smooth closed path that is homotopic to zero in D, then
Z
f (z) dz = 0.
γ
Theorem 4.47 (Deformation Theorem) Let D be an open nonempty set in C. If
f is holomorphic in D and γ1 and γ2 are closed piecewise regular paths that are
homotopic in D, then
Z
γ1
f (z) dz =
Z
f (z) dz.
γ2
Theorem 4.48 Any function f which is holomorphic in a simply connected domain
D possesses an anti-derivative in this domain.
If we consider the function f = 1/z, which is holomorphic in the annulus U =
{z ∈ C | 0 &lt; |z| &lt; 2}, we can see immediately that, in Theorem 4.44, the assumption
that D is simply connected is essential: the global existence theorem of an anti-
derivative does not hold in general for multiply connected domains. The same
example shows that the integral of a holomorphic function over a closed path in a
multiply connected domain might not vanish.
74
COMPLEX ANALYSIS
Proposition 4.49 Let D be a domain in C and f : D → C be a continuous function.
Then, if
Z
f (z) dz = 0
γ
for any closed path γ in D, f admits an anti-derivative on D.
Remark
4.50 (Morera) a) If f is a continuous complex function on a domain D
Z
f (z) dz = 0 for every closed contour γ in D, then f is holomorphic on D.
and
γ
b) If a complex function f possesses an anti-derivative on a domain D, then f
is holomorphic on D.
To summarize, if D is a domain in C and f is a continuous function on D, then
the following three statements are equivalent:
a) f has an antiderivative in D;
b) The integral
Z
f (z) dz vanishes for all piecewise smooth closed paths γ
γ
contained in D;
Z
c) The integral
f (z) dz is path-independent.
γ
Remark 4.51 (Extended Cauchy-Goursat Theorem) Let Γ and γ1 , . . . , γn be simple
closed positively oriented contours such that all the paths γi lie in the interior of Γ.
Assume that the interiors of γi and γj are disjoint sets, for all i 6= j. If f is
holomorphic on a domain containing the paths Γ and γi , for i = 1, n, and the points
between them, then
Z
Γ
f (z) dz =
n Z
X
f (z) dz.
i=1 γi
Theorem 4.52 (Cauchy’s Integral Formula) Let f be a holomorphic function in
a simply connected domain D and γ a positively oriented simple closed path in D.
75
COMPLEX INTEGRATION
Then, for any point z in the interior of the domain bounded by γ, we have
f (z) =
1
2πi
Z
f (ξ)
dξ.
ξ−z
γ
(4.4)
Let us notice that (4.4) is a kind of a ”holographic” formula, stating that a
holomorphic function in a simply connected domain is determined by its behaviour
on the boundary of the domain. As a consequence, it follows that if two holomorphic
functions coincide on the boundary of a simply connected domain, they coincide
everywhere in the domain.
Remark 4.53 Formula (4.4) gives a representation of a holomorphic function in
the closure of a domain in terms of an integral over the boundary of the domain. In
other words, if f is holomorphic in the closure of a domain Ω that is bounded by a
finite number of continuous paths, then, at any point z ∈ Ω, f can be represented as
f (z) =
1
2πi
Z
∂Ω
f (ξ)
dξ,
ξ−z
where ∂Ω is the positively oriented boundary of Ω. If the point z lies outside Ω, the
f (ξ)
function
is holomorphic in Ω and we have
ξ−z
1
2πi
Z
∂Ω
f (ξ)
dξ = 0.
ξ−z
Remark 4.54 It is not difficult to see that we can extend Cauchy’s formula (5.4)
for the case in which the closed contour γ is the oriented boundary of a multiple
connected domain D. For instance, if ∂D = γ1 ∨ γ2 , then
f (z) =
1
2πi
Z
γ1
f (ξ)
1
dξ −
ξ−z
2πi
Z
γ2
f (ξ)
dξ,
ξ−z
z ∈ D.
Proposition 4.55 If f is holomorphic in a simply connected domain D ⊆ C and γ
is a positively oriented simple closed path in D, then f is indefinitely differentiable
76
COMPLEX ANALYSIS
in D and, for any z ∈ D \ {γ} and n ∈ N, we have
Z
n!
f (ξ)
f (n) (z) =
dξ.
2πi
(ξ − z)n+1
(4.5)
γ
Definition 4.56 Let γ be a piecewise smooth path in C. If z0 ∈ C \ {γ}, we can
define the so-called index of γ with respect to z0 as being
Z
1
dξ
n(γ, z0 ) =
.
2πi
ξ − z0
γ
It can be proven that if γ is a closed path in C, its index with respect to a point
z0 ∈ C \ {γ} is an integer. This integer is also known as the winding number of the
closed path γ about the point z0 . Intuitively, the index gives the number of times
a closed contour γ winds counterclockwise around a point z0 . The sign of the index
is determined by the orientation we choose on the path.
It is not difficult to see that the index n(γ, z0 ) has the following properties:
1) if γ1 and γ2 are homotopic paths in C \ {z0 }, then n(γ1 , z0 ) = n(γ2 , z0 );
2) n(γ −1 , z0 ) = −n(γ, z0 );
3) if γ = γ1 ∨ γ2 , then n(γ, z0 ) = n(γ1 , z0 ) + n(γ2 , z0 ).
Theorem 4.57 Let f be a holomorphic function on an open set D ⊆ C and γ a
closed contour which is homotopic to zero in D. Then, f is indefinitely derivable on
D and, for any z ∈ D \ {γ} and n ∈ N, we have
Z
n!
f (ξ)
(n)
n(γ, z) f (z) =
dξ.
2πi
(ξ − z)n+1
(4.6)
γ
Theorem 4.58 (Generalized Cauchy’s Theorem) Let D be a domain in C and let
f be holomorphic on D. Let γ1 , . . . , γn be n closed paths in D. If
n(γ1 , z) + &middot; &middot; &middot; + n(γn , z) = 0,
77
COMPLEX INTEGRATION
for all z ∈
/ D, then
Z
γ1
f (z) dz + &middot; &middot; &middot; +
Z
f (z) dz = 0.
γn
Proposition 4.59 (Cauchy’s Inequalities) Let z0 ∈ C, r ∈ R+ and f holomorphic
on an open set G ⊇ B(z0 , r). If
M = sup{|f (z)| | z ∈ B(z0 , r)},
then, for any n ∈ N, we have
f (n) (z0 ) ≤ n!
M
.
rn
(4.7)
Theorem 4.60 (Liouville) If f is holomorphic on C and there exists a positive
number M such that |f (z)| ≤ M for any z ∈ C, then f is constant on C.
Theorem 4.61 (The Fundamental Theorem of Algebra) Any polynomial
P (z) = a0 + a1 z + &middot; &middot; &middot; + an z n ,
with ai ∈ C, an 6= 0 and n ≥ 1, has at least one root in C.
Exercise 4.62 Show that if γ is a simple closed path in C and a is a point not lying
on γ, then
Z
γ

 2πi, a is inside {γ},
1
dz =

z−a
0, a is outside {γ}.
Solution. Note that f (z) = 1/(z − a) is holomorphic everywhere, except at z = a,
where it has, as we shall see later, a simple pole. If a lies outside {γ}, then, from
Cauchy’s theorem, the integral is zero. If a is inside {γ}, then the integral is equal to
the integral along a suitable chosen circle centered at z = a, which can be computed
directly by using the parametrization γ(t) = a + reit , with t ∈ [0, 2π] and r &gt; 0. As
we saw in a previous example, the value of the integral in this case is 2πi.
78
COMPLEX ANALYSIS
Exercise 4.63 Prove that the integral
Z
sin z dz = 0, for any Jordan curve γ in
γ
the complex plane.
Solution. Since the function f (z) = sin z is entire and γ is a Jordan path, using
Cauchy’s theorem, it follows immediately that our integral is equal to zero.
Exercise 4.64 Show that
Z
γ
(z 2
1
dz = 0,
+ 4)(z 2 + 9)
where γ is the positively oriented circle |z| = 1.
Solution. The zeros of the polynomial (z 2 + 4)(z 2 + 9) are &plusmn;2i and, respectively, &plusmn;3i
and they lie outside the disk D = B(0, 1). Therefore, the function
f (z) =
(z 2
1
+ 4)(z 2 + 9)
is holomorphic on D and, since the path γ is smooth and closed, it follows that the
integral is zero.
Exercise 4.65 Show that if f is holomorphic in a domain D and its derivative is
zero on D, then f is constant on D.
Exercise 4.66 Prove that if f is holomorphic in the annulus
U = {z ∈ C | r &lt; |z − a| &lt; R},
then the integral
I=
Z
f (z) dz
|z−a|&lt;ρ
has the same value for all ρ with r &lt; ρ &lt; R.
79
COMPLEX INTEGRATION
Exercise 4.67 Compute the value of the integral
I=
Z
γ
where γ is the circle |z − 2| =
√
z+1
dz,
− 1)
z 2 (z
2, traversed counterclockwise.
Solution. We note that
z+1
f (z)
=
,
− 1)
z−1
z 2 (z
where
z+1
.
z2
√
The function f is holomorphic in the disk B(2, 2), its only singularity, at z = 0,
f (z) =
lying outside the closed contour γ. Therefore, using Cauchy’s integral formula (...),
we get
I = 2πi f (1) = 4πi.
Exercise 4.68 Evaluate the integral
I=
Z
γ
z−1
dz,
z 2 (z − 2)
where γ is the circle |z| = 1, traveled counterclockwise.
Solution. We remark that
z−1
f (z)
= 2 ,
z 2 (z − 2)
z
where
f (z) =
z−1
.
z−2
The function f is holomorphic in the disk B(0, 1). Using Cauchy’s integral formula
(5.5), we get
′
I = 2πif (0) = −
πi
.
2
80
COMPLEX ANALYSIS
Exercise 4.69 Compute the value of the integral
I=
Z
sinh z
dz,
(z − πi)4
γ
where γ is the positively oriented circle |z − 2i| = 3.
Solution. Since
f (z) = sinh z
is holomorphic on the disk B(2i, 3), using Cauchy’s integral formula (5.5), we get
I=
πi
2πi ′′′
f (πi) = − .
3!
3
Exercise 4.70 Compute the integral
I=
Z
γ
ez
dz,
z 2 − 2z
where γ is the positively oriented circle C(0, 4).
Solution. Since
I=
1
2
Z
γ
ez
1
dz −
z−2
2
Z
γ
ez
dz,
z
it follows immediately that
I=
1
1
2π ie2 − 2π ie0 = π i e2 − 1 .
2
2
Exercise 4.71 Find the length of the following paths:
a)
the positively oriented circle C(1 + 2 i, 1);
b)
the line segment from −1 + 2 i to 3 i;
c)
the rectangle with vertices &plusmn;2 &plusmn; 3 i.
81
COMPLEX INTEGRATION
Exercise 4.72 Evaluate the integral I =
Z
e4z dz for each of the following paths:
γ
a)
γ is the line segment from 2 to 3 i;
b)
γ is the positively oriented circle C(0, 2).
Exercise 4.73 Show that
Z
2
z 3 ez dz = 0
γ
for any closed piecewise smooth path γ.
Exercise 4.74 Compute the following integrals:
a)
b)
I=
I=
Z
C(0,2)
Z
C(0,8)
ez
dz;
z (z − 7)
ez
dz.
z (z − 7)
Exercise 4.75 Show that
Z
C(0,3)
z2
1
dz = 0.
+1
82
Chapter 5
Taylor and Laurent Series
5.1
Sequences and Series of Functions. Power Series
Definition 5.1 Let us consider a sequence of complex functions (fn )n≥0 , defined on
a nonempty set D ⊆ C. The sequence (fn )n≥0 is said to be pointwise convergent at
the point z0 ∈ D if the number sequence (fn (z0 ))n≥0 is convergent.
Definition 5.2 The sequence (fn )n≥0 is said to be pointwise convergent on the set
D if (fn )n≥0 is convergent at any point z ∈ D. In this case, we can define a function
f : D → C, by putting, for z ∈ D,
f (z) = lim fn (z).
n→∞
The function f is called the limit of (fn )n≥0 on D. We shall use the notation
fn → f
to denote the fact that f is the limit of the sequence (fn )n≥0 .
Definition 5.3 The sequence (fn )n≥0 is said to be uniformly convergent on D to f
if for any ε &gt; 0 there exists Nε ∈ N such that, for any z ∈ D,
|fn (z) − f (z)| &lt; ε,
83
84
COMPLEX ANALYSIS
for any n ≥ Nε . We shall write
u
fn −→ f
to denote the fact that the sequence (fn )n≥0 is uniformly convergent to f .
Remark 5.4 Of course, uniform convergence implies pointwise convergence.
Proposition 5.5 Let fn : D ⊆ C → C, with n ≥ 0, be a pointwise convergent
sequence and let f (z) = lim fn (z). If
n→∞
an = sup |fn (z) − f (z)|
z∈D
and
lim an = 0,
n→∞
then (fn )n≥0 is uniformly convergent to f on D.
Proposition 5.6 If the sequence (fn )n is uniformly convergent to f on D and all
the maps fn are continuous at z0 ∈ D, then f is continuous at z0 .
Remark 5.7 Let us notice that if the sequence (fn )n is uniformly convergent to f
on D and all the maps fn are continuous at z0 ∈ D, then
lim lim fn (z) = lim lim fn (z).
z→z0 n→∞
n→∞ z→z0
Proposition 5.8 Let (fn )n≥0 be a sequence of continuous functions defined on a
domain D containing a contour γ. If (fn )n≥0 converges uniformly to f on D, then
Z
Z
Z
lim
fn (z) dz =
lim fn (z) dz = f (z) dz.
n→∞
n→∞
γ
γ
γ
Definition 5.9 Let fn : D → C, with n ≥ 0, and Sn : D → C given by
Sn (z) =
n
X
fi (z).
(5.1)
i=0
The pair ((fn )n≥0 , (Sn )n≥0 ) is said to be a series of complex functions (the series
associated to the sequence (fn )n≥0 ).
85
TAYLOR AND LAURENT SERIES
We shall denote such a series by
X
X
fn . Sn is called the nth partial sum of the series
n≥0
fn and fn is said to be its general term.
n≥0
Definition 5.10 The series
X
n≥0
fn is said to be convergent at z ∈ D if the sequence
(Sn )n≥0 is convergent at z ∈ D. The limit S(z) of (Sn (z))n≥0 is called the sum of
X
the convergent series
fn at the point z. We shall use the following symbol to
n≥0
denote the sum of a convergent series:
S=
∞
X
fn .
n=0
Definition 5.11 A series which is not convergent is called divergent.
Definition 5.12 A series
number series
X
n≥0
X
n≥0
fn is called absolutely convergent at z0 ∈ D if the
|fn (z0 )| is convergent.
Definition 5.13 The series
X
fn is called uniformly convergent on the set D if
n≥0
(Sn )n≥0 is uniformly convergent on D.
Theorem 5.14 (Weierstrass Test) Let fn : D ⊆ C → C, with n ≥ 0. If there exists
an ≥ 0 such that
|fn (z)| ≤ an ,
for any n ≥ 0 and for any z ∈ D and if the series
is absolutely and uniformly convergent on D.
X
an is convergent, then
n≥0
Theorem 5.15 Let fn : D → C, with n ≥ 0. If the series
fn
n≥0
X
fn is uniformly
n≥0
convergent on D and fn are continuous at z0 ∈ D, then the sum of the series
is continuous at z0 ∈ D.
X
X
n≥0
fn
86
COMPLEX ANALYSIS
Proposition 5.16 Let
X
fn be a series of continuous functions which converges
n≥0
uniformly to f on a domain D. If γ is a contour lying entirely in D, then
Z X
∞
∞ Z
X
fn (z) dz =
fn (z) dz.
γ n=0
n=0 γ
Proposition 5.17 a) Let (fn )n≥0 be a sequence of holomorphic functions on a domain D ⊆ C. If the sequence (fn )n≥0 converges uniformly to f on D, then f is
holomorphic on D.
b) If (fn )n≥0 is a sequence of holomorphic functions that converges uniformly to
a function f on every compact subset of a domain D, then f is holomorphic on D.
(k)
Moreover, the sequence of derivatives (fn )n≥0 converges uniformly to f (k) on every
compact subset of D, for any k ≥ 1.
Proposition 5.18 a) Let (fn )n≥0 be a sequence of holomorphic functions on a doX
main D. If the series
fn converges to f uniformly on D, then f is holomorphic
n≥0
on D.
b) Let (fn )n≥0 be a sequence of holomorphic functions on a domain D. If
X
fn
n≥0
converges to f uniformly on every compact subset of D, then for any k ≥ 1 and for
all z ∈ D, we have
i.e. the series
X
n≥0
∞
X
fn(k)(z) = f (k)(z),
n=0
fn may be differentiated term-by-term. Moreover, for any k ≥ 1,
the above convergence is uniform on any compact subset of D.
Definition 5.19 Let z0 ∈ C and an ∈ C, with n ≥ 0. A series of functions for
which
fn (z) = an (z − z0 )n , for n ≥ 0,
is called a power series on C. We shall denote such a power series by
X
an (z − z0 )n .
n≥0
(5.2)
87
TAYLOR AND LAURENT SERIES
Let us notice that the above series is convergent for z = z0 .
Theorem 5.20 (Cauchy-Hadamard) Let R ≥ 0 be defined as follows:
R=
lim
n→∞
with the convention that R = 0 if
lim
n→∞
Then:
1
p
n
|an |
,
p
n
|an | = ∞ and R = ∞ if
lim
n→∞
p
n
|an | = 0.
1) if R &gt; 0, the power series (5.2) is absolutely convergent for all the points z
with |z − z0 | &lt; R and divergent for any z with |z − z0 | &gt; R;
2) if R &gt; 0, the power series (5.2) is uniformly convergent over any compact
disk |z − z0 | ≤ r, with 0 &lt; r &lt; R;
3) if R = 0, the series (5.2) is convergent only for z = z0 .
Remark 5.21 The non-negative quantity R is called the radius of convergence for
the series (5.2).
The set
Dc = {z ∈ C | |z − z0 | &lt; R}
is called the disk or the domain of convergence for the series (5.2). On the boundary
of this disk, i.e. at the points |z − z0 | = R, the power series (5.2) may be either
convergent or divergent. Therefore, we have to examine separately what happens
at these points. Alternatively, we can apply other theorems (see ,  and ) to
deal with this case (for instance, we can use Abel’s theorem).
Remark 5.22 If there exists lim
n→∞
an
, then
an+1
R = lim
n→∞
Moreover, if there exists lim
n→∞
an
.
an+1
an
, then
an+1
R = lim
n→∞
an
.
an+1
88
COMPLEX ANALYSIS
Remark 5.23 A power series about the point z0 = 0, i.e. a series of the form
X
an z n ,
n≥0
is called a Maclaurin series.
Remark 5.24 If the power series (5.2), having the radius of convergence R, is
uniformly convergent over |z − z0 | &lt; r, with 0 &lt; r &lt; R, then its sum defines a
continuous function at any point z with |z − z0 | &lt; R. Moreover, in |z − z0 | &lt; R, the
sum of the series (5.2) is a holomorphic function and its derivative can be obtained by
X
n an (z − z0 )n−1 , obtained by differentiating
termwise differentiation. The series
n≥1
(6.2) with respect to z term by term, has the same radius of convergence as the series
(5.2).
Remark 5.25 A power series is infinitely complex differentiable in its disk of convergence and all its derivatives are power series that can be obtained by termwise
differentiation.
Using power series, we can rigorously define the most commonly used elementary
functions of a complex argument.
I. A polynomial functions P = P (z) can be defined as being a power series with
a finite number of nonzero coefficients:
P (z) = a0 + a1 z + &middot; &middot; &middot; + an z n ,
with ai ∈ C. The radius of convergence of this power series is ∞ and the polynomial
function is indefinitely derivable on C.
II. Rational functions are complex functions of the form
R(z) =
P (z)
,
Q(z)
where P and Q are polynomial functions in z and Q is not the zero polynomial. The
domain of R is the set of all the points z for which the denominator Q(z) is not zero.
89
TAYLOR AND LAURENT SERIES
III. The exponential function exp : C → C is defined by
exp (z) =
X zn
n≥0
n!
.
This function is also denoted by ez and its radius of convergence is ∞.
We have already seen that this function is an entire one and its derivative is
d z
(e ) = ez .
dz
We shall give now other properties of the exponential function.
a) ez+w = ez ew , for any z, w ∈ C;
′
b) (ez ) = ez , for any z ∈ C;
c) eiz = cos z + i sin z, for any z ∈ C (Euler’s formula);
d) The exponential function is periodic, with complex period 2πi, i.e.
ez+2kπi = ez ,
for any z ∈ C, k ∈ Z.
e) The complex exponential is not a one-to-one map. As already mentioned, we
can divide the complex plane into horizontal strips of height 2π in such a way that in
each strip the exponential function is one-to-one. More precisely, for z = x + iy ∈ C,
we set
Sk = {z = x + iy ∈ C | x ∈ R, (2k − 1)π &lt; y ≤ (2k + 1)π},
k ∈ Z.
The sets Sk are called fundamental strips for the complex exponential function. It
follows that if z1 and z2 belong to the same strip Sk , then ez1 = ez2 implies that
z1 = z2 .
IV. The sine function sin : C → C is defined by
sin z =
X
(−1)n
n≥0
z 2n+1
.
(2n + 1)!
90
COMPLEX ANALYSIS
Its radius of convergence is ∞.
V. The cosine function cos : C → C, defined by
cos z =
X
(−1)n
n≥0
z 2n
,
(2n)!
has the radius of convergence R = ∞.
VI. The hyperbolic sine function sinh : C → C, defined by
sinh z =
X
n≥0
z 2n+1
.
(2n + 1)!
Its radius of convergence is ∞.
VII. The hyperbolic cosine function cosh : C → C, defined by
cosh z =
X z 2n
.
(2n)!
n≥0
Its radius of convergence is ∞.
Properties
Let z, w ∈ C. It is not difficult to see that the following properties hold true:
eiz − e−iz
,
2i
1.
sin z =
2.
sinh z =
3.
sin2 z + cos2 z = 1,
4.
(sin z) = cos z,
5.
(sinh z) = cosh z,
ez − e−z
,
2
′
′
cos z =
eiz + e−iz
;
2
cosh z =
ez + e−z
;
2
cosh2 z − sinh2 z = 1;
′
(cos z) = − sin z;
′
(cosh z) = sinh z;
91
TAYLOR AND LAURENT SERIES
6.
sin (−z) = − sin z,
7.
sinh (−z) = − sinh z,
8.
sin (z + w) = sin z cos w + cos z sin w;
9.
cos (z + w) = cos z cos w − sin z sin w;
cos (−z) = cos z;
cosh (−z) = cosh z.
10.
sinh (z + w) = sinh z cosh w + cosh z sinh w;
11.
cosh (z + w) = cosh z cosh w + sinh z sinh w.
Exercise 5.26 Solve the equation cos z = 0.
Exercise 5.27 Solve the equation sin z = 5.
Exercise 5.28 Solve the equation sin z = 1 + i.
Example 5.29 Let D = {z ∈ C | |z| &lt; 1} and fn : D → C, with n ≥ 1, defined by
fn (z) = z n .
Then, for any z ∈ D,
lim fn (z) = 0,
n→∞
i.e. (fn )n is convergent on D.
Exercise 5.30 Let fn (z) = z n , with n ≥ 1, and f (z) = 0. Prove that:
(a) (fn )n≥1 is pointwise convergent to f on the open disk B(0, 1);
(b) (fn )n≥1 converges uniformly to f on any closed disk B(0, ρ), with 0 &lt; ρ &lt; 1;
(c) (fn )n≥1 is not uniformly convergent to f on the closed disk B(0, 1).
Exercise 5.31 Compute the radius and the set of convergence for the series
X
n≥0
n! z n .
92
COMPLEX ANALYSIS
Solution. It is easy to see that the radius of convergence for this series is equal to
zero. Therefore, the series is convergent only for z = 0, i.e. Dc = {0}.
Exercise 5.32 Find the radius of convergence for the series
X zn
n≥0
n!
.
Solution. The radius of convergence for the above series is equal to ∞. So, the
domain of convergence is Dc = C.
Exercise 5.33 Prove that the series
X zn
n≥0
n2
is absolutely convergent on |z| &lt; 1.
Exercise 5.34 Find the radius of convergence for the series
X nn z n
n≥1
n!
.
Exercise 5.35 Compute the radius of convergence for the series
X
2
2−n z n .
n≥1
Exercise 5.36 Determine the radius of convergence for the series
X zn
.
ln n
n≥2
Exercise 5.37 Find the radius of convergence for the series
2
X
1 n n
1+
z .
n
n≥1
Exercise 5.38 Compute the radius of convergence for the series
X z n2
n≥1
n2
.
93
TAYLOR AND LAURENT SERIES
Exercise 5.39 Find the radius of convergence of the hypergeometric series
F (α, β, γ; z) = 1 +
∞
X
α (α + 1) &middot; &middot; &middot; (α + n − 1) β (β + 1) &middot; &middot; &middot; (β + n − 1)
n! γ (γ + 1) &middot; &middot; &middot; (γ + n − 1)
n=1
zn,
for α, β ∈ C and γ 6= 0, −1, . . . .
Exercise 5.40 Prove that ez = 1 if and only if z = 2kπi, with k ∈ Z.
Exercise 5.41 Prove that ez1 = ez2 if and only if z1 = z2 + 2kπi, with k ∈ Z.
Exercise 5.42 Show that
ez = ez ,
for any z ∈ C.
Exercise 5.43 Prove that

 sin (2z) = 2 sin z cos z;
cos (2z) = cos2 z − sin2 z;
Exercise 5.44 Show that

2
 | sin z|2 = sin2 x + sinh y;
| cos z|2 = cos2 x + sinh2 y;


sinh (2z) = 2 sinh z cosh z.
| sinh z|2 = sinh2 x + sin2 y;
Exercise 5.45 Prove that

 cos (iz) = cosh z;
5.2

| cosh z|2 = sinh2 x + cos2 y.
sin (iz) = i sinh z;
sinh (iz) = i sin z;
cosh (iz) = cos z.
Taylor Series
Theorem 5.46 (Taylor’s Theorem) Let z0 ∈ C and r &gt; 0. If f : B(z0 , r) → C
∞
X
is holomorphic on B(z0 , r), then there exists a unique power series
an (z − z0 )n
n=0
having the radius of convergence R ≥ r such that, for any z ∈ B(z0 , r),
f (z) =
∞
X
n=0
an (z − z0 )n .
(5.3)
94
COMPLEX ANALYSIS
The coefficients an in (5.3), called the Taylor coefficients, are given by
Z
f (n) (z0 )
1
f (ξ)
an =
=
dξ,
n!
2πi
(ξ − z0 )n+1
(5.4)
γ
where γ = ∂B(z0 , ρ), with 0 &lt; ρ &lt; r.
Corollary 5.47 If a complex function f is holomorphic on a domain D, then, for
any n ≥ 0, its n-th derivative f (n) exists and is holomorphic on D.
Remark 5.48 If z0 = 0, the Taylor series (5.3) becomes
′
f (z) = f (0) +
′′
f (0)
f (0) 2
z+
z + &middot;&middot;&middot; .
1!
2!
(5.5)
This series is said to be the Maclaurin series of f .
Definition 5.49 Let D be an open nonempty set in C. We say that the complex
function f : D → C is expandable in a Taylor series about the point z0 ∈ D if there
∞
X
exists r &gt; 0 such that B(z0 , r) ⊆ D and there exists a power series
an (z − z0 )n
n=0
convergent in B(z0 , r) such that, for any z ∈ B(z0 , r), we have:
f (z) =
∞
X
n=0
an (z − z0 )n .
(5.6)
Definition 5.50 A function f : D → C is called analytic on the open set D if it is
expandable in a convergent Taylor series about every point z0 in D.
Remark 5.51 This definition characterizes the class of the functions which can be
locally approximated by convergent power series.
So, a function f defined on an open set D is said to be analytic (or to possess a power
∞
X
series expansion) at a point z0 ∈ D if there exists a power series
an (z − z0 )n ,
with a positive radius of convergence, such that
f (z) =
∞
X
n=0
for any z in a neighbourhood of z0 .
an (z − z0 )n ,
n=0
95
TAYLOR AND LAURENT SERIES
Remark 5.52 Let us note that any power series with a positive radius of convergence defines an analytic function on the interior of its region of convergence.
Elementary operations with analytic functions can be performed. If f and g are
analytic at a point z0 , then f &plusmn; g, c f (where c is a constant) and f g are analytic
functions at z0 , too. Also, if h is analytic at f (z0 ), then (h ◦ f )(z) is analytic at z0 .
Theorem 5.53 A function f : D → C is holomorphic on the open set D if and only
if f is analytic on D.
So, the terms holomorphic and analytic can be used in an interchangeable manner.
Definition 5.54 Let f : D ⊆ C be holomorphic on a domain D. A point z0 ∈ D
is called a zero of the function f if f (z0 ) = 0. If there exists n ∈ N∗ such that
′
f (z0 ) = f (z0 ) = &middot; &middot; &middot; = f (n−1) (z0 ) = 0 and f (n) (z0 ) 6= 0, then z0 is called a zero of
order n for f .
Example 5.55 For the function f (z) = sin z − z, the point z0 = 0 is a third order
′
′′
′′′
zero. Indeed, it is not difficult to see that f (0) = f (0) = f (0) = 0 and f (0) 6= 0.
Remark 5.56 If f is holomorphic on a domain D ⊆ C and z0 is a zero of order
n for f , then there exists a function g, holomorphic on D, such that g(z0 ) 6= 0 and
f (z) = (z − z0 )n g(z), for all z ∈ D.
Remark 5.57 It is not difficult to see that if f is holomorphic on a domain D ⊆ C,
has a zero at the point z0 ∈ D and does not vanish identically in D, then there exists
a neighbourhood U ∈ V(z0 ), U ⊆ D, a non-vanishing holomorphic function g on U
and a unique n ∈ N∗ such that
f (z) = (z − z0 )n g(z),
for any z ∈ U.
Proposition 5.58 Let f and g be holomorphic on a domain D ⊆ C. Then, f ≡ g
if and only if there exists z0 ∈ D such that f (n) (z0 ) = g(n) (z0 ), for any n ∈ N.
96
COMPLEX ANALYSIS
Remark 5.59 If f and g are two holomorphic functions on a domain D ⊆ C and
f (z) = g(z) for all z in a nonempty open subset of D, then f (z) = g(z), for any
z ∈ D.
Theorem 5.60 (The Maximum Modulus Principle) If f : D → C is holomorphic
on a domain D ⊆ C and there exists z0 ∈ D such that
|f (z)| ≤ |f (z0 )|,
for any z ∈ D, then f is constant on D.
Remark 5.61 If f is holomorphic on a bounded domain D and continuous on D,
then
max |f (z)| = max |f (z)| .
z∈D
z∈∂D
Theorem 5.62 (The Analytic Continuation Principle) Let f : D → C be holomore ⊇ D and a function fe,
phic on a domain D. Assume that there exist a domain D
e and such that fe | = f . Then, fe is unique. The function fe is
holomorphic on D
D
said to be the analytic continuation of f into fe.
e with D ⊆ D
e and
So, if f and fe are analytic on the domains D and, respectively, D,
if they agree on D, then fe is said to be an analytic continuation of f into the domain
e The above theorem states that there is only one such analytic continuation, i.e.
D.
fe is uniquely determined by f .
Exercise 5.63 Prove that if f : D → C is holomorphic on a domain D ⊆ C and
there exists a point z0 ∈ D such that f (n) (z0 ) = 0, for any n ∈ N, then f (z) = 0, for
any z ∈ D.
Exercise 5.64 Prove that if f : D → C is holomorphic on a domain D ⊆ C and the
set {z ∈ D | f (z) = 0} has an accumulation point contained in D, then f (z) = 0,
for any z ∈ D.
Exercise 5.65 Prove that
1
= 1 + z + z2 + &middot; &middot; &middot; ,
1−z
for |z| &lt; 1.
97
TAYLOR AND LAURENT SERIES
Exercise 5.66 Show that
1
= 1 − z + z2 − &middot; &middot; &middot; ,
1+z
for |z| &lt; 1.
Exercise 5.67 Prove that
and
1
= 1 + 2z + 3z 2 + &middot; &middot; &middot; ,
(1 − z)2
for |z| &lt; 1
1
= 1 − 2z + 3z 2 − &middot; &middot; &middot; ,
(1 + z)2
for |z| &lt; 1.
Exercise 5.68 Let f : C \ {0, 1} → C, defined by
f (z) =
1
.
z(z − 1)
Expand f in a Taylor series about the point z = 0.
Exercise 5.69 Find the Maclaurin series for the error function erf (z), defined by
2
erf (z) = √
π
Zz
2
e−t dt.
0
Exercise 5.70 Find a power series representation about the origin for the following
functions:
a)
f (z) = cos(z 2 );
b)
f (z) = z 3 sin(z 2 );
c)
f (z) = (sin z)2 .
Exercise 5.71 Find the Taylor series expansion of the function f : C \ {&plusmn;i} → C
given by
1
1 + z2
about the point z = 0 and compute its radius of convergence.
f (z) =
98
COMPLEX ANALYSIS
Solution. We have
f (z) =
∞
∞
k=0
k=0
X
k X
1
=
−z 2 =
(−1)k z 2k .
2
1 − (−z )
So, the radius of convergence of the above power series is R = 1.
....
Solution Manual -desktop .......
5.3
Laurent Series
As we saw, Taylor series are suited for representing holomorphic functions on a disk
centered at a given point z0 . Now, we shall consider more general power series, containing both positive and negative powers of (z−z0 ), and characterizing holomorphic
functions in annuli, i.e. in sets of the form
D = {z ∈ C | r &lt; |z − z0 | &lt; R},
with z0 ∈ C and 0 ≤ r &lt; R ≤ +∞. Using such a generalized expansion, we shall be
able to describe the behavior of a function having a singularity at the point z0 .
Definition 5.72 Let z0 ∈ C and an ∈ C. A series of the form
∞
X
n=−∞
an (z − z0 )n
(5.7)
is called a Laurent series.
So, a Laurent series is a series of functions of the form
∞
X
n=−∞
an (z − z0 )n = &middot; &middot; &middot; +
a−n
a−1
+&middot;&middot;&middot;
+ a0 + &middot; &middot; &middot; + an (z − z0 )n + &middot; &middot; &middot; . (5.8)
n
(z − z0 )
z − z0
In fact, a Laurent series about a given point z0 is a sum of two independent power
series, one consisting of positive powers of (z − z0 ) and the other one of negative
powers of (z − z0 ):
∞
X
n=0
n
an (z − z0 ) +
∞
X
n=1
a−n (z − z0 )−n .
99
TAYLOR AND LAURENT SERIES
Still, we shall often use the short notation (5.8), despite the fact that, conceptually,
a Laurent series is the sum of two distinct power series.
Definition 5.73 The series
∞
X
n=1
a−n (z − z0 )−n is called the principal part or the
singular part of the Laurent series (5.8). The series
∞
X
n=0
Taylor part or the regular part of the series (5.8).
an (z − z0 )n is said to be the
Remark 5.74 Any power series is a Laurent series, with an = 0, for n &lt; 0.
∞
X
Definition 5.75 The Laurent series
n=−∞
an (z − z0 )n converges on a set D ⊆
C \ {z0 } if both its principal part and its Taylor part are convergent on D. Moreover,
if we denote by Π(z) and T (z) the sums of the principal part and, respectively, the
Taylor part of the series (5.8), then the sum
S(z) =
∞
X
n=−∞
an (z − z0 )n
of the Laurent series (5.8) is defined as being
∀z ∈ D ⊆ C \ {z0 }.
S(z) = Π(z) + T (z),
Theorem 5.76 (Laurent) For the Laurent series (6.8), let
r = lim
n→∞
and
R=
lim
p
n
n→∞
with the convention that R = 0 if
lim
n→∞
|a−n |
1
p
n
|an |
(5.9)
,
p
n
|an | = ∞ and R = ∞ if
(5.10)
lim
n→∞
p
n
|an | = 0.
a) If r &lt; R, the Laurent series (5.8) is absolutely and uniformly convergent over
compact sets in the annulus U (z0 ; r, R) and divergent in C \ U (z0 ; r, R).
100
COMPLEX ANALYSIS
b) If R &lt; r, the Laurent series (5.8) is divergent on C.
c) If R = r, we may have points of convergence only on the circle ∂B(z0 , r).
It is not difficult to see that the sum
S(z) =
∞
X
n=−∞
an (z − z0 )n
is a holomorphic function in the annulus U (z0 ; r, R).
Theorem 5.77 If f is holomorphic in the annulus D = U (z0 ; r, R), with 0 ≤ r &lt;
R ≤ ∞, then f can be expanded uniquely as
f (z) =
∞
X
n=−∞
an (z − z0 )n ,
for any z ∈ D,
(5.11)
with the coefficients an given, for n ∈ Z, by
1
an =
2πi
Z
γ
f (ξ)
dξ.
(ξ − z0 )n+1
(5.12)
Here, γ = ∂B(z0 , ρ), with r &lt; ρ &lt; R.
Remark 5.78 The Laurent expansion of a function f which is holomorphic on a
whole disk D centered at a point z0 has no principal part and its regular part coincides
with the power expansion of f on the disk D.
Definition 5.79 Let f be a single-valued holomorphic function defined on an open
set D ⊆ C. A point z0 ∈ C is called an isolated singular point for the function f if
z0 ∈
/ D, but there exists a punctured neighbourhood of z0 contained in D, i.e there
exists R &gt; 0 such that U̇ (z0 ; R) ⊆ D.
So, a function f has an isolated singularity at a point z0 if f is differentiable in some
punctured disk centered at z0 and f is not defined at z0 .
101
TAYLOR AND LAURENT SERIES
Example 5.80 For the function f : C \ {0, &plusmn;4i} → C, defined by
f (z) =
z+3
,
+ 16)
z 3 (z 2
the points z = 0 and z = &plusmn;4i are isolated singularities.
Remark 5.81 Let us notice that if z0 is an isolated singularity for f , then z0 belongs
to the boundary of D and D ∪ {z0 } is an open set. Moreover, if D is a domain, then
D ∪ {z0 } is a domain, too.
Definition 5.82 Let f : D ⊆ C → C be a holomorphic function on an open set D
and z0 ∈ C be an isolated singular point for f .
a)
The point z0 is said to be a removable singularity if f is holomorphically exe = D ∪ {z0 }.
tendable to D
b) The point z0 is called a pole if there exists lim f (z) = ∞.
z→z0
c) The point z0 is said to be an essential singularity if the limit lim f (z) does not
z→z0
exist.
Remark 5.83 The points at which f is holomorphic, together with those at which
f has a removable singularity, are said to be regular points for f .
Remark 5.84 The point z0 is a removable singular point for f if and only if there
exists lim f (z) and it is finite. We shall often use the same notation f for the
z→z0
e = D ∪ {z0 }, by putting f (z0 ) = lim f (z).
extension of f to D
z→z0
Remark 5.85 Let D ⊆ C be an open subset of the complex plane, f ∈ H(D) and
z0 an isolated singular point of f . There are several ways to decide if an isolated
singularity is a removable one. More precisely, it is not difficult to see that the
following statements are equivalent:
(i) f is holomorphically extendable to z0 ;
(ii) f is continuously extendable to z0 ;
102
COMPLEX ANALYSIS
(iii) there exists a deleted neighbourhood of z0 on which f is bounded;
(iv) lim (z − z0 ) f (z) = 0.
z→z0
Remark 5.86 Let f ∈ H(D) and z0 an isolated singular point for f . The point z0
1
is a pole of order n for f if z0 is a removable singularity for
and a zero of order
f
n for the holomorphic extension to z0 of this function.
Remark 5.87 Let us notice that if z0 is a pole for f , then there exist a unique
n ∈ N∗ and a unique function g ∈ H(D ∪ {z0 }) such that g(z0 ) 6= 0 and
f (z) = (z − z0 )−n g(z),
for any z ∈ D.
In fact, f has a pole of order n at z0 if n is the smallest positive integer for which
(z − z0 )n f (z) is holomorphic at z0 .
An important tool for analyzing the behavior of a holomorphic function f about an
isolated singularity z0 is offered by its Laurent series around z0 . Let f ∈ H(D), z0
an isolated singularity for f and R &gt; 0 such that U (z0 ; 0, R) ⊆ D. Then, in this
annulus, we have
f (z) =
∞
X
n=−∞
with
1
an =
2πi
Z
γ
f (ξ)
dξ,
(ξ − z0 )n+1
an (z − z0 )n ,
γ = ∂B(z0 , r), 0 &lt; r &lt; R.
(5.13)
(5.14)
Theorem 5.88 Let f ∈ H(D) and z0 ∈ C be an isolated singular point for f .
a) The point z0 is a removable singularity for f if and only if the principal part
of the Laurent series of f about z0 in a punctured disk centered at z0 is identically
zero.
b) The point z0 is a pole for f if and only if the principal part of the Laurent
series of f about z0 in a punctured disk centered at z0 consists only of a finite number
103
TAYLOR AND LAURENT SERIES
of terms, i.e. there exists a unique n ∈ N∗ such that, in a punctured disk centered at
z0 , we have
f (z) =
a−n
a−1
+ &middot;&middot;&middot; +
+ a0 + a1 (z − z0 ) + &middot; &middot; &middot; ,
n
(z − z0 )
z − z0
(5.15)
with a−n 6= 0. The integer n, called the order or the multiplicity of the pole, describes
the rate at which the function grows near z0 .
c) The point z0 is an essential singularity if and only if the principal part of
the Laurent series of f about z0 in a punctured disk centered at z0 has an infinite
number of terms.
Remark 5.89 Let us notice that there are functions that have non-isolated singular
points. For example, the function
f (z) =
1
sin
1
z
1
, for k = &plusmn;1, &plusmn;2, . . . . The
kπ
points zk are simple poles that accumulate in z = 0. Thus, z = 0 is a non-isolated
has singular points at z = 0 and, respectively, at zk =
singular point for f (it is an accumulation point of poles).
Definition 5.90 Let D be an open nonempty subset of C. A function f is called
meromorphic on D if there exists a set E ⊆ D such that f ∈ H(D \ E) and E
contains only removable singularities or poles for f .
If we denote by M(D) the set of all the meromorphic functions on D, it follows that
H(D) ⊆ M(D).
The above definition holds true also for D ⊆ C∞ .
Definition 5.91 Let f ∈ H(D). If there exists r &gt; 0 such that {z ∈ C | |z| &gt; r} ⊆
D, i.e. f is holomorphic on the domain r &lt; |z| &lt; ∞, then the point at infinity
z = ∞ is said to be an isolated singular point for f .
104
COMPLEX ANALYSIS
In order to study the behavior of the function f at ∞, we shall consider the function
1
g(ξ) = f
,
ξ
which is holomorphic on U̇
for g.
1
0;
. In other words, ξ = 0 is an isolated singularity
r
We shall say that z = ∞ is a removable singularity, a pole of order n or, respec-
tively, an essential isolated singular point for f if ξ = 0 is a removable singularity, a
pole of order n or, respectively, an essential isolated singular point for g.
Example 5.92 Let us analyze the behavior of the function
f (z) =
1
z(z + 1)
in a neighbourhood of the point at infinity. Let
z=
and
1
ξ
1
g(ξ) = f
.
ξ
Expanding g for small |ξ|, we get
g(ξ) = ξ 2 − ξ 3 + ξ 4 − &middot; &middot; &middot; .
Therefore, the point z = ∞ is a regular point for the function f .
Exercise 5.93 Expand in a Laurent series about the point z0 = −1 the function
f (z) =
1
1 − z2
in the annulus D = {z ∈ C | 0 &lt; |z + 1| &lt; 2}.
105
TAYLOR AND LAURENT SERIES
Solution. The function f is holomorphic in the annulus D. Thus, on D, it can be
expanded in a convergent Laurent series:
f (z) =
∞
X
an (z + 1)n .
n=−∞
We have
f (z) =
1
=
1 − z2
1
z+1
2(z + 1) 1 −
2
.
Therefore,
f (z) =
X (z + 1)n
1
1 1
1
=
+ + (z + 1) + &middot; &middot; &middot; .
n
2(z + 1)
2
2(z + 1) 4 8
n≥0
This expansion is valid provided that z ∈ D.
Exercise 5.94 Let D = {z ∈ C | 0 &lt; |z − i| &lt; ∞}. Expand the holomorphic
function f : D → C, defined by
f (z) =
e2z
,
(z − i)3
in the annulus D, in a convergent Laurent series about the point z = i.
Solution. We have
e2z
e2i
e2i
2(z−i)
=
e
=
3
3
(z − i)
(z − i)
(z − i)3
So,
f (z) =
1 + 2(z − i) +
(2(z − i))2
+ &middot;&middot;&middot; .
2!
e2i
2e2i
2e2i
4e2i 2e2i
+
+
+
+
(z − i) + &middot; &middot; &middot; .
(z − i)3 (z − i)2 (z − i)
3
3
Exercise 5.95 Expand in a Laurent series about the point z0 = 1 the function
f (z) =
1
1 − z2
in the annulus D = {z ∈ C | 2 &lt; |z − 1| &lt; ∞}.
106
COMPLEX ANALYSIS
Exercise 5.96 Develop in a Laurent series in the punctured disk D = {z ∈ C | 0 &lt;
|z| &lt; 1} the function
f (z) =
1
.
z(z − 1)
Exercise 5.97 Find the Laurent series about the point z0 = 1 for the function
f (z) = sin
1
z−1
in the annulus D = {z ∈ C | 0 &lt; |z − 1| &lt; ∞}.
Exercise 5.98 Let z0 ∈ C, an ∈ C and 0 ≤ r &lt; R ≤ ∞. Prove that if the Laurent
∞
X
an (z − z0 )n is convergent in the annulus D = U (z0 ; r, R) and its sum is
series
n=−∞
zero in D, then an = 0, for any n ∈ Z.
Exercise 5.99 Prove that, in the annulus D = {z ∈ C | 3 &lt; |z| &lt; ∞}, the function
f (z) =
1
z 4 + 9z 2
can be expanded in the following Laurent series:
f (z) =
n
∞
1 X
9
n
(−1)
.
z 4 n=0
z2
Exercise 5.100 Let z0 ∈ C, an , bn ∈ C and 0 ≤ r &lt; R ≤ ∞. Prove that if the
∞
∞
X
X
Laurent series
an (z − z0 )n and
bn (z − z0 )n are convergent in the annulus
n=−∞
n=−∞
D = U (z0 ; r, R) and their sums are equal in D, then an = bn , for any n ∈ Z.
Exercise 5.101 Let f : C \ {0} → C be defined by
f (z) =
sin z
.
z
Show that the point z0 = 0 is a removable singularity for f .
107
TAYLOR AND LAURENT SERIES
Solution Due to the fact that
lim f (z) = 1,
z→0
we can extend f to C by defining the function f at origin as being f (0) = 1.
Exercise 5.102 Let f : C \ {1} → C be defined by
f (z) =
z
.
z−1
Prove that the point z0 = 1 is a pole for f .
Solution. We have
lim f (z) = ∞.
z→1
Hence, the point z0 = 1 is a pole for f .
Exercise 5.103 Let us consider f : C \ {0} → C, defined by
1
2
f (z) = e z .
Show that z0 = 0 is an essential singularity for the function f .
Solution. Since the limit lim f (z) doesn’t exist, the point z0 = 0 is an essential
z→0
singularity for the function f .
Exercise 5.104 Show that the function f : C \ {0} → C, defined by
f (z) =
1 − cos z
,
z5
has a third order pole at z0 = 0.
Solution. In the annulus D = {z ∈ C | 0 &lt; |z| &lt; ∞}, the function f is holomorphic
and can be expanded in the following Laurent series:
f (z) =
1
1
z
−
+ − &middot;&middot;&middot; ,
3
2! z
4! z 6!
which implies that z0 = 0 is a pole of order three for f .
108
COMPLEX ANALYSIS
Exercise 5.105 Prove that the function f : C \ {0} → C, defined by
1
f (z) = sin ,
z
has an essential singularity at z0 = 0.
Solution. In the annulus D = {z ∈ C | 0 &lt; |z| &lt; ∞}, the function f is holomorphic
and can be expanded in the following Laurent series:
f (z) =
1
1
1
+
− &middot;&middot;&middot; ,
−
3
z 3! z
5! z 5
which implies that z0 = 0 is an essential singularity for f .
Exercise 5.106 Find the zeros and their multiplicities of the following functions:
a)
4
f (z) = 1 + z 2 ;
b)
f (z) = z 4 cos z;
c)
f (z) = sin2 z.
Exercise 5.107 Determine the power series of f (z) = ez centered at the point
z = −2.
Exercise 5.108 Find the poles of the following functions and determine their orders:
a)
f (z) =
b)
f (z) =
c)
f (z) =
z
(z 2
+ 1) (z − 2)2
sin z
;
z6
ez
4
z
.
−1
;
109
TAYLOR AND LAURENT SERIES
5.4
Residues
We shall discuss now the concept of residue of a single-valued holomorphic function
f at an isolated singular point z0 . Let us recall that if we consider f ∈ H(D), z0 an
isolated singularity for f and R &gt; 0 such that U (z0 ; 0, R) ⊆ D, then, in this annulus,
the function f can be expanded in a Laurent series of the form
f (z) =
∞
X
n=−∞
with
1
an =
2πi
Z
γ
f (ξ)
dξ,
(ξ − z0 )n+1
an (z − z0 )n ,
(5.16)
γ = ∂B(z0 , r), 0 &lt; r &lt; R.
(5.17)
Definition 5.109 Let z0 ∈ C and f be a regular function in the punctured disk
D = {z ∈ C | 0 &lt; |z − z0 | &lt; R}. The coefficient a−1 of the Laurent series (6.19) of
f about the point z0 in D is called the residue of the function f at the point z0 and
is denoted by Res (f, z0 ).
So,
1
Res (f, z0 ) =
2πi
Z
f (z) dz,
(5.18)
γ
where γ = ∂B(z0 , r), with 0 &lt; r &lt; R. Rearranging this formula, we get
Z
f (z) dz = 2πi Res (f, z0 )
(5.19)
γ
and, hence, the value of the residue of f at the point z0 is instrumental for evaluating
complex integrals. In fact, formula (6.22) holds true for any simple closed contour
in D having the index n(γ, z0 ) = 1. Let us notice that the only term in the Laurent
series expansion for f that has a nonzero contribution to the integral in (6.22) is
a−1 and this justifies the use or the name residue for this coefficient.
Example 5.110 If D = {z ∈ C | 0 &lt; |z| &lt; 1} and f (z) =
f (z) =
1
, then
− z)
z 3 (1
1
1
1
+
+ + 1 + z + z2 + &middot; &middot; &middot; .
z3 z2 z
110
COMPLEX ANALYSIS
In this case, the residue of the function f at the point z0 = 0 is a−1 = 1.
Example 5.111 If D = {z ∈ C | 0 &lt; |z| &lt; 1} and f (z) =
f (z) =
ez
, then
z3
1
1
1
1
z
+ 2+
+ + + &middot;&middot;&middot;
3
z
z
2! z 3! 4!
So, the residue of the function f at the point z0 = 0 is a−1 =
1
.
2
1/z
Example 5.112 If D = {z ∈ C | 0 &lt; |z| &lt; +∞} and f (z) = e
f (z) = &middot; &middot; &middot; +
, then
1
1
1
+
+ +1
3
2
3! z
2! z
z
The residue of the function f at the point z0 = 0 is a−1 = 1.
Let us see now how we can compute the residue of a given function f at an
isolated singular point z0 . As we saw, such a point can be a removable singularity,
a pole or an essential singularity.
I. If z0 is a regular point for f , then
Res (f, z0 ) = 0.
(5.20)
II. If z0 is a first order pole for f , then
Res (f, z0 ) = lim (z − z0 ) f (z).
z→z0
(5.21)
In particular, if
f (z) =
g(z)
,
h(z)
′
with g, h ∈ H(D), g(z0 ) 6= 0, h(z0 ) = 0, h (z0 ) 6= 0, then
Res (f, z0 ) =
g(z0 )
.
h′ (z0 )
Indeed, since in a punctured neighbourhood of z0 ,
f (z) =
a−1
+ a0 + a1 (z − z0 ) + &middot; &middot; &middot; ,
(z − z0 )
(5.22)
111
TAYLOR AND LAURENT SERIES
it follows that
(z − z0 ) f (z) = a−1 + a0 (z − z0 ) + a1 (z − z0 )2 + &middot; &middot; &middot; ,
which gives immediately (6.24).
If f (z) =
g(z)
, then
h(z)
lim (z − z0 )f (z) = lim
z→z0
z→z0
g(z)
g(z0 )
= ′
.
h(z) − h(z0 )
h (z0 )
z − z0
III. If z0 is a pole of order m, then
Res (f, z0 ) = lim
z→z0
1
dm−1
[(z − z0 )m f (z)] .
(m − 1)! dz m−1
(5.23)
Indeed, since in a punctured neighbourhood of z0 ,
f (z) =
a−m
a−1
+ &middot;&middot;&middot;
+ a0 + a1 (z − z0 ) + &middot; &middot; &middot; ,
m
(z − z0 )
z − z0
it follows that
[(z − z0 )m f (z)](m−1) = (m − 1)! a−1 + m! a0 (z − z0 ) + &middot; &middot; &middot; ,
which implies that
lim [(z − z0 )m f (z)](m−1) = (m − 1)! a−1 .
z→z0
IV. If z0 is an essential singularity, then
Res (f, z0 ) = a−1 .
So, in this case, the residue of f can be computed only as being the coefficient a−1
in the Laurent series of f about z0 .
112
COMPLEX ANALYSIS
Example 5.113 For the function f : C \ {0, 1} → C, defined by
f (z) =
1
,
z 2 (1 − z)
the point z = 0 is a second order pole and the point z = 1 is a simple pole. Then,
it is not difficult to see that
Res (f, 0) = 1,
Res (f, 1) = −1.
Example 5.114 Let f : C \ {&plusmn;i} → C, defined by
f (z) =
Then,
1
Res (f, i) = ,
2
z
.
z2 + 1
1
Res(f, −i) = .
2
Example 5.115 For the function f : C \ {(2k + 1)πi | k ∈ Z} → C, defined by
f (z) =
ez
1
,
+1
the residue at the point z0 = πi, which is a simple pole, can be computed using
formula (6.25). We get
Res (f, πi) = −1.
Example 5.116 Let f : C \ {1} → C, defined by
1
1
z−1
f (z) =
e
.
z−1
The point z0 = 1 is an essential singularity for f . Indeed, in the annulus {0 &lt;
|z − 1| &lt; ∞}, the function f can be expanded in a Laurent series of the form
f (z) =
1
1
1
+
+
+ &middot;&middot;&middot; .
z − 1 (z − 1)2 2! (z − 1)3
Thus, we obtain
Res (f, 1) = 1.
113
TAYLOR AND LAURENT SERIES
Exercise 5.117 Locate and classify the singularities of the following functions:
a) f : C \ {0, 2} → C,
f (z) =
b) g : C∗ → C,
g(z) = z 2 e1/z ;
c) h : C∗ → C,
h(z) =
z3 + 1
;
z 4 (z − 2)
sin z − z
.
z2
Exercise 5.118 Find and classify the singularities of the function f : C \{1, 4} → C,
given by
f (z) =
z ez
.
(z − 1)2 (z − 4)3
Exercise 5.119 Compute the residue of the function f : C∗ → C, defined by
1
f (z) = z 3 cos ,
z
at the point z = 0.
Exercise 5.120 Evaluate the residue of the function f : C \ {i} → C, defined by
1
f (z) = z sin
,
iz + 1
at the point z = i.
Exercise 5.121 Compute the residue at the point z = 0 of the function
f (z) =
sin z
.
z3
Solution. The function f has a double pole at z = 0 and
Res (f, 0) = 0.
Exercise 5.122 Compute the residue at the point z = 0 of the functions:
a)
f (z) =
cos z
;
z4
b)
f (z) =
ez − 1
.
sin2 z
114
COMPLEX ANALYSIS
Let us introduce now the notion of residue at the point at infinity.
Definition 5.123 If z = ∞ is an isolated singular point for f , i.e. f is holomorphic
in a punctured neighbourhood D = U (0; r, +∞) of ∞, then f can be expanded in D
in a Laurent series
f (z) =
∞
X
cn z n
in r &lt; |z| &lt; ∞.
n=−∞
(5.24)
The residue of the function f at ∞ is defined as being the coefficient −c−1 in the
expansion (6.27), i.e.
Res (f, ∞) = −c−1 .
Remark 5.124 Let us notice that if we compute, for z = ∞, the integral of f along
a closed path lying in the punctured domain of holomorphy and having the index 1
with respect to z = ∞, we get
1
2πi
Z
γ
f (z) dz = −c−1 ,
(5.25)
where γ = γ1−1 , γ1 = ∂B(0, R), for r &lt; R, and this justifies the above definition.
Example 5.125 Let f : C \ {&plusmn;i} → C, defined by
f (z) =
z2
z
.
+1
For |z| &gt; 1, we can expand f in the following Laurent series:
f (z) =
1
1
1
−
+
− &middot;&middot;&middot;
z z3 z5
So,
Res (f, ∞) = −c−1 = −1.
Chapter 6
The Residue Theorem.
Applications
6.1
The Residue Theorem
Theorem 6.1 (Cauchy’s Residue Theorem) Let γ be a positively oriented simple
closed path in a simply connected domain D. We assume that f is analytic everywhere on D, except at a finite number of isolated singularities z1 , . . . , zn lying in the
interior of γ. Then,
Z
f (z) dz = 2πi
n
X
Res (f, zk ).
(6.1)
k=1
γ
Theorem 6.2 If f is holomorphic on an open set D \ {zi }i∈I , where zi are isolated
singularities for f , then for any compact set K ⊂ D with γ = ∂K being a piecewise
smooth path which doesn’t contain singular points for f , there exists a finite number
of points zi ∈ K and
Z
γ
f (z) dz = 2πi
X
Res (f, zi ).
(6.2)
zi ∈K
Theorem 6.3 Let f : D ⊆ C → C be holomorphic on the open set D and let S be
the set of the isolated singular points of f . If γ is a closed contour in D which is
115
116
COMPLEX ANALYSIS
homotopic to zero in D ∪ S, then the set {z ∈ S | n(γ, z) Res(f, z) 6= 0} is finite and
Z
X
f (z) dz = 2πi
n(γ, z) Res (f, z).
(6.3)
z∈S
γ
Remark 6.4 In what follows, in all the applications of Cauchy’s residue theorem,
we shall work only with paths γ having the winding numbers around all the involved
singularities equal to one or zero.
Proposition 6.5 Let f be holomorphic in D = C \ {z1 , z2 , . . . , zn } and let us denote
zn+1 = ∞. Then,
n+1
X
Res (f, zk ) = 0.
k=1
Example 6.6 Let us look at the integral
Z
z+1
I=
dz,
z (z 2 + 9)2
γ
where γ = ∂B(0, 7).
The function f has isolated singularities at the points 0 and &plusmn;3i and they belong to
the domain bounded by the contour γ. So, one way to compute the integral I is to
write
I = 2πi [Res (f, 0) + Res (f, 3i) + Res (f, −3i)] ,
which gives
I = 0.
On the other hand, we can evaluate the residue of the function f at z = ∞ and, then,
using Proposition 7.5, we can easily compute the value of the integral I. Indeed, for
|z| &gt; 3, we have
f (z) =
i.e.
f (z) =

z+1
z+1
=

z (z 2 + 9)2
z5
1
1
+ 5
4
z
z
1
2

,
9 
1+ 2
z
9
81
1 − 2 + 4 − &middot;&middot;&middot;
z
z
2
,
117
THE RESIDUE THEOREM. APPLICATIONS
which implies that
Res (f, ∞) = 0
and, hence,
I = 0.
Exercise 6.7 Compute the integral
Z
I=
|z|=2
4
dz.
(z 2 + 1)(z − 3)2
Solution. The function f has isolated singularities at the points z = &plusmn;i and z = 3.
Since only the singular points &plusmn;i, which are simple poles, belong to the domain
bounded by the closed contour γ, it follows that
I = 2πi [Res (f, −i) + Res (f, i)] =
12
πi.
25
Exercise 6.8 Compute the integral
I=
Z
γ
sin z
dz,
z6
where γ = ∂B(0, 3).
Solution. The function f can be expanded, in the annulus 0 &lt; |z| &lt; ∞, in a Laurent
series about the point z = 0 of the form
1
1
z3 z5
1
1
f (z) = 6 z −
+
− &middot;&middot;&middot; = 5 −
+
− &middot;&middot;&middot; .
3
z
3!
5!
z
3! z
5! z
So, z = 0 is a pole of order five and we obtain
I = 2πi Res (f, 0) =
Exercise 6.9 Evaluate the integral
Z
I=
γ
where γ = ∂B(0, 5).
2πi
.
5!
z+1
dz,
z (z 2 + 4)2
118
COMPLEX ANALYSIS
Solution. Since the singular points 0 and &plusmn;2i belong to the domain bounded by the
contour γ, we get
I = 2πi [Res (f, 0) + Res (f, 2i) + Res (f, −2i)] ,
i.e.
I = 0.
Exercise 6.10 Compute the integral
Z
I=
γ
1
dz,
ez + 1
where γ = ∂B(2i, 2).
Solution. Since z0 = πi is a first order pole for f which belongs to B(2i, 2) and all
the other singular points of f lie outside the domain bounded by γ, we obtain
I = 2πi Res (f, πi),
i.e.
I = −2πi.
Exercise 6.11 Compute the value of the integral
Z
z2 + 1
I=
dz,
ln z − πi
γ
where γ is the positively oriented circle |z + 1| =
on its principal branch.
1
and the logarithm is considered
2
Solution. We notice that z0 = −1 is a first order pole for f . Therefore,
I = 2πi Res (f, −1),
i.e.
I = −4πi.
119
THE RESIDUE THEOREM. APPLICATIONS
6.2
Evaluation of Real Integrals
Cauchy’s residue theorem proves to be a powerful tool for computing a wide range
of definite integrals. As we shall see, using this theorem, we shall be able to evaluate
some trigonometric integrals of the form
Z2π
I=
R (sin t, cos t) dt,
(6.4)
0
where R is a suitable real rational function, or important types of improper real
integrals of the form
I=
Z∞
f (x) dx.
(6.5)
−∞
I. Trigonometric Integrals
Let us consider trigonometric integrals of the form
I=
Z2π
R (sin t, cos t) dt,
0
where R is a real rational function of its arguments.
Theorem 6.12 Let R = R(u, v) be a real rational function of its arguments, which
has no poles on the circle u2 + v 2 = 1. Then,
Z2π
R (sin t, cos t) dt = 2πi
The notation
1
R
g(z) =
iz
X
Res (g, z),
z − z −1 z + z −1
,
2i
2
.
Res (g, z) means the sum of the residues of g at all the
|z|&lt;1
poles lying inside the disk |z| &lt; 1.
(6.6)
|z|&lt;1
0
where
X
(6.7)
120
COMPLEX ANALYSIS
Exercise 6.13 Compute
I=
Zπ
0
1
dt.
2 − cos t
Solution. We notice that the integral is defined only over [0, π]. Since cos (2π − t) =
cos t, we get
Z2π
π
1
dt =
2 − cos t
Zπ
0
1
dt,
2 − cos t
which implies that
1
I=
2
Z2π
0
1
dt.
2 − cos t
Therefore,
1
I=
2
Z
γ
1
dz
1
=−
1
1 iz
2
2−
z+
2
z
Z
γ
1
2
dz.
2
i z − 4z + 1
The function
1
1
g(z) = −
i z 2 − 4z + 1
√
√
has simple poles at z = 2 &plusmn; 3. Hence, since only the pole z = 2 − 3 lies inside
the unit disk and
Res (g, 2 −
we get
√
3) =
1
√ ,
2i 3
π
I=√ .
3
Remark 6.14 In a similar manner, we can treat integrals of the form
In =
Z2π
cos nt R (sin t, cos t) dt
Z2π
sin nt R (sin t, cos t) dt,
0
or
Jn =
0
121
THE RESIDUE THEOREM. APPLICATIONS
for n ∈ N. In this case, we consider the integral In + iJn and, using the same change
of variables and de Moivre’s formula, we can easily compute the above integrals.
Exercise 6.15 For a &gt; 1, show that
Z2π
0
1
2π
dt = √
.
a + cos t
a2 − 1
Exercise 6.16 Prove that
Z2π
0
1
2π
dt = √
,
1 + a cos t
1 − a2
for 0 &lt; |a| &lt; 1.
Exercise 6.17 Show that
Zπ
0
1
aπ
dt = 2
,
2
(a + cos t)
(a − 1)3/2
for a &gt; 1.
Exercise 6.18 For n ∈ N, prove that
Z2π
(1 + 2 cos t)n cos nt dt = 2π.
0
II. Improper Integrals
If the improper integral
I=
Z∞
f (x) dx
(6.8)
−∞
is convergent, the main idea for computing its value by using Cauchy’s residue
theorem will be to extend f to the complex plane and to choose a suitable family of
contours of integration γr such that
lim
r→∞
Z
γr
f (z) dz =
Z∞
f (x) dx.
−∞
In order to deal with integrals of the form (6.8), we shall make use of a famous
lemma due to Camille Jordan.
122
COMPLEX ANALYSIS
Lemma 6.19 (Jordan’s Lemma) Let D = {z ∈ C | Im z ≥ 0} and f : D → C be a
continuous function. If, for r &gt; 0, the path γr : [0, π] → C is defined by
γr (t) = r eit
and
lim f (z) = 0,
z→∞
then
lim
r→∞
Z
f (z) eiz dz = 0.
(6.9)
(6.10)
γr
Moreover, if f is an analytic function in the upper-half complex plane such that f
tends to zero, uniformly on any upper semi-circle γr centered at the origin and with
radius r → ∞, then, for a &gt; 0,
lim
r→∞
Z
eiaz f (z) dz = 0.
γr
Let us remark that if a &lt; 0, a similar result holds for the semicircle γr− lying in the
lower half-plane.
Lemma 6.20 Let f be a continuous function on the closed sector S0 [θ1 , θ2 ] and
γr (t) = r ei[θ1 +t(θ2 −θ1 )] ,
(i) If lim z f (z) = 0, then lim
z→∞
r→∞
Z
with t ∈ [0, 1].
f (z) dz = 0.
γr
(ii) If lim z f (z) = 0, then lim
z→0
r→0
Z
f (z) dz = 0.
γr
(iii) If θ1 = 0, θ2 = π/p and lim z 1−p f (z) = 0, then
r→∞
lim
r→∞
Z
γr
p
f (z) eiz dz = 0.
123
THE RESIDUE THEOREM. APPLICATIONS
(iv) If f is holomorphic on the sector S0 [θ1 , θ2 ] \ {0} and z = 0 is a simple pole for
f , then
lim
r→0
Z
γr
f (z) dz = (θ2 − θ1 ) i Res(f, 0).
Remark 6.21 The Jordan’s lemma is obtained from (iii) if we take p = 1.
II.1. Improper integrals of rational functions
We intent to evaluate now integrals of the form
I=
Z∞
R(x) dx,
−∞
with R a real rational function. We shall consider here only integrals for which
the integrand is a rational function with no real poles and such that the degree
of the denominator is at least two units higher than the degree of the numerator.
Therefore, the integrals will be convergent.
Theorem 6.22 Let R = P/Q be a real rational function, with P and Q polynomials
of degree n and, respectively, m. We suppose that Q has no zeros on the real axis
Z∞
and lim z R(z) = 0, i.e. n ≤ m − 2. Then, the integral
R(x) dx is convergent
|z|→∞
−∞
and
Z∞
R(x) dx = 2πi
X
Res (R, z).
(6.11)
Im z&gt;0
−∞
Let us notice that in (6.11) we consider the sum of the residues of the function
R = R(z) at all its poles that are situated in the upper half-plane.
Exercise 6.23 Evaluate the integral
I=
Z∞
0
2x2
dx.
(x2 + 1)(x2 + 4)
124
COMPLEX ANALYSIS
y
γr+
−r
x
r
O
Figure 6.1: The closed contour γ
Solution. Obviously,
1
I=
2
Z∞
−∞
2x2
dx.
(x2 + 1)(x2 + 4)
Let R : C \ {&plusmn;i, &plusmn;2i} → C, defined by
R(z) =
2z 2
.
(z 2 + 1)(z 2 + 4)
The function R has simple poles at the points &plusmn;i and &plusmn;2i. We define the closed
contour γ = γr+ ∨[−r, r], where r &gt; 2 and γr+ is the semicircle going counterclockwise
from (r, 0) to (−r, 0) (see Figure 6.1). The function R satisfies the conditions of
Theorem 6.22. Its singular points in the upper half-plane are the simple poles z1 = i
and z2 = 2i. Thus,
I = πi [Res (R, z1 ) + Res (R, z2 )] = πi
i
2i
−
3
3
=
Exercise 6.24 Show that
Z∞
0
(x2
1
π
dx = 3 ,
2
2
+a )
4a
whenever a &gt; 0.
π
.
3
125
THE RESIDUE THEOREM. APPLICATIONS
Exercise 6.25 Prove that
Z∞
−∞
x4
1
π
dx = √
,
4
+a
2a3
for a &gt; 0.
Exercise 6.26 Evaluate the integral
I=
Z∞
−∞
x2
dx.
(x2 + 1)2 (x2 + 9)
II.2. Improper integrals of products of rational and trigonometric functions
We shall consider now integrals of the following form:
I=
Z∞
R(x) eiax dx;
I=
−∞
Z∞
R(x) cos (ax) dx;
−∞
I=
Z∞
R(x) sin (ax) dx.
−∞
In order to deal with such integrals, we state the following result.
Theorem 6.27 Let R = P/Q be a rational real function, with P and Q polynomials
of degree n and, respectively, m. We assume that Q has no zeros on the real axis
Z∞
and lim R(z) = 0, i.e. n ≤ m − 1. Then, for a &gt; 0,
R(x) eiax dx is convergent
|z|→∞
−∞
and
Z∞
−∞
R(x) eiax dx = 2πi
X
Im z&gt;0
where
g(z) = R(z) eiaz .
Res (g, z),
(6.12)
126
COMPLEX ANALYSIS
Exercise 6.28 Compute the integral
Z∞
I1 =
−∞
x−1
cos 5x dx.
x2 − 2x + 5
Solution. We consider also the integral
I2 =
Z∞
−∞
x2
x−1
sin 5x dx.
− 2x + 5
We compute
I = I1 + iI2 ,
i.e.
I=
Z∞
−∞
x2
x−1
ei5x dx.
− 2x + 5
The rational function
z−1
z 2 − 2z + 5
is holomorphic on C \{1&plusmn;2i} and has simple poles at z = 1&plusmn;2i. Since the polynomial
R(z) =
z 2 − 2z + 5 has no roots on the real axis and lim R(z) = 0, using Theorem 6.27, we
z→∞
get
Z∞
R(x) ei5x dx = 2πi
X
Res (g, z),
(6.13)
Im z&gt;0
−∞
where
g(z) = R(z) ei5z .
Since in the upper half-plane lies only the simple pole z = 1 + 2i, computing the
residue of g at this pole, we get
i.e.
I = πi e−10 cos 5 + ie−10 sin 5 ,
I1 = Re I = −πe−10 sin 5
and
I2 = Im I = πe−10 cos 5.
127
THE RESIDUE THEOREM. APPLICATIONS
Exercise 6.29 Prove that
I=
Z∞
π
x sin αx
dx = e−α ,
2
1+x
2
for α &gt; 0.
0
Exercise 6.30 Show that
Z∞
−∞
e−a
cos x
dx
=
π
,
x2 + a2
a
for a &gt; 0.
x sin x
dx = π e−a ,
x2 + a2
for a &gt; 0.
Exercise 6.31 Prove that
Z∞
−∞
Exercise 6.32 Evaluate the following improper integrals:
a) I =
Z∞
x sin 2x
dx.
(x2 + 4)(x2 + 1)2
0
b) I =
Z∞
−∞
x2
x+2
sin 3x dx.
+ 2x + 5
III. The Poisson Integral
Using Cauchy’s residue theorem, we can prove that
Z∞
sin x
π
dx = .
x
2
0
Consider the function f : C∗ → C, defined by
f (z) =
eiz
,
z
128
COMPLEX ANALYSIS
which is holomorphic on C∗ . This function has a simple pole at z = 0 and an
essential singularity at z = ∞. Let 0 &lt; r &lt; R and the paths γR : [0, π] → C, defined
by γR (t) = R eit , and, respectively, γr : [0, π] → C, given by γr (t) = r ei(π−t) (see
Figure 6.2). Let
γ = γR ∨ [−R, −r] ∨ γr ∨ [r, R].
y
γR
γr
−R
−r
O
r
R
x
Figure 6.2: The contour of integration for the Poisson integral
So, using this indented semicircle and Cauchy’s residue theorem, it follows that
Z
γR
eiz
dz +
z
Z−r
eix
dx +
x
−R
Z
γr
eiz
dz +
z
ZR
r
eix
dx = 0.
x
(6.14)
ZR
(6.15)
Since
eix = cos x + i sin x,
it follows that
Z−r
−R
eix
dx +
x
ZR
r
eix
dx =
x
ZR
r
eix − e−ix
dx = 2i
x
r
sin x
dx.
x
THE RESIDUE THEOREM. APPLICATIONS
129
Using Jordan’s lemma, we get
lim
Z
eiz
dz = 0.
z
(6.16)
eiz
dz = −iπ.
z
(6.17)
R→∞
γR
Also,
lim
r→0
Z
γr
Hence, for r → 0 and R → ∞, from (7.20)-(7.23), we obtain
Z∞
sin x
π
dx = .
x
2
0
IV. Other Improper Integrals
Using the same techniques, we can compute other important types of real integrals.
Exercise 6.33 Show that
Z∞
−∞
eax
π
dx =
,
1 + ex
sin πa
for 0 &lt; a &lt; 1.
Solution. Let us consider the function
f (z) =
eaz
,
1 + ez
which is holomorphic on C \ {(2k + 1)πi | k ∈ Z} and a contour consisting of a
rectangle in the upper half-plane with one side lying on the real axis and another
one parallel with the real axis and situated at the level of the line Im z = 2π (see
Figure 6.3). More precisely, if R ∈ (0, ∞), we take
ΓR = [−R, R] ∨ [R, R + 2πi] ∨ [R + 2πi, −R + 2πi] ∨ [−R + 2πi, −R].
130
COMPLEX ANALYSIS
y
−R + 2π i
R + 2π i
πi
−R
O
R
x
Figure 6.3: The contour ΓR
The only singularity of f in this rectangle is z0 = πi, which proves to be a simple
pole. Then,
Res (f, z0 ) = −eaπi
and, using the residue theorem, we get
Z
f (z) dz = −2πi eaπi .
ΓR
Let
IR =
ZR
f (x) dx,
Z∞
eax
.
1 + ex
−R
which tends, when R → ∞, to
I=
−∞
Let us evaluate now the integrals of f over each side of the rectangle. It is not
difficult to see that the integrals over the vertical sides tend to zero as R → ∞, while
the integral on the top side of the contour is equal to −e2πia IR .
Therefore, when R tends to infinity, we get
I − e2aπi I = −2πi eaπi ,
131
THE RESIDUE THEOREM. APPLICATIONS
i.e.
I=
π
.
sin πa
In a similar manner, we can compute the Fresnel’s integrals, occurring in the theory
of diffraction. We have
Z∞
2
cos x dx =
0
Z∞
1
sin x dx =
2
2
0
r
π
.
2
Exercise 6.34 Show that
Z∞
−∞
x sin x
dx = πe−1 .
x2 + 1
Exercise 6.35 Prove that
Z∞ √
0
π x ln x
π
√
dx
=
ln
a
+
,
x2 + a2
2
2a
for a &gt; 0.
Exercise 6.36 Show that
Z∞
0
x ln x
1
dx = − .
(x2 + 1)3
8
Exercise 6.37 Evaluate the following improper integrals:
a) I =
Z∞
sin ax
dx,
x(x2 + b2 )2
0
b) I =
Z∞ √
0
x ln x
dx.
(1 − x)2
for a, b &gt; 0.
132
COMPLEX ANALYSIS
Remark 6.38 Let f be a continuous real function. An improper integral of the
form (6.8) is convergent if there exists and it is finite
lim
R→∞
r → −∞
ZR
f (x) dx.
r
In this case,
Z∞
f (x) dx =
−∞
lim
R→∞
r → −∞
ZR
f (x) dx.
r
If there exists
lim
ZR
R→∞
−R
f (x) dx
and it is finite, the integral (6.8) is said to be convergent in the sense of Cauchy
principal value. In this case, the limit is called the Cauchy principal value of the
integral I. We use the notation
p.v.
Z∞
−∞
f (x) dx = lim
ZR
R→∞
−R
f (x) dx.
If the integral (6.8) is convergent, then it is convergent also in the Cauchy sense,
the converse implication being, in general, false. In the case in which the integral
(6.8) is not convergent and exists only as an improper Lebesgue integral, we have
to consider its Cauchy principal value, i.e. the value we must assign to such an
improper integral is the principal value integral.
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