A FRIENDLY APPROACH TO
COMPLEX ANALYSIS
Sara Maad Sasane • Amal Sasane
,
►world Scientific
A FRIENDLY APPROACH TO
COMPLEX ANALYSIS
A FRIENDLY APPROACH TO
COMPLEX ANALYSIS
Sara Maad Sasane
Lund University, Sweden
Amol Sasane
London School of Economics, UK
1!) World Scientific
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A FRIENDLY APPROACH TO COMPLEX ANALYSIS
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Preface
We give an overview of what complex analysis is about and why it is im­
portant. As the student must have learnt the notion of a complex number
at some point, we will use that familiarity in our discussion here. Later on,
starting from Chapter 1 onwards, we will start things from scratch again.
So the reader should not worry about being lost in this preface!
What is Complex Analysis?
In real analysis, one studies (rigorously) calculus in the setting of real num­
bers. Thus one studies concepts such as the convergence of real sequences,
continuity of real-valued functions, differentiation and integration. Based
on this, one might guess that in complex analysis, one studies similar con­
cepts in the setting of complex numbers. This is partly true, but it turns out
that up to the point of studying differentiation, there are no new features
in complex analysis as compared to the real analysis counterparts. But
the subject of complex analysis departs radically from real analysis when
one studies differentiation. Thus, complex analysis is not merely about
doing analysis in the setting of complex numbers, but rather, much more
specialized:
Complex analysis is the study of "complex differentiable" functions.
Recall that in real analysis, we say that a function f : JR: -+ JR: is differentiable
at x0 E JR: if there exists a real number L such that
lim f(x) - f(xo) = L,
x➔xo
X - Xo
V
A Friendly Approach to Complex Analysis
vi
that is, for every E > 0, there is a 8 > 0 such that whenever 0 < lx-xol < 8,
there holds that
x - x o) _
I f( ) f(
x-xo
LI <
E.
In other words, given any distance E, we can make the difference quotient
f(x) - f(xo)
x-xo
lie within a distance of E from the real number L for all x sufficiently close
to, but distinct from, xo.
In the same way, we say that a function f : (C --+ (C is complex differen­
tiable at z0 E (C if there exists a complex number L such that
z - zo
lim f ( ) f( )
z - Zo
z-+zo
= L,
that is, for every E > 0, there is a 8 > 0 such that whenever O < lz -zol < 8,
there holds that
I
f(z) - J(zo) _
z-zo
LI
<
E.
The only change from the previous definition is that now the distances are
measured with the complex absolute value, and so this is a straightforward
looking generalization.
But we will see that this innocent looking generalization is actually .
quite deep, and the class of complex differentiable functions looks radically
different from real differentiable functions. Here is an instance of this.
x2 if X > 0,
.
Example 0.1. Let f: Ill--+ Ill be given by J(x) = {
2
-X
I"f X < 0.
f
0
Fig. 0.1
Graphs of the functions
f and its derivative f'.
vii
Preface
Then f is differentiable everywhere, and
f'(x)
=
{
2x �fx:::::o,
-2X IfX < 0.
(0.1)
Indeed, the above expressions for f'(x) are immediate when x =/- 0, and
J'(0) = 0 can be seen as follows. For x =/- 0,
I
f(x) -f(O) - o = f(x) = El:_= =
lxl Ix - 0I,
I
J I
x -0
x
lxl
and so given E > 0, we can take J =
o < Ix -01 < o,
I
E
(> 0) and then we have that whenever
f(x) - f(O) oJ I - I
- = x 0 < J = E.
x-0
However, it can be shown that f' is not differentiable at 0; see Exercise 0.1.
This is visually obvious since f' has a corner at x = 0.
Summarizing, we gave an example of an f : IR?. -+ IR?., which is differ­
entiable everywhere in IR?., but whose derivative f' is not differentiable on
R
In contrast, we will later learn that ifF : (C -+ (C is complex differen­
tiable function in (C, then it is infinitely many times complex differentiable!
In particular, its complex derivative F' is also complex differentiable in C.
Clearly this is an unexpected result ifall we are used to is real analysis. We
will later learn that the reason this miracle takes place in complex analysis
is that complex differentiability imposes some "rigidity" on the function
which enables this phenomenon to occur. We will also see that this rigid­
ity is a consequence of the special geometric meaning of multiplication of
complex numbers.
◊
a
Exercise 0.1. Prove that J' : R ➔ R given by (0.1) is not differentiable at 0.
Why study complex analysis?
Although it might seem that complex analysis is just an exotic generaliza­
tion ofreal analysis, this is not so. Complex analysis is fundamental in all
ofmathematics. In fact real analysis is actually inseparable with complex
analysis, as we shall see, and complex analysis plays an important role in
the applied sciences as well. Here is a list ofa few reasons to study complex
analysis:
viii
A Friendly Approach to Complex Analysis
(1) PDEs. Iff : (C-+ (C is a complex differentiable function in C, then we
have two associated real-valued functions u, v : IR2 -+ IR, namely the
real and imaginary parts off: for (x,y) E IR2, u(x,y) := Re(f(x,y))
and v(x,y) := Im(f(x,y)).
(x,y)
� v(x,y)
- -y
f(x,y)
u(x,y)
The real and imaginary parts u, v off.
Fig. 0.2
It turns out that real and imaginary parts u, v satisfy an important
basic PDE, called the Laplace equation:
{P u
Dou := fJ 2
x
P
+ {fJ u2 = 0.
y
Similarly Dov = 0 in IR as well. The Laplace equation itselfis important
because many problems in applications, for example, in physics, give
rise to this equation. It occurs for instance in electrostatics, steady-.
state heat conduction, incompressible fluid flow, Brownian motion, etc.
(2) Real analysis. Using complex analysis, we can calculate some integrals in real analysis, for example
2
1
00
cosx
---2 dx or
-oo 1 + x
Note that the problem is set in the reals, but one can solve it using
complex analysis.
Moreover, sometimes complex analysis helps to clarify some matters
in real analysis. Here is an example ofthis. Consider
1
f(x) := - -2 , x E IR\ {-1,1}.
1 -x
Then f has a "singularity" at x = ±1, by which we mean that it is
not defined there. It is, however defined in particular in the interval
(-1,1). The geometric series
1 + x2
+ x + x6 + ...
4
ix
Preface
convergesfor lx 2 1 < 1, or equivalentlyfor lxl < 1, and we have
1
=f(x)forxE(-1,1).
l+x 2+x 4+x 6+-··=1 -x 2
From theformulafor f, it is not a surprise that the power series rep­
resentation of the function f is valid onlyfor xE(-1,1), since f itself
has singularities atx =
1 and atx =
-1. But now let us consider the
new function g given by
g(x) :=
1
,
l+ x2
xER
The geometric series 1- x2+ x4 -x 6+- . . . convergesfor I- x2 I < 1,
or equivalentlyfor lxl < 1, and we have
1
= g(x)forxE(-1,l)
.
2
+x
l
So the power series representation of the function g is again valid only
for xE(-1,1), despite there being no obvious reason from theformula
for gfor the series to break down at the points x =
-1 and x = +1.
The mystery will be resolved later on in this book, and we need to look
at the complex functions
1-x 2+x 4-x 6 +-···=
1
1
and G(z) =
F(z) =
2
1_z
1 + z2
( whose restriction to � are the functions f and g, respectively). In
particular, G now has singularities at z = ±i, and we will see that
what mattersfor the power series expansion to be valid is the biggest
size of the disk we can consider with center at z =
0 which does not
contain any singularity of G.
G
F
Fig. 0.3
Singularities of F and G.
x
A Friendly Approach to Complex Analysis
(3) Applications. Many tools used for solving problems in applications,
such as the Fourier/Laplace/ z-transform, rely on complex function the­
ory. These tools in turn are useful for example to solve differential
equations which arise from applications. Complex analysis plays an
important in applied subjects such as mathematical physics and engi­
neering, for example in control theory, signal processing and so on.
(4) Analytic number theory. Perhaps surprisingly, many questions
about the natural numbers can be answered using complex analytic
tools. For example, consider the Prime Number Theorem, which gives
an asymptotic estimate on the number 1r(n) of primes less than n for
large n:
71
(n)
Theorem 0.1. (Prime Number Theorem) lim
n ➔ oo n 1 og n)
= 1.
It turns out that one can give a proof of the Prime Number Theorem
using complex analytic computations with a certain complex differ­
entiable function called the Riemann zeta function. Associated with
the Riemann zeta function is also a famous unsolved problem in an­
alystic number theory, namely the Riemann Hypothesis, saying that
all the "nontrivial" zeros of the Riemann zeta function lie on the line
Re(s) = ½ in the complex plane. We will meet the Riemann zeta func­
tion in Exercise 4.5 later on.
What will we learn in Complex Analysis
The central object of study in this course will be
I holomorphic functions in a domain I
that is, complex differentiable functions f : D ➔ <C, where D is a "do­
main" (the precise meaning of what we mean by a domain will be given in
Subsection 1.3.4).
The bulk of the book is then in Chapters 2, 3 and 4, where we construct
the following three lanterns to shed light on our central object of study,
namely holomorphic functions in a domain:
(1) The Cauchy-Riemann equations,
(2) The Cauchy Integral Theorem,
(3) Taylor series.
xi
Preface
The core content of the book can be summarized in the following Main
Theorem 1:
Theorem 0.2. Let D be an op_en path connected set and let f : D --+ C.
Then the following are equivalent:
(1) For all z ED, f'(z) exists.
(2) For all z ED and all n 2: 0, f( n )(z) exists.
fJu
fJv
· ble an d
·
(3) u := Re(f), v := Im(f) are continuous
ly d:Fr
=
,
iJJerentia
fJx
{)y
8u
8v .
inD.
8x
{)y
(4) For each simply connected subdomain S ofD, there exists a holomorphic F: S--+ <C such that F'(z) = f(z) for all z ES.
(5) f is continuous on D and for all piecewise smooth closed paths , in
each simply connected subdomain ofD, we have
-=--
1
f(z)dz
= 0.
(6) If {z E <C: lz - zol ::S; r} CD, then there is a unique sequence (cn )n ::::o
in <C such that for all z with lz - zol
CX)
< r,
f(z) = :�::::Cn(z - zot-
1
n=O
( )
f(
= -.
d(
1 i 1(-zol=r (1"'> - zo )n+l
21r
------------Furthermore, Cn
and Cn
=
f( n )(zo)
.
n.1
1 Don't worry about the unfamiliar terms/notation here: that is what we will learn,
besides the proof!
xii
A Friendly Approach to Complex Analysis
Complex Analysis is not complex analysis!
Indeed, it is not very complicated, and there isn't much analysis. The
analysis is "softer" than real analysis: there are fewer deltas and epsilons
and difficult estimates, once a few key properties of complex differentiable
functions are established. The Main Theorem above tells us that the subject
is radically different from Real Analysis. Indeed, we have seen that a real­
valued differentiable function on an open interval ( a, b) need not have a
continuous derivative. In contrast, a complex differentiable function on
an open subset of (C is infinitely many times differentiable! This happens
because the special geometric meaning of complex multiplication implies
that complex differentiable functions behave in a rather controlled manner
locally infinitesimally, and aren't allowed to map points willy nilly. This
controlled behaviour makes these functions rigid and we will see this in
Section 2.3. Nevertheless there are enough of them to make the subject
nontrivial and interesting!
The intended audience
These notes constitute a basic course in Complex Analysis, for students
who have studied calculus in one and in several variables. The title of
the book is meant to indicate that we aim to cover the bare bones of the
subject with minimal prerequisites. The notes originated as lecture notes
when the second author gave this course for third year students of the BSc
programme in Mathematics and/with Economics.
Acknowledgements
Thanks are due to Raymond Mortini, Adam Ostaszewski and Rudolf Rupp
for many useful comments. This book relies heavily on some of the sources
mentioned in the bibliography. This applies also to the exercises. At some
instances we have given detailed references in the section on notes at the
end of each chapter, but no claim to originality is made in case there is a
missing reference.
Sara Maad Sasane and Amol Sasane,
London and Lund, 2013
Contents
Preface
V
.
1 Complex numbers and their geometry
1.1
1.2
.
1 3
1.4
1.5
The field of complex numbers .
Geometric representation of complex numbers .
Topology of (C ••.•.•••.•......•..
1.3.1
Metric on (C .••....•....•..
Open discs, open sets, closed sets and compact
1.3.2
sets ..............
1.3.3
Convergence and continuity
Domains ..........
1.3.4
The exponential function and kith
The exponential exp z ..
1.4.1
1.4.2
Trigonometric functions
1.4.3
Logarithm function
Notes
2. Complex differentiability
2.1
2.2
2.3
2.4
2.5
1
5
1 1
1 2
1 3
1 4
1 5
17
18
2 2
2 3
27
9
2
Complex differentiability .
Cauchy-Riemann equations ..........
Geometric meaning of the complex derivative
The d-bar operator .
Notes ..................
.
3 Cauchy Integral Theorem and consequences
.
3 1
1
Definition of the contour integral
xiii
0
3
36
49
56
58
59
59
xiv
A Friendly Approach to Complex Analysis
3..
1 1 An important integral ..........
Properties of contour integration ........
Fundamental Theorem of Contour Integration .
The Cauchy Integral Theorem........ . .
3.4.1 Special case: simply connected domains .
3.4.2
What happens with nonholomorphic functions?
3.5
Existence of a primitive ...............
3.6 The Cauchy Integral Formula ...............
3.7 Holomorphic functions are infinitely differentiable....
3.8 Liouville's Theorem; Fundamental Theorem of Algebra.
3.9 Morera's Theorem: converse to Cauchy's Integral
Theorem.
3.10 Notes .......
3.2
3.3
3.4
4. Taylor and Laurent series ·
Series .......
Power series .......................
4.2.1 Power series and their region of convergence
4.2.2
Power series are holomorphic .
4.3
Taylor series......
Classification of zeros .......
4.4
4.5
The Identity Theorem .......
4.6 The Maximum Modulus Theorem.
4.7 Laurent series ............
4.8 Classification of singularities....
.
4 8.1 Wild behaviour near essential singularities
4.9 Residue Theorem .
4.10 Notes ....
4.1
4.2
5. Harmonic functions
5.1
5.2
5.3
What is a harmonic function?
What is the link between harmonic functions and holomorphic functions? .....................
Consequences of the two way traffic: holomorphic ++
harmonic .................
5.3.1 Harmonic functions are smooth
5.3.2
Mean value property
5.3.3
Maximum Principle .......
67
69
73
77
82
85
88
91
96
99
101
103
105
106
107
107
113
117
12 2
12 6
12 9
13 1
14 1
15 1
15 3
162
163
163
165
170
171
172
172
Contents
5.4
5.5
Uniqueness of solution for the Dirichlet problem ,
Notes . . . . . . . . . . . . . . . . . . . . . . . .
xv
173
176
Solutions
177
Bibliography
269
Index
271
Chapter 1
Complex numbers and their geometry
In this chapter, we set the stage for doing complex analysis. We study three
main topics:
(1) We will introduce the set of complex numbers, and their arithmetic,
making <C into a field, "extending" the usual field of real numbers.
(2) Points in <C can be depicted in the plane JR2, and we will see that the
arithmetic in <C has geometric meaning in the plane. This correspon­
dence between <C and points in the plane also allows one to endow <C
with the usual Euclidean topology of the plane.
(3) Finally we will study a fundamental function in complex analysis,
namely the exponential function. We also look at some elementary
functions related to the exponential function, namely trigonometric
functions and the logarithm.
1.1
The field of complex numbers
By definition, a complex number is an ordered pair of real numbers. For
example,
(1, 0), (0, 1), (0, 0),
(-¾, v2)
are all complex numbers. The set JR x JR of all complex numbers is denoted
by <C. Thus
<C = { z = ( x, y) : x E JR and y E JR}.
For a complex number z = (x, y) E <C, where x, y E JR, the real number xis
called the real part of z, and y is called the imaginary part of z.
1
2
A Friendly Approach to Complex Analysis
We define the operations of addition "+" and multiplication "•" on C by:
(x1,Y1) + (x2,Y2) = (x1 +x2,Y1 + Y2),
(x1,Y1) · (x2,Y2) = (x1x2- Y1Y2,X1Y2 + X2Y1),
for complex numbers (x1,YI), (x2,Y2). With these operations,C is a field,
that is,
(Fl) (C,+) is an "Abelian group",
(F2) (C \ {0}, •) is an Abelian group, and
(F3) the distributive law holds: for a, b, c EC, (a+ b) · c =a· c + b · c.
In (Fl), Abelian group just means that the operation + on C is associative,
commutative, there exists an "identity element" (0,0), such that
(x,y) + (0,0) = (x,y) = (0,0) + (x,y)
for all (x,y), and every element (x,y) has an "additive inverse" (-x,-y):
(x,y) + (-x,-y) = (0,0) = (-x, -y) + (x,y).
Similarly, in (F2), the multiplicative identity is (1,0), and the multiplicative
inverse of a complex number (x, y) EC \ { (0,0)} is given by
(
X
, -y )
x2 + y2 x2 + y2 ·
Exercise 1.1. Check that (1.1) is indeed the inverse of
(1.1)
(x, y)
EC\ {(O, 0)}.
We have:
Proposition 1.1. (C, +, ·) is a field.
� is "contained" in C. In fact, we can embed� inside C, and view� as
a "subfield" of C, that is, one can show that the map
1-7 (x,0)
X
sending the real number x to the complex number (x,0) is an injective
field homomorphism. This just means that the operations of addition and
multiplication are preserved by this map, and distinct real numbers are sent
to distinct complex numbers.
�
X
X1 +x2
X1 · X2
1
0
I-+
I-+
I-+
I-+
I-+
C
(x,0)
(x1 + x2,0) = (x1,0) + (x2,0)
(x1 · x2,0) = (x1,0) · (x2,0)
(1,0)
(0,0)
3
Complex numbers
Thus we can view real numbers as if they are complex numbers via this
identification. For example, the real number v'2 can be viewed as the
complex number ( v'2, 0). If this makes one uneasy, one should note that
we have been doing such identifications right from elementary school, where
for instance, we identified integers with rational numbers, for example,
Z 3 3 = l E Q,
and we didn't lose sleep over it!
But the advantage of working with (C is that while in IR there was no
solution x E IR to the equation
x 2 + 1 = 0,
now with complex numbers we have
(0,1) · (0,1) + (1,0) = (-1,0) + (1,0) = (0,0).
If we give a special symbol, say i, to the number (0,1), then the above says
that
i2 + 1 = 0,
where we have made the usual identification of the real numbers 1 and 0
with their corresponding complex numbers (1,0) and (0,0).
Henceforth, for the complex number (x, y), where x, y are real, we write
x + yi, since
(x, y) = (x, 0) + (y, 0) • (0, 1) = x + yi .
._.,_,
.._,_,
.._,_,
=Y =i
=X
As complex multiplication is commutative, in particular, yi = iy, and so
we have x + yi = x + iy.
Exercise 1.2. Let 0 E
x, y
are real.
( - 27r, 27r) .
Express
1
+ i tan 0
1 _ i tan 0
.
m the form x
+ yi,
where
Historical development of complex numbers. Contrary to popular
belief, historically, it wasn't the need for solving quadratic equations, but
rather cubic equations, that led mathematicians to take complex numbers
seriously. The gist of this is the following. Around the sixteenth century,
one viewed solving equations like
ax 2 +bx+c = 0
as the geometric problem of finding the intersection point of the parabola
y = x2 with the line y = -bx - c. Based on this geometric interpretation,
it was easy to dismiss the lack of solvability in reals of a quadratic such as
x 2 + 1 = 0, since that just reflected the geometric fact that parabola y = x 2
did not meet the line y = -1. See the picture on the left in Figure 1. 1.
4
A Friendly Approach to Complex Analysis
0
-1
Fig. 1.1 Lack of solvability in reals of x2 = -1 versus the fact that x 3 =
always has a real solution x.
3px
+ 2q
Meanwhile, Cardano (1501-1576) gave a formula for solving the cubic x3
3px + 2q, namely,
X = {/q + ✓q2 _ p3 + {/q _
=
✓q2 _ p3.
For example, one can check that for the equation x3 = 6x + 6, with p = 2
and q = 3, this yields one solution to be x = ,W + �- However, note that
by the Intermediate Value Theorem, the cubic y = x 3 always intersects
the line y = 3px + 2q. See the picture on the right in Figure 1.1. But for
an equation like x 3 = 15x + 4, that is when p = 5 and q = 2, we have
q2 - p3 = -121 < 0, and so Cardano's formula fails with real arithmetic,
but we do have a real root, namely x = 4:
43
= 64 = 60 + 4 = 15 · 4 + 4.
Three decades after the appearance of Cardano's work, Bombelli suggested
that maybe with the use of complex arithmetic, Cardano's formula would
give the desired real root. So we may ask if
x = {12 + lli + {/2 - lli � 4.
One can check that (2+i) 3 = 2+11i and (2-i) 3 = 2-lli, so that the above
does work with these values of the cube root. Thus Bombelli's work estab­
lished that even for real problems, complex arithmetic might be relevant.
From then on, complex numbers entered mainstream mathematics.
Exercise 1.3. A field IF is called ordered if there is a subset P c IF, called the set
of positive elements of IF, satisfying the following:
(Pl) For all x,y E P, x+y E P.
(P2) For all x, y E P, x · y E P.
(P3) For each x E P, one and only one of the following three cases is true:
.3,0 -XE P.
1° X = 0.
£0 XE P.
5
Complex numbers
For example, the field of real numbers IR! is ordered, since P := (0, oo) is a set of
positive elements of R (Once one has an ordered set of elements in a field, one
can compare the elements of F by defining a relation > p in F by setting y > p x
for x, y E F if y - x E P.) Show that C is not an ordered field.
Hint: Consider x := i, and first look at x · x.
Geometric representation of complex numbers
1.2
Since (C = JR2, we can identify complex numbers with points in the plane.
See Figure l. 2.
Y ,____.(x,y)
(0, 0)
Fig. 1.2
X
The complex number x + iy in the complex plane.
The complex plane is sometimes called the Argand1 plane.
Exercise 1.4. Locate the following points in the complex plane: 0,
cos
1, -;, i, -v'2i,
3 +ism 3.
7r
. •
7r
So we can identify (Cas a set with the points in the plane JR2. Do the field
operations in (Chave some geometric meaning in the plane? We see below
that this is indeed the case: addition in (C is vector addition in the plane,
and multiplication in (Chas a special geometric meaning in the plane, which
is explained below as well.
Geometric meaning of complex addition. With this identification of
complex numbers with points in the plane, it is clear that complex addi­
tion is just addition of vectors in JR2. By addition of vectors, we mean the
1 It is named after Jean-Robert Argand (1768-1822), although it was used earlier by
Caspar Wessel (1745-1818).
A Friendly Approach to Complex Analysis
6
usual way of combining two vectors, that is, by completing the parallelo­
gram formed by the line segments joining (0,0) to each of the two complex
numbers as sides, and then taking the endpoint of the diagonal from (0,0)
as the sum of the two given complex numbers. See Figure 1.3.
0
Fig. 1.3
Addition of complex numbers is vector addition in JR. 2 •
Indeed, the middle picture shows that addition of z1 and z2 as vectors in
the plane yields the correct x-coordinate of their sum as complex numbers
by looking at the two congruent blue triangles. Similarly, by the rightmost
picture shows that the y-coordinate is also correct.
Geometric meaning of complex multiplication. We will now see the
special geometric meaning of complex multiplication. In order to do this, it
will be convenient to use polar coordinates. Thus, let the point (x, y) E JR2
have polar coordinates r:::: 0 and 0 E
(-1r, 1r].
This means that the distance
of the point (x,y) to (0,0) is r (:::: 0), and the angle made by the ray from
(0,0) to (x, y) makes an angle of 0 with the positive real axis (the x-axis).
(If (x,y) is itself (0,0), we set 0 = 0.)
Y i------�(x, y)
r
0
Fig. 1.4
X
Polar coordinates (r, 0) of (x, y) E JR. 2 •
7
Complex numbers
Then from the right-angled triangle shown in Figure 1.4, we have
x = rcos0,
y = rsin0.
Thus we can express the complex number in terms of the polar coordinates
(r,0):
x + yi = rcos0 + ( rsin0)i = r(cos0 + isin0).
Now we give the geometric interpretation of complex multiplication. For
two complex numbers expressed in polar coordinates as
z1 = r1(cos01 + isin01),
z2 = r2(cos02 + isin02),
we have that
z1 · z2 = r1(cos01 + isin01) · r2(cos02 + isin02)
= r1r2(cos01cos02 - sin01sin02 + i(cos01sin02 + cos02sin01) )
= r1r2(cos (01 + 02) + isin (01 + 02) ),
using the trigonometric identities for angle addition. Thus z1 · z2 has the
polar coordinates ( r1r2, 01+02). In other words, the angles z1 and z2 make
with the positive real axis are added in order to get the angle z1 • z2 makes
with the positive real axis, and the distances to the origin are multiplied to
get the distance z1 · z2 has to the origin. See Figure 1.5.
Fig. 1.5 Geometric meaning of complex multiplication: angles get added, distances to
the origin get multiplied.
As a special case, consider multiplication by cos a:: + isin a::, which is at a
distance of 1 from the origin. Then from the above, we see that if z E <C,
then z • (cos a:: + isin a::) is obtained by rotating the line joining O to z
anticlockwise through an angle of a::. In particular, multiplying z by
i.
= 0 + i. • 1 = cos
2 °+ i sm 2
1r
produces a counterclockwise rotation of 90 .
•
•
1r
8
A Friendly Approach to Complex Analysis
______ -------: · (coso:+isino:)
0
Fig. 1.6
iz
0
Multiplication by cos a+ i sin a produces an anticlockwise rotation through a.
De Moivre's formula and nth roots. We have for all n E N
(cos 0+isin0)n = cos(n0) +isin(n0).
This is called de Moivre's formula.
Exercise 1.5. Recover the trigonometric equality cos(30) = 4(cos 0)3 -3 cos 0 using
de Moivre's formula.
Exercise 1.6. Express (1+i) 10 in the form x+iy with real x, y without expanding!
Exercise 1.7. By considering (2+i)(3+i), show that�= tan- 1 � +tan- 1
i.
Exercise 1.8. Gaussian integers are complex numbers of the form m+in, where
m, n are integers. Thus they are integral lattice points in the complex plane.
Show that it is impossible to draw an equilateral triangle such that all vertices
are Gaussian integers.
Hint: Rotation of one of the sides should give the other. Recall that v'3 r/. Q.
De Moivre's formula gives an easy way of finding the nth roots of a complex
number z, that is,complex numbers w that satisfy w n = z. Indeed,we first
write z = r(cos0+isin0) for some r:?: 0 and 0 E [0,21r). Now ifwn = z,
where w = p(coso:+isino:),then
w n = pn (cos(na) +isin(no:)) = r(cos0+isin0) = z,
and so by equating the distance to the origin on both sides, we obtain
pn = r. Hence p = y'r, as both p and r are nonnegative. On the other
hand,the angle that wn makes with the positive real axis is no:, which is in
the set{·•• , 0 - 41r, 0 - 21r, 0, 0+ 21r, 0+ 41r, 0+ 61r, •··},because the
angle made by a nonzero z with the positive real axis is unique only up to
integral multiples of 21r, that is, instead of 0, we could just as well have
used 0+ 21rk for any integer k. See Figure 1.7.
9
Complex numbers
z
0 + 21r etc.
Fig. 1. 7
The angle made by z with the positive real axis.
Thus we get that a E { �
+ 2: k : k E Z} , and this gives distinct w for
0 0 27r 0
21r
0
21r
aE {-, -+-, -+2·-, ... , -+(n-1)·-} .
nn n n
n
n
n
In particular, if z = 1, we get the
nth roots of unity, which are located at the
vertices of an
n-sided regular polygon inscribed in a circle. See Figure 1.8.
Fig. 1.8
The six 6th roots of unity.
Exercise 1.9. Find all complex numbers w such that w4
the complex plane.
=
-1. Depict these in
Exercise 1.10. Find all complex numbers z that satisfy z 6 - z 3 - 2
= 0.
Exercise 1.11. If a, b, c are real numbers such that a + b + c = ab+ be+ ca,
then they must be equal. Indeed, doubling both sides and rearranging gives
(a - b) 2 + (b - c)2 + (c - a) 2 = 0, and since each summand is nonnegative, it
must be the case that each is 0. On the other hand, now show that if a, b, c are
complex numbers such that a2 + b2 + c2 = ab + be + ca, then they must lie on
the vertices of an equilateral triangle in the complex plane. Explain the real case
result in light of this fact.
Hint: Calculate ((b - a)w + (b - c)) • ((b - a)w 2 + (b - c)), where w is a nonreal
cube root of unity.
2
2
2
10
A Friendly Approach to Complex Analysis
Exercise 1.12. The Binomial Theorem says that if a, bare real numbers and n EN,
then
are the binomial coefficients. The algebraic reasoning leading to this is equally
valid if a, b are complex numbers. Show that
(
3n
0
)
+
(
Hint: Find (1 + 1) 3n
root of unity.
3n
3
)
+
3n
( )
6
+ ... + (3n
3n)
+ (1 + w) + (1 + w2 )
3n
3n
,
= 23n
+ 2 . (-l) n
3
where w denotes a nonreal cube
Exercise 1.13. Show, using the geometry of complex numbers, that the line seg­
ments joining the centers of opposite external squares described on sides of an
arbitrary convex quadrilateral are perpendicular and have equal lengths.
Absolute value and complex conjugate. The absolute value lzl of the
complex number z = x + iy, where x, y E JR:, is defined by
lzl
= Jx2 + y2 .
Note that by Pythagoras' Theorem, this is the distance of the complex
number z to O in the complex plane. See the picture on the left of Figure 1.9.
By expressing z1, z2 E C in terms of polar coordinates, or by a direct
calculation, it is clear that lz1z2 I = lz1I · lz2 IExercise 1.14. Verify the property lz1z2 I
Cartesian coordinates.
iy
Z
lzl
lxl
Z
= X +iy
= X +iy
X
I YI
-iy
Z=
X
- iy
Fig. 1.9 The absolute value of z is the distance of z to the origin, and the complex
conjugate is obtained by reflecting z in the real axis.
11
Complex numbers
The complex conjugate z of z
= x + iy where x, y E IR?.,
is defined by
z= x-iy.
In the complex plane, z is obtained by reflecting the point corresponding
to
z
in the real axis. See the picture on the right of Figure 1.9. From this
geometric interpretation, convince yourself that for all z1, z2 E
z1 + z2
= z1 + z2
and
z1 · z2
The following properties are easy to check:
Re(z)
z=z,
= --,
z+z
2
= z1 · zz.
Im(z)
Exercise 1.15. Verify that the four equalities above hold.
Exercise 1.16. Prove that for all z EC, lzl
Give geometric interpretations of each.
= lzl,
re,
z-z
2i
= -.-.
IRe(z)I :S: lzl and IIm(z)I :S: lzl.
Exercise 1.17. If a, z E (C satisfy lal < 1 and lzl :S: 1, then prove that
It�
:S: 1.
;z I
Exercise 1.18. Consider the polynomial p given by p(z) = eo + c1z + • • • + Cd Zd ,
where eo, c1, ..., Cd E � and Cd -I 0. Show that if w EC is such that p(w) = 0,
then also p(w) = 0.
l
Exercise 1.19. Show that the area of the triangle formed by 0, a, b EC is
1 z1 z1
Exercise 1.20. Prove for any complex z1, z2 , zs that i <let [ 1 z2 z2
1 Z3 Z3
I I·
Im ab)
;
is real.
Exercise 1.2 1. Show that for any two complex numbers z1, z2, there holds that
lz1 + z2 1 2 + lz1 - z2 1 2 = 2(lz11 2 + lz2 / 2 ). What is the geometric interpretation of
this equality?
1.3
Topology of
re
The concepts in ordinary calculus in the setting of IR?., like convergence of
sequences, or continuity and differentiability of functions, all rely on the
notion of closeness of points in IR?.. For example, when we talk about the
convergence of a real sequence (an ) n EN to its limit L E IR?., we mean that
given any positive E, there is a large enough index N such that beyond that
index, the corresponding terms an all have a distance to L which is at most
12
A Friendly Approach to Complex Analysis
E. This "distance of an to L" is taken as Ian - LI, and this is the length of
the line segment joining the numbers an and L on the real number line.
Now in order to do calculus with complex numbers, we need a notion
of distance d(z1,z2 ) between for pairs of complex numbers (z1,z2 ), and the
first order of business is to explain what this notion is.
1.3.1
Metric on
C
Since C is just JR.2, we use the usual Euclidean distance in JR.2 as the metric
in C. Thus, for complex numbers z1 = x1 +iy1 and z2 = x2 +iy 2 , we have
✓
d(z1,z2 ) = (x1 - x 2 )2 +(Y1 - Y2 )2 = lz1 - z2 IBy Pythagoras's Theorem, this is the length of the line segment joining the
points (x1,Y1),(x 2 ,Y 2 ) in lR.2 ; see Figure 1.10.
I�
Z2 = (x 2 ,Y2 )
lx1-x2 I
d(z1,z2 ) :=lz1-z2 I
....._______.....,,.z1 = (x1,Y1)
IY1 - Y2 I
Fig. 1.10
The distance between z1 and z2 is the length of the segment joining z1 to z2.
Using the geometric meaning of addition of complex numbers, and the well
known result from Euclidean geometry that the sum of the lengths of any
two sides of a triangle is at least as big as the length of the third side, we
obtain the following triangle inequality for the absolute value:
lz1 +z2 I::; lz1I +lz 2 I for z1,z2 EC.
See Figure 1.11. The triangle inequality above can also be verified analyti­
cally by using the Cauchy-Schwarz inequality for real numbers x1,x2 ,y1,Y 2 :
(xr +YI)(x� +y�) :2'.: (x1x2 +Y1Y2 ) 2 .
0
Fig. 1.11
Triangle inequality.
13
Complex numbers
Exercise 1.22. Show that for all z1,z2EC, lz1-z2l 2: llz1I -lz2II­
Exercise 1.23. Sketch the following sets'in the complex plane:
(1) {zEC:lz-(1-i)l=2}.
(2) {zEC:lz-(1-i)I < 2}.
(3)
{zEC: 1 < lz-(1-i)I <
2}.
(4)
{zEC: Re(z-(1-i)) = 3}.
(5)
{zEC: IIm(z-(1-i))I
(6)
{zEC: lz-(1-i)I =lz-(1 + i)l}-
(7)
{zEC: lz-(1-i)I +
(8) {zEC: lz-(1-i)I
1.3.2
< 3}.
lz-(1 + i)I = 2}.
+ lz-(1 + i)I
< 3}.
Open discs, open sets, closed sets and compact sets
In order to talk about sets of points near a given point, it will be convenient
to introduce the following definitions.
An open ball/disc D(zo,r) with center zo and radius r
D(zo,r) := {z Ere: lz - zol < r}.
> 0 is defined by
U ofreis called open if for every z E U, there exists an Tz > 0 such
D(z,rz ) C U. In other words, no matter what point we choose in U,
A subset
that
there is always some "room" around that point comprising points only from
U.
D(z0, r) is an open set.
D(z0, r) makes sense. Here are
annulus Ar := { z Ere: r < lzl < 1}
For example, it can be checked that an open disc
So using the adjective
open in
the name for
some more examples of open sets. The
and the right half-plane IHI:= {z Ere: Re(z)
> O}
are open sets.
It is also convenient to give a special name to sets whose complement is
open, and these are called
closed sets.
One can also give a characterization
of closed sets in terms of sequential convergence: A set
and only if for every sequence (zn ) nE N in
there holds that the limit LE F.
F
F
c re is closed if
which is convergent in re to
L,
14
A subset
for all
z
S
of
A Friendly Approach to Complex Analysis
re is called
bounded if there exists a M
>
0 such that
E S, JzJ ::; M. Thus S is contained in a big enough disc in the
complex plane. A subset K c
re is called compact if it is both closed and
bounded. We will often use the known fact from real analysis that a real
valued continuous function on a compact set possesses a maximizer and a
minimizer.
1.3.3
Convergence and continuity
We can also talk about convergent sequences in
E
re.
A sequence (zn )n EN is said to be convergent with limit L if for every
>
0, there exists an index
holds that Jzn - LJ
<
N E N such that for every n > N, there
E. It follows from the triangle inequality that for a
convergent sequence the limit is unique, and we write
lim Zn
n➔oo
Example 1.1. Let
=
L.
z be a complex number with JzJ <
1. Then the sequence
(zn )n EN converges to 0. Indeed, Jzn - OJ= Jzn J= JzJn = JJzJn - OJ, but as
JzJ
<
◊
1, we know that Jzl n -+ 0 as n-+ oo.
Exercise 1.24. Consider the polynomial p given by p(z) =co+ c1z+ • • • + cd zd ,
where eo, c1, ..., Cd E C and Cd =/- 0. Show that there exist M, R > 0 such that
jp(z)I � Mlzl d for all z EC such that lzl > R.
Exercise 1.25. Show that a sequence (zn ) nE N of complex numbers is convergent
to L if and only if the two real sequences (Re(zn ))nE N and (Im(zn ))nE N are con­
vergent respectively to Re(L) and Im(L).
Exercise 1.26. Show that a sequence (zn ) nE N of complex numbers is convergent
to L if and only if (zn ) nE N converges to L.
Exercise 1.27. Prove that C is complete, that is, every Cauchy sequence in C
converges in C. (A sequence (zn )nE N is called a Cauchy sequence if for every
E > 0, there is an index N E N such that for all indices m, n > N, there holds
that Jzn - Zml < E.)
re,
re.
Let S be a subset of
z0 E S and f : S -+
continuous at z0 if for every E > 0, there exists a 8
Then f is said to be
> 0 such that whenever
< E. f is said to
z ES satisfies Jz - zol < 8, there holds that lf(z) - f(zo)I
be continuous if for every z E S, f is continuous at z.
One can also give a characterization of continuity at a point in terms
of convergent sequences: f : S -+
re is continuous at z0
E S if and only
Complex numbers
15
if for every sequence (zn )nE N in S convergent to z0, there holds that the
sequence (f(zn))nE N is convergent to f(zo).
Example 1.2. Complex conjugation is continuous, that is, z 1-t z : (C --+ (C
is continuous. Indeed, we have lz-zol = lz - zol = lz-zol for all z, zo E <C.
This shows that complex conjugation is continuous at each z0 E <C, and so
it is a continuous mapping. This is geometrically obvious, since complex
conjugation is just reflection in the real axis, and so the image stays close
to the reflected point if we are close to the point!
Since (z) = z for all z E <C, complex conjugation is its own inverse.
So complex conjugation is invertible with a continuous inverse. Thus com­
plex conjugation gives a homeomorphism (that is, a continuous bijective
mapping with a continuous inverse) from (C to <C.
◊
Exercise 1.28. Prove that the map z
1.3.4
i-+
Re(z) : (('.-+ JR is continuous.
Domains
In the sequel, the notion of a path-connected open set will play an important
role. By this we mean that we will prove results about our central object
of study, namely functions f : D --+ (C that are complex differentiable at
every point of the set D ( C <C), and it will turn out that for the validity
of many of these theorems, we will need D to be a "nice" subset of <C, and
not just any old subset of <C. Sets which satisfy this "niceness" assumption,
stipulated precisely below, will be what we call a domain.
We will call an open path-connected subset of, (C a domain. We already
know what "open" means. Now let us explain what we mean by "path­
connectedness" .
Definition 1. 1.
(1) A path (or curve) in (C is a continuous function 1: [a, b] --+ <C.
a
b
16
A Friendly Approach to Complex Analysis
(2) A stepwise path is a path 'Y : [a, b] -+ C such that there are points
to = a < ti < · · · < tn < tn+l = b such that for each k = 0, l, ..., n,
the restriction 'Y: [tk, tk+d -+ C is a path with either a constant real
part or constant imaginary part.
(3) An open set U C C is called path-connected if for every z1, z2 E U,
there is a stepwise path 'Y: [a, b] -+ C such that 'Y(a) = z1, "f(b) = z2,
and for all t E [a, bl, 'Y(t) E U.
Actually in the above definition of path-connected open sets, the restriction
that the paths being stepwise paths can be relaxed, that is, if we look at
those open sets in which any two points can be joined merely by a path,
then this class of open sets coincides with our path-connected sets. But
this is an unnecessary diversion for us. So we just live with the definition
we have given instead.
Example 1.3.
(1) The open unit disc lI)) := {z EC: lzl < 1} is a domain.
(2) For r E (0, 1), the annulus Ar := {z EC: r < lzl < 1} is a domain.
(3) The right half-plane IHI:= {z EC: Re(z) > 0} is a domain.
IHI
Fig. 1.12
The domains lill, Ar and IHI.
On the other hand, the set S := {z E C : lzl -=/=- 1} =: C \ '][' is not a
domain, since although it is open, it is not path-connected. Indeed, there
17
Complex numbers
is no path joining, say O and 2: for if there were one such path 1, then by
the Intermediate Value Theorem applied to the map t H l,(t)I : [ab]--+
,
R,
we see that since l,(a)I = 0 < 1 < 2 = l,(b)I, there must be at* E [ab]
,
such that l,(t*)I = 1, but then ,(t*) fj. S.
◊
Exercise 1.29. Show that the set
domain.
Exercise 1.30. Let
also a domain.
1.4
D
{z
E (('.: Re(z) • Im(z) > 1} is open, but not a
be a domain. Set
D* := {z
E (('.: z ED}. Show that
D*
is
The exponential function and kith
In this last section, we discuss some basic complex functions:
the exponential function z H exp z,
the trigonometric functions z H sin z, cos z,
and the logarithm z H Logz.
They will serve as counterparts to the familiar functions from calculus, to
which they reduce when restricted to the real axis. In other words, when
we restrict our functions to the argument z = x E R, then we get the usual
real-valued functions
xHe x ,
x H sinx, cosx,
x H logx.
So our definitions provide extensions of the usual real-valued counterparts;
see Figure 1.13.
Fig. 1.13 The real valued functions map points on the real line to the real line, but our
definitions will give extensions of these to the shaded region, the complex plane.
18
A Friendly Approach to Complex Analysis
We will see that these extensions have new and interesting properties in
the complex domain that are not possessed by them when the argument
is only allowed to be real. Also they will serve as important examples of
complex differentiable functions: we will see later on that the exponential
and trigonometric functions are complex differentiable everywhere in the
complex plane, and the logarithm function is complex differentiable where
it happens to be continuous.
Let's begin with the exponential function.
1.4.1
The exponential exp z
Definition 1.2 (The complex exponential). For z = x+iy E <C, where
x, y are real, we define the complex exponential, denoted by exp z, as fol­
lows:
expz = e x (cos y + i sin y).
First we note that when y = 0 , the right hand side is simply e x . So our
definition extends the usual exponential function (IR 3) x r-+ e x ( E IR).
But this definition does appears to be mysterious. After all, z r-+ e Re(z)
also gives an extension of the usual real exponential function. So why
not use this simple definition instead? We define our exp function in the
way we do, because we will see later on that this is the unique extension
of the real exponential to the whole complex plane having the property
that the extension is complex differentiable everywhere; see Example 4 .8
on page 1 28. In fact, just like the real counterpart, where we have that
d x
e = e x for all x E IR ,
dx
we will later see that
d
expz = expz for all z E <C.
dz
So eventually we will learn that our mysterious looking definition is actually
quite natural! Right now, let us check the following elementary properties:
Proposition 1.2.
(1)
(2)
(3)
(4)
(5)
exp0 = e 0 (cos0 + i sin0) = 1 · (1 + i0) = 1 .
For z1 , z2 E <C, exp(z1 + z2) = (expz1)(expz2).
1
For z E <C, expz-=/=- 0 , and (expz)- = exp(-z).
For z E <C, exp(z + 21ri) = expz.
For z E <C, I expzl = e Re(z) .
19
Complex numbers
Proof.
(2) If z1=x1 + iy1 and z2=x2 + iy2, then
exp(z1 + z2)
i
= e(x1+x2 )+ (y1+v2 ) =ex1 +x2 (cos(y1 + Y2) + i sin(y1 + Y2))
= ex1 ex 2 (cos Y1 cos Y2 - sin Y1 sin Y2 + i (sin Y1 cos Y2 + cos Y1 sin Y2))
= ex1(cosy1 + i siny1)ex2 (cosy2 + i siny2)=(expz1)(expz2).
(3) From the previous part, we see that
1=exp O=exp(z - z) =(expz)(exp(-z)),
showing that expz =I- 0 and (expz)-1=exp(-z). Thus exp maps (C to the
"punctured" plane (C \ { 0}.
(4) We have
exp(z + 21ri )=(expz)(exp(21ri )) =(expz) • e0 (cos(21r) + i sin(21r))
=(expz)·l·(l + i ·O)=expz.
This shows that exp is "periodic in the y-direction" in the complex plane,
with a period of 21r; see Figure 1.14.
'
'
I
I
Z3
W2
-�
W1
expzi
exp
21r
-�
Wo
z2
Z1
0
zo
w_1
expwi
''
'
''
'
Fig. 1. 14
21r-periodicity of exp in the y-direction.
This phenomenon is not present in the x-direction, where just like in the
real setting, the function x r--+ exp(x+iyo) (with Yo E IR fixed) is one-to-one.
See Figure 1.15.
20
A Friendly Approach to Complex Analysis
Fig. 1.15
x >-+ ex is one-to-one.
J
(5) For x, y E �, le x cosy+ ie x sin YI = e 2x ((cosy)2 + (siny)2 ) = e x . So
I exp(x+iy)I = e x . This implies that exp maps vertical lines in the complex
plane ( that is all points having a common real part) into circles ( that is all
points having the same absolute value, in other words same distance to the
origin).
□
Proposition 1.2.(3) above shows that the map exp is not one-to-one, but
rather, it is periodic with period 21ri. Figure 1.16 shows the effect of the
mapping z f--t exp z on horizontal (fixed imaginary part y) and vertical
lines (fixed real part x). This picture is arrived at by putting together the
observations displayed in Figure 1.17.
,
21r
37f
I
,
I�
I
,
-1
Fig. 1.16
0
1
�
z ,-+ expz
2
7f
7f
2
0
2
The image of horizontal and vertical lines under the exponential map.
21
Complex numbers
-oo-----+---+-__,,__+_oo
exp
�
"e-(X)"
"e+= "
---------6----'T---
e"(cos 0o+i sin0o)
· +i0o
exp
�
Xo+i·
Xo
Fig. 1.17
Bo
exo (cos • +i sin·)
exp
�
The image of horizontal and vertical lines under the exponential map.
In Figure 1.16, we note that exp preserves the angle between the curves we
have considered in its domain. Namely, the horizontal and vertical lines,
which are mutually perpendicular, are mapped to circles and radial rays,
which are also mutually perpendicular. We will see later on that this is no
coincidence, and this property of "conformality", that is, of the preservation
of angles between curves in the domain together with their "orientation"
is something which is possessed by all complex differentiable functions in
domains.
Euler's formula. Note that for z =iy, where y is real, we have
exp(iy) =cosy+i siny.
This is the so-called Euler's formula. Hence the polar form of a complex
number can now be rewritten as z =r(cos0+i sin 0) =r exp(i0).
22
A Friendly Approach to Complex Analysis
9
Exercise 1.31. Compute exp z for the following values of z: i ;, 3 + 1ri.
Exercise 1.32. Find all z E (('. that satisfy exp z = 1ri.
Exercise 1.33. Plot the curve ti-+ exp(it) : [O, 21r] -+ <C.
Exercise 1.34. Describe the image of the line y = x under the exponential map
z = x + iy i-+ exp z. Proceed as follows: Start with the parametric form x = t,
y = t, and get an expression for the image curve in parametric form. Plot this
curve, explaining what happens when t increases, and when t-+ ±oo.
Exercise 1.35. Find the modulus and the real and imaginary parts of exp(z 2) and
of exp (1/ z) in terms of the real and imaginary parts x, y of z = x + iy.
1.4.2
Trigonometric functions
Just as we extended the real exponential function, we now extend the famil­
iar real trigonometric functions to complex trigonometric functions. From
the Euler formula we established earlier, we have for real x that
exp(ix) = cosx + i sinx and
which gives
exp(ix)
exp(-ix) = cosx - i sinx,
+ exp( -ix)
exp(ix) - exp(-ix)
.
and sm x =
.
2
2i
This prompts the following definitions: for z E C, we define
exp(iz) + exp( -iz)
exp(iz) - exp( -iz)
.
cosz =
and smz =
.
2
2i
Clearly these definitions give extensions of the usual real trigonometric
functions because when we put z = x, we get cos z = cos x and sin z = sin x,
as we had seen above from Euler's formula.
Several trigonometric identities continue to hold in the complex setting.
For instance, cos(z1 +z2) = (cosz1)(cosz2) - (sinz1)(sinz2). Indeed,
cos x =
(cosz1)(cosz2) - (sinz1)(sinz2)
exp(iz1) + exp(-iz1)
exp(iz2) + exp(-iz2)
= (
) (
)
_
(
exp(iz1)
2
�t
xp(-iz2)
) (
exp(iz2)
2
�t
xp(-iz2)
)
2 exp(i (z1 +z2)) + 2 exp(-i (z1 +z2))
= COS(Z1 + Z2 ) •
4
Exercise 1.36. Show that sin(z1 + z2) = (sinz1)(cosz2) + (cosz1)(sinz2) for all
z1, z2 E <C.
=
23
Complex numbers
Also, (sinz)2 + (cosz)2
= 1, since
2
2
exp(iz) - exp(- iz)
exp(iz)+exp(-iz)
.
(sm z )2 + (cos z )2 = (
)
)+ (
2i
2
exp(2iz) - 2+ exp(- 2iz) exp(2iz)+ 2+ exp(- 2iz)
+
-4
4
= 1.
However, as opposed to the real trigonom etric functions which satisfy
I sin xi ::; 1 and I cos xi ::; 1 for real x, z r-+ sinz and z r-+ cosz are not
bounded. Indeed, for z = iy, where y is real, we have
exp( ( y) )+ exp(- ( y)) exp(- y)+ exp(y)
cos(iy) = _ _ _ z_· z_· _ _ _ _ _ _ _z·_ z·_ _ = _ _ _ _ _ _ _ _ _ _
2
2
and so cos(iy) ➔+oo as y ➔ ±oo. Sim ilarly, since
sin(iy) =
2
e-Y - eY
2i
we also have that I sin(iy)I ➔ oo as y ➔ ±oo.
We will see later that z r-+ cos z and z r-+ sin z are complex differentiable
everywhere in the complex plane.
Exercise 1.37. Show that
cos z = ( cosx)(cosh y) - i(sinx)(sinh y)
and
for z = x+iy, where x, y are real, and cosh y
I cos zl 2 = ( cosh y) 2 - (sinx)2 ,
eY-e-Y
.
:= eY+e-Y
and smh y :=
2
2
Exercise 1.38. We know that the equation cosx = 3 has no real solution x. How­
ever, show that there are complex z that satisfy cosz = 3, and find them all.
1.4.3
Logarithm function
In the real setting, given a positive y, logy E IR.is the unique real num ber
such that e 10gy = y. Thus log : (0, oo) ➔IR.serves as the inverse of the
function x r-+ e x : IR.➔ (0, oo ). See Figure 1.18.
�
e
---------------0---�--
logy
�
Fig. 1.18
0
The maps x ,-, e x : JR➔ (0, oo) and y ,-, logy: (0, oo) ➔ R
A Friendly Approach to Complex Analysis
24
In the complex case, we know that exp : C--+ C\ {0}, and we now wonder
if there is a "complex logarithm function" mapping C\ {0} to Cthat serves
as an inverse to the complex exponential function. Given a z -/=- 0, we
seek a complex number w such that exp w = z, and we would like to
call this w the "complex logarithm of z". However, we have seen that
the exponential function exp is 21r-periodic in the y-direction, and so the
moment we find one w such that e w = z, we know that there are infinitely
many others, since exp(w + 21rin) = exp w = z for all n E Z. Given this
infinite choice, which w must we call the complex logarithm of z? We
remedy this problem of nonuniqueness by just choosing a w that lies in a
fixed particular horizontal strip of width 21r. Indeed, all possible nonzero
complex numbers can be obtained as the exp of something lying in any
such strip, and now for the purpose of defining the complex logarithm, we
choose (somewhat arbitrarily), the strip§:= {z EC: -1r < Im(z) :S 1r}.
See Figure 1.19.
7r
-Jr
Fig. 1.19
The strip § := JR x (-1r, 1r] is mapped by exp onto C \ {0}.
This will give, as we shall show below, a unique w in the strip such that
expw =:= z, and we will call this unique w the "principal logarithm of z",
denoted by Log z. In order to do this, we will first introduce the notion of
the principal argument of a nonzero complex number.
Principal argument of a nonzero complex number. For z-/=- 0, let 0
be the unique real number in the interval (-1r, 1r] such that
z = lzl(cos0+i sin0).
This value of 0 is called the principal argument of z, and is denoted by
Arg z. Here are a few examples:
Arg(-1) = 1r,
Arg(i) =
i,
.
7r
Arg(-i) = --.
2
We note that if we start at a point on the positive real axis in the complex
plane, and go around anticlockwise in a circle, then there is a sudden jump
Arg(3) = 0,
25
Complex numbers
in the value of the principal argument as we cross the negative real axis:
on the negative real axis, the value of the principal argument is 1r, while
just below the negative real axis, the principal argument is close to -1r.
Exercise 1.39. Depict { z E
(C:
z =/ 0, � < IArg(z)I <
i} in the complex plane.
Now we are ready to define -the principal logarithm of nonzero complex
numbers.
Definition 1.3. The principal logarithm Log z of z -/- 0 is defined by
Log z = log lzl +iArg z.
First of all, let us observe that
exp(Log z) =e 10glzl (cos(Arg z) +i sin(Arg z))
= lzl (cos(Arg z) +i sin(Arg z)) = z.
This shows that exp :§ H C \ {0} is onto. Also, it is one-to-one, because
if z1,z2 E § are such that expz1 = expz2, then exp(z1 -z2) = 1, and so
z1 -z2 = 21rni, for some n E Z. But as z1,z2 E §, their imaginary parts
must differ by a number< 21r. This implies that n = 0, and so z1 =z2.
So the two maps exp :§➔ C \ {0} and Log : C \ {0}➔§ are inverses
of each other.
Of course, had we chosen to define the principal argument 0 of a nonzero
z = lzle i0 to lie in a different interval (a, 21r+a] or [a, 21r+a) for some other
a, we would have obtained a different well-defined notion of the logarithm
(which would also be equally legitimate). But when we talk about the
principal logarithm of z, in this book we will always mean log lzl +iArg z,
with the principal argument Arg z E (-1r,1r]. Here is an example:
Log(-i) = log I
- ii+iArg(-i) = log1 - �2 i = 0 - �2 i = -�2 i.
Continuity of Log in C \ (-oo, 0]. First of all, we remark that owing
to the lack of continuity of Arg : C \ {0} ➔ (-1r,1r] across the negative
real axis, also the function Log is not continuous on C \ {0}. It fails to be
continuous at each point in (-oo, 0). Let us show the lack of continuity at
-1. To this end, consider the sequence
which converges to -1:
=
\
=
lim exp (i(- 1r+�))= lim (-1) (cos�+i sin�)= -1(1 +i0) =-1.
n➔
n
n➔
n
n
26
A Friendly Approach to Complex Analysis
Also,
Thus we have
}11! Log (exp (i( - 1r + ¾))) = i(O- 1r) = -i1r
Log
(}11! exp (i( - 1r + ;) ) ) = Log(-1) =
il--
i1r
showing that Log is not continuous at -1 EC\ {O}.
On the other hand, Log is continuous on the smaller set C\ (-oo, OJ. This
just follows by observing that the principal argument Arg(z) is continuous
in C\ (-oo, OJ. Indeed, the key thing is that if we take any complex number
z0 not lying on (-oo, O], then there is some room around z0 not touching
the negative real axis: we can always find a small enough r such that the
disc D(z0 ,r) does not touch the line (-oo,OJ. Thus, given an E > 0, by
shrinking r further if necessary, we can ensure that the points z in D ( zo,r)
satisfy IArg(z) -Arg(zo)I < E. See Figure 1.20.
. Arg
�
7r
Fig. 1.20
Arg(zo) -1r
Continuity of the principal argument Arg in C \ ( -oo, O].
As both z f-t log lzl and Arg are continuous in C \ (-oo, OJ, it follows
that Log is continuous there too.
Using the continuity of Log on C\ (-oo, O], we will see later on that Log
is complex differentiable in C\ ( -oo,O].
ab for a EC\ {O} and b EC. One can now also talk about a b , where a, b
are complex numbers and a f=. 0, and we define the principal value of a b as
a b := e bLog(a).
Complex numbers
27
For example, the principal value of ii is
exp(i • Log(i)) = exp(i(log Iii+iArg i)) = exp,(i (0+ii)) = e-i.
Exercise 1.40. Find Log(l + i).
Exercise 1.41. Find Log(-1) and Log(l). Show that Log(z2 ) isn't always equal
to 2 • Log(z).
Exercise 1.42. Find the image of the annulus { z E (C : 1 < izl < e} under the
principal logarithm.
Exercise 1.43. Find the principal value of (1 + i) 1 -i.
1.5
Notes
The remark on the historical development of complex numbers is taken from
[Needham (1997)]. Exercise 1.2 is taken from [Shastri (2000)]. Exercises 1.7,
1.7, 1.13, 1.19 are taken from [Needham (1997)]. Exercises 1.23 and 1.34
are taken from [Beck, Marchesi, Pixton, Sabalka (2008)].
Chapter 2
Complex differentiability
In this chapter we will learn three main things:
(1) The definition of complex differentiability, that is, given f : U -+ <C,
where U is an open subset <C, and zo EU, we will learn the meaning of
the statement "f is complex differentiable at z0 with complex derivative
f'(zo)".
.
.
8u 8v
8u
8v
(2) The Cauchy-Riemann equations:
and
By = - ox•
ox = By
These are PDEs that are satisfied by the real and imaginary parts u, v
of a complex differentiable function f : U -+ <C wherever it is complex
differentiable.
(x,y)
u
� v(x,y) - - - f(x,y)
;
0
u(x,y)
Fig. 2.1
The real and imaginary parts u, v off.
Vice versa, we will learn that if these Cauchy-Riemann equations are
satisfied everywhere in an open set and u, v are 0 1 , then f = u + iv is
complex differentiable in U.
(3) The geometric meaning of the complex derivative f'(z0): infinitesi­
mally the map f is an amplification by lf(zo)I together with a twist (a
counterclockwise rotation) through Arg(f'(z0)).
29
30
A Friendly Approach to Complex Analysis
The central result in this chapter is the necessity and (under mild condi­
tions) sufficiency of the Cauchy-Riemann equations for the complex differ­
entiability of a function in an open set.
2.1
Complex differentiability
Definition 2.1.
(1) Let Ube an open subset of (C,
f:
U --t (C and z0 E U. Then f is said
to be complex differentiable at z0 if there exists a complex number L
such that
z - zo
lim f( ) f( ) = L
z - zo
that is, for every E > 0, there is a J > 0 such that whenever z EU and
0 < Jz - zol < J, we have
z---+zo
I
f(z) - f(zo)_
z-zo
LI
<
.
E
We denote this L ( which can be shown to be unique) by
(2)
A
f'(zo)or
dz(zo).
function f : U --t (C which is complex differentiable at all points of
the open set U is called holomorphic 1 in U.
(3) A function that is holomorphic in (C is called entire,
that is, the domain
of f is understood to be the whole of (C and moreover, f is holomorphic
in (C.
Let us look at a simple example of an entire function.
Example 2.1. Consider the function
f : (C --t (C defined by f(z)
(z E q. We show that f is entire. Note that
2
- z5
f(z)- f(zo) z
---=z -zo
z -zo
= z + zo :::::: 2zo
z2
for z near z0, and so we guess that f'(zo)= 2z0. Let us show this now. For
z -/- zo, we have
f(z)- f(zo)
l _ _ _ _ _ _ _ _ - 2zol
z - z0
=
z2 - z
l- _ _ _
5 - 2zol = Jz + zo - 2zol = Jz - zol-
z-zo
1The word "holomorphic" is derived from the Greek "holos" meaning "entire", and
"morphe" meaning "form" or "appearance".
Complex differentiability
31
So we see that the left hand side can be made as small as we please when
z is close enough to z0. Let E > 0. Set 15 := E
satisfies O < lz-zol < 15, we have
I
f(z)-f(zo)
z-zo
>
0. Then whenever z EC
-2zol = lz-zol < 15
= E.
Hence f'(zo) = 2z0. As z0 EC was arbitrary, f is holomorphic in C, that
is, f is entire, and from the above it follows that
d zz
dz
= 2z,
z EC.
◊
On the other hand, here is an example of a natural mapping which is not
complex differentiable.
Example 2.2. Consider the function g : C ➔ C defined by g(z) = z
(z E q. We show that g is differentiable nowhere. Suppose that g is
differentiable at z0 EC. Let E := ½ > 0. Then there exists 15 > 0 such that
whenever z satisfies O < lz- zol < 15, we have
l
g(z)-g(zo)
z-zo
-zo
-g'( Zo) - -- -g'( Zo < E.
z
)I
1 1:Z-zo
See the picture on the left of Figure 2.2. The above says that whenever z
is in the punctured disc of radius 15 with center z0, we are guaranteed that
this inequality holds. We will now make special choices of the z as the blue
and the red point shown in the figure, and show that the special cases of
the inequality above yield that g'(zo) must lie in discs of radius 1/2 with
centers at -1 and 1. See the picture on the right of Figure 2.2. But these
discs have no intersection, and this will be our contradiction. We give the
details below.
(9
•
Takmg z
Fig. 2.2
15
Non complex differentiability of complex conjugation
= zo + 2,
we have O < lz-zol < 15, and so
1: = :: -
g'(zo)I
=I:;� - g'(zo)I = II - g'(zo)I < E.
(2. 1)
32
A Friendly Approach to Complex Analysis
Also, taking z = zo + i�, we have 0 < lz - zol < J, and so
l
z-zo
- g'(zo
z _ zo
)I = ,-iiJJ/22 -g'(zo )I = I 1 + g'(zo)I <
1
It follows from (2.1) and (2.2) that
E.
2 = 11-g'(zo)+l+g'(zo)I::; 11-g'(zo)l+ll+g'(zo)I < E+E = 2E = 2-�
a contradiction. So g is not differentiable at z0.
(2.2)
= 1,
◊
2
Exercise 2.1. Show that f : (C ➔ (C defined by f(z) = lzl for z E C, is complex
differentiable at 0 and that f' (0) = 0. We will see later (in Exercise 2.9) that f
is not complex differentiable at any nonzero complex number.
Exercise 2.2. Let D be a domain, and f : D ➔ (C be holomorphic in D. Set
D* := {z E (C: z ED}, and define f*: D* ➔ (C by f*(z) = f(z) (z ED*). Prove
that f* is holomorphic in D*.
The following reformulation of complex differentiability is useful to prove el­
ementary facts about complex differentiation. Roughly speaking, the result
says that for a complex differentiable function f with complex derivative L
at z 0, f(z) -f(zo) - L · (z -zo) goes to 0 "faster than z - zo".
Lemma 2.1. Let U be an open set in C, z 0 EU, and f : U ➔ C. Then
the following are equivalent:
(1) f is complex differentiable at zo with f'(zo) = L.
(2) There exists an r > 0, and a function h : D(zo,r) ➔ C, where
D(zo,r) := {z EC: lz - zol < r} , such that
(a) f(z) = f(zo) + (L + h(z))(z -zo) for lz - zol < r, and
(b) lim h(z) = 0.
z➔zo
Proof.
(2)::::}(1): For z E D(z 0, r) \ {zo}, we have, upon rearranging, that
f(z) - f(zo)
Z - Zo
_ L = h(z) z.2+° 0
and so f is complex differentiable at zo, and f'(zo) = L.
(1)*(2): Now suppose that f is complex differentiable at zo. Then there
is a J1 > 0 such that whenever 0 < lz - zol < J1, z EU and
I
f(z)
-
f(zo)
Z - Zo
_ f'(zo)I < 1.
33
Complex differentiability
Set r := 81, and define h: D(zo, r) ---+ C by
h(z)= {
f(z) - f(zo) _
f'(zo) if Z-=/=- Zo,
z-zo
if z = zo,
0
Then f(z) = f(zo) + (f'(zo) + h(z))(z - zo) holds whenever lz - zol < r. If
E > 0, then there is a 8 > 0 (which can be chosen smaller than r) such that
whenever O < lz - zol < 8, we have
I
f(z) - f(zo)
- f'(zo)I (= lh(z) - OI) <
z - zo
This completes the proof.
E.
□
For example, using this lemma, we see that holomorphic functions must be
continuous.
Exercise 2.3. Let D be a domain in C. Show that if f : D ➔ (C is complex
differentiable at zo E D, then f is continuous at zo. Later on, we will see that if
f is holomorphic in D, then in fact it is infinitely many times differentiable in D!
Using Lemma 2.1, it is also easy to show the following.
Proposition 2.1. Let U be an open subset of C. Let f, g : U ---+ C be
complex differentiable functions at z 0 EU. Then:
(1) f + g is complex differentiable at zo and (f + g)'(zo) = f'(zo) + g'(zo).
(Here f + g: U---+ C is defined by (f + g)(z) = f(z) + g(z) for z EU.)
(2) If a EC, then a· f is complex differentiable and (a· f)'(zo) = af'(zo).
(Here a· f: U---+ C is defined by (a· f)(z) = af(z) for z EU.)
(3) fg is complex differentiable at zo and moreover, there holds that
(fg)'(zo) = f'(zo)g(zo) + f(zo)g'(zo).
(Here Jg: U---+ C is defined by (fg)(z) = f(z)g(z) for z EU.)
Remark 2.1. Let U be an open subset of C, and let Hol(U) denote the
set of all holomorphic functions in U. Then it follows from the above that
Hol(U) is a complex vector space with pointwise operations. On the other
hand, the third statement above shows that the pointwise product of two
holomorphic functions is again holomorphic, and so Hol(U) also has the
structure of a ring with pointwise addition and multiplication.
Example 2.3. It is easy to see that if f(z) := z (z Eq, then f'(z) = 1.
Using the rule for complex differentiation of a pointwise product of two
A Friendly Approach to Complex Analysis
34
holomorphic functions, it follows by induction that for all n E N, z f-t z n is
entire, and
d n
Z
dz
= nz n -1 _
◊
In particular all polynomials are entire.
Exercise 2.4. Prove Proposition 2.1.
Exercise 2.5. Let D = {z E (C : lzl < 1} and Hol(D) denote the complex vector
space of all holomorphic functions in D with pointwise operations. Is Hol(D) finite
dimensional?
Exercise 2.6. Let U be an open subset of C, and let f : U ➔ (C be such that
f (z) =I= 0 for z E U and f is holomorphic in U. Prove that the function
J:
U ➔ C, defined by
is holomorphic, and that
(y)'
Exercise 2.7. Show that in
(C \
(y)
(z) = ftz for all z E U,
)
(z) = - :��)�2
(
(z E U).
{O}, for each m E Z, !zm
= mzm -l
Just like one has the chain rule for the derivative of the composition of
maps in the real setting, there is an analogous chain rule in the context of
composition of holomorphic functions.
Proposition 2.2. (Chain rule) Let
(1)
(2)
(3)
(4)
DJ, D9 be domains,
f: DJ➔ C be holomorphic in DJ,
g: D9➔ C be holomorphic in D9 , and
f(DJ) C D9 •
Then their compos ition go f : DJ ➔ C, defined by (go f)(z) = g(f(z)),
z E Dt, is holomorphic in Dt and (gof)'(z) = g'(f(z))f'(z) for all z E Dt.
f
g
Fig. 2.3 The chain rule: (go f)'(z) = g'(f(z))f'(z), z E Dt ·
Complex differentiability
35
Proof. Let z0 E DJ. Then f(z0) E D9. From the complex differentiabil­
ity off at z0 and that of g at f (z0), we know that there are functions h f
and h9, defined in the discs D(z0, TJ) C DJ and D(f(z0), r9) C D9 such
that
f(z) - f(zo) = (f'(zo) + h1(z))(z - zo),
g(w) - g(f(zo)) = (g'(f(zo)) + h9(w))(w - f(zo)),
and
lim h (w) = 0.
lim h z = 0,
z-+zo 1( )
w➔ f(zo) 9
But it follows from the continuity of f at z0 that when z is close to z0,
w := f(z) is close to f(zo), and so if z-=/= zo, but close to zo, we have
(g O f)(z) - (g O f)(zo) =
(g'(f(zo)) + h9(f(z)))(f'(zo) + h1(z)),
Z - Zo
□
and so the claim follows.
Example 2.4. From Exercise 2.7, we know that
(�)
= - 2 z EC \{0},
:,
:z
but this is also easy to see from the definition, because
1
1
Zo-Z
-1 z-+zo 1
Z Zo
=
zzo(z - zo) zzo ----+ - z3
Z-Zo
for zo EC \{0}.
1
f := 1 + z 2
g := z
i
.,,,.-------.._
-i
DJ :=C \{-i,i}
Fig. 2.4
D9 :=C \{0}
Application of the chain rule.
Now consider the functions f := l+z 2 on DJ :=C\{-i,i} and g := 1/ z
on D9 :=C \{0}. Clearly, f (DJ) c D9, and so, by the Chain rule,
1
2z
d
1
• 2z = (
) =(1 + z2)2
(1 + z2)2
dz 1 + z2
inC \{-i,i}.
◊
36
A Friendly Approach to Complex Analysis
Exercise 2.8. Assuming that expz is entire and that exp' z
this later), show that
1 +z)
z r-+ exp (1-z
is holomorphic in the unit disc I[)):=
2.2
{z
E (C: [z[
<
= expz (we will prove
1}, and find its derivative.
Cauchy-Riemann equations
We now prove the main result in this chapter, which says roughly that a
= u + iv is holomorphic if and only if its real and imaginary
u, v (viewed as real valued functions living in an open subset of IR.2 )
function f
parts
satisfy a pair of partial differential equations, called the Cauchy-Riemann
equations.
Cauchy-Riemann
Let U be an open subset of <C, and let f : U ➔ <C be a function. Then
taking any point (x,y) EU, we have f(x +iy) E <C, and we can look at the
real part
u(x,y) of f(x + iy), and the imaginary part v(x,y) of f(x + iy).
See Figure 2.5.
� v(x,y) - -,f(x, y)
u(x,y)
Fig. 2.5
The real and imaginary parts u, v off.
37
Complex differentiability
If one changes the point (x,y), then f(x + iy) changes, and so do u(x,y)
and v(x,y). In this manner, associated with f , we obtain two real-valued
functions
u: U---+ IR., U 3 (x,y) f-t Re(f(x + iy)) =: u(x,y),
v: U---+ IR., U 3 (x,y) f-t Im(f(x + iy)) =: v(x,y).
Our first result in this section is the necessity of the Cauchy-Riemann
equations for complex differentiability, and we will prove this result in
Theorem 2.1 below. The result says that if f is complex differentiable
at (xo,Yo) E U, then
OU
OX
av
o
y
and
OU
o
y
av
at (xo,Yo),
OX
and these two equations are called the Cauchy-Riemann equations. So for a
function to be complex differentiable, it has got to satisfy these equations.
In o,ther words, if these equations aren't satisfied by the real and imaginary
part of some complex valued function at some point, then at that point we
know that the function can't be complex differentiable. Here is an example.
We had seen earlier (by "brute force", using the E-8 definition of complex
differentiability), that z f---t z is not complex differentiable any where in the
complex plane. Let us revisit Example 2.2.
Example 2.5. For the function g: C---+ C defined by g(z) = z (z EC), we
have that
u(x,y) = Re(g(x + iy)) = Re(x - iy) = x,
Thus
v(x,y) = Im(g(x + iy)) = Im(x - iy) = -y.
OU
o (x,y) = 1 i- -1
x
av
=o
(x,y).
y
This shows that the Cauchy-Riemann equations can't hold at any point.
So we recover our previous observation that g is complex differentiable
nowhere.
◊
Before proving the necessity of the Cauchy-Riemann equations for complex
differentiability, let us also mention the second important result we will
show in this section, namely the sufficiency of the Cauchy-Riemann equa­
tions for holomorphicity in an open set. More precisely, we will show in
38
A Friendly Approach to Complex Analysis
ThE:Jorem 2.2 below that if u,v : U ➔ IR are two real-valued functions in an
open set U that are continuously differentiable in U (as functions of two
real variables), and the Cauchy-Riemann equations are satisfied by u and v
everywhere in U, then the new complex-valued function f: U ➔ C, created
from u,v by setting f(x+iy) := u(x,y)+iv(x,y), (x,y) EU, (so that u,v
turn out to be the real and imaginary parts, respectively, of the f we have
constructed), is holomorphic in U. This important result will enable us to
establish the holomorphicity of important functions without having to go
through the rigmarole of verifying the E-8 definition. Here is an example.
Example 2.6. Consider the functions u,v defined in the punctured plane
IR2 \ {(0, 0)} as follows:
X
u(x,y) := 2
v(x,y):= 2-y 2' (x, y) =/- (0, 0).
X +y 2'
X +y
Then we have
y 2 -x 2
1 • (x 2 + y 2 ) - x • 2 x
&u
2
x
2
+
y
2
x
)
&x
(
( 2 + y2 )2 '
&u
-x-2y
-2xy
2
y
2
x
2
x
2
&y
( + )
( + y 2 )2 '
y2 - x2
&v
2xy
&v
and
.
=
&x
(X 2 +y 2 )2
(x 2 + y 2 ) 2 '
y
8
Clearly (x,y) M (x 2 +y 2 ) 2 , y 2 -x 2 , ±2xy are continuous in IR2 and (x 2 +y 2 ) 2
is nonzero in IR2 \ {(0, 0)}. So it follows that each of these partial derivatives
is continuous in IR2 \ {(0, 0)}. So u,v are continuously differentiable in
IR2 \ {(0, 0)}. Also the Cauchy-Riemann equations hold. Thus f := u + iv
is holomorphic in (C \ {0}.
In fact, the f above is just the reciprocation map z M 1/ z:
X
-y
X
iy
z
z I z =/- 0.
f = u + iv = x 2 y 2 + i x2 y 2 ) = x2 + y 2 =
=
=
+
+
zz -;; '
•
•(
-
W
◊
Theorem 2.1. Let U be an open subset of (C and let f: U ➔ (C be complex
differentiable at zo = xo + iyo E U. Then the functions
(x,y) M u(x,y) := Re( f(x + iy)) : U ➔ IR and
(x,y) M v(x,y) := Im(f(x + iy)) : U ➔ IR
are differentiable at (xo, Yo) and
&u
&v
(xo, Yo) =
(xo, Yo) and
&x
&y
(2.3)
39
Complex differentiability
Proof. (The idea of the proof is easy, we just let (x,y ) tend to (xo,Yo)
by first keeping y fixed at y0, and then by keeping x fixed at x 0, and look
at what this gives us. See Figure 2.6.)
(xo,Y)
(xo,Yo)
(x,yo)
Fig. 2.6 Proof of the necessity of the Cauchy-Riemann (CR) equations for complex
differentiability.
Let z0 = (x0, y0) EU. Let E > 0. Then there is /j > 0 such that whenever
0 < [z - z0[ < 8, we have z EU and
I
f(z ) - f(zo)
Z - Zo
- J'(zo)I < E.
(2.4)
Step 1. We will show that �: (xo,Yo) exists and equals Re(f'(z0)).
Let z := x + iy0, where x E � is such that O < [x - x0[ < 8. Then
z - zo = x - xo, and so O < [z - zo[ = [x - xo[ < 8. Thus
l
u(x,yo) - u(xo,Yo)
_ Re(f'(zo))I
x-xo
f(x + iyo
= I Re (
l
z
=
=
;�
xo + iyo)
) - Re(f'(zo))I
= IRe ( f( ; �?o)) - Re(f'(zo))I
::::; I
f(z ) - f(zo) - ' z
f ( o)I < E,
z-zo
using (2.4). Thus the partial derivative
ou
. u(x,Yo) - u(xo,Yo)
- (xo,Yo) = 11m
OX
x➔xa
= Re(f'(zo)).
X - Xo
40
A Friendly Approach to Complex Analysis
Step 2. We show that i:(xo,Yo) =Im(f'(zo)).
Proceeding in a manner similar to Step 1 above, we also have with the same
notation that
v(x,yo)-v(xo,Yo) _
Im(f'(zo))I
l
X-Xo
(x + iyo
xo + iyo)
= IIm ( f
; ��
) - Im(f'(zo))I
(z
= IIm ( f ;
:s; I
=
o)
:? ) - Im(f'(zo))I
f(z) - f(zo)
Z - Zo
=
- f'(zo)I < E,
ov
. v(x,yo)-v(xo,Yo)
=Im(!'(zo )). Thus
and so -;:,- (Xo,Yo ) = 11m
uX
x----+xo
X - Xo
.av
OU
I
f (zo) = ox (xo,Yo)+ i o (xo,Yo)x
(2.5)
Step 3. We show that :: (xo,Yo) = -Im(f'(zo)).
Now let z := xo + iy, where y E JR is such that O < IY - Yol < 8. Then
z - zo = i(y - Yo) and so O < lz - zol = IY - Yol < 8. Thus using the fact
that for a, b E JR there holds that Re( a+ ib) =Im(i(a+ ib)), we obtain that
Im( (f
u(xo, y) - u(xo,Yo)
+ Im(!'(zo))I = I i (z) - f(zo)) + Im(f'(zo))I
I
y-�
y-�
(z
= Im (-f ; :?o) + f'(zo))
:s;
I
If
=
(z) - f(zo)
Z - Zo
J'(zo)
I
< E.
I
Thus the partial derivative
ou
_ . u(xo,y) - u(xo,Yo) _ -Im(!'( )) .
zo
-;:,- (xo,Yo ) - 11m
y----+yo
uy
y-yo
Recall that in Step 2, we had obtained
i:(xo,Yo) =Im(f'(zo)),
and so, we have established one of the two Cauchy-Riemann equations,
namely that
Complex differentiability
41
Step 4. We show that ::(xo,Yo) = Re(f'(zo)).
Proceeding as in Step 3 above, with the same notation , and using the fact
that for a, b E JR there holds that Im(a+ib) = -Re(i(a+ib)), we have that
I
v(xo,y)- v(xo,Yo) R !
(z) - f(zo)
- e( '(zo))I = I-Re (J
) - Re(f'(zo) I
y-yo
y-yo
z)
:S 1-J( - f(zo) - f'(zo)I
y-yo
f(z)- f(zo)
= I
- f'(zo)I < E.
Thus the partial derivative
Hence
Z- Zo
av
_ Re(!'(zo )).
_ 1.1m v(xo,y) - v(xo,Yo) � (xo,Yo ) uy
Y-+Yo
y-yo
J'(zo) = :: (xo,Yo) - i :: (xo , Yo).
(2.6)
From (2.5) and (2.6), it follows that
av
av
au
au
=
(xo,Yo) = - (xo,Yo)(xo,Yo) and
ax
ax (xo,Yo) ay
ay
So we have got both Cauchy-Riemann equations.
Finally, we show that u, v are differentiable (as real valued functions of
two real variables) at (xo,Yo). For z = (x,y) satisfying O < l z - zol < 8,
lu(x,y)-u(xo,Yo)-(�(xo,Yo)) (x-xo)-(�(xo,Yo)) (y-yo)I
ll(x,y)- (xo,Yo)ll2
_
-
lu(x,y) - u(xo,Yo)- (�(xo, Yo)) (x - xo) + (�(xo, Yo)) (y -Yo)I
ll(x,y)- (xo,Yo)ll2
IRe(f(z) - f(zo) - f'(zo)(z - zo))I
=���-��--��-�<€
lz- zol
Thus u is differentiable at (x0, Yo). Similarly, v is also differentiable at
(xo,Yo).
□
Remark 2.2. We will see later on that in fact the real and imaginary parts
of a holomorphic function are infinitely many times differentiable.
42
A Friendly Approach to Complex Analysis
Exercise 2.9. Consider Exercise 2.1 again. Show that J is not differentiable at
any point of the open set (C \ {0}.
One also has the following converse to Theorem 2.1. This is a very useful
result to check the holomorphicity of functions.
Theorem 2.2. Let
(1) U be an open subset of C,
(2) u,v : U ➔ IR. be continuously differentiable, and
(3) u,v satisfy the Cauchy-Riemann equations:
for all (x,y) EU,
au
av
au
av
(x,y) = (x,y) and
(x,y) = - (x,y).
ay
ax
ax
ay
Then f := u + iv : U ➔ C is holomorphic in U and
J' (X + iy) = ;: (X' y) + i ;: (X' y) for X + iy E u.
Proof. Let z0 = x 0 + iy0 E U. Let E > 0. Let o > 0 be such that
whenever z = x +iy belongs to the disc D(z0, o) := { w EC: lw- z0 1 < o},
we have z EU ,
l
au
au
(x,y )- (xo,Yo) I
ax
ax
av
av
< E, and 1ax(x,y )- ax(xo,Yo) I < E.
(2.7)
(That this is possible follows from the fact that u,v are continuously dif­
ferentiable.)
Let z = x + iy be a fixed point in the punctured disc D(z0, o) \ { z0},
and consider the part of the line in D (z0, o) joining z0 to z. A point on this
line can be represented by
p(t) = (1-t)zo + tz = ((1- t)xo + tx, (1- t)yo + ty )­
In particular, p(O) = z0 and p(l) = z. See Figure 2.7.
z = (x,y)
p(t)
zo = (xo,Yo)
Fig. 2.7
= (1 - t)z0 + tz
= ((1 - t)xo + tx, (1 - t)yo + ty)
Points p(t), on the line joining zo to z.
43
Complex differentiability
Define r.p1,r.p2 : (-1, 1) ➔ JR.by
[r.p1(t)] · = u(p(t))
]
[
v(p(t))
r.p2(t) ·
Then using the chain rule we obtain
·
- xo)+ �; (p(t)) · (y - Yo)
]
· - xo)+ -y (aavp(t)) · (y - Yo)
�: (p(t)) · (x
['Pi(t)] ·[
.
av
'P2I (t)
-a(p(t)) (x
x
Let
0
u
a
av
(p(t)) =
(p(t)),
ya
xa
au
av
B(t) := - (p(t)) =
(p(t)),
ya
xa
where we have used the Cauchy-Riemann equations to obtain the rightmost
equalities. Thus with this notation, we have
['Pi(t)]= A(t)(x-xo) B(t)(y-yo)]=[Re((A(t)+iB(t))(z zo))
[
]
r.p�(t)
B(t)(x x0) + A(t)(y-y0)
lm((A(t) + iB(t))(z zo))
A(t) :=
-
-
So
f(z)
-
·
- f(zo) = u(x,y) - u(xo,Yo)+i(v(x,y) - v(xo,Yo))
= r.p1(l) - r.p1(0) + i(r.p2(l) - r.p2(0))
=
=
Thus
11
(11
11
1
1
+
'Pi(t)dt+i
A(t)dt
r.p;(t)dt
i
· - zo)­
B(t)dt) (z
u
a(
) . av( ))
- f(zo) - (
-a Xo,Yo + i-a Xo,Yo
u
a
u
a
) . [ av
((p(t)) (p(O)) dt+ l ((p(t)) -
f(z)
Z-Zo
X
X
)
1
av
[
(p(O)) dt.
i
l
= o
a
a
a
a
x
x
x
x
o
By (2.7), it follows that
v
a
u
a
f(z) f(zo)
((xo,Yo)+i(xo,Yo) <E+€=2€.
a
a
z zo
x
x
This holds for all z satisfying O < lz - zol < 8, and so f is complex differ­
entiable at z0 and
1
I
-
-
.
)
I
f'(zo) = �: (xo, Yo)+i �: (xo,Yo).
This completes the proof.
□
44
A Friendly Approach to Complex Analysis
Let us revisit Example 2 .1 again, but now instead of using the E-8 def­
inition of complex differentiability, we will use the above result to check
holomorphicity of the squaring map.
Example 2. 7. For the function f : (C -+ C defined by f(z) =z2 (z E q
we have that
u(x,y)= Re(f(x+ iy))=Re(x 2
-y
v(x,y)=Im(f(x + iy))=Im(x 2
-y
2
+ 2xyi)=x 2 - y2,
2
+ 2xyi)=2xy.
Thus
au
2 x=
(x,y)
ax
au
(x,y) = -2y =
ay
which shows that the Cauchy-Riemann equations hold in C. So we recover
our previous observation that f is entire, and since
J'(z)= �:(x,y)+i::(x,y)=2 x +2yi =2z,
◊
we also obtain that f'(z)=2z for z EC.
Example 2.8. (exp, sin, cos are entire.) For the function g: (C-+ C defined
by g(z)=expz (z E q we have that
u(x,y)= Re(g(x+ iy))=Re(e x (cosy+i siny))=e x cosy,
Thus
v(x,y)= Im(g(x+ iy))=Im(e x (cosy+i sin y))=e x sin y.
au
av
e "' cosy=
(x,y),
(x,y)
ay
ax
au
av
x
(x,y) = -e siny = - (x,
y),
ax
ay
which shows that the Cauchy-Riemann equations hold in C. So we arrive
at the important result that exp is entire, and since
"'
x
g'(z)= �:(x,y)+i;:(x,y) =e cosy+ie sin y=expz,
we also obtain that : exp z=exp z for z EC. Hence from Proposition 2 .2 ,
z
also the trigonometric functions
exp(iz) - exp(-iz)
.
smz=
2i
and
cosz=
exp(iz)+exp(-iz)
2
Complex differentiability
45
are entire functions, and moreover,
d l.
S nz
�
p( iz)- (-i) exp(-iz) -----iz)
i ex-------- = exp( iz) +exp(-= cosz ' and
=-
d
=
- COS Z
�
�
exp(-iz)
i exp( iz) +(-i)
--------2
=
2
exp( iz)- exp(-iz)
- ------�
= - Sl. n Z .
◊
Example 2.9. (Holomorphicity of Log.) We will show that the principal
logarithm is holomorphic in the open set C\ (-oo, O]. Note that the principal
logarithm is defined in the bigger set (C \ { 0}, but we had seen earlier that it
is not continuous in this bigger set ( because at each negative real number,
it is discontinuous). We had also seen that in the smaller set (C \ (-oo, O],
the principal logarithm is continuous. We will now use this continuity to
show that Log is in fact holomorphic in (C \ (-oo, O], and that
d
l
- Log(z) = - for z E (C \ (-oo, O].
dz
z
First note that when z , zo EC\ (-oo, O] are distinct, then Log(z) -/- Log(zo)­
(Why?) Let E > 0. Set
Since expw is differentiable at wo := Log(zo), there is a 81 > 0 such that
whenever w satisfies O < lw -wol = lw- Log(zo)I < 81, there holds that
I
I
1-1
expw-zo
expw - exp Wo
-zo <
- expwo w- Log(zo)
w-wo
E1-
But by the continuity and injectivity of Log in (C \ (-oo, O], there exists a
o > 0 such that whenever O < lz -zol < o, we have
0 < I Log z - Log zo I < 81.
Thus with w := Log z, and O < lz -zol < o, we have O < lw -wol < 81,
and so
I
I
z -zo
----- --zo <
Log z - Log zo
E1.
I
46
But then
I
we have
A Friendly Approach to Complex Analysis
I
z - zo
lzol
whenever 0 <lz-zol <J,
�lzol-E1 �-.So
Log z - Log zo
2
Log z-Log z0
z-z0
1
- zo
I (
I
=
zo -
z-zo
Log z-Log z0 )
= zo Log
z-zo
z-Log z0
1
1
<El· -- • lzol/2 lzol
I
·
1
1
z
-z
o
·
Log z - Log zo
1
z-zo
Log z-Log zo
2E1
= --2 <E.
1
Zo
1 ·�
1
lzol
Thus Log is holomorphic in (C \ (-oo, 0] and moreover, d Log z = �- ◊
z
dz
We now consider an example illustrating the fact that the assumption of
differentiability of u, v in Theorem 2.2 (as opposed to mere existence of
partial derivatives of u, v satisfying the Cauchy-Riemann equations), is not
superfluous.
Example 2.10. Consider the function f: (C--+ (C given by
xy(x+iy)
.
f( X +iy )X 2 +y 2
if x + iy-/- 0, and f (0) = 0. We have that for nonzero (x, y) E IR2,
x2 y
u(x, y) = Re(f(x + iy)) = 2
X +y 2,
xy2
v(x, y) = Im(f(x + iy)) = 2
X +y 2,
and u(0, 0) = v(0, 0) = 0. Thus
av
au
au
av
(0, 0) = 0 = - (0, 0),
(0, 0), and
(0, 0) = 0 =
ax
ay
ay
ax
which shows that the Cauchy-Riemann equations hold at the point (0, 0).
However, the function is not complex differentiable at 0, since if it were, we
would have
au
.av
I
f (0) = ax (0, 0) + i ax (0, 0) = 0 + iO = 0,
and for E = 1/4, there would exist a corresponding J such that whenever
0 <lz- 0I = Ix+ iyl <J, we would have
xy
f(z) - f(O) = 2 2 <E,
x +y
z-0
I
j'(O)I I
I
47
Complex differentiability
but taking x + iy=
2 + i 2, we arrive at the contradiction that
J
J
1
I I
I
xy
< f. =
4·
x 2 + y2
This shows that f is not complex differentiable at 0.
We note that there is no contradiction to Theorem 2.2, since for example
u is not differentiable at (0,0). It it were, its derivative at (0,0) would have
to be the linear transformation
2=
[:] � [ :: (0,0) :� (0,0)] [:] = [ 0 0] [:] = 0.
But then with f. := 1/3 > 0, there must exist a J > 0 such that whenever
0 < ll(x,y)-(0,0)11 2 < J, we would have
1
x2
lu(x,y)-u(0,0)-0((x,y)-(0,0))1
-------------= 2 y 2 3 < f.=3
ll(x,y)-(0,0)11 2
(x +y )2
.
Then
with
(x, y) = ( �'�), we have ll(x,y) -(0,0)11 2= � < J, and so
J2 J
4 · 2 -=x y
1
1
----���
<E=-1 ==
'
2
2
2
(x +y ) !
J !
(�+ :)
J8
a contradiction. So u is not differentiable at (0,0).
3
v'§
◊
The Cauchy-Riemann equations can also be used to prove some interest­
ing facts, for example the following one, which highlights the "rigidity" of
holomorphic functions alluded to earlier. See also Exercise 2.12 below.
Example 2.11. (Holomorphic function with constant modulus on a disc
is a constant.) Consider the disc D= {z EC: lz-zol < r}. We will show
using the Cauchy-Riemann equations that if f : D -+ C is holomorphic
in D, with the property that there is a c E JR such that IJ(z)I= c for all
z E D, then f is constant. (We will use this fact in later when we learn
about a result called the "Maximum Modulus Theorem".) See the picture
below.
f
,-------------.
48
A Friendly Approach to Complex Analysis
Let u, v denote the real and imaginary parts of f. By assumption, we have
c2= 1/1 2= u2 + v2, and so by differentiation,
au
u
ax
v
+
ay
v
+
au
u
va
ax
va
ay
=
o,
= 0.
va
u.
. va
.
= - a Ill thefirst equatIOn and =
Vsmg
ax
ay
ay
au
x
a
.
.
Ill the second equat10n,
au
au
u- -v- =0,
ax
au
u
To remove
To remove
au
ay
au
ax
a
y
(2.8)
a
y
+
v
au
ax
(2.9)
= 0.
au
.
, multiply (2.8)
by u, (2.9)
by v, and add: (u 2 + v 2) = 0.
ax
au
.
, multiply (2.8by
) -v, (2.9by
) u, and add: (u 2 + v 2) = 0.
ay
If c= 0 so that u + v = c2= 0, then u= v= 0 and so f= 0 in D.
If C-/=- 0, then
2
2
au
ax
=
au
ay
va
va
. .
=
= 0.
= 0, and the CR equat10ns give also
ax
By the Fundamental Theorem of Integral Calculus, it follows that
x
ay
1 u (�, Yo)d�= 0,
Y u
) - u( , Yo)= 1 � ( , rt)drt= 0.
y
u(x, Yo) - u(xo, Yo)=
u(x, y
xo
a
aX
x
Yo
x
So the value of u at any (x, y) is the same as the value of u at zo= (xo, Yo).
Thus u is constant in D.
(x y)
(xo, Yo ) J
(x, Yo)
Similarly, v is constant in D.
Consequently, f= u + iv is constant in D.
◊
Exercise 2.10. Show that
z i---+ z3
is entire using the Cauchy-Riemann equations.
49
Complex differentiability
Exercise 2.11. Show that z I-+ Re(z) is complex differentiable nowhere.
Exercise 2.12. Let DC (C be a domain. Show, using the Cauchy-Riemann equa­
tions, that if f : D ➔ (C is holomorphic in D, with the property that f(z) E R
for all z E D, then f is constant in D.
Exercise 2.13. Let D C (C be a domain. Show that if f : D ➔ (C is holomorphic
in D, with the property that J' (z) = 0 for all z E D, then f is constant in D.
Exercise 2.14. Suppose that f : (C ➔ (C is an entire function, and let u := Re(!),
v := Im(!). Suppose, moreover, that there exists a differentiable h: R ➔ R such
that u =ho v. Prove that f must be a constant.
Exercise 2.15. Let k ER be a fixed, and let f be defined by f(z) = (x 2 -y 2 )+kxyi
for z = x + iy, x, y ER. Show that f is entire if and only if k = 2.
2.3
Geometric meaning of the complex derivative
In ordinary calculus we have learnt the geometric meaning of the derivative
of a real valued function f:�➔�at a point xo E�: f'(x0) is the slope
of the tangent to the graph off at x0• See Figure 2.8.
f
slope= f'(xa)-------..._
Xo
Fig. 2.8 Meaning off' (xo)lim
:z:-+:z:o
J(x) - f(xo)
J(x) - J(xo)
=f'(x0) implies
R::-j'(xo) for x near xo, i.e.
X - Xo
X - Xo
f(x) - J(xo) R::- J'(xo)(x - xo)-
This means that locally, around x0, J(x) - J(x0) looks like the action of
the linear map h rt f'(x0 )h: �➔�on x - Xo- Visually this means that
near x 0, there is very little difference between the (tangent) line with slope
f'(x0) passing through (x0, f(x0)) and the graph off. That is, if we zoom
into the graph of the function around the point (x0, f(x0)), then the graph
looks like a straight line.
50
A Friendly Approach to Complex Analysis
In this section, we pose an analogous question for a complex valued
function map f : U ➔ C defined on an open set U, that happens to be
complex differentiable at a point z0:
What is the geometric meaning of the complex number f'(z0)?
We can't draw a graph off, because z as well as f(z) belong to C = �2,
and so (z, f(z)) would be a point in �2 x �2 = �4 ! But we can draw a copy
of U in the plane on the left hand side, and on the right hand side, we can
imagine a copy of C, with f mapping points from U on the left to points
on the right, as shown below.
f
�
C
·f (z)
We will now show that the complex number f'(z0) describes the action of
the complex differentiable function locally infinitesimally around z0 by a
anticlockwise rotation through the angle Arg(f'(zo)) together with a scal­
ing/ magnification by lf'(zo)lf(z) - f(zo) =
(z) f(zo)
�f'(zo) for z near zo, i.e.
f'(z0), implies f lim
z➔zo
Z - Zo
Z - Zo
f(z) - f(zo) � f'(zo)(z - zo).
But from the geometric meaning of complex multiplication, we know that
when we multiply z-zo by f'(z0), z- zo gets rotated anticlockwise through
the angle Arg(f'(zo)), and the length of z - z0 gets multiplied by the length
of f'(z0), namely we get a magnification by the factor lf'(zo)I. In order to
further understand this geometrically, look at Figure 2.9.
z
zo
Fig. 2.9
f
�
f(zo)
Geometric local meaning of the complex derivative.
Complex differentiability
v'3
51
Suppose that f'(zo) =
+ i, so that lf'(zo)I = 2 and Arg(f'(zo)) = ?T/6.
First look at z - z0 shown in the domain U as the solid line segment between
z and z0. On the right hand side, we have shown a translated version of
this line segment as a dashed line, emanating from f (z0). In order to find
out where f(z) is, we just use the fact that f(z) - f(z0) is approximately
equal to f'(zo) multiplied by z - z0• So the solid line denoting f(z) f(z0) on the right hand side is obtained by rotating the rightmost dashed
line anticlockwise through an angle of Arg(f'(zo)) ( assumed to be 30 ° in
this picture), and magnifying the length of the dashed line by lf'(zo)I =
2. Note that if want to find the image of another point z which is near
z0, we have to repeat the same procedure. Namely we first look at the
line segment joining z0 to z, which is the solid line on the left. We have
shown a translated version of this as a dashed line, emanating from f (z0),
in the picture on right hand side. In order to find the position of f(z),
we first rotate the leftmost dashed line anticlockwise through an angle of
f' ( z0), that is 30 ° , and magnifying the length of the leftmost dashed line
by lf'(zo)I = 2. In this manner, we obtain the solid line on the right hand
side representing f(z) - f(z0). Placing one end at f(z0), the other end is
the point f(z) (approximately!). So we see that locally the action off is
as follows. Imagine the domain as a rubber sheet, and look at a point z0
on this rubber sheet. Tear out a small portion of this rubber sheet around
z0. The function f takes the point z0 on this rubber sheet to a point f(zo)
somewhere in the complex plane. If we want to know how the rest of the
points on our little torn rubber sheet are mapped by f, one follows this
procedure. We place our rubber sheet such that zo on our rubber sheet is
lying over the point f(zo) in the complex plane. (Imagine pinning it on the
plane with the pin passing through the point marked z0 on our little torn
rubber sheet.) Then we stretch out our rubber sheet about the point z0 by a
factor of lf'(zo)I, and then rotate this stretched rubber sheet anticlockwise
by an angle of Arg(f'(z0)) around the point z0 on the rubber sheet.
In order to stress this geometric interpretation, let us revisit Example 2.1
yet again, where we considered the squaring map z i-+ z 2. See Figure 2.10
and Example 2.12.
52
A Friendly Approach to Complex Analysis
.
2,5
;
'
1
:
;
.
2·····L ....l±J.------1 .....
� LL_I
, !_J.. L
I
l'
:
.. J! ... � .. � ...... ) ..
l
Q.8
Q.6
o.4
0.2-
0
Q,2
Q,4
0,6.
0,8
-1
-1
Fig. 2.10
-(),5
0
�.5
1
The image of a square and a square grid under the mapping z 2 •
Example 2.12. Suppose that we assume complex differentiability of the
squaring map f, z H z 2 , at a point z0 E C. Let us then try to prove that
the complex derivative of the squaring map at z0 must be 2z0 by figuring
out geometrically the local amplification at z0 and the local rotation at z0
produced by the squaring map.
We first ask: what is the local rotation produced? In order to find this
out, consider a point z close to z0 along the ray joining z0 to z. Look
at Figure 2.11, which shows the effect of the squaring map: the angle is
doubled, and the distance to O is squared. Thus z 2 lies on the ray joining
0 to z5, which makes an angle 2Arg(z0) with the positive real axis. Hence
the line segment joining z5 to z 2 is obtained by rotating the line segment
joining z0 to z anticlockwise through the angle Arg(z0). Consequently,
Arg(f'(zo)) = Arg(zo)-
53
Complex differentiability
Z H z2
z
�o)
0
Fig. 2.11
Calculation of the amount of local rotation produced by the squaring map.
Next we ask: what is the local amplification produced? In order to find this
out, consider a point z close to z0 which is at the same distance from O as
z0, but it makes an slightly bigger angle 0 + d0 with the positive real axis.
Look at Figure 2.12, which shows the effect of the squaring map. Since d0
is tiny, the length lz-zol is approximately lzol ·d0, while the length lz 2 - z5I
is approximately lzol 2 • 2d0. Consequently the magnification factor lf'(zo)I
must be equal to (lzol 2 · 2d0)/(lzol · d0) = 2lzol-
Z
H z2
z
0
Fig. 2.12
0
Finding the amount of local magnification produced by the squaring map.
A Friendly Approach to Complex Analysis
54
Putting it all together, we have
J'(zo) = lf'(zo)I · ( cos(Arg(J'(zo))) +isin(Arg(J'(zo))))
= 2lzol · ( cos(Arg(zo)) +isin(Arg(zo)))
= 2z0 ..
So by investigating the local behaviour of the squaring map f around the
point z0, we could find out the complex derivative f'(z0).
◊
Example 2.13. (Complex conjugation is complex differentiable nowhere.)
See Figure 2.13.
w
w
Fig. 2.13
z t-+ z is not holomorphic.
Suppose that z r-+ z is complex differentiable at z0. Then the local be­
haviour of the map around zo should be a rotation followed by an am­
plification. Consider the point z near z0 obtained by a tiny horizontal
translation. From the picture, by looking at the images z0 and z, we see
that since z - z0 = z - z0, no rotation is produced. On the other hand,
if we look at w which is near z0 obtained by a tiny vertical displacement,
then from the picture, we see that w - z0 = -(z - z0), and so there is a
rotation through 180 ° . But this means that locally the map is not a rota­
tion (because if it were, all infinitesimal vectors emanating at z0 would be
rotated by f by a same fixed amount!).
◊
Exercise 2.16. We know that the power function z r-+ z n , n EN, is entire. Find
its complex derivative by investigating its local behaviour.
Hint: Since the rotation and magnification is the same for all tiny arrows ema­
nating from zo, one can simply the argument given in Example 2.12 by looking
at what happens to a tiny vector perpendicular to the ray joining Oto zo.
55
Complex differentiability
Exercise 2.17. We know that the exponential function z >---+ e z is entire. Find its
complex derivative by investigating its local behaviour.
Hint: Take a point zo, and move it vertically up through a distance of o. Look
at the image to determine the amplification. Similarly, by moving zo horizontally
through o, determine the amount of rotation produced locally.
Exercise 2.18. Give a visual argument to show that the map z >---+ Re(z) is not
complex differentiable any where in <C.
Conformality. Look at Figure 1.16 on page 20 again, which shows the
action of the entire mapping exp. In this picture, we see that just like in the
domain, also the images of the vertical and horizontal lines are mutually
perpendicular. We had mentioned that this is the property of conformality,
that is, of the preservation of angles between curves in the domain together
with their "orientation" is something which is possessed by all complex
differentiable functions in domains. Moreover, we had mentioned that this
property is possessed by every holomorphic function in its domain. Let us
now see why holomorphic functions possess this property, based on what
we have learnt about the local action of complex differentiable functions.
f
f(p)
Fig. 2.14
Let f :
U ---+
Conformality of a holomorphic
(C be holomorphic in
ing at a point p E
U.
U.
f.
Here cp
= Arg(f'(p)).
Imagine two smooth curves intersect­
Since the curves are smooth, they have tangents at
p, say T1 and T2. Look at Figure 2.14. Near the point p, there is very little
difference between the curve and its tangent line at p, so we may assume
that the curves are replaced by their tangent lines. These tangent lines
make a certain angle. Now let us look at what f does to these lines. Each
of the curves is mapped to new curves in (C by f, and these intersect at
f of p, that is f(p). They are smooth too, and they possess new tangent
lines. But since the local action of f around p is rotation clockwise by
Arg(f' (p)) followed by magnification, the new tangent lines are obtained
by just rotating clockwise the old tangent lines, and magnifying the image.
So it is obvious that the angle will be the same, and so will the orientation.
So the conformality of holomorphic maps isn't a mystery anymore!
56
A Friendly Approach to Complex Analysis
Relation to real differentiability, and the Cauchy-Riemann equa­
tions revisited. Consider a map f : U -+ C which is complex differentiable
at z0 = (x 0, y0) in the open set U. If u, v denote the real and imaginary
part of the function f, then we know that u, v : U -+ IR are differentiable
at (xo,Yo)- Thus f, that is, the map (x,y) H (u(x,y),v(x,y)): U-+ IR2 , is
differentiable (in the real variable sense), and we know that its derivative
is the linear transformation
l
�:(xo,Yo) �� (xo,Yo)
·- [
A.
av
av
(xo,Yo)
(xo,Yo)
ay
ax
which describes the local action. But since f is complex differentiable, we
know that its local action is a anticlockwise rotation by 0 := Arg(f'(zo)),
followed by a magnification by r := lf'(zo)I, and so it is the linear trans­
formation described by the matrix
cos0 -sin0
r[
sm
· 0 cos 0] ·
Since A must equal this, we obtain that
au
av
( Xo,Yo) = r cos0 =
( Xo,Yo),
ax
ay
au
av
(Xo,Yo) = r sin0 = - ay ( Xo,Yo).
ax
Moreover, f'(zo) = r(cos0+isin0)
= �:(xo,Yo) +i��(xo,Yo)-
Summarizing, if f is complex differentiable at a point zo, then it is
real differentiable (as a mapping from U c IR2 to IR2 ), but what distin­
guishes complex differentiability from mere real differentiability is that the
real derivative for a complex differentiable mapping is not just any old lin­
ear transformation, but a very special one: it is an anticlockwise rotation
through an angle 0 followed by a scaling by r.
2.4
The d-bar operator
The two Cauchy-Riemann equations can be written as a single equation by
introducing what is called the "d-bar operator"
{)
az·
57
Complex differentiability
Let us define the differential operators
�
:=
!
az
2
(
�
� -i�
:=
) and
ax
!
az
ay
2
(
� +i �
ax
ay
).
By a differential operator, we mean something which can act on functions
and produce new functions. For example, the above two differential opera­
tors can act on smooth functions cp
,
: U -+ JR, where U is an open subset of
JR2 , and by the smoothness of cp we mean that it possesses at least its first
partial derivatives with respect to x and y everywhere in U. Then
acp
:= !
az
2
Also, for
u, v
( acpx a
/'P)
ay
and
acp
!
:=
az
2
a
a
( acpx +i acpy ).
smooth real-valued functions in U, we define
.
.
a
a
av
(u+iv) := aauz +i. az
(u+iv) := aauz+i. aavz and az
.
az
With this notation, we have for a holomorphic f =
UC
<C that
a
az
u + iv in
the open set
a
. =au
. av
+i­
-(u+iv)
az
az
az
.1
1
av
. av
au
.
au
=- (-+i-) +i- (-+i- )
2 ax
ay
2 ax
ay
= ! ( au_ av) +i ( au+ av)
ax
2 a
2 ax
a
-f =
!
y
y
= O+iO = 0,
using the Cauchy-Riemann equations for the real and imaginary parts
off. Also, we have
�f
fu
= �(u+iv)= au+i av
fu
fu
fu
av _ av
( a x i ay )
= ! ( au+ av) + i (- au+ av)
2 ax
ay
2
ay
ax
1
au
.1
av
au
. av
= - · 2- + i- · 2- = - + i2 ax
2
ax
ax
ax
au
au
( _ i ay ) +i!2
2 ax
= !
!
= J'.
Summarizing, for f holomorphic in
U that : f = 0,
z
and
:zf =
J'.
u, v
58
A F'riendly Approach to Complex Analysis
So philosophically, we ought to think of holomorphic functions as "func­
tions of
z, z
which are independent of z".
Example 2.14. z is not holomorphic because
8
_
{)(
")
OZZ = OZ X - zy =
1
2
8
OX+z ay
1
. 1 .
= - -z · - • z =1 =/-
2
.
(8
)
.
. 1 (8
X- 2
2
8
OX+z Oy
)
y
o.
2
◊
Example 2.15. lzl = zz is not holomorphic in (C \ {O} because
2
8
-
8
1
--=(zz) = --=(x +y ) =2
oz
2
oz
2
.
(8
8
-+zax
ay
)
(x +y )
2
2
=½(2x+i2y) =x+iy =z =f. 0 in (C \ {O}.
◊
Example 2.16. z 2 is entire because
(
!(z 2 ) = :z
1
=2
(
x2
-
8
2
y +2xyi)
.
8
-+zax
ay
)( 2
2)
.1 (
X -y +z2
8
.
8
)(
8
.
8
)
-+zax
ay
)
2xy
=½(2x -i2y)+i½(2y+i2x) =0.
Moreover,
:z (z 2 )
2
2
= !(x - y +2xyi)
1
=2
1
(
8
.
8
--iax
ay
)
2
2)
.1 (
(x -y +z2
1
--iax
ay
(2xy)
= (2x+i2y)+i (2y -i2x) =2(x+iy) =2z.
2
Exercise 2.19. Show that 4
2.5
2
!!
= .6., where .6. := ::2
◊
+
::2 is the Laplacian.
Notes
The section on the geometric meaning of the complex derivative follows
closely the exposition in [Needham (1997)]. Exercises 2.16 and 2.17 are
taken from [Needham (1997)].
Chapter 3
Cauchy Integral Theorem and
consequences
Having become familiar with complex differentiation, we now turn to inte­
gration. In this chapter we will learn a very important theorem in complex
analysis, called
I The Cauchy Integral Theorem, I
We will begin by defining "contour integration". And then we will show
the Cauchy Integral Theorem. One might ask: Why is this contour inte­
gration and the Cauchy Integral Theorem so important? The importance
of integration in the complex plane stems from the fact that it will lead to
a greater understanding of holomorphic functions, for example, the funda­
mental fact that holomorphic functions are infinitely many times complex
differentiable! In this chapter we will learn the following main topics:
(1)
(2)
(3)
(4)
Contour integration and its properties.
The Fundamental Theorem of Contour Integration.
The Cauchy Integral Theorem.
Consequences of the Cauchy Integral Theorem:
(a)
(b)
(c)
(d)
3.1
Existence of a primitive.
Infinite differentiability of holomorphic functions.
Liouville's Theorem and the Fundamental Theorem of Algebra.
Morera's Theorem.
Definition of the contour integral
In ordinary calculus, given a continuous function f : [a, b] ➔ JR.,
l
b
f(x)dx
59
(3.1)
60
A Friendly Approach to Complex Analysis
has a clear meaning. Now suppose we wish to generalize this in the complex
setting: given z, w complex numbers, want to give meaning to something
like
1
w
f(() d(.
Then a first question is: How do we get from z to w?
In JR, if a< b, then there is just one way of going from the real number
a to the real number b, and so our data in the real case is just:
(1) a< b, and
(2) a continuous function f: [a, b] -t R
But now z and w are points in the complex plane, and so there are many
possible connecting paths along which we could integrate. See Figure 3.1.
- - - - o----e:::,,.---00
a
b
Fig. 3.1
JR
- - - -
z
c,;r
Which path to go from z to w?
So in the complex setting, besides specifying the end points z and w, we
will also specify the path 'Y taken to go from z to w, and we will replace
the above expression (3.1) in the real case by an expression which looks like
this in the complex setting:
1
f(z)dz.
We call such an expression a "contour" integral, for the computation of
which we need the following data:
(1) A domain D (c IC), and z,w ED.
(2) A continuous function f : D-+ C.
(3) A smooth path 'Y : [a, b] -t D joining z to w.
Cauchy Integral Theorem
61
We note that we need not merely a path, but a smooth path joining z to
w. What do we mean by "smooth" here? Recall that a path,: [a, b]-+ D
is a continuous function. We can decompose, into its real and imaginary
parts:
,(t) = x(t) + iy(t),
where x, y : [a, b] -+ R The path, is called
differentiable. Here are some examples.
t E [a, b],
smooth if x, y are continuously
re be given by,(t) = t(l+i), t E [O, l]. Then
Example 3.1. Let,: [O, 1]-+
the real and imaginary parts x, y : [O, 1] -+ IR of, are given by x(t) = t,
y(t) = t, t E [O, l]. Since x, y are continuously differentiable on [O, 1],, is a
smooth path. See Figure 3.2.
l+i
'Y
0
Fig. 3.2
The smooth path 1.
Similarly consider the two paths,1, , 2 : [O, 21r] -+ IR, given by
where
t
,1 (t)
= exp(it)
and ,2(t)
= exp(2it),
E [O, 21r]. Then since the real and imaginary parts of these paths
are cos t, sin t, cos(2t), sin(2t), all of which are continuously differentiable, it
follows that,1, , 2 are smooth paths. See Figure 3.3.
Fig. 3.3
The smooth path 1.
So the set of image points (the ranges of,1 and ,2) is the same, that is:
{,1 (t) :
t
E [O, 21r]}
=
{,2(t) :
t
E [O, 21r]}
= {z
circle with center 0). However ,1 and ,2 are
functions are not the same: for example, ,1(1r)
E
re :
lzl
= 1}, (the
unit
different paths, because the
= -1-/=- 1 = , 2(1r).
◊
62
A Friendly Approach to Complex Analysis
Remark 3.1. It is very common and convenient to refer to the
t E [a, bl}
path/curve itself.
range
{'y(t) :
of a path
'Y
: [a, b] --+ C as the
W ith this usage, a path
becomes a concrete geometric object (as opposed to being a mapping), such
as a circle or a straight line segment in the complex plane and hence can
be easily visualized. The difficulty with this abuse of terminology is that
several
different paths can have the same image, and so it causes ambiguity.
The precise definition of the contour integral is given below.
Definition 3.1. Given
(1) a domain
D,
:
(2) a continuous function f D --+ C (with real and imaginary parts de­
noted by u, v: D ➔ �), and
(3) a smooth path 'Y : [a, b] --+
by x, y : [a, b] --+ �),
we define
i
f(z)dz
:=
:=
:=
D
(with real and imaginary parts denoted
b
1 f('Y(t))'Y'(t)dt
b
1
·(x'(t)+iy'(t))dt
b
1 (u('Y(t)) x'(t) -v('Y(t)) y'(t))dt
b
1 (v('Y(t)) x'(t) u('Y(t)) y'(t))dt.
(3.2)
(u('Y(t))+ iv('Y(t)))
·
·
•
+i
•
+
We note that the two integrals in the last line above are just the usual
Riemann integrals of real-valued continuous functions.
The contour integral can be interpreted geometrically as follows. The term
'Y'(t)dt = x'(t)dt + iy'(t)dt
can be viewed as an infinitesimal incremental piece of the contour.
multiply this by the (almost constant) value f('Y(t))
We
off on this incremental
piece. Finally, we add up all these contributions along the contour to get
the total as the integral
b
1 f('Y(t))'Y'(t)dt.
See Figure 3.4.
63
Cauchy Integral Theorem
D
f('Y(t))
f
'Y(b)
�
iy'(t)dt
\
'Y(a)
x'(t)dt
Fig. 3.4
Geometric meaning of the contour integral.
Here is an example.
Example 3.2. Let
(l)
D
= C,
(2) 'Y be the smooth path given by
(3) f = (z Hz).
l
Then
f(z)dz =
=
11
11
t(l + i) • (1 + i)dt
t(l - i) · (1 + i)dt =
= 2 r1 tdt = 2 .
Jo
'Y(t) = t(l + i), t E
t2 1
l
2o
=2
.
11
t(l 2
-
[O,
i2 )dt =
1],
and
11
t(l + l)dt
1
2 = 1.
◊
Exercise 3.1. Consider the three paths 11,,2,13: [0,27r]--+ <C defined by
= exp(it),
12(t) = exp(2it),
13(t) = exp(-it),
11(t)
for t E [O,27r].
integrals
are all different.
Show that their images are the same, but the three contour
64
A Friendly Approach to Complex Analysis
Exercise 3.2. Let f be holomorphic in a domain and let 'Y: [O, 1] ➔ D be a smooth
path. Show that
:/('Y(t))
= J'('y(t)) · "/(t) for all t E
[O, l].
In the sequel, we will often assume that our smooth paths are parameterized
by [O, 1] (rather than some more general interval [a, bl). Let us explain why
we may assume this.
Suppose that 'Y : [a, b] ➔ and -;:;; : [c, d] ---+
are two smooth paths,
such that there is a continuously differentiable function cp : [c, d] ➔ [a, b]
such that a= cp(c), b = cp(d), and -;:;;(t) = "f(cp(t)) for t E [a, b]. We call such
smooth paths "equivalent". (Imagine going from 'Y(a) = -;::;(c) to 'Y(b) = -;::;(d)
along the same route, but with possibly different speeds.) See Figure 3.5.
We now show the following.
re
b
cp(t)
1
''
: cp
a
C
d
t
re,
'(b)�',(a)
�
J
'\') � ,(,p(t))
)
�
Fig. 3.5
'Y(a) = -;::;(c)
Equivalent paths.
Equivalent paths give the same integral. By the chain rule it follows
that
b
1 J(-;::;(t))-;::;'(t)dt = 1 f('Y(cp(t))h'(cp(t))cp'(t)dt
b
hf(z)dz
=
(-r=cp(t))
id J('Y(T)h'(T)dT =
c
In particular, given any 'Y: [a, b] ➔
cp(t) =
1
"Y
J(z)dz.
re, we can define cp : [O, 1] ---+ [a, b] by
t)a + tb,
t E [a, b].
Then cp is continuously differentiable, and cp(O) = a, cp(l) = b. So with
c := 0, d := 1 in the above, and with -;:;; : [O, 1] ➔ re defined by -;:;; = 'Y o cp,
(1-
we see that
h J(z)dz =
i
f(z)dz,
Cauchy Integral Theorem
65
and so there is no loss of generality (when it comes to statements about
contour integrals) in assuming that the smooth path is parameterized by
[O, l].
Contour integrals along piecewise smooth paths. We extend the
definition above to paths with "corners". A path , : [a, b] -+ C is called a
piecewise smooth path/curve if there exist points
a < c1 < · · · <
and, is continuously differentiable on
For such a path, we define
1
7
f(z)dz :=
l
a
+
Here is an example.
ci
1
<b
Cn
[a, c1l, [c1, c2l, ..., [cn -1, cn l, [en, b].
f(,(t)),'(t)dt +
1::
J
c2
q
f(,(t)),'(t)dt +
Example 3.3. Let 7y be the path from O to 1
_
,(t)= {
See Figure 3.6.
t
c1, ..., Cn such that
f(,(t)),'(t)dt + · · ·
1:
f(,(t)),'(t)dt.
+ i given by
ift E [O, 1]
l+(t-l)i iftE(l,2].
l+i
'
Fig. 3.6
Then we have
J2
12
{ zdz = f\1dt+
1�
lo
=
=
fo 1 tdt +
1
0
The piecewise smooth path
(1 + (t- l)i)idt= (
(i
lo
+ (t- l))dt
-1 = 1 +i.
!2 +i+ 4-l
2
1
'i-
tdt+
12
1
(1 -
(t- l)i)idt
66
A Friendly Approach to Complex Analysis
Let us look at the answers obtained in Examples 3.2 and 3.3. We see that
z f---+ z), the end
+ i), and the paths joining
the integrand is the same ( the nonholomorphic function
points of the paths are the same (namely O and 1
these two points are different ('-y and ::Y). See Figure 3.7.
l+i
'
Fig. 3.7
Iz
0
'
The two paths "Y and-;;;.
hz
The answer we obtain in each case is different too:
Thus the integral
z
f---+ z.
depends
dz = l -=/- 1
+i=
dz.
on the path for the nonholomorphic integrand
This is not strange, because from the definition of the contour
integral of course we expect the value of the contour integral to depend
on the route chosen. The main goal in this chapter is to show that the
contour integration of a holomorphic function along two paths joining the
same points from
z to w
is the
same provided
that the map is holomorphic
everywhere in the region between the two paths!
It
turns out that this
is a fundamental result (called the Cauchy Integral Theorem) in complex
analysis because many further results follow from this. Let us check this in
an example.
Example 3.4. Consider the same two contours 1, ::Y considered in Exam­
ples 3.2 and 3.3 above. But instead of taking the nonholomorphic map
z f---+ z,
I
1 + i)t(l + i)dt = 1 1 2itdt = i,
zdz = 1
1 t ldt + 12 + (t l)i)idt
consider now the
=
·
1
[ tdt +
lo
function
(l
h 1
zdz =
entire
(1
j\1
z:
Then we have
and
-
(t - l))dt
=
!2 - !2 + i = i.
We note that this time the answer is the same for I and for :::;;.
◊
Cauchy Integral Theorem
Exercise 3.3. Integrate the following functions over the circle
anticlockwise:
67
lzl =
2, oriented
(1) z+z.
(2) z 2 - 2z + 3.
(3) xy, where z
= x + iy, x, y E JR.
Exercise 3.4. Evaluate
i
Re(z)dz, where 'Y is:
(1) The straight line segment from Oto 1 + i.
(2) The short circular arc with center i and radius 1 joining Oto 1 + i.
(3) The part of the parabola y
3.1.1
= x 2 from x = 0 to x = l joining Oto 1 + i.
An important integral
Let us now calculate a very important simple contour integral, which will
recur again and again in the course. We fix a useful convention: through­
out the book, unless otherwise stated, a circular path with center z0 and
radius r > 0 traversed in the anticlockwise direction will mean the path
C: [O, 21r] -+ <C given by C(t) = z0 + rexp(it), t E [O, 21r]. (Thus the circle
is traversed once.) See Figure 3.8.
C
zo
r
Fig. 3.8 The circular path C with center zo and radius r traversed in the anticlockwise
direction.
We will now calculate the integrals
L
(z - zotdz, n E Z.
This calculation will prove to be very useful later on.
Theorem 3.1. Let C be a circular path with center z0 and radius r
traversed in the anticlockwise direction. Then
{ ( _ )nd = { 21ri if n = -1,
Jc z zo z
O
if n -/- -1.
>0
68
A Friendly Approach to Complex Analysis
We note that the answer is independent of r.
Proof. We have C(t) = zo+ rexp(it) = zo+ rcost+ irsint, t E [0,27!"],
and so C'(t) = -rsint+ircost= ir(cost+i sint) = irexp(it) , t E [0,27!"].
We now consider the two cases:
1 ° When n = -l, we have
{ (z - zotd z = { (z - zo)-1d z =
le
le
=
fo
21r
f 21r
lo
1
· irexp(it)dt
it
rexp(")
idt= 27l"i.
2° When n =/- -l, we have
fc (z - zo) d z= fo21r r exp(nit) • irexp(it)dt
2
= fo ir +l exp(i (n+ l)t)dt
2
2
= -r fo :in((n+ l)t)dt+ ir fo :os((n+ l)t)dt
n
n
n
1r
n+ 1
n+1
= 0+0= 0.
□
This completes the proof.
We will see later that this has significant consequences. For instance,
suppose that we have an f that has a "expansion in terms of integral powers
of z" (whatever that means) , in an annulus A. := { z E <C : r < lz - zol < R}
with center z0, inner radius r, and outer radius R like this:
f(z)
= L an (z - zot , z EA..
nEZ
We will give precise meaning to the (infinite) "sum" above later on, but
for now, one may just imagine a finite sum (so that all but finitely many
an s are zeros). Then by multiplying both sides by (z - z0 )-(m +l) for some
m E Z, we obtain
__f_(_z)_...,.. = " an (z - zo) n-m-1 ,
L..i
(z - zo)m +l
and so
l
-.
1
nEZ
1
f(z)
" an (z - zo) n-m-ld z = a .
d z = L..i
m
27ft e (Z - Zo )m +l
e
,,,
nEa..
Cauchy Integral Theorem
69
In the above, we assumed that the sum passes through integration over C,
which for finite sums, follows from the definition of the integral, and we
will see this in the next section. When the sum is not finite, we will make
precise the details later, but things work out essentially as suggested by
this calculation. The upshot of it all is that the coefficients are expressible
in terms of a contour integral, and we will see later that any function
holomorphic in an annulus will have such an expansion.
Exercise 3.5. Let C be the circular path with center O and radius 1 traversed in
the anticlockwise direction. Show that for O :S k :S n,
(n)
k
3.2
z) n
= � { (lz+
dz.
k
+1
27ri }0
Properties of contour integration
In this section we will show some useful properties of contour integration.
The following result follows in a straightforward manner from the definition
of contour integration.
Proposition 3.1. Let D be a domain in <C and "I: [a, b] ---+ D be a piecewise
smooth path. Then the following hold:
(1) For all continuous f, g : D---+ <C,
l (f + g)(z)dz= l f(z)dz+ l g(z)dz.
(2) For all continuous f : D ---+ <C and all a E <C,
l (af)(z)dz =al f(z)dz.
Let C(D; <C) denote the vector space over <C of all complex-valued continu­
ous functions on D with pointwise operations. Then the above result says
that each piecewise smooth path "/ in D induces a linear transformation
from C(D;<C) to <C, namely
f f--t l f(z)dz: C(D;<C)---+ <C.
Exercise 3.6. Prove Proposition 3.1.
70
A Friendly Approach to Complex Analysis
Opposite paths. Given a smooth path 'Y: [a, b] ---+Din a domain D, its
oppos ite path,-"(: [a, b] ---+ D, is defined by (-'Y)(t) = 'Y(a+b-t),t E [a, b].
Then (-'Y)(a) = 'Y(b) and (-'Y)(b) = 'Y(a), and so-"( starts where 'Y ends,
and ends at the starting point of 'Y while traversing the same path of 'Y but
in the opposite direction. See Figure 3.9.
Fig. 3.9
The opposite path --y to the path 'Y·
But why do we denote the opposite path by -"(? Here's why.
Proposition 3.2. Let 'Y : [a, b]---+ D be a smooth path in a doma in D an d
f : D ---+ C be a continuous function . Then
1
-"!
Proof.
['Y
We have
f(z)dz =
-1
'Y
f(z)dz.
b
f(z)dz
1 f( (-'Y)(t)) · (-'Y)'(t)dt
b
1 f('Y(a+b- t)) · ('Y'(a + b- t)) · (-l)dt
(r= a� b -t) l
a
-1
f('Y(T)). 'Y'(T)dT =
f(z)dz.
-1
b
f('Y(T)). 'Y'(T)dT
□
Exercise 3.7. Show that-(-')')= 'Y, where 'Y: [a,b]-+ Dis a smooth path in a
domain D.
Concatenation of paths. Let D be a domain and let
"(1 : [a1, b1] ---+ D and
'Y2 : [a2, b2]---+ D
be two paths such that
71
Cauchy Integral Theorem
(so that
'Y2
starts where
'Yl
their "concatenation" by:
('Yi+'Y2)(t)
=
{
ends). Define
to be
'Yl (t)
for a 1 ::; t ::; b 1 ,
b
'Y2(t - 1 +a2 ) for b 1 ::; t::; b 1 +b2 - a2 .
'Yl
'Yl +'Y2
Concatenation ')'1
Fig. 3.10
'Yl + 'Y2 : [a 1 ,b 1 +b2 - a2 ]
'Y2
+ ')'2 of two paths ')'1, ')'2.
Proposition 3.3. Let D be a domain in <C and let 'Yl : [a 1 ,b 1 ] ➔ D and
'Y2: [a2,b2] ➔ D be two paths such that 'Y 1 (b1) ="f2 (a2 ). Then
1
Proof.
')'1 +"12
1
"/1 +"12
We have
f(z)dz
=1
f(z)dz =
bl+bra
1�1
1
1�1
1
1
1
2
a1
=
+
=
=
=
+
"/1
"/1
1
f(z) dz +
"/1
1
"/2
f(z)dz.
f(('Y 1 +'Y2)(t))('Y 1 +'Yd(t)dt
f(('Y 1 +'Y2)(t))('Y 1 +'Y2)'(t)dt
b 1 +b 2- a 2
f(('Y 1 +'Y2)(t))('Y 1 +'Y2 )'(t)dt
b1
f('Y 1 (t))'Y�(t)dt
b 1 +b 2- a 2
f('Y2 (r - b 1 +a2)h�(r - b 1 +a2 )dr
b1
f(z) dz+
f(z) dz+
j
1
b2
a2
"/2
f('Y2 (s))'Y�(s)dt
f(z) dz.
(s
= r - b 1 + a2 )
□
Exercise 3.8. If')': [a, b] ➔ Dis a smooth path in a domain D, and f: D ➔ IC is
continuous, then show that
f
Ir+<--r)
f(z)dz = 0.
A useful estimate. We now prove an inequality for the size of the contour
integral in terms of the size of
lfl along the contour, and the length of the
contour. This will prove to be indispensable in the sequel.
72
A Friendly Approach to Complex Analysis
Proposition 3.4.
Let
be a domain in <C,
1 : [a, b] ---+ D be a piecewise smooth path and
(3) f : D ---+ <C be a continuous function.
(l)
(2)
D
11
Then
'Y
f(z)dzl::; ( max lf('Y(t))I) · (length of 1).
(3.3)
tE[a,b]
Recall that the length of I is given by
1
b
J(x'(t))2 + (y'(t))2,
where x, y : [a, b] ---+ IR are the real and imaginary parts of 1. See Fig­
ure 3.11.
1(b)
y'(t)dt
1(a)
Fig. 3.11 The arc length of the path 'Y is the sum of the incremental arc lengths ds,
where ds = (x'(t)dt)2 + (y'(t)dt)2 = (x'(t))2 + (y'(t)) 2 dt.
✓
✓
Proof.
Consider first a curve
To see this, let
11 cp(t)dtl
b
b
1 <p(t)dt = r
=
· exp(i0), where
r = exp(-i0) • r • exp(i0)
= exp(-i0) ·
=
<p : [a, b] ---+ <C, for which we prove
1
b
b
1 cp(t)dt = 1
Re (exp(-i0) ·
b
r?: 0 and 0 E (-1r, 1r]. T hen
exp(-i0) •
<p(t)) dt + i
1
b
<p(t)dt
Im (exp(-i0) ·
cp(t)) dt.
73
Cauchy Integral Theorem
But the left hand side is real, and so the integral of the imaginary part on
the right hand side must be zero. Consequently,
11 cp(t)dtl
b
=1
b
Re (exp(-i0) • cp(t)) dt
:::; 1 [Re (exp(-i0)·cp(t))[ dt
b
:::; 1 [exp(-i0)·cp(t)[ dt
b
= 1 [cp(t)[ dt.
b
The proposition now follows, since with cp(t):= f('Y(t))·'Y'(t), t E
II
If 'Y(t)
=
f(z) dzl
x(t)
11 f('Y(t))'Y'(t)dtl
::::: 1 lf('Y(t))'Y'(t)[dt
b
+
=1
b
[f('Y(t))ll'Y'(t)[dt
:::; ( max [f('Y(t))I) J ['Y'(t)[dt.
tE[a,b]
b
iy(t), where x, y are real-valued, then
1 ['Y'(t)[dt
b
=
[a, b],
b
=1
b
This completes the proof.
a
J(x'(t))2
+ (y'(t))2 dt = length of 1'·
□
Exercise 3.9. Calculate the upper bound given by (3.3) on the absolute value of
the integral
i
z2 dz,
where 'Y is the straight line path from O to 1
find its absolute value.
+ i.
Also, compute the integral and
Exercise 3.10. Using the calculation done in Exercise 3.5, deduce that (:) ::; 4 .
n
3.3
Fundamental Theorem of Contour Integration
Let us recall the Fundamental Theorem of Calculus in the real setting:
Theorem 3.2. (Fundamental Theorem of Calculus) If F: [a, b]--+ JR.
is continuously differentiable and F' =:f on [a, bl, then
1 f(x)dx
b
= F(b) - F(a).
74
A Friendly Approach to Complex Analysis
This is an important result, because it facilitates the computation of the
Riemann integral. Indeed, if we know that a function is the derivative of
something, then it is easy to calculate its integral. For example,
b3 _a3
d
because x 2 =
x 2dx =
( 3) .
dx
:
3
ab
Analogously, we will now see that if f is the derivative of a holomorphic
function, then the calculation of the contour integral
l
I
fz
( )dz
is easy, since we have (similar to the Fundamental Theorem of Calculus in
the real setting):
Theorem 3.3. (Fundamental 1 Theorem of Contour Integration)
Let
(l) Dbe a d omain in C,
(2) 'Y : [a, b] ➔ Dbe a piecewise smooth path,
(3) f: D ➔ C be a continuous function in D,
(4) F: D ➔ C be a holomorphic function such that F' =fin D.
Then
I
fz
( (a)).
( b)) - F,
( )dz = F,(
How does this theorem help? One can now calculate some contour integrals
very easily (just like in ordinary calculus). Here is an example.
Example 3.5. Since
1
zdz =
'Y
d z
( :)
dz
=z z( E q, for
(l + i)2 _ 0 2
2
2
1
+ 2i + i2
any 'Y joining O to 1
2
1
+ i,
+ 2i - 1 = i,
2
and so in particular, we recover the answer obtained in Example 3.4.
◊
I
In particular, as we have also seen in the previous example,
fz
( )dz = Fw
( ) - Fz
( )
is independent of the path 'Y joining the points z tow, when f possesses an
"antiderivative" or "primitive" Fin D.
1The naming of this result is done just to highlight the similarity with the real analysis
analogue. However, in complex analysis, this isn't all that "fundamental". We will soon
learn about Cauchy's Integral Theorem, which is certainly more fundamental!
75
Cauchy Integral Theorem
= z for
all z E C. Indeed, the calculation in Examples 3.2 and 3.3 shows that the
contour integral along paths joining O to 1 + i does depend on the path
chosen.
◊
Example 3.6. There is no function F : (C ➔ (C such that F'(z)
Proof. (of Theorem 3.3.) For z = x + iy ED, where x,y are real, define
the real-valued functions U,V,u,v by
F(x + iy) = U(x,y) + iV(x,y),
f(x + iy) = u(x,y) + iv(x,y).
Also, set 'Y(t) = x(t) + iy(t) (t E [a, bl), where x,y are real-valued. Then by
the Cauchy-Riemann equations, we have
u(x,y) + iv(x,y) = f(x + iy) = F'(x + iy)
au
av
.a v
.a u
= a (x,y) + i a (x,y) = a (x,y) - i a (x,y).
x
x
y
y
By the chain rule and the above, we have
!
U(x(t),y(t)) =
!�
(x(t),y(t)) · x'(t) +
!�
(x(t),y(t)) · y'(t)
= u(x(t),y(t)) · x'(t) - v(x(t),y(t)) · y'(t).
!
Similarly,
i
V(x(t),y(t)) =
!:
(x(t),y(t)) • x'(t) +
!;
(x(t),y(t)) • y'(t)
= v(x(t),y(t)) · x'(t) + u(x(t),y(t)) · y'(t).
Thus
J(z)dz =
l
b
J('Y(t))'Y'(t)dt
(
b
= l u(x(t),y(t)) + iv(x(t),y(t))) (x' (t) + iy'(t) )dt
=lb !
U(x(t),y(t))dt + i
lb !
V(x(t),y(t))dt
= U(x(b),y(b))-U(x(a),y(a)) +i(V(x(b),y(b))-V(x(a),y(a)))
= F('Y(b)) - F('Y(a)).
This completes the proof.
□
76
A Friendly Approach to Complex Analysis
Exercise 3.11. Show, using the Cauchy-Riemann equations, that z 1--t
primitive in <C.
z has no
Exercise 3.12. (Integration by Parts Formula.) Let f, g be holomorphic functions
defined in a domain D, such that J', g' are continuous in D, and let 'Y be a
piecewise smooth path in D from w E D to z E D. Show that
i
f(()g'(()d( = f(z)g(z) - J(w)g(w) -
Exercise 3.13. Evaluate
i
i
J'(()g(()d(.
cos zdz, where 'Y is any path joining -i to i.
Definition 3.2. A path 'Y : [a, b] ---+ C is said to be closed if 'Y(a)
= "f(b).
Corollary 3.1. Let
(1)
(2)
(3)
(4)
D be a domain in C,
'Y : [a, b] ---+ D be a closed piecewise smooth path,
f: D ---+ C is a continuous function in D,
F: D---+ C be a holomorphic function such that F' = f in D.
Then
1
Proof.
f(z) dz = 0.
1
f(z)dz = F('Y(b)) - F('Y(a)) = 0 since "f(b) = 'Y(a).
1
Example 3.7. Form E Z \ {0}, z ED:= C \ {0},
□
� ( ::) = zm -1; so
d
zm -ldz = 0
for any closed path 'Y in D. What if m = 0? Note that Log' z = I/ z for
z E jj := C \ (-oo, 0), and so for any path 7 in D, we do have
f !dz= 0.
}-:y z
However, in D,
!z doesn't have a primitive; see Exercise 3.16.
◊
Cauchy Integral Theorem
77
i
Exercise 3.14. Use the Fundamental Theorem of Contour Integration to write
down the value of
expzdz
where I is a path joining O and a + ib. Equate the answer obtained with the
parametric evaluation along the straight line from O to a+ ib, and deduce that
1
1
_ a( e a cosb-l)+be a sinb
e a x cos (bx)d x•
a 2 + b2
o
Exercise 3.15. Applying the Fundamental Theorem of Contour Integration to
expz and integrating round a circular path, show that for all r > 0,
[211"
J
o
e rcosO cos(rsin0 + 0)d0 = 0.
Exercise 3.16. Show that 1/z has no primitive in the punctured plane <C \ {O}.
The Cauchy Integral Theorem
3.4
We will now show one of the main results in complex analysis, called the
Cauchy Integral Theorem.
Theorem 3.4. (The Cauchy Integral Theorem) Let
(1)
(2)
(3)
D be a domain in <C,
f: D --+ <C be holomorphic in D, and
'Yo, 'Yi : [O, 1] --+ D be two closed, piecewise smooth, D-homotopic paths.
Then
1
')'O
f(z)dz =
1
f(z)dz.
')'1
Before we go further, let us try to understand the statement. Notice, first
of all, that the two paths in D are closed. Secondly, what do we mean
by saying that the two closed paths are "D-homotopic"? Intuitively, this
means the following. Look at Figure 3.12, where we have depicted the two
paths in the domain. Imagine you have placed a rubber band along 'YO·
For 'Yo to be D-homotopic to 'Yi, we should be able to deform this rubber
band so as to get 'Yl, with the condition that each intermediate position
of the rubber band lies in D. Clearly this is not possible sometimes, for
example if the domain has holes. See for example the picture on the right of
Figure 3.12, we expect the two paths in the domain D taken as punctured
complex plane D = <C \ {O} to be not D-homotopic.
78
A Friendly Approach to Complex Analysis
intermediate position
of the rubber band
obstruction
Fig. 3.12 ')'O, ')'1 in D :=CareC-homotopic in the picture on the left, but in the picture
on the right, the two paths ')'O, ')'l in D = C\ {O} are not C\ {O}-homotopic.
We have the following precise definition.
Definition 3.3. Let D be a domain in C and ')'o, ')'1 : [0, 1] -+ D be closed
paths. Then 'Yo is said to be D-homotopic to 'Yl if there is a continuous
function H: [0, 1] x [0, 1] -+ D such that the following hold:
(Hl) For all t E [0, 1], H(t,0) = 1'o(t).
(H2) For all t E [0, 1], H(t, 1) = 'Yi(t).
(H3) For alls E [0, 1], H(O,s)= H(l,s).
We can think of the Has a family of closed paths from [0, 1] to D, param­
eterized by "time", the s-variable. You may think of the closed path at
time s as the position of the rubber band at time s, Initially, when s = 0,
H(·, 0) is the path 'Yo, while finally, when the times= 1, we end up with
H(·, 1), which is the path 'Yl· So far, this is what (Hl) and (H2) say. The
requirement (H3) just says that at each point of time s, the intermediate
path 'Ys := H(·,s) is closed too. Continuity of H means that the rubber
band never breaks, and the deformation takes place smoothly. The picture
below illustrates this.
H(·,s)
H(-,0) = 'Yo
79
Cauchy Integral Theorem
Example 3.8. Let D = C, and 'Yo, 11 : [0, 1] --+ C be the two circular paths
given by 'Yo= 4exp(27rit), and 11 = 2i + exp(27rit), fort E [0, 1]. Then 'Yo
is C-homotopic to 'Yl· Indeed, we can define H by just taking a "convex
combination" of the points 1o(t) and 11(t). See Figure 3.8.
Set H: [0, 1] x [0, 1]--+ C by H(t,s)= (1- s) · 1o(t) + s · 11(t)= (4 - 3s) ·
exp(27rit) + s • 2i , 0 ::::; t, s ::::; 1. Then His clearly continuous, and moreover,
(Hl) for all t E [0, 1], H(t,0) = (4- 0) · exp(27rit) + 0 · 2i = 10 (t),
(H2) for all t E [0, 1], H(t, 1) = (4- 3) · exp(27rit) + 1- 2i = 11(t), and
(H3) for each s E [0, 1], H(0,s) = (4- 3s) · 1 + s · 2i = H(l,s).
Hence (Hl), (H2), (H3) are satisfied and so 10 is C-homotopic to 11.
On the other hand, the same two paths are not C \ {0}-homotopic.
Why is that? If they were C \ {0}-homotopic, then by the Cauchy Integral
Theorem, the contour integral of the holomorphic function 1/z in C \ {0}
along the two paths would be the same. But we have
1!
dz= 27l"i =I= 0=
1
!dz,
')'1 Z
'Yo Z
where the last equality follows from the Fundamental Theorem of Contour
Integration, because 1/ z has the primitive Logz in C \ (-oo, 0].
◊
Exercise 3.17. Let D be a domain in <C. Show that D-homotopy is an equivalence
relation on the set of all closed paths in D. In particular, we can say ",o, 11 are
D-homotopic" instead of saying that ",o is D-homotopic to 11".
Proof. ( of Theorem 3.4.) We will make the simplifying assumption that
the homotopy H is twice continuously differentiable. This smoothness con­
dition can be omitted (see for example [Conway (1978)]), but then the proof
becomes technical. Moreover, the assumption of twice continuous differen­
tiability is mild, and we will invoke this below when we will exchange the
order of partial differentiation:
fP H
8s8t
82 H
8t8s ·
80
A Friendly Approach to Complex Analysis
Idea of the proof: Let 'Ys
:= H(·, s) be the intermediate curve at times.
Define
I( s )
:=
1
f(z)dz,
s
E [O, l].
/S
(We will use differentiation under the integral sign with respect tos to show
that
d
/(s) 0,
d
showing thats H I( s) is constant, and in particular
=
1
f(z)dz = I(0) = J(l) =
10
1
1
f(z)dz,
'Yl
which is the desired conclusion.) We have
8H
dI
d
d [1
dz
s
z
( )=
f( ) =
l f(H(t, s)) 8t (t, s)dt
ds
ds o
ds ,.
=
1(
1
1
d
s
s
:s (f(H(t, ))�� (t, )) t
82 H
8H
8H
(
s
s
s
s
))
[J (t, s )) dt
)
+
f(H(t,
))
[J
f'(H(t,
(t,
(t,
)
l
= o
s t
88
8t
(1
=l
o
=
1
( f'(H(t, s)) 8H (t, s) 8H (t, s) + f(H(t, s)) [J82[JH (t, s)) dt
1!
8t
88
t
s
(f(H(t, s )) �� (t, s)) dt,
and so by the Fundamental Theorem of Integral Calculus,
(1 d
8H
dI
(t, s)) dt
f(H(t, s))
(s) = l
(
88
ds
o dt
8H
8H
= f(H(l, s)) 88 (l, s) - f(H(0, s )) 88 (o, s)
H(l, a) -H(l, s ) -f H s))lim H(O, a)- H(O, s)
( (l,
o---+s
a - s
a - s
a
a
s
)
,
)
,
H(
,
H(
H(
l
l )-H(l, s )
-f(H(l,s))lim
- l
=f(H(l,s))lim
o---+s
o---+s
a - s
a - s
=f(H(l,s))lim
o---+s
=
0.
1
1
Hence the maps H I( s) : [0, 1] -+ (C is constant. In particular,
11
f(z) dz = J(l) = J(0) =
12
f(z) dz.
81
Cauchy Integral Theorem
This completes the proof. But where did we use the holomorphicity of
f?
We used this in order to get the equality in the third line above. Indeed,
we claim that
:/f (H (t,
Let f
=
s))) = f' (H (t, s)) · �� (t,
s
).
=
X
u + iv, and H
+ iY, where u, v, X, Y are real valued. We
suppress the arguments (t, s) below. Then we have
:/f (H (t,
s
)))
=
=
:/u (X, Y ) + iv (X, Y ))
u
a
,
(X Y
ax
+i (
=
u
a
(
= f'
v
ax
(X
ax
+i (
=
a
u
a
ax
(X
ax
aX
as
(X, Y ) ·
Y
'
av
).
).
aX
as
(X, Y ) ·
Y
(X
'
)+
+
8s +
aX
_
v
(X
as
v
(X, Y ).
y
a
u
aY
as
(X, Y ) ·
ax
)). (
+iY )
8s )
aY
a
Y
(X
'
aY
a
(X ' Y ) .
8s +
i ax
s
a
ax
aX
. av
+iY ) · :
u
.
(X, Y )
ay
a
= f'
a
x
as
(
8s )
aY
. aY
+ia
H (t ,
s
s
)
)) · �� (t,s).
The careful reader would have also noticed that we also assumed that f' is
continuous when we differentiated under the integral sign. Again, the result
holds without this assumption, but we will not do this here. The interested
student can find the complete proof of the Cauchy Integral Theorem for
example in [Conway (1978)].
D
Exercise 3.18. We have seen that if C is the circular path with center O and radius
1 traversed in the anticlockwise direction, then
{ !dz = 21ri.
la z
Now consider the path S, comprising the four line segments which are the sides
of the square with vertices ±1 ± i, traversed anticlockwise. Draw a picture to
convince yourself that Sis (C \ {O}-homotopic to C. Evaluate parametrically the
integral
{ !dz,
ls z
and confirm that the answer is indeed 21ri.
82
A Friendly Approach to Complex Analysis
Exercise 3.19. Let a> 0, b > 0, and E: [O,27r] ➔ <C be the elliptic path
By considering
3.4.1
= acost+ibsint,
jEz-1 dz, show that 12
E(t)
71"
O
t E [O,27r].
1
271"
• 0 2 d0 = a2(cos 0)2 + b2( sm
ab •
)
Special case: simply connected domains
Consider a "degenerate" closed path, which is constant. That is, if D is
a domain and p E D, then consider "/p : [0,1] ➔ D given by "/p (t) = p,
t E [0, l]. Then "/p is closed, because "/p(O) = p = "/p (l). For any continuous
f: D ➔ C, what is
1
IP
f(z)dz?
0, because 'Y�(t) = 0 for each t E [a, b] and
1
1p
1
f(z)dz = [ f('Yp (t)) · "f�(t)dt = 0.
lo
In light of this, we see that an important special case of the Cauchy Integral
Theorem is obtained when we know that a closed path 'Y is D-homotopic
to a point (that is, the constant path "/p (t) = p for all t E [0,1]). In this
case we say that 'Y is D-contractible. Imagine placing a rubber band along
'Y, and then shrinking it to a point such that each intermediate position
of the rubber band is in D. Indeed for a D-contractible path 'Y (which is
D-homotopic to a point p E D), and for a holomorphic function f: D ➔ C,
we have by the Cauchy Integral Theorem that
1
1
J(z)dz =
1
IP
J(z)dz = 0.
A domain in which every closed path is D-contractible is called simply
connected.
For example, the domains C, II)):= {z EC: lzl < 1}, C \ (-oo,0], are
all simply connected. For example if we take any p in the domains in first
two cases, then the homotopy given by
H(t, s):= (1- s)'Y(t) + sp,
t, s E [0,1]
does the job. In the case when D = C\ (-oo,0], given any 'Y: [0,1] ➔ D, we
first choose any real p > 0 for example p = 1, and then use the same Has
above. Note that none of the above domains have any "holes" in them. On
Cauchy Integral Theorem
83
the other hand, domains with holes are not simply connected. For example,
the punctured complex plane (C \ { 0} is not simply connected. For example
consider the circular path with center O and any positive radius r traversed
once in the anticlockwise direction. We have seen that
1
!dz= 27ri.
cZ
But if C were (C \ { 0}-contractible to a point in the punctured plane, we
should have had
1
!dz= 0
cZ
by Cauchy's Integral Formula. So this means that C is not (C \ {O}­
contractible to a point in (C \ {O} and so (C \ {O} is not simply connected.
Similarly, one can show that the annulus
{ z E (C : 1 < lzl < 2}
is not simply connected. See Figure 3.13. The obstruction of the hole
can be thought of as a nail or a pillar emanating from the plane, which
prevents a rubber band encircling it from being shrunk to a point in the
domain while always staying in the plane.
simply connected
not simply connected
Fig. 3.13 The domains IC, ll)) := {z E IC : JzJ < 1} and IC\ (-oo, OJ are simply connected,
while the annulus A:= {z E IC: 1 < JzJ < 2} and the punctured plane IC\ {O} aren't.
84
A Friendly Approach to Complex Analysis
We have the following corollary of the Cauchy Integral Theorem.
Corollary 3.2.
Let
(1) D be a simply connected domain,
(2) 'Y
be a closed piecewise smooth path in D and
(3) f: D ➔ C be holomorphic in D.
Then
i
f(z)dz = 0.
This corollary itself is also sometimes called the Cauchy Integral Theorem.
Example 3.9. For any closed path 'Y, since exp is entire, and (C is simply
i
connected, we have
In fact for any entire function
It can happen for a
f: D ➔
function
non
(C that
expzdz = 0.
f :=
1/z • Then
f(z)dz = 0,
for any closed path 'Y·
simply connected domain
i
for every closed smooth path 'Y in
2
i
f,
f(z)dz = 0
1
'Y
D.
D
◊
and a holomorphic
Here is an example: take
D=
C\ {0},
�dz=O
z
for every closed path 'Y in the punctured plane because 1/z 2 possesses a
primitive in (C \ {0}:
Exercise 3.20. Integrate the following functions over the circular path given by
Jzl = 3 traversed in the anticlockwise direction:
(1) Log(z - 4i).
(2)
1
z-1
(3) Principal value of iz -3.
85
Cauchy Integml Theorem
Exercise 3.21. (Winding number of a curve.) Suppose that 1 : [0, 1] ➔ C is a
smooth closed path that does not pass through 0. We define the winding number
of 1 (about 0) to be
1
1
1
w (1) .·- - / -dz - - 21ri 'Y z
21ri
11
O
'(t)
,(t)
']__dt.
(1) Using the observation that exp(21ria) = 1 if and only if a E Z, show that
w(,) E Z by proceeding as follows. Define cp: [0, 1] ➔ C by
To show that w(,) E Z, it suffices to show that cp(l) = 1. To this end,
calculate cp' (t), and use this expression to show that cph is constant in [0, 1].
Use this fact to conclude that cp(l) = 1.
(2) Calculate the winding number of r1: [0, 1] ➔ C given by r1(t)
(t E [0, l]).
= exp(21rit)
(3) Prove that if ,1, 12: [0, 1] ➔ C are smooth closed paths not passing through
0, and 11 · 12 is their pointwise product, then w(,1 · 12) = w(,1) + w(,2).
(4) Let m E N. Calculate the winding number of the curve rm: [0, 1] ➔ C given
by rm(t) = exp(21rimt) (t E [0, 1]).
(5) Show that the winding number function I M w(,) is "locally constant", by
which we mean that if ,o: [0, 1] ➔ C \ {0} is a smooth closed path, then there
is a 8 > 0 such that for every smooth closed path 1 : [0, 1] ➔ C \ {0} such
that II,- ,olloo := max{l,(t)- ,o(t)I : t E [0, 1]} < 8, we have w(,) = w(,o)­
(In other words, if we equip the set of curves with the uniform topology, and
equip Z with the discrete topology, then IM w(,) is continuous.)
3.4.2
What happens with nonholomorphic functions?
We now highlight the fact that the Cauchy Integral Theorem may fail if
one drops the assumption of holomorphicity of f. Let us see what happens
when we consider our favourite nonholomorphic function, the complex con­
jugation map z H z. We will show that rather than the integral around
the closed loop 'Y being 0, the contour integral of z around 'Y yields the
area enclosed by 'Y, which is of course very much dependent on 'Y, and
two <C-homotopic paths can enclose widely different areas (just imagine
two concentric circles with different radii). We will only give a plausibility
argument by resorting to a specific picture, as shown in the picture below.
86
A Friendly Approach to Complex Analysis
(x(t), y(t))
(x(a), y(a))
II
(x(b), y(b))
For the smooth path 'Y : [a, b] -+ <C, we have
j zdz
'Y
=
1
b
ab
=1
=
(x(t) - iy(t))(x'(t) + iy'(t))dt
x(t)x'(t) + y(t)y'(t) + i(x(t)y'(t)-y(t)x'(t))dt
(x(b))2 -(x(a))2
b
+ (y(b))2 -(y(a))2 + i
2
= O+i 1 (x(t)y'(t)-y(t)x'(t))dt.
_, ' '
x'(t)dt
Fig. 3.14
l
b
f (x(t)y'(t)-y(t)x'(t))dt
la
' '
�I
b
x(t)y'(t)dt and
b
l
l�
-x'(t)dt
x'(t)y(t)dt.
87
Cauchy Integml Theorem
Look at Figure 3.14. From the two pictures on the top, we see that
b
l x(t)y'(t)dt = (Area enclosed by "f).
From the two pictures on the bottom, we see that
Thus
l
b
l x'(t)y(t)dt = -(Area enclosed by "f).
b
z dz = i l (x(t)y'(t) - y(t)x'(t))dt = 2i • (Area enclosed by "f).
So in contrast to the Cauchy Integral Theorem for holomorphic functions,
we see that for this nonholomorphic function, the integral along a closed
contour is not zero, but yields the area of the contour!
Exercise 3.22. Suppose a coin of radius r rolls around a fixed bigger coin of radius
R. Then the path traced by a point on the rim of the rolling coin is called an
epicycloid, and it is a closed curve if R
nr, for some n EN. See Figure 3.15.
=
Fig. 3.15
The epicycloid with n = 6.
(1) With the center of the fixed coin at the origin, show that the epicycloid can
be represented parametrically as z(t)
r((n
1) exp(it) - exp(i(n
l)t)),
t E [O, 21r].
=
+
+
(2) By evaluating the integral of z along the epicycloid, show that the area en2
closed by the epicycloid is equal to 1rr (n
l)(n 2).
+
+
In the rest of this chapter we will learn about several consequences of the
Cauchy Integral Theorem. In particular, we will learn
(1) that in simply connected domains every holomorphic function possesses
a primitive;
(2) · that holomorphic functions are infinitely many times differentiable;
(3) that bounded entire functions are constants (Liouville's Theorem) (and
also use this to prove the Fundamental Theorem of Algebra);
(4) a result called Morera's theorem, which is a sort of a converse to the
Cauchy Integral Theorem.
A Friendly Approach to Complex Analysis
88
3.5
Existence of a primitive
We will show that on a simply connected domain, every holomorphic func­
tion is the derivative of some holomorphic function.
Theorem 3.5. If
(1) D is a simply connected domain and
(2) f: D-+ C is holomorphic,
then there is a holomorphic function F : D -+ C such that for all z E D,
F'(z) = f(z).
Proof.
Fix a point p E D. Define F : D -+ C by
F(z)=i f(()d(,
zED,
where 'Y is any smooth path in D joining p to z. Why is this F well-defined,
that is, why does the F(z) not depend on which path we take joining p to
z? If:=:; is another smooth path in D that joins p to z, then 'Y -:=:; is a closed
smooth path in the simply connected domain D. See Figure 3.16.
7z
z
'Y
p
Fig. 3.16 'Y - ::Y forms a closed path.
1_
Cauchy Integral Theorem gives
so that
1
7z
0=
f ( () d(
7-7
f(z)dz =
1
b
f(()d( - � f(()d(,
h
= � f ( () d(, and F is well-defined.
Ir
Next we show the holomorphicity of F and that F' = f in D. Since
f is holomorphic in D, it is also continuous there, and so given a z E D
and an E > 0, there is a 8 > 0 such that whenever lw - zl < 8, we have
lf(w) - f(z)I < f.. Thus if we take aw such that O < lw - zl < 8, then
F(
(z)
= W�Z
w2=:
f(()d().
f(()d(-
(Jw
Jz
89
Cauchy Integral Theorem
If 'Yzw is a straight line path joining z to w, then the concatenation of 'Yz
with the concatenation of 'Yzw with -,w is a closed path, and so by the
Cauchy Integral Theorem, we obtain
0
=
1
'Y.z+'Yzw-")'w
1
f(()d( =
f(()d( +
'Yz
1
'Yzw
f(()d(-1 f(()d(.
'Yw
/1;::\)
'Y
-
,
,'
',
p
1
1
1
The Fundamental Theorem of Contour Integration gives
and so
F(w
I
and so
F( w
2
=
2
=
Id(=
'Yzw
:
'Yzw
(z) - z =
f( ) w � z
z
:( ) - f(z)I =
('d( = w - z,
'Yzw
f(()d( - w � z
= w � 1 (!(() -f(z))d(,
z 'Yzw
I i.w
W �
z
= w�
zl
l
1
1
'Yzw
f(z)d(
(!(() -f(z))d(I
li.w
(!(() -f(z))d(I
max If(() - f(z)I) · (length of 'Yzw )
:::; -1 -- (1 (E"/zw
W - Z
:::; ---1 1:lw-zl =
I w-z
Thus F'(z) = f(z), and Fis holomorphic.
1
€.
□
Remark 3.2. A primitive for a holomorphic function f in a simply con­
nected domain is unique up to a constant. Indeed, if F, F are both primi­
tives for F, then F' = f = F' in D, and so
d
(F - F) = F' - F' = f - f = O in D.
dz
By Exercise 2.13, it follows that there is a constant C such that F-F = C
in D. So F = F + C in D.
90
A Friendly Approach to Complex Analysis
Example 3.10. exp(-z 2 ) is entire. So there exists an F, which is also
entire, such that for all z EC, F'(z) = exp(-z 2 ). (But one cannot express
F in terms of elementary functions. One primitive is given by
F(z) =
1
2
e-C d(
'Yz
for z E C, where 'Yz is the straight line path joining O to z. Then in
particular, for real x,
and it turns out that this ( and so any other primitive too) can't be expressed
in terms of elementary functions.)
◊
Exercise 3.23. Suppose that Dis a domain. If f is holomorphic in D, and there
is no F holomorphic in D such that F' = f in D, then we know that D cannot
be simply connected. Give a concrete example of such a Dand f.
Corollary 3.3. If
( 1) D is a simply connected domain,
(2) f : D ---+ C is holomorphic,
(3) 'Y : [a, b] ---+ D and -::Y : [c, d] ---+ D are two smooth paths such that
they have the same start and end points, that is, 'Y(a) = -::Y(c) and
"f(b) = -::Y(d),
then
1
Proof.
1
f(z) dz=
/4
f(z) dz.
f has a primitive F and so
f(z)dz = F('Y(l))- F('Y(O))
= F(-::Y(l)) - F(-::Y(O)) = /4J(z).
□
91
Cauchy Integral Theorem
3.6
The Cauchy Integral Formula
We will now learn about a result, called the Cauchy Integral Formula,
which says, roughly speaking that if we have a closed path I without self­
intersections, and f is a function which is holomorphic inside 1, then the
value of f at any point inside I is determined by the values of the function
on 1! This illustrates the "rigidity" of holomorphic functions.
Later on, in the next chapter, we will study a more general Cauchy
Integral Formula, which will allow us to even express all the derivatives of
f at any point inside I in terms of the values of the function on 1. So we
can consider the basic result in this section as the "n = 0 case" of the more
general result to follow.
Theorem 3.6. (The Cauchy Integral Formula for circular paths)
Let
(1) D be a domain,
(2) f: D--+ C be holomorphic in D,
(3) r > 0, zo ED and the disc�:= {z EC: lz-zol � r} CD.
Then
f(w)
=�
21ri
r
f(z) dz,
zw
lcr
lw-zol
< r,
where Cr is the circular path Cr(t) = zo + r exp(it), t E [O, 21r], with center
zo and radius r > 0 traversed in the anticlockwise direction.
•
In order to prove this result, we will first prove the following technical fact,
which will also prove to be useful later on.
92
A Friendly Approach to Complex Analysis
Proposition 3.5. Let
(1) D be a domain, zo ED,
(2) f: D---+ <C be holomorphic in D \ {zo}, and continuous on D,
(3) r > 0 and the disc Ll := {z E <C: lz - zol::::; r} be contained in D.
Then
f f(z) dz,
f(zo)= �
2m lc r z-zo
where Cr is the circular path Cr (t) = zo +rexp(it), t E [0,21r], with center
zo and radius r > 0 traversed in the anticlockwise direction.
Proof. Let E > 0. Then there is a 8 > 0 (which we can arrange to be
smaller than r) such that whenever O < !z - zol ::::; 8, lf(z) - f(zo)I < E.
Consider the circular path C0 , with center zo and radius 8 traversed in
the anticlockwise direction. But C0 and Cr are easily seen to be D \ {zo}­
homotopic.
/�
('•·----Qqo
·. i .•' _____\cJ r
: '//
"�-
\_o'
Indeed the homotopy H can be obtained by just taking the convex combi­
nation of the points on Cr and C0: H(·, s) := (l-s)Cr(·)+sC0(·), s E [O, l].
Thus by the Cauchy Integral Theorem, we have
f f(z) dz= f f(z) dz.
lc 0 z-zo
Jc r z-zo
Hence,
I� r
1� r
I� /
r
_l dz
JJi)_dz-f(zo)�
JJi)_dz-f(zo)I =
21rilcrz-zo
21rilc 8 Z-Zo
21rilc 8 Z-Zo l
f(z) - J(zo)
=
dzl
21rilc 0
z-zo
(z)
(zo)
<
max lf - f I). 27r8
- zECo
21rlz - zol
(
f
< -·27rD=E.
21r8
Since
E
> 0 was arbitrary, the claim follows.
□
93
Cauchy Integral Theorem
The following is an immediate corollary.
Corollary 3.4. Let
(1) D be a domain,
(2) f: D----+ C be holomorphic in D,
(3) r > 0, zo ED and the disc Ll := {z EC: lz - zol::; r} c D.
1
Then
f(z)
1
.
--dz,
f(zo) = 2 7fZ Cr Z - Zo
where Cr is the circular path Cr (t ) = zo +rexp(it), t E [0,21r], with center
zo and radius r > 0 traversed in the anticlockwise direction.
We now prove the basic version of the Cauchy Integral Formula, namely
Theorem 3.6. Note that as opposed to the previous Corollary 3.4, now the
w can be any point inside the circle with center z0 and radius r , and not
necessarily the center zo as in the corollary above.
Proof. (of Theorem 3.6.) Let w be such that lw - zo I
< r . Choose a o > 0
small enough so that the circular path C0 with center w and radius o is
contained in the interior of Cr . But now Cr and C0 are D \ { w}-homotopic,
and this can be seen in the same manner as in Example 3.8.
Since
!(·)
· -W
is holomorphic in D \ {w}, it follows from the Cauchy Integral Theorem
that the second equality holds below:
1
1
f(z) dz.
f(z) dz=�
21rz c8 Z - W
21rz Cr Z - W
This completes the proof.
f(w) = �
D
94
A Friendly Approach to Complex Analysis
Exercise 3.24. Let O < a < l, and let 'Y be the unit circle with center O traversed
anticlockwise. Show that
{
J,
Y
21r
1
i
{
dz=
dt.
(z-a)(az-l)
}0
l+a2-2acost
Use Cauchy's Integral Formula to deduce that
1
2
,r
0
1
dt
1 +a2-2acost
21r
= -1-a
-- •
2
Exercise 3.25. Fill in the blanks.
(l)
(2)
{ exp zdz=---, where 'Y is the circle lzl
},y z-1
wise direction.
r
z +1
:
dz
J, z -1
= --- )
=
2 traversed in the anticlock­
where 'Y is the circle lz -11
1 traversed in the
where 'Y IS
· 1e I z -i·1
· the circ
.
1 traversed m the
anticlockwise direction.
(3 )
(4 )
(5 )
z 2 +l
_ dz = ___
1
z2
1
anticlockwise direction.
jf
,
1
z 2 +l
· the circ
· 1e I z + 11 = 1 traversed m
. the
-dz = ___ , where 'Y IS
-2
, z -1
anticlockwise direction.
z: +l
dz = ---1
clockwise direction.
{
}, z
Exercise 3.26. Does
z
1--t
,
where 'Y is the circle lzl
1
2)
zl-z
(
=
3 traversed in the anti­
have a primitive in {z E (C: 0
< lzl <
1}?
Corollary 3.5. {Cauchy's Integral Formula for general paths)
Let
(1) D be a domain,
(2) f: D-+ C be holomorphic in D,
(3) zo ED, and
(4) 'Y be a closed path in D which is D \ {zo}-homotopic to a circular path
C centered at z0, such that C and its interior is contained in D.
Then we have f(zo)
Proof.
=�
27rZ
1
f(z)
'Y Z - Zo
dz.
1
By the Cauchy Integral Formula for circular paths, it follows that
f(zo)
1
=.
2 11"Z
f(z)
--dz.
CZ - Zo
1
95
Cauchy Integral Theorem
1
But since 'Y is D \ {z0 }-homotopic to C, by the Cauchy Integral Theorem,
f(z) dz = _1_
f(z ) dz.
_1_
21ri c z - Zo
21ri 'Y z - Zo
This completes the proof.
□
This result highlights the "rigidity" associated with holomorphic functions
mentioned earlier. By this we mean that their highly structured nature
(everywhere locally infinitesimally a rotation followed by a magnification)
enables one to pin down their precise behaviour from very limited infor­
mation. That is, even if we know the effect of a holomorphic function in a
small portion of the plane (for example the values along a closed path), its
values can be inferred at other far away points in a unique manner. The
picture below illustrates this in the case of the Cauchy Integral Formula,
where knowing the values of f on the curve 'Y enables one to determine the
values at all points in the shaded region!
exp(iz)
. 3.27. Let F be defined by F(z) = Exercise
-- and let R > 1.
z2 + 1
(1) Let u be the closed semicircular path formed by the segment S of the real
axis from - R to R, followed by the circular arc T of radius R in the upper
half plane from R to - R. Show that
{ F(z)dz = �e
}"
(2) Prove that I exp(iz)I � 1 for z in the upper half plane, and conclude that for
large enough lzl, IF(z)I � 2/Jzl2•
(3) Show that lim { F(z)dz = 0, and so lim { F(z)dz = �-
�ooh
�ooh
e
(4) Conclude, by parameterizing the integral over Sin terms of x, that
r cosx dx := lim { cosx dx = �R ➔ oo }_ R l+x 2
J_ oo l+x 2
e
oo
Exercise 3.28. Evaluate
R
1
2
O ,,-
ecos 9 cos(sin 0)d0. Hint: Consider exp(exp(i0)).
96
A Friendly Approach to Complex Analysis
3. 7
Holomorphic functions are infinitely differentiable
In this section we prove the fundamental property of holomorphic func­
tions in a domain, namely that they are infinitely many times complex
differentiable.
Let us contrast this with the situation in Real Analysis. We have already
seen in Example 0.1 that the derivative may fail to be differentiable at
isolated points. There are even more extreme examples of this phenomenon,
and there exist functions f : � ➔ � which are differentiable everywhere,
but f' is differentiable nowhere! We refer the interested reader to §3.8,
[Gelbaum and Olmsted (1964)], where one can find an example of a function
g : � ➔ � that is continuous everywhere, but differentiable nowhere; the
integral
f(x)
= fo
x
g(�)d�,
x E �,
of this g, then gives our sought for f : � ➔ � that is differentiable every­
where, but whose derivative (g!) is differentiable nowhere.
Corollary 3.6. Let
(1) D be a domain, and
(2) f: D ➔ <C be holomorphic in D.
Then f' is holomorphic in D.
Note that the above gives the following chain of implications:
I
f E Hol(D)
I⇒I
f' E Hol(D)
I⇒I
f" E Hol(D)
I⇒
···
So whenever f is holomorphic in a domain D (that is, f E Hol(D)), it is
infinitely many times complex differentiable.
Here is a plan of how we will show this.
Formula, we know that
where
f(z)
=�
2m
1
From the Cauchy Integral
f(()
d ,
Cr (- Z (
Cr is a circle centered at z with radius r. If we were to formally
differentiate under the integral sign, we would get an expression for the
derivative of
f:
f'(z)
=
_1
21ri
1
f(()
d(,
Cr ((-z)2
Cauchy Integral Theorem
97
Having shown this formula, we will show that
. f'(w)- f'(z)
1Im----w➔z
Z-W
exists by using the above expression for the derivative at z and w.
Let zo E D. Let g be defined by
f(z)- f(zo) I·r z ---1r zo,
g(z)= {
z-zo
if z= zo.
f'(zo)
Clearly g is holomorphic in D \ {zo} and continuous in D. We will now
apply the technical fact we had shown in Proposition 3.5 tog. Choose an
r > 0 small enough so that the disc {z EC: lz-zol � r} is contained in D,
and let Cr denote the circular path with center zo and radius r traversed
anticlockwise. Then
1 f g(z)
dz
.
J'(zo) = g(zo) = 2 7rZ lcr z- Zo
= _1 f f(z)- f(zo) dz
( 3 .4 )
21ri lcr (z-zo) 2
1
f(z) dz_ f(zo) {
= _1_ f
dz
21ri lcr (z-zo) 2
21ri lcr (z - zo) 2
1 f
f(z) dz- O ·
=( 3·5 )
21ri lcr (z-zo) 2
Thus for w inside Cr, but with w f z0, we have that
f(z) dz
f(z)
= _1 f
_1 f
J'(w) = 21ri } z-w 2 dz 21ri J
z
-w
)
)2 '
cr (
00 (
where C0 is a small circular path with center w and radius 8 that lies inside
Cr. The second equality above follows from the Cauchy Integral Theorem,
because
J(-)
(· - w)2
is holomorphic in D \ {w} and the paths Cr, C0 are D \ {w}-homotopic.
Proof.
Cr,___-�
98
A Friendly Approach to Complex Analysis
So we have for w -/ zo inside Cr that
f'(w) - f'(zo) = _1 _ (-1 f
f(z) dz)
f(z) dz __
l f
2
21ri lcr (z - zo)2
w - Zo
w - zo 21ri lcr (z - w)
__1_ f f(z)(2z - Zo - w) dz
- 21ri Jcr (z - w)2 (z - zo)2
What does this look like when w � zo? The numerator looks like
f(z)(2z - zo - zo) = 2f(z)(z - zo),
while the denominator looks like (z - zo)2 (z - zo)2 . So we guess that
(w) - f'(zo) = -2_ f
f(z) dz
lim f'
w-tzo
w - Zo
21ri lcr (z - zo) 3 '
and we prove this claim below. So let us calculate
f'(w) -f'(zo) _ -2_ f f(z) dz= (w _ zo _l_ f (3z -zo -2w)f(z) dz
)
w -zo
21ri Jc/z -zo)3
21ri Jcr (z - w) 2 (z -zo)3
Of course if we manage to show that the integral is bounded by some
constant for all w close to zo, then we see that since this is being multiplied
by w - zo, as w ➔ zo, we can make the overall expression as small as we
please. To this end, let us consider a disc with center z0 and having radius
smaller than r , say r/2, and we will confine w (which is anyway supposed
to be near z0) to lie within this disc. So from now on, w will lie in the
(compact set)
{ w E <C: lw - zo I
Consider the continuous map <p,
(
) 'P. (3z - zo - 2w)f(z)
z, w H
(z - w)2 (z - zo)3
I
I
: Cr x { w E <C: lw - zol:::;
:::; i} .
i} (=:KC <C
2
= JR4 ) ➔ R
But K is a compact set in JR4 because it is closed and bounded. Thus the
continuous function <p: K ➔ JR has some maximum value M � 0 on the
compact set K. Hence,
f (�z - z) ( 2w)fj:) dzl :::; _!_M(length of Cr )= _!__M21rr = Mr,
21f
21f
21ri jCr Z - W 2 Z - Zo
and so
f'(w) - f'(zo) - �
f(z) dzl :::; lw - zolM
r W�o 0.
w - zo
21ri Jcr (z - zo) 3
This shows that f' is differentiable at zo. As the choice of zo was arbitrary,
f' is holomorphic in D.
I�
I
r
□
Exercise 3.29. Suppose f is holomorphic in a domain D. Is it clear that if n EN,
then J<n) has a continuous complex derivative?
Cauchy Integral Theorem
3.8
99
Liouville's Theorem; Fundamental Theorem of Algebra
Here is one more instance of the rigidity associated with holomorphicity.
Theorem 3.7. (Liouville's Theorem) Every bounded entire function is
constant.
Let us again contrast this with the situation in Real Analysis. There we
know that for example x H sin x is differentiable everywhere on �, and it is
bounded too: I sin xi :::; 1 for all x ER But sin is not a constant function.
On the other hand, in light of the above Liouville's Theorem, the entire
function z H sin z, being nonconstant, must necessarily be unbounded on
(C! We had checked this by brute force using the definition earlier on page
23, and proved that I sin(iy)I ➔ oo as y ➔ ±oo.
Proof. Let M;:::: 0 be such that for all z EC, IJ(z)I :::; M. Suppose that
w E C, and let I be the circular path with center w and radius R, where
R is any positive number. Then (from the proof of Corollary 3.6, see in
particular (3.5))
f(z)
,
1
2 dz,
f (w) = -.
27ri '"Y (z - w )
and so
1
f(z)
IJ'(w)I = 1dz < _..!__. M . 21rR = M_
R
21ri '"Y (z-w)2 l - 21r R2
1
1
But since R > 0 was arbitrary, it follows that f'(w) = 0. So f' (w) = 0 for
all w EC, and hence f is constant. We had seen this in Exercise 2.13, but
here is another way to see this. If z E C, by considering the straight line
path 'Yz joining Oto z, we have
f(z) - f(0) =
This completes the proof.
1
'"Y
z
J'(()d( = 0.
□
This result can be used to give a short proof of the Fundamental Theorem
of Algebra2 •
Corollary 3.7. (Fundamental Theorem of Algebra) Every polynomial
of degree ;:::: 1 has a root in C.
2Despite its name, there is no purely algebraic proof of the theorem, since any proof
must use the completeness of the reals, which is not an algebraic concept. Additionally,
it is not really "fundamental" for modern algebra; its name was given at a time when
the study of algebra was mainly concerned with the solutions of polynomial equations
with real or complex coefficients.
A Friendly Approach to Complex Analysis
100
For a polynomial p: (C--+ C, given by p(z) =co+ c1z + · · · + cd zd , z EC,
where co, c1, · · · , cd EC, Cd -/ 0, the number dis called the degree of p. A
number z0 EC such that p(zo) = 0 is called a zero/root of p.
Proof.
Suppose p(z) =co+ c1z + · · · + cd z d is a polynomial with d � 1,
and such that it has no root in C. That is, for all z E C, p(z) -I 0. But
since p is entire and nonzero, its reciprocal, namely the function f given by
f(z) = l/p(z) (z EC), is entire; see Exercise 2.6. In Exercise 1.24, we had
shown an estimate on the growth of a polynomial:
there exist M, R > 0 such that jp(z)I � Mlzld whenever lzl > R.
In the compact set {z EC : lzl ::::; R}, the continuous function z r-+ jp(z)I
has a positive (because pis never 0) minimum m. Thus
lf(z)I::::; min{ �d '
M
!} ,
z EC.
By Liouville's Theorem, f must be constant, and sop must be a constant,
a contradiction to the fact that d � l.
For example, the polynomial p given by p(z) = z
- 3z
guaranteed to have a root somewhere in the complex plane.
1976
28
□
+ v'399 is
Exercise 3.30. Let f be an entire function such that f is bounded away from 0,
that is, there is a 8 > 0 such that for all z E <C, lf(z)I 2: 8. Show that f is a
constant.
Exercise 3.31. Show that an entire function whose range of values avoids a disc
{w E <C: lw - wol < r} must be a constant.
Exercise 3.32. Asstime that f is an entire function that is periodic in both the
real and in the imaginary direction, that is, there exist Ti, T2 in JR such that
f(z) = f(z +Ti)= f(z + iT2) for all z E <C. Prove that f is constant.
Exercise 3.33. A classical theme in the theory of entire functions is to try to
characterize the entire function f based on the way Iii grows for large lzl. Here
is one instance of this.
( 1) Show that if f is entire and lf(z)I � I expzl, for all z E <C, then in fact f
is equal to c • expz for some complex constant c with lei � 1. (Thus if a
nonconstant entire function "grows" no faster than the exponential function,
it is an exponential function.)
( 2) One may be tempted to argue that this can't be right on the grounds that
"polynomials grow more slowly than the exponential function", but surely
p =f. exp z. Find the flaw in this reasoning by showing that if p is a polynomial
satisfying IP( z) I � I exp zI for all z E <C, then p 0.
Hint: Look at z = x < 0.
=
101
Cauchy Integral Theorem
Exercise 3.34. Let f : (C -+ (C be an entire function. Suppose that a1, a2 are
complex numbers such that a1 =/ a2, and such that a1, a2 are contained in the
interior of the circular path C with radius R > 0 and center 0, traversed once in
the counterclockwise direction.
Ii
z
f( )
c ( z - a1)( z - a2)
(1) Prove that
dz
21rR
max IJ(z)lI :S (R - I a1 l)(R - I a2 I) zEC
a
(2) F ind a, /3 E (C such that for all z E C, -,------,--,---....,.. - -z - a1
(z - a1)(z - a2)
(3) Express
1
/3
+ -z - a2 ·
z
z
z
dz in terms of { f( ) dz and { f( ) dz.
{
f( )
}0 (z-a1)(z-a2)
}0 z-a1
}0 z-a2
Use the Cauchy Integral Formula to simplify these latter expressions.
(4) Deduce Liouville's Theorem.
3.9
Morera's Theorem: converse to Cauchy's Integral
Theorem
Recall the Cauchy Integral Theorem, where we have learnt that if
(1)
(2)
(3)
(4)
Dis a domain,
f : D-+ <C is holomorphic,
b. is any disc such that b. C D,
'Y is a closed piecewise smooth path in b.,
then
i
f(z)dz
= 0.
Now we will see that the following converse to the above result holds.
Theorem 3.8. (Morera's Theorem) If
(1) D is a domain,
(2) f : D-+ <C is a continuous function such that
(3) for every closed rectangular path 'Y in every disc contained in D,
i
f(z)dz
= 0,
then f is holomorphic in D.
In other words if the contour integral is zero for some special paths, then
we are allowed to conclude that f is holomorphic!
A Friendly Approach to Complex Analysis
102
Proof. Let zo E D, and let b. be a disc with center z0 such that b. C D,
and 'Yzo ,z is the path joining zo to z by first moving horizontally and then
moving vertically.
Zo 'Yzo, z
Define F: b.---+ <C by F(z) = 1
f(()d(,
z Eb..
We will show that F is holomorphic in b., and its derivative is f. This
shows that f is holomorphic in b.! Why? f (being the derivative of a
holomorphic function) is itselfholomorphic in b..
Let z E b.. Suppose E > 0. Since f is continuous, there exists o > 0
such that whenever lw-zl < o and w Eb., it follows that lf(w)-f(z)I < E.
We have
1'zo,z
F(w) -F(z) = 1
1'zo,w
f(()d(-1
'"t'zo,z
f(()d(.
Using the fact that the integral off on closed rectangular paths is zero,
F(w) - F(z) = 1
"Yz,w
f(()d(,
where 'Yz,w is the path joining z to w by again first moving horizontally and
then moving vertically. See the picture below which shows one particular
case, and we have (suppressing the integrand):
F(w) - F(z) = lz !(()d(- l. .!(()d( =
o,
0
=
1 1 1 1 1
B
+
C
+
-"'(z,w
=0
+
-D
+
l l fc-(l 1J
"Yz,w
+
+
f(()d(.
+
103
Cauchy Integral Theorem
..... - - ...
jj\n �
Zo
Thus for 0
w
F( 2
< lw - zl < 8,
=
:
(z)
- f(z)
= w �z
= w �z
1
1
'Yz,w
"Yz,w
':'/z,w
D
f(()d( - f(z) w � z
(!(() - f(z))d(,
1
'Yz,w
ld(
where we have used the Fundamental Theorem of Contour Integration for
the holomorphic function 1 to obtain
1
"Yz,w
ld(
= w-z.
Consequently, from the above, and using the fact that the length of ':'fz,w is
we have
IRe(w - z)I + IIm(w - z)I 21w - zl,
:::;
E
This completes the proof.
3.10
:::; -1w-z
--1
(IRe(w - z)I + IIm(w - z)I)
< 2E.
□
Notes
The proof of Theorem 3.4 follows closely the exposition in [Beck, Marchesi,
Pixton, Sabalka (2008)]. Exercises 3.3, 3.4, 3.12, 3.20, 3.27 are taken from
[Beck, Marchesi, Pixton, Sabalka (2008)]. Exercises 3.10, 3.14, 3.18, 3.19,
3.22, 3.24 are taken from [Needham (1997)]. Exercise 3.21 is taken from
[Rudin (1987)]. Exercise 3.26, 3.33, 3.34 is taken from [Flanigan (1972)].
Exercise 3.28 is taken from [Howie (2003)].
Chapter 4
Taylor and Laurent series
In this chapter we will first learn about the fundamental result which says
that a holomorphic function f has a power series expansion around any
point in the domain D where it lives. See the picture on the left below.
L c (z - zo)
00
Taylor series:
n
n =O
n
Laurent series:
L c (z - zot
nEZ
n
That is, for each z0 E D, there exists an R > 0 such that
J(z)
=
L=O c (z - zot,
00
n
lz - zol < R.
n
Vice versa, every power series
n =O
that converges for at least two points converges in lz - zol < R for some R
and is holomorphic there. En route we will also prove further fundamental
results on holomorphic functions:
(1) The (general) Cauchy Integral Formula and the Cauchy inequality.
(2) The classification of zeros and the Identity Theorem.
(3) The Maximum Modulus Theorem.
105
106
A Friendly Approach to Complex Analysis
In the second part of the chapter, we will learn about Laurent series, which
are like power series, except that negative integer powers of the terms z - zo
also occur in the expansion. This will be useful to study functions that are
holomorphic in annuli (and in particular punctured discs). See the picture
on the right above. They are also useful to classify "singularities" , and to
evaluate some real integrals, as we will see at the end of this chapter.
4.1
Series
Just like with real series, given a sequence (an)nE J\I of complex numbers,
one can form a new sequence (sn)nE J\I of its partial sums:
s2 := a1
s3 := a1
+ a 2,
+ a2 + a3,
Definition 4.1.
=
=
a := lim S n .
(1) The series"
a converges if (s n)nE J\I converges, and"
L..,,; n
L..,,; n
n-+ oo
n=l
n=l
=
(2) The series Lan diverges if (sn)nE J\I diverges.
n=l
=
=
L
(3) Lan converges absolutely if the real series
Jan i converges.
n=l
n=l
From the result in Exercise 1.25, which says that a complex sequence con­
verges if and only if the sequences of its real and imaginary parts converge,
=
Lan converges
¢?
n=l
the real series
=
=
L Re(a ) and L Im(a ) converge.
n
n
n=l
n=l
Thus the results from real analysis lend themselves for use in testing the
convergence of complex series. For example, it is easy to prove the following
two facts, which we leave as exercises.
00
Exercise 4.1. If
Lan
n=l
converges, then lim
n--+oo
an
= 0.
00
Exercise 4.2. If
Lan
n=l
00
converges absolutely, then
Lan
n=l
converges.
107
Taylor and Laurent series
Exercise 4.3. Show that if lzl
< 1, then
f:
Exercise 4.5. Show that the series 1-•
satisfying Re(s) > 1. Thus
s 1-t
t
n=O
Exercise 4.4. Show that if Jzl < 1, then
z
n
converges and that
nz -l
n
+ 2-• +
f:
n=O
z
n
=
1�
z.
= l � 2•
z)
(
3-•
+···
converges for all
s
EC
00
1
((s) := " �n•
n=l
is a well-defined map in the half-plane given by Re(s) > 1, and is called the
Riemann zeta function. The link of the Riemann zeta function with the number
theoretic world of primes is brought out by the Euler Product Formula, which
says that if p1 := 2 < p2 := 3 < p3 := 5 < · · · is the infinite list of primes in
increasing order, then
K
1
IT----.,
K --+oo k=l 1 - Pk
((s) = lim
Re(s) > 1.
Bernhard Riemann (1826-1866) showed that the function ( can be extended
holomorphically to C \ {1 }. It can be shown that the function ( has zeros at
-2, - 4, -6, ..., called "trivial zeros", but it also has other zeros. All the non­
trivial zeros Riemann computed turned out to lie on the line Re(s) = 1/2. This
led him to formulate the following conjecture, which is a famous unsolved problem
in Mathematics.
Conjecture 4.1. (Riemann Hypothesis)
zeta function lie on the line Re(s) = 1/2.
4.2
4.2.1
All non-trivial zeros of the Riemann
Power series
Power series and their region of convergence
Let (cn)nE N be a complex sequence (thought of as a sequence of "coeffi­
cients"). An expression of the type
00
LCn Z
n
n =O
is called a power series in the complex variable z. Thus we imagine putting
in specific values of z in the above series. Then for some z E C, the power
series will converge, while for other values of z it may diverge.
108
A Friendly Approach to Complex Analysis
Example 4.1. All polynomial expressions are power series, with only
finitely many nonzero coefficients. Polynomials converge for all z E C.
The power series
00
n =O
converges whenever I z I
<
1. It diverges if I z I � 1 since , ( lim z n
n--+oo
= o). ◊
A fundamental question is:
L
00
For what values of z E (C does the power series
n =O
Cn Z n
converge?
The following result gives the answer to this question.
L
00
Theorem 4.1. For
n =O
Cn Z n ,
exactly one of the following hold:
(1) Either it is absolutely convergent for all z EC.
(2) Or there is a unique nonnegative real number R such that
L
L
00
(a)
n =O
Cn Z n
is absolutely convergent for all z EC with lzl
Cn Z n
is divergent for all z EC with lzl
00
(b)
n =O
< R,
and
> R.
(The unique R > 0 in the above theorem is called the radius of convergence
of the power series, and if the power series converges for all z E C, we
say that the power series has an infinite radius of convergence, and "'.rite
"R = oo".)
diverge
R
zo
converge
Fig. 4.1
Convergence region of a power series in C.
109
Taylor and Laurent series
What happens on the circle lzl = R? Complex power series may diverge at
every point on the boundary (given by lzl = R), or diverge on some points
of the boundary and converge at other points of the boundary, or converge
at all the points on the boundary. There is no general result answering
what happens at each point on the circle, and one just has to look at the
specific power series at hand to find out the behaviour.
Proof. (of Theorem 4.1.) Let
S := { y E [0, oo) : 3z EC such that y = lzl and � Cn Z n converges} .
Clearly OE S. Only two cases are possible:
l S is not bounded above. In this case we will show that the radius of
convergence is infinite. Given z EC, there exists a y ES such that lzl < y.
But as y ES, there exists a zo EC such that y = lzol and
O
converges. It follows that its terms tend to 0 as n ➔ oo, and in particular,
they are bounded: lcnzol :=::; M. Then with r := lzl/lzol ( < 1), we have
lcn zn l
L Mr
= lcnzol
00
But
n =O
n
converges (r
<
c��,)
n
:=:; Mrn
(n EN).
1!), and so by the Comparison Test,
00
LCn Z n
n =O
is absolutely convergent. Since z was arbitrary, the claim follows.
2_0 S is bounded above. In this case, we will show that the radius of con­
vergence is sup S, that is,
(a) if lzl
(b) if lzl
< sup S,
L enz
00
then
> supS, then
n =O
n
converges absolutely, and
00
Lcnzn diverges.
n =O
110
A Friendly Approach to Comple$ Analysis
If z E <C and lzl < sup S, then by the definition of supremum, it follows
that there exists a y E S such that lzl < y. Then we repeat the proof in 1 °
as follows. Since y E S, there exists a zo E <C such that y = lzol and
converges. It follows that its terms tend to O as n ➔ ao, and in particular,
they are bounded: lcn zol::; M. Then with r := lzl/lzol (< 1), we have
lcn zn l
= lcn zol
C�'i)
n
::; Mrn
00
00
n=O
n=O
(n EN).
But L Mrn converges (r < 1!), and so L Cn Z n is absolutely convergent.
Finally, if z E <C and I z I > sup S, then setting y .- I z I, we see that
y ¢ S, and by the definition of S,
diverges (otherwise y E S).
•
z
The uniqueness of R can be seen as follows. If R, R have the property
described in the theorem and R < R, then
R+R
R<r:=--<R.
2
00
00
□
As r < R, Lcn rn converges. As R < r, Lcn rn diverges, a contradiction.
=l
�
The calculation of the radius of convergence is facilitated in some cases by
the following result.
111
Taylor and Laurent series
Theorem 4.2. Consider the power series
l
.
Cn + I exists,
.
If L := 11m I -then
n--+oo
Cn
(1) the radius of convergence is 1/L if L-/=- 0.
(2) the radius of convergence is infinite if L = 0.
Proof. Let L -/=- 0. We have that for all nonzero z such that lzl
that there exists a q < 1 and an N large enough such that
lc
for all n
> N.
1 z:
n+
lcnZ
+l l
I
(This is because
I
c 1
;:
=
I
Cn + jlzl '-5, q
l
Cn
<
l
<
1/ L
zl n� Llzl < 1.
So we may take q = (Llzl 1)/2 < 1.) Thus by the Ratio Test, the power
series converges absolutely for such z.
+
If L = 0, then for any nonzero z E <C, we can guarantee that there exists a
q < 1 such that
lc +1 z: + 1 1
n
= Cn +l jlzl '-5, q < l
for all n
lcnZ
> N.
I
I
(This is because
Cn
I I
n--+oo
Cn+l
--z
--+ 0IzI
Cn
= 0 < 1.
So we may take q = 1/2 < 1.) Thus again by the Ratio Test, the power
series converges absolutely for such z.
On the other hand, if L -/=- 0 and lzl
enough such that
lc
l
+ z:
l l t
CnZ
c 1
as I : zl n� Llzl
>
=
>
1/ L, then there exists an N large
c
l n+l jlzl > 1 for all n > N,
Cn
1. By the Ratio Test, the power series diverges.
□
112
A Friendly Approach to Complex Analysis
Example 4.2. Consider the power series
Lz
oo
n=l
1
lim
(n+ l) 2
n2
•
We have
= 1'
1
n--+oo
n
n2
and so the power series converges for lzl < 1 and diverges for lzl > 1. Note
that if lzl = 1, then
and since
L n1
00
n=l
2
converges, it follows that
I:
oo
n=l
n
:2
converges absolutely. Thus at every point of the circle lzl = 1, the power
series converges. We see that this situation is in contrast to the case of the
geometric series
00
where we had convergence at no point of the circle
Exercise 4.6. Consider the power series
izl = 1.
L enxn. If L :=
00
n=O
lim
n--+oo
◊
vTcJ exists, then
(1) the radius of convergence is 1/ L if L-:/ 0.
(2) the radius of convergence is infinite if L = 0.
Exercise 4.7. Show that
L nnzn converges only when z = 0.
00
n=l
zn
Exercise 4.8. Show that �
0nn converges for all z E C.
n=l
Exercise 4.9. Find the radius of convergence of the following complex power series:
oo (-l)n n
L-n-z'
n=l
00
""'
2012
wn
n=O
n
z '
113
Taylor and Laurent series
4.2.2
Power series are holomorphic
We have seen that polynomials are power series with an infinite radius of
convergence, that is, they converge in the whole of <C. They are of course
also holomorphic there. This is not a coincidence. Now we will see, more
generally, that a power series
f(z) :=
that converges for lzl
there holds that
L CnZn
00
n=O
< R is actually holomorphic there,
and for lzl
J'(z) = d: (co+ c1z + c 2 z2 + · · ·) = c1 + 2c 2 z + 3c3z 2 + · · · =
f
n =l
< R,
Cnnzn-l,
(as expected, if one imagines differentiating the series termwise, as we do
in the case of finite sums, that is, polynomials).
Theorem 4.3. Let R > 0 and f(z) :=
f'(z) =
Proof.
L ncnzn-l
00
n =l
for lzl
L CnZn converge for lzl < R. Then
00
n=O
< R.
Step 1. First we show that the power series
g(z) :=
L nCnZn-l = C1 + 2c z + · · ·+ nCnZn-l + ...
00
n =l
is absolutely convergent for lzl
By hypothesis
2
< R.
Fix z and let r satisfy lzl
< r < R.
converges, and so there is some positive number M such that lcn rn l
for all n. Let p := lzl/r. Then O ::=; p < 1, and
1
z n -1 Mnpn-1
lncn zn -ll = lcn rn l · - · nl-l
:C:::: --r • r
r
<M
L npn-I converges (to 1/(1-p) Exercise 4.4). By the Comparison Test,
=l
L ncnzn-I converges absolutely.
00
2
n
00
n =l
;
A Friendly Approach to Complex Analysis
114
Step 2. Now we show that f'(zo) = g(zo) for lzol < R, that is,
lim (f(z) - f(zo) - g(zo)) = 0.
z--+zo
Z - Zo
As before, let r be such that lzol < r < R and since z ---+ zo, we may also
restrict z so that lzl < r.
Let
E
> 0. As
L nc r -l converges absolutely, there is an N such that
00
n
n =l
n
Keep N fixed. We have f(z) - f(zo)=
f
f
LC (
00
n =l
n z
n
- zo), and so for z -=I- zo,
n
f(z) - f(zo) =
Cn Z - Zo =
Cn (zn -1 + zn -2 ZO
z - zo
z - zo
n =l
n =l
+ ... + Zo- 1).
f
Thus
f(z) - j(zo)
Cn (zn -1 + zn -2 ZQ + ... + Zo- 1 -nzo- 1).
- g(zo) =
Z - ZQ
n =l
We let S1 be the sum of the first N - 1 terms of this series (that is, from
n= l ton= N -1) and S2 be the sum of the remaining terms. Then since
izl, izol < r, it follows that
!S2 1 s;
L ic l ( r -1 + r -1 + ... + r -1 +nr -1) = L 2nlc lr 00
00
n =N
Also,
S1 =
n
n
n
n terms
LC
N
n =l
n l
n (z -
n
n
n =N
n
n 1
+ zn-2 zo + · · · + zz0- 2 + z0- 1 -nz0-1)
< �-
115
Taylor and Laurent series
is a polynomial in z and by the algebra of limits,
.
� Cn (z0n- 1 + z0n-2 zo + · · · + zoz0n-2 + z0n- 1
11m S1 = �
N
z➔zo
-
n=l
=
LC
N
n=l
n
(nz�-l - nz�- 1) = 0.
nz0n- 1 )
So there is a 8 > 0 such that whenever lz - zol < 8, we have 1S11 < E/2.
Thus for lzl < r and 0 < lz - zol < 8, we have
I
f(z) - f(zo)
z-�
- g(zo)/ � IS1 I + IS2 I < :_
2
This means that f'(zo) = g(zo), as claimed.
+ :_ = E.
2
D
Remark 4.1. If (cn )nE N is a sequence of real numbers, then consider the
real power series
From Real Analysis, we know that such a power series converges in an in­
terval of the form (- R, R) and diverges in JR \ [- R, R] for some R ;:::: 0.
The two results in Theorems 4.1 and 4.3 imply that if we replace the real
variable x by a complex variable z, then we can "extend/continue" the real
power series to a holomorphic function in the disc given by I z I < R in the
complex plane. So we can view real analytic functions (namely functions of
a real variable having a local power series expansion) as restrictions of holo­
morphic functions. This again highlights the interplay between the worlds
of real analysis and complex analysis. (We have seen a previous instance
of this interaction when we studied the Cauchy-Riemann equations.)
By a repeated application of the previous result, we have the following.
Corollary 4.1. Let R > 0 and let f(z) :=
Then fork;:::: 1,
LC Z
00
n
n=O
L n(n - l)(n - 2) · · · (n -k+ l)c z -k
00
j{k)(z) =
.
n
n=k
1
In particular, for n;:::: 0, Cn = JC n)(0).
1
n.
n
n
converge for lzl
for lzl
< R.
< R. (4.1)
116
A Friendly Approach to Complex Analysis
Proof. This is straightforward, and the last claim follows by setting z = 0
in (4.1):
f
n(n-1) · · · (n - k + l)c n zn -k-ll = k!ck.
n=k+l
z=0
jCkl(O)=k(k-1) · · · lck + z
Also, f (O) = co.
□
There is nothing special about taking power series centered at 0. One can
also consider
Lc n (z-zot,
n=0
00
where zo is a fixed complex number. The following results follow immediately from Theorems 4.1 and 4.3.
Corollary 4.2. For
L C (z - zo) , exactly one of the following hold:
00
n=0
n
n
(1) Either it is absolutely convergent for all z E C.
(2) Or there is a unique nonnegative real number R such that
(a)
L c (z - zot is absolutely convergent for lz - zol < R, and
=0
L c (z - zot is divergent for lz - zol > R.
00
n
n
00
(b)
n=0
n
Corollary 4.3. Let zo E <C, R > 0 and f(z) :=
for lz-zol
jCkl(z)
=
< R. Then
L c (z - zot converge
00
n=0
n
L n(n -1) · · · (n - k + l)c (Z - zor-k for lz-zol < R,
00
n=k
n
n
In particular, for n � 0, Cn = .!,f( l(zo).
n.
k � 1.
Remark 4.2. (Uniqueness of coefficients.) Suppose that
L C (Z - zot and I:cn(Z - zot
00
00
n=0
n=0
are two power series which both converge to the same function f in an open
disk with center z0 and radius R > 0. Then from the above, for n � 0, we
have
f( n )(zo) = Cn ·
Cn =
n.I
n
117
Taylor and Laurent series
Exercise 4.10. For
lzl < 1, what is 12 + 22 z + 32 z2 + 42z3 +. • •?
LC Z .
00
Exercise 4.11. True or false? All statements refer to power series
n
n
n =O
(1) The set of points z for which the power series converges equals either the
singleton set {O} or some open disc of finite positive radius or the entire
complex plane, but no other type of set.
(2) If the power series converges for z = 1, then it converges for all z with lzl < 1.
(3) If the power series converges for z = 1, then it converges for all z with lzl = 1.
(4) If the power series converges for z
= 1,
then it converges for z
= -1.
(5) Some power series converge at all points of an open disc with center O of some
positive radius, and also at certain points on the boundary of the disc (that
is the circle bounding the disc), and at no other points.
(6) There are power series that converge on a set of points which is exactly equal
to the closed disc given by lzl :S 1.
(7) If the power series diverges at z
4.3
= i, then it diverges at z = 1 + i as
well.
Taylor series
We have seen in the last section that complex power series
00
Lcn(z-zot
n=O
are holomorphic in their region of convergence lz - zol < R, where R is the
radius of convergence. In this section, we will show that conversely, if f is
holomorphic in the disc lz - zo I < R, then
f(z)
=
L cn(z - zot whenever lz - zol < R,
00
n=O
where the coefficients can be determined from the f. Thus every holomorphic function f defined in a domain D possesses a power series expansion
in a: disc around any point zo E D.
Theorem 4.4. If f is holomorphic in D(zo,R) := {z E (C: lz - zol < R},
then f(z) = co +c1(z - zo) + c2(z - zo)2 + c3(z - zo)3 + · · · for z E D(zo,R),
where for n � 0,
d(
- _1 f
f(()
Cn - 21ri Jc (( - zo)n+l '
and C is the circular path with center z0 and radius r, where O < r < R
traversed in the anticlockwise direction.
, -Q
_
A Friendly Approach to Complex Analysis
118
'
'
'
',
,
:
:,
'
'
c---,,,
''
'
r
zo R
- -- ............
'
--�
',
,'
'
'
'
'
,
:
,:
'
,,,."
Proof. Let z E D(zo, R). Initially, let r be such that lz-zol < r < R.
Then by Cauchy's Integral Formula,
f(() d( = _1
f(()
d(
f(z) = _1
2ni c (-z
2ni c (-zo+ zo-z
f(()
d(
=
2ni c
l z-zo
((-zo) ( (-zo)
1
1
�1
z-zo
lz-zol
Then lwl = -'------'- < . Thus
Set w := ---.
1
1
.,, -zo
r
Wn
1
1
2
3
n 1
----=--=
l+w+w
+w
+···+w
+-­
l z-zo
-w
-w
1
1
(- Zo
z-zo
(z-zot
(z-zor-1
= +--+···+----+-----1 (-zo
((-zo)n-1(( - z)'
((-zo)n-l
and so plugging this in the above, we obtain
1
(z-zot-1
(z-zot
d(
(() (-1- + ...+
+
f(z) = _1
f
((-zo)n((-z))
((-zo)n
(-zo
2ni c
where
=co+c1(z-zo)+ · · ·+ Cn-1(z-zo)n-l + Rn(z),
1
f(()(z-zor d(.
Rn(z) := _1
2ni c ((-zo)n((-z)
So we would be done ifwe manage to show that Rn(z) goes to O as n--+ oo.
We note that IJI is bounded on the circle, since it is a continuous real valued
function on the compact set C, that is, there is an M > 0 such that for all
(EC, lf(()I < M. (Here with a slight abuse, we think ofthe path C, and
the set ofpoints C(t), t E [O, 2n], as being the same.) Also, for (EC,
lz-zol nn
(z-zot
l ((- nl = ( r ) �O.
zo)
Taylor and Laurent series
119
But what about 1/1( - zl for ( E C? Is this bounded by something? The
picture below shows that indeed this term is bounded by the reciprocal of
the "distance between the circle C and z" .
We have 1(-zl = 1(-zo-(z-zo)I 2: 1(-zol-lz-zol = r- lz-zol- Thus
n
M
lz - zol
n
�O.
)
IRn (z)l:::;(
r -lz - zol
r
Thus the series co+ c1 (z - zo)+ c2 (z - zo)2 + c3 (z - zo)3 + · · · converges
to f(z). Note that we have only shown the expression
Cn
= _l
1
(
f( )
d
21ri c ((- zo)n +l (
where r is such that Iz-zoI < r < R. But by the Cauchy Integral Theorem,
we see that this integral is independent of r, and any value of r E (0, R)
can be chosen here:
(
! -� +l is holomorphic in the punctured disc D*(z0, R) given by
(1)
(• - ZQ n
0 < I z - zo I <__R;
(2) besides C, if C is another circular path with center zo and some other
radius E (0, R) then C, C are D*(zo, R)-homotopic.
r
D
This completes the proof.
But we had learnt earlier in Theorem 4.3 that whenever we have a
J(z) =
L c (z - zot for lz - zol < R,
n=O
n
we know that
JC n )(zo)
n!
for n 2'. 0. And in the above result, we found different expressions for the
coefficients en , given in terms of integrals. But the coefficients of the power
Cn
=
120
A Friendly Approach to Complex Analysis
series expansion are unique in any disc, and so these have to be the same.
With this observation in mind, we obtain the following result.
Corollary 4.4. (Taylor 1 Series) If
(1) D be a domain,
(2) f: D -+ (C is holomorphic, and
(3) zo ED,
then
f(z)
=
f'(zo)
f"(zo)
- zo) + --, -(z - zo) 2 +... ,
f(zo) + --(z
1
1.
2.
lz - zol
< R,
where R is the radius of the largest open disk with center z0 contained in
D. Also,
J< n)(zo)
=
_2:!_
1
f(z)
21ri c (z - z0) n +l
dz '
(4.2)
where C is the circular path with center zo and radius r , where O
traversed in the anticlockwise direction.
<r<R
(4.2) is called the (general) Cauchy Integral formula. Earlier we had just
seen the cases when n = 0 (Theorem 3.4) and n = 1 (Proof of Theorem 3.6).
Also, we can obtain from here that
J< n )(w)
=
_2:!_
21ri
1
f(z)
dz '
c (z - w) n +l
for any point w Eb.:= {z E (C: lz-zol < R}. This follows from the Cauchy
Integral Theorem, since we can first consider a small circle C& centered at
w for which the above formula holds, and then note that the paths C, C&
are b. \ {w}-homotopic, while the function f(-)/(· - w) n +i is holomorphic
in b. \ {w}.
Proof.
f(z)
From Theorem 4.4, we have
= co+c1(z-zo)+c2(z-zo) 2+c3(z-zo) 3+... ,
z E D(zo, R), (4.3)
where D(zo, R) := {z E (C: lz - zol < R} and R is the radius of the largest
open disk with center zo contained in D. Also, for n :::: 0,
1
f(z)
_ 1
d
Cn - 21ri C (z - zo) n +l z,
-------------
1 Named after Brook Taylor (1685-1731) who, among others, studied this expansion in
the context of real analytic functions.
121
Taylor and Laurent series
and C is the circular path with center z0 and radius r, where 0 < r < R
traversed in the anticlockwise direction. But from Corollary 4.3, the power
series above is infinitely many times differentiable, and also, for n � 0,
1 n)
JC (zo) = Cn.
1
n.
Thus the result follows.
□
Summarizing, if f is holomorphic in iz - zol < R, then
f(z) =
00
00
1
f(()
f( )(zo)
n
((z - zo)n = L
L
+l
d(
·
.
_
n
1 (z - zo) ,
)
2
(
ni
C
zo
n.
)
1
(
n=0
n=0
n
for lz-zol < R, where C(t) = zo+r exp(it), t E [0, 2n], and r is any number
such that 0 < r < R.
Example 4.3. The exponential function f, z 1---t f(z) := expz, is entire.
So we know that we can write
exp z =
00
CnZn for z EC,
L
n=0
for some coefficients Cns. What are these coefficients? They are given by
f(n)(zo)
, n � 0.
Cn =
n.1
Since
!
exp z = exp z, it follows that j(nl(o) = 1, and so
f(z) =
f(n)( 0)
'°'-(z -Ot = '°' �zn,
oo
oo
�
n=0
n!
� n!
n=0
◊
for all z EC.
Example 4.4. The function f defined by f(z) = Log(z) is holomorphic in
C \ (-oo, 0]. The largest open disc with center zo = 1 in this cut plane is
D = {z EC: lz - 11 < 1}. Since
JCnl(zo) =
we have Log(l + w) = w -
(-lt (n - 1)!
z[j
w2
2
+···+
= (-lt(n - 1)!,
(-ltwn
n
+ · · · for lwl < 1.
Exercise 4.12. Show that for z EC:
3
z
sin z = z - 3!
5
- ···
+5! +
z
z2
and cos z = 1 - 2!
4
z
-+···
+4!
.
◊
122
A Friendly Approach to Complex Analysis
Exercise 4.13. Find the Taylor series of the polynomial z 6- z4 + z2 - 1 with the
point zo =1 taken as the center.
Exercise 4.14. Find en 's in the Taylor series
(1) f(z) =
1
1'0z
LC Z
n=O
n
n
around Ooff given by
exp((2 )d(, z EC, where ')'oz is the straight line path from Oto z.
Hint: J'(z) =exp( z2 ).
z2
(2) J(z)= ( z l)2,
+
z
EC\{-1}. Hint: Fori z l<l,
1
z +l
=1- z+z2 -+···.
Here is a consequence of the general Cauchy Integral Formula.
Corollary 4.5. (Cauchy's inequality) If
(1) f is holomorphic in D(zo,R) := {z EC: lz - zol < R} and
(2) 1/(z)I :SM for all z E D(zo, R),
n1M
then for n 2'. 0, lf( n ) (zo)I :S �n •
Proof.
1� r
Let C be the circle with center zo and radius r < R. Then
1/Cnl(zo)I
=
21ri
f(z)
Jc (z- zo)n+l dzl
I
I
n! M
n!M
n!
f(z)
<-max----- -21rr= ---21rr= --.
- 21r zEC (z- zo) n +l
21r r n +l
rn
The claim now follows by passing the limit r /' R.
□
Exercise 4.15. Suppose that f is an entire function for which there is an M > 0
and an integer n 2: 0 such that for all z E C, IJ( z)I :S Ml z ln - Use Cauchy's
inequality to prove that J<n+i) (z) = 0 for all z and show that f is a polynomial
of degree at most n. What happens when n =O?
Exercise 4.16. Evaluate
{
s
;�1: dz, where C is the circular path with center 0
la z
and radius 1 traversed in the anticlockwise direction.
4.4
Classification of zeros
Suppose f : D-+ C is holomorphic in a domain D. We ask the question:
What do the zeros of a nonzero f look like? (A point z0 ED is a zero off if
f(zo) = 0.) We will learn in this section that the answer to this question is,
that the zeros are "isolated". Such a thing doesn't happen with continuous
123
Taylor and Laurent series
functions. Zeros of nonzero continuous functions, which aren't as rigid as
holomorphic functions needn't be isolated.
Example 4.5.
(1) expz has no zeros in <C. Indeed, I exp(z)I = e Re(z) > 0 for all z E <C.
(2) cos z-3 has infinitely many zeros in <Cat 2nn±ilog(3+2v'2), n E Z, all
of which lie on a horizontal line, and they are isolated, with a distance
of 2n between any two isolated adjacent zeros. We had seen this in
Exercise 1.38.
(3) The polynomial p, p(z) = (z + l)3 z 9 (z - 1)9, has zeros at -1, 0, 1. ◊
If p is a nonzero polynomial such that p(zo) = 0, then by the Division
Algorithm there exists a polynomial q (the quotient) such that
p(z) = (z - zo)q(z)
(that is, the remainder is 0). Now we have two possible cases:
1° q(zo)-/= 0. Then zo is an isolated zero of p.
2 ° q(z0) = 0. Then we repeat the above procedure with p replaced by q.
Eventually, we obtain p(z) = (z - z0 )m q(z) for some m 2: 1 and q(zo)-/= 0.
Then we we call m the multiplicity/order of zo (as a zero of p). We will
now see in Theorem 4.1 below that the same sort of a thing holds for
holomorphic functions f ( replacing the polynomial p), except that we end
up with another holomorphic function (g instead of the polynomial q). This
is not completely surprising, since we know that power series are analogues
of polynomials, and every holomorphic function has a local power series
expansion.
But let us first give the following definitions.
Definition 4.2. Let D be a domain and f: D--+ <C be holomorphic in D.
A point zo ED is called a zero off if f(zo) = 0.
If there is a smallest m E N such that
(1) f(m )(zo)-/= 0 and
(2) f(zo) = · · · = f(m-l)(zo)
= 0,
then z0 is said to be a zero of f of order m.
that f(o) := f.)
(We adopt the convention
We have the following result on the classification of zeros of a holomorphic
function.
124
A Friendly Approach to Complex Analysis
Proposition 4. 1.
( Classification of zeros) Let
(1) D be a domain,
(2) f: D-+ <C be holomorphic in D and
(3) zo ED be a zero of f.
Then there are exactly two possibilities:
1° There is a positive R such that f(z)=0 for all z satisfying lz-zol < R.
2_0 There exists an m E N such that zo is a zero off oforder m, and
there exists a holomorphic function g: D-+ <C such that g(zo)-/= 0 and
f(z)=(z - zo)m g(z) for all z ED.
We note that ifwe are in case 2 ° , then since g is continuous and as g(zo)-/= 0,
gis not zero in a small disc b. centered at z0, and so f is nonzero in b.\ {0}
using the fact that for z Eb.\ {zo}, f(z)=(z - zorg(z)-/= 0. Thus zo is
the only zero of f in b., that is, z0 is isolated.
Also, we will soon learn about something called the Identity Theorem,
and from that result, it will follow that in fact in case 1 °, we can conclude
that f 0 in the whole of the domain D.
Proof.
=
We have a power series expansion for f in a disc with some radius
R > 0 and center zo:
f(z)=co+ c1(z - zo)+ c2 (z - zo) 2 + · · · for lz - zol < R.
Since f(zo)=0, we know that co=0. Now there are exactly two possibil­
ities:
1° All the Cn are zero. Then f(z)=0 whenever lz - zol < R.
2_0 There is a smallest m ;:::: 1 such that Cm -/= 0. Then we have that
co=c1 = · · · =Cm -1=0, and so using the fact that
f( n l(zo)
, n;:::: 0,
Cn =
n.1
it follows that zo is a zero of order m. Moreover, from the power series
expansion, we have
+
Cm +k(z-zot
f(z)=Cm(z-zor+cm +l(z-zo)m l+. · · =(z-zor
I:
(X)
k=O
for lz - zol < R. Thus, if we define g: D-+ <C by
f(z)
for z -/= zo,
z
( - zo ) m
g(z)= {
k
� Cm +k(z - zo) for lz - zol < R,
(X)
(4.4)
125
Taylor and Laurent series
From (4.4), the two definitions give the same value whenever both are
applicable. So g is well-defined. Moreover, we have:
is holomorphic in D. For z =/- z0, this follows by observing that
both f and 1/(· -zor are holomorphic in D\ {O}. For lz -zol < R,
this follows since g is given by a power series!
(2) g(zo) = Cm =/- 0. (Definition of m!)
(3) f(z) = (z -zo) m g(z) for z =/- D\ {zo} for z ED\ {z0} follows from
(4.4). On the other hand if z = z0, then both sides are zero. Thus
for all z E D, f(z) = (z -zo)m g(z).
(4) z0 is a zero of order m. Indeed, Cn = fn (z0 )/n! for all n's, and so
the claim follows using Cm =/- 0, while co = c1 = · · · = Cm-1 = 0.
(1)
g
□
This completes the proof.
Example 4.6.
(1) mr is a zero of sin z for each n E Z. We know that sin z is not identically
zero in any neighbourhood of mr by just looking at the restriction of
sin z to the real axis. We ask: what is the order of mr as a zero of sin z?
Since sin' z = cos z and cos zl z =mr = (-l) n =/- 0, it follows that mr is a
zero of sin z of order 1.
(2) exp(z2 ) -1 has a zero at O since exp(02 ) - 1 = 1 - 1 = 0. What is its
order? We have
exp(z 2 ) = 1 +
z2
z4
- + - + · · · ' z E <C,
1!
2!
1
z2
and so exp(z 2 ) - 1 = z 2 g (z) (z E q, where g(z) := + 1 + · · ·
1
1.
2.
g is given by a power series that converges in <C, and so g is entire.
Also, g(O) = 1 =/- 0. Thus the order of O as a zero of exp(z 2 ) - 1 is 2.
Alternately, we could have observed that
2
2
d� (exp(z ) -1{= = (exp(z )) · 2zl =O
0
z
=
0,
::2 (exp(z 2 ) -1{= = (exp(z 2 )) · 2 + (exp(z 2 )) · (2z) 2 1 =O
0
z
= 2 =/- 0,
and so the order of O as a zero of exp(z 2 ) - 1 is 2.
◊
Exercise 4.17. Let D be a domain, m E N, R > 0 and zo E D. Let f, g : D ➔ <C be
m
holomorphic such that g(zo) -/= 0 and whenever lz-zol < R, J(z) = (z-zo) g(zo).
Prove that zo is a zero of f of order m.
126
A Friendly Approach to Complex Analysis
Exercise 4.18. Find the order of the zero zo for the function fin each case:
(1) zo
= i and f(z) = (1 + z2 )
4
•
(2) zo = 2mri, where n is an integer, and f(z) = expz -1.
(3) zo = 0 and f(z) = cosz -1 + �(sinz) 2 .
Exercise 4.19. Let fbe holomorphic in a disc that contains a circle, in its interior.
Suppose there is exactly one zero zo of order 1 off, which lies in the interior of
,. Prove that
1
zj'(z)
zo =
dz.
27ri , J(z)
1
Exercise 4.20. Let D be a domain and f be holomorphic in D such that f has a
zero of order m > 1 at zo ED. Prove that the function z >-+ (f(z))2 has a zero of
order 2m at zo, and that J' has a zero of order m - 1 at zo.
Exercise 4.21. A complex valued continuous function on a domain needn't exhibit
the dichotomy of behaviour of its zeros as in Theorem 4.1 for holomorphic func­
tions. Give an example of a function f: <C ➔ <C, such that O is neither an isolated
zero nor is it the case that fis identically zero in a small disc around O centered
at 0.
4.5
The Identity Theorem
In this section, we will learn the Identity Theorem, which once again high­
lights the rigidity of holomorphic functions. It says roughly that a nonzero
holomorphic function cannot have an accumulation of zeros in its domain.
Theorem 4.5. Let
(1) D be a domain,
(2) f: D ➔ <C be a holomorphic function in D,
(3) (zn) nE N be a sequence of distinct zeros off which converges to z* ED.
Thenf is identically zero in D.
Proof.
First let us note that z* is itself a zero of f because by the con­
tinuity off, we have
f(z*)
=
= f ( n➔oo
lim Zn ) = lim f(zn) = lim 0 = 0.
n➔oo
n➔oo
We claim thatf 0 in some disc� with center z* and radius r > 0. If not,
then z* is a zero of some order m, and f(z) = (z -z*) m g(z) for z in�, and
Taylor and Laurent series
127
g(z) -=f. 0 in �- But then for all large n, 0 = f(zn ) = (zn - z*) m g(zn ) -=f. 0,
a contradiction.
Next we will show that f
0 in all of D. Suppose that f(w) -=f. 0
for some w E D. Then there is a path, : [O, 1] ---+ D that joins z* tow:
1(0) = z* and 1(1) = w. Let S := {t E [0, 1] : f(,(T)) = 0 for 0:::; T:::; t},
and T := supS. We note that supS exists since S is nonempty (0 belongs
to S!) and S is bounded above (by 1). W hat is T? If we think of t as time
as we travel from z* (time t = 0) tow (time t = 1) along,, then T is the
largest time such that f has been O along the path covered so far. If T = 1,
then we are done (since then by continuity, f(w) = f(T) = 0). So let us
suppose that T < 1. Then f(,(T)) = 0, by continuity. But then f(z) = 0
for z's in a disc around,(T) of radius say 8, since,(T) can't be an isolated
zero! This implies that f(,(t)) = 0 for t's that are bigger than T (because
for all t's close enough to T, l,(t) -,(T)I < r).
=
\ ,(T) /
''
,,'
This contradicts the definition of T. Thus T can't be less than 1. This
D
completes the proof.
Example 4.7. We had shown that for all z E re, (cosz)2 + (sinz)2 = 1
using the definitions of cosz and sinz. Here is a proof using the above
result. Consider f: re➔ re, f(z) = (cos z) 2 + (sinz)2 -1, z Ere. Then f is
entire. Also, for all x E �, f( x) = (cosx)2 +(sinx)2 -1 = 0 by Pythagoras's
Theorem. So by the result above, f 0 in re!
◊
=
The following is an immediate consequence of this result.
Corollary 4.6. (Identity Theorem) Let
(1) D be a domain,
(2) f, g: D ➔ re be holomorphic in D,
(3) (zn ) n EN be a sequence of distinct points in D which converges to z* E D,
and such that for all n EN, f(zn ) = g(zn )Then f(z)
= g(z)
for all z ED.
Proof. Define h : D ---+ re by h(z) = f(z) -g(z), z E D, and note that
the ZnS are zeros of the holomorphic function h. By the result above, h
128
A Friendly Approach to Complex Analysis
□
must be identically zero in D, and so the claim follows.
Example 4.8. We know that exp: <C---+ <C defined by
expz = exp(x
+ iy):= ex(cosy + i siny),
z=x
+ iy E <C,
is an entire function such that exp x = ex for x E IR. In other words, exp
is an entire extension of the usual real exponential function. Is there any
other entire extension possible? We show that the answer is no! Suppose
that g: <C---+ <C is entire and g(x) = ex for real x. But then
expx = g(x) for all x ER
In particular,
and 1/n ➔ 0 E <C. So by the Identity Theorem, expz = g(z) for all
z E <C. So there is only one entire function whose restriction to IR is e x .
This explains the naturalness of the definition of the complex exponential
in Section 1.4.1.
◊
Exercise 4.22. Show, using the Identity Theorem, that for all z1, z2 EC,
cos(z1
+ z2) = (cosz1)(cosz2) - (sinzi)(sinz2),
by appealing to the corresponding identity when z1, z2 are real numbers.
Exercise 4.23. Let D be a domain and let Hol(D) be the set of all functions
holomorphic in D. Then it is easy to check that Hol(D) is a commutative ring
with the pointwise operations
(f + g)(z)
(f · g)(z)
= f(z) + g(z),
= f(z)g(z),
for z E D and f,g E Hol(D). (By a commutative ring R, we mean a set R
with two laws of composition + and· such that (R, +) is an Abelian group, · is
associative, commutative and has an identity, and the distributive law holds: for
a,b,c E R, (a+ b)· c = a· c + b· c.)
Check that Hol(D) is an integral domain, that is, a nonzero ring having no
zero divisors. In other words, if f • g =0 for f,g E Hol(D), then either f =0 or
g =0.
If instead of Hol(D), we consider the set C(D) of all complex-valued con­
tinuous functions on D, then C(D) is again a commutative ring with pointwise
operations. Is C(D) an integral domain? (This shows that continuous functions
are not as "rigid" as holomorphic functions.)
129
Taylor and Laurent series
Exercise 4.24. Let f, g be holomorphic functions in a domain D. Which of the
following conditions imply f = g identically in D?
(1) There is a sequence (zn ) nE N of distinct points in D such that J(zn ) = g(zn)
for all n EN.
(2) There is a convergent sequence (zn ) nE N of distinct points in D with its limit
in D such that f(zn ) = g(zn) for all n EN.
(3) 'Y is a smooth path in D joining distinct points a, b E D and f
(4) w ED is such that J <nl(w)
= g <n)(w)
= g on 'Y­
for all n 2: 0.
Exercise 4.25. Suppose that f is an entire function, and that in every power series
(that is, for every zo E q f(z)
Prove that f is a polynomial.
4.6
=
L c (z - zot, at least one coefficient is 0.
n=O
n
The Maximum Modulus Theorem
In this section, we prove an important result, known as the Maximum
Modulus Theorem which says that for a nonconstant holomorphic function
f : D ➔ C, Ill can't have a maximizer in the domain D.
Theorem 4.6. (Maximum Modulus Theorem) Let
(l) D be a domain,
(2) f: D ➔ (C be holomorphic in D,
(3) z0 ED be such that for a ll z ED, lf(zo)I 2:: lf(z)IThen f is constant on D.
Proof. Let r > 0 be such that the disc with center z0 and radius 2r
is contained in D. Let Cr be the circular path Cr(t) = z0 + rexp(it),
t E [O, 21r]. Then by the Cauchy Integral Formula,
1
!(zo ) = 21ri
1
=27r
1
f(
l 1
-dz = z)
z - zo
21ri
271' ---'-----'---'-'-irexp
'
f(zo +rexp(it)).
(it )dt
1271' f(zo + rexp(it))dt,
7
0
O
rexp(it)
130
A Friendly Approach to Complex Analysis
where the last expression can be viewed as the "average" of the values of f
on Cr. Since lf(zo + rexp(it))I::; lf(zo)I for all t, the above yields
lf(zo)I = 1 �
2
1
f(zo + rexp(it))dtl ::; �
2
211:
211:
1211:lf(zo + rexp(it))ldt
=
fz
1
::; �
l ( o)ldt lf(zo)I2
So in the above, all the inequalities ::; have to be equalities, and by rear­
ranging, we get
1211:
( lf(zo)l - lf(zo + rexp(it))I)dt = 0.
;:,:o
But the integrand is pointwise nonnegative, and so in light of the above, we
can conclude that lf(zo + rexp(it))I = lf(zo)I for all t. But by replacing r
by any smaller number, the same conclusion would hold. Thus f maps the
disc�:= {z E <C: lz-zol ::; r} into the circle {w E <C: lwl = lf(zo)I}- This
implies by Example 2.11 that f is constant on �- The Identity Theorem
now implies that f must be constant on the whole of D.
2�
□
Example 4.9. Let lHI := {z E <C : Re(z) :2 O} denote the right half plane,
and consider f : lHI --+ <C given by
e
)
f(z) = ;l-/ , z E lHI.
Then it can be shown that
llfll= := max lf(z)I
zEIH[
exists. Without worrying about the existence of this maximum, let us
instead see how, assuming its existence, the Maximum Modulus Theorem
enables us to calculate its value. Suppose that zo E lHI is a maximizer. Then
this maximizer z0 can't have a real part which is positive, by the Maximum
Modulus Theorem. So zo E iffi?., that is, z0 = iy0 for some Yo ER But
1
e p( iy)
, y ER
lf(iy)I = l �i - I=
Jy + 1
y+1
1
= 1. ◊
Thus llfll= = max lf(z)I = maxlf(iy)I = max
zEIH[
yEIR
yEIR
l
J0
+1
+
y
2
H+i
2
2
Exercise 4.26. Let D be a domain and let f: D ➔ <C be a nonconstant holomor­
phic map. Prove that there is no maximizer for the map z >-+ IJ(z)J on D.
Exercise 4.27. (Minimum Modulus Theorem) Let D be a domain and let
f: D ➔ <C be holomorphic in D. Suppose that there is a zo ED such that for all
z ED, IJ(zo)I :S: Jf(z)J. Then prove that either f(zo) = 0 or f is constant on D.
Exercise 4.28. Consider the function f defined by
mum and minimum value of Jf(z)J on {z E <C: JzJ
f(z) = z 2
:S:
1}.
-
2. Find the maxi­
131
Taylor and Laurent series
4. 7
Laurent series
Laurent series generalize Taylor series Indeed, while a Taylor series
.
I:cn (z-zor
n=O
has nongenative powers of the term z-z0, and converges in a disc, a Laurent
series is an expression of the type
Lcn (Z-zor
nEZ
= · · · +C- 1 (z-Zo)-l +Co+C1 (z-zo) 1 + · · · ,
which has negative powers of z - z0 too
.
We will see that
(1) Laurent series "converge" in an annulus {z E <C : r < lz - zol < R}
with center z0 and gives a holomorphic function there, and
(2) conversely, if we have a holomorphic function in an annulus with center
z0 and it has singularities that lie in the "hole" inside the annulus,
then the function has a Laurent series expansion in the annulus For
.
example, we know that for all z E <C,
z z 2 z3
+ + +· ,
l! 2! 3!
··
and so for z =/- 0, we have the "Laurent series expansion"
exp z = 1+
exp
1
1
1 1
1 1
= 1+ +
+
+
� 2! z2 3! 3! ... .
�
Note that exp(l/z) is holomorphic in <C \ {O}, which is a degenerate
annulus centered at O with inner radius r = 0 and outer radius R = +oo!
Let us first define what we mean by the convergence of
L c (z-z ) .
nEZ
n
0
n
132
A Friendly Approach to Complex Analysis
Definition 4.3.
The Laurent series
L c (z - z0)
nEZ
L C- (z - zo)00
If
n=l
n
L e (z - zo)
nEZ
n
n
n
n
n
converges (for z) if
converges and
L e (z - zor converges.
00
n=O
n
converges, then we write
nEZ
00
00
n=l
n=O
and call this the sum of the Laurent series.
Example 4.10. For what z E <C does the Laurent series
1
4z
1
8z
... + _ _3 + _ _2 + _!_ + 1 + z + z 2 + z 3 + ...
converge? We have:
2z
(1) 1 + z + z 2 + z 3 + · · · converges for zl < 1, and it diverges for lzl > 1.
1
1
< 1 and diverges for 1 � I > 1,
+ :3 + :5 + · · · converges for
(2)
2
2zI
2z 4
8
that is, it converges for lzl > 1/2 and diverges for lzl < 1/2.
Hence the given Laurent series converges when lzl < 1 and lzl > 1/2, that
is, it converges inside the annulus {z E <C: 1/2 < lzl < 1}, and it diverges
when lzl > 1 or when lzl < 1/2.
◊
For what z does
L C (Z - zor converge?
nEZ
n
(1) From Theorem 4.1, for
L C (Z - zor, there is some R such that it
00
n =O
n
converges for lz - zol <Rand diverges for lz - zol > R.
(2) What about the series
L c_
00
n (z - zo)-
n
? The power series
n=l
also converges for lwl <Rand diverges for lwl >
Then
L C- (z - zo)-
R.
L C- W
00
n
n=l
Set w := (z-zo)-1.
n
converges when 1/lz - zol < R, that is for
n=l
lz - zol > 1/.R =: r, and diverges for lz - zol < r.
00
n
n
133
Taylor and Laurent series
Hence the Laurent series converges in the annulus {z Ere: r
and diverges if either lz - zol < r or lz - zol > R.
< lz-zol < R}
Is it holomorphic in the annulus where it converges?
L c (z - zot is holomorphic in {z Ere: lz - zol < R} and so in
00
(1) z I-+
n
n=O
particular, also in {z Ere: r
< lz - zol < R}.
(2) The map
00
n
�
9
W i-----=--t
�C-nW
n=l
is holomorphic in {w E re : lwl < .R}. Also we see that the mapping
z � (z - z0)-1 : re\ {zo} -+ re is holomorphic. So their composition
go f is holomorphic in {z Ere: lz - zol > r}, that is,
00
)-n
g f �
zi-=---:+
�C-n (z - zo
o
n=l
is holomorphic in {z E re : r < lz - zol}, and so also in particular in
the annulus {z Ere: r < lz - zol < R}.
L
+ L c_
00
00
Hence the sum z I-+
en (z - zot
n (z - zo)-
n=l
is holomorphic in the annulus {z Ere: r < lz - zol
n=O
n
=
< R}.
L c (z - zot
nEZ
n
Summarizing, we have learnt that any Laurent series
L C (z - zo)
nEZ
n
2
converges in an annulus {z Ere: r
map
z I-+
n
< lz - zol < R} for some r, R, and the
L C (z - zot
n
nEZ
is holomorphic in {z Ere: r
< lz - zol < R}.
That conversely, a function holomorphic in an annulus has a Laurent
series expansion is the content of the following theorem.
2which may be empty!
134
A Friendly Approach to Complex Analysis
Theorem 4.7. If f is holomorphic in A:= {z E (C: r
then
f(z) =
where
(1) Cn
l
.
=2
1f'I,
1
C
L C (z - zo)
nEZ
n
f(()
,
(( - zo)n+l d(
(2) C is the circular path given by C(t)
(3) p is any number such that r
n
<
lz -z0/
for z EA,
= z0 + pexp(it),
< p < R.
<
R},
(4.5)
t E [O, 21r],
Moreover, the coefficients are unique in (4.5).
Example 4.11. Define f : (C \ {O} --+ (C by f (z) = z3 exp(l/z), z =/- 0.
Then f is holomorphic in A:= {z E (C: 0 < /zl < +oo}. So by the above
result, f must have a Laurent expansion
LCn Z n .
nEZ
In this case we can find the coefficients just by inspection: since
1
1
1
exp; = 1 + ; +
+ · · · , z =/- 0,
21z2
2
z
l
1
3
1
3
we have f(z) = z exp;
z + z + ! + ! + ! + · · ·, for z =/- 0.
4z
2
3
Consequently,
1
1
1
· · · , c_1 = , co = , c1 = , c2 = 1, c3 = 1,
1
1
2.
41.
3.
and Cn = 0 for n 2: 4.
◊
Proof. (of Theorem 4.7.) (Existence.) Fix z EA. Chooser and R such
that r <r < lz -zol < R < R. Let ,1 and ,2 be the circular paths
+ rexp(it),
,2(t) = zo + Rexp(it),
and 0:= Arg(z) + 1r/2. Let 13
')'1(t) =
fort E [0, 21r + 0],
ZQ
. z -zo
.
,3 (t) = ti I
z -zo I
:
[r, R] --+A be the path
(This is just a straight line path joining 11 and ,2, and the peculiar mul­
tiplication by i produces a rotation by 90 ° , ensuring that this path avoids
z!) See Figure 4.2.
135
Taylor and Laurent series
Fig. 4.2
Laurent series.
Clearly the path 'Y := "/2 -"/3 - "(1 + "/3 is A \ {z}-homotopic to a small
()
circle C8 centered at z. Also, f - is holomorphic in A\ {z}, and so
. -z
1
(
(
f( ) d( = f f( ) d( = (z) · 21ri,
f
z
z
(
(
J
c
0
7
where the first equality follows from the Cauchy Integral Theorem, and the
second equality follows from the Cauchy Integral Formula. Thus since the
contour integral along "/3 cancels with that along --y3, we obtain
1
(
!( ) d( = �
(z) = �
f
27ri "( ( - Z
27ri
=
1 - fl - 1
72
/73
(
(
f( ) d(-�
f( ) d(.
2m 171 ( - z
2m 172 (-z
�
"fl
+ f/ f(()
_
/73 (-Z
d(
(II)
(I)
We will see below that the integral (I) gives the term
(X)
Lcn(z-zot,
n=O
while the integral (II) gives the term
n
L C-n(z-zo)- ,
(X)
n=l
and these put together we will yield the desired Laurent series expansion
of .
f
136
A Friendly Approach to Complex Analysis
1
Step 1. In this step we will show that - .
21ri
We have for ( E "/2 that
!(()
( -z
!(()
( -zo +
zo-z
1 f�) f
"/2 <,
Z
d( =
n=O
cn(z-z0 )n.
!(()
!(()
((-zo)( 1-w) '
z-zo
((-zo) (1---)
(- Zo
z-zo
lz-zol
lz-zol
where w := = -- < 1, and so
--· We have lwl =
r
-zo
-zo
I(
.,
R
I
n
W
l
2
3
n-1
--=l + w + w + w + ···+ w
--.
+ l-w
l -w
Using this, we obtain
f(() = f(() (1 + w + ···+ wn+ _:!!:_)
l-w
( -zo
( z-zo
= !(() + ...+ !(() n(z _ zor-1 + f ( )( n t .
((-zo)
(-zo
((-zo) ( ( -z)
( -z
�1
Thus
f(() d(
21ri "12 (-z
where
=�1
�1
f(() d( - (z-zot-1
f(() d( + ...+
21ri "12 ((-zo) n
21ri "12 (-zo
l
n
f(()
.
+ 21ri "12 (( - zo)n(( -z) d( (z-zo)
=Co+ c1(z-zo) + ··· + Cn-1(z-zor-1 + Rn(z),
1
1
Rn(z) := _
21ri
1
f( ( )(z-zot d(.
zo)n( ( -z)
Here we have used the fact that "/2 is A-homotopic to any circle C with
center z0 and radius p where r < p < R, and so by Theorem 3.4,
1
"12 ((-
1
!(() d
!(() d
(
(
zo)k = c ((-zo)k
fork=l, ..., n -l. We would be done and obtain
"12 ((-
�1
21ri
f(()
"12 ( - Z
d(
=
f
n=O
= Ck
Cn(z-zot
if we show that lim Rn = 0. There is an M > 0 such that for all ( E "/2,
n-+oo
lf(()I < M. Moreover, I(-zol =Rand
I(-zl = 1(-zo-(z-zo)I 2: I(-zol - lz-zol = R
- lz-zol,
137
Taylor and Laurent series
and so
n
M
IR (z)I::; lz - zol
� O.
n
)
(
R-lz - zol
R
n
1
3
2
Thus co+ c 1(z - zo)+ c2 (z - zo) + c3 (z -zo) +· ·· = -.
1ri2
1
1
�1
l.
Step 2. We will show that-21ri
We have
__l
21ri
1
f(() d(
z
'°Yl (-
=
( -zo
Set w :=--. Then lwl =
z -zo
1
1
L
(() d
', 12
z
(.
f(()
oo
C- n (z - zo)-n.
-r- d( =
z
n=l
'°Yl ', -
= _1
1f
f(()
d(
z - Zo) - (( - Zo)
21ri
'°Yl (
21ri
'°Y1
f(()
(z - zo) (1-
r
< .1 Thus
z
zo I
I
( -zo
Z - Zo)
d(
W
n
----=--=l+w+w2 +w3 +···+wn-1 +--,
l
and so
l
27ri
_ ( - zo
z - zo
1
f(() d(
=
Z
1
�1
1
11 ( -
=
l-w
1- w
1
( (-zo)n
..+ ((-zor- +
(-1
·
+
) d(
f(() z -zo
(z-zo)n (z-()
(z-zo)n
11
1
l
d(·
f(()
d ·-- +... +
21ri 11 ((-zo)-n+l
21ri 11 f(() ( z-zo
(z-zo)n
_1
21ri
-l
+
21ri
11
/(()((- zo) d
(
(z- zo)n (z - ()
�1
n
= C-1(z - zo)-1+···+ C-( n -l)(z - zo)-(n -l)+ Rn (z),
1
where
!(()(( - zor d .
Rn (z) := _1
(
21ri 11 (z - zo)n (z - ()
Here we have used the fact that 'Yl is A-homotopic to any circle C with
center z0 and radius p where r < p < R, and so by Theorem 3.4,
1
1
f(() d
!(() d = C
k
( =
c ((- zo)k (
zo)k
'°Y1 ((-
A Friendly Approach to Complex Analysis
138
for k =-l, ..., -(n- 1). There is an M > 0 such that for all ( E 'Yi,
lf(()I < M. Moreover, 1(- zol =rand
iz-(I= l(z-zo )- ((-zo)I 2:: lz-zol-I(-zol= lz-zol -
_
and so
(
IRn(z)I::;
r
n
Iz-zo
I)
M
--+<X>
n
--+
Iz-zo I-r
o.
Thus C-1(z-zo)-1 + · · · + C-(n-I)(z-zo )-( n-l) + · · ·=-�
211'i
1
r,
f(() d .
(
z
"/1 ( -
This completes the proof of the part on the existence of the Laurent ex­
pansion.
Uniqueness of coefficients. Cauchy's Integral Formula allows us to show
that the Laurent expansion is unique, that is, if
f (z)= 2:cn(Z-zo r for r < lz-zol < R,
nEZ
d
(z-zo )n+l
then for all n, Cn = Cn , If n-/- l, then (z-zo )n= - ( ----). So
n+ 1
dz
l
(z-zo rdz
= 0 (n-/- 1),
where C is given by C(t)= zo +pexp(it), t E [O, 211']. By a direct calculation,
f
_l_ dz= f
lo
lc z-zo
21r
l ' irexp(it)dt= 211'i.
rexp(it)
Hence, if term-by-term integration is justified in the annulus, we would
have
{e: (z-zo )
lC
n
-m-1 �� Cn(z-Zo ) dz
nEZ
� Cn l{e: (z-Zo )n-m-l dz = 211'iC·-m ,
= �nEZ
C
and the claim about the uniqueness of coefficients would be proved. This
term-by-term integration can be justified as follows. We have
(
�
Cn(z-zo )n-m-l = .. ·
�-
nEZ
+
Cm-2
-
(z-zo) 3
+
Cm-1
-
(z-zo )2
)
+ (cm+l + Cm+2(z-zo ) + · · ·)
+�
z-z
o
Cm
= fi(z) +-+ fz(z).
Z-Zo
139
Taylor and Laurent series
We need only show that Ji, h have a primitive in the annulus and then
l
{ Lcn(Z -zor-m- dz
jC nEZ
=
r
j
C
(1i(z) + Cm
Z -Zo
+ h(z)) dz
= 0 + 21ricm + 0 = 21ricm,
00
as required. But h(z) = LCm+n (z -zor-1 for lz -zol < R, and so if
n=l
L Cm+ (z -zot
00
F2(z) :=
-
n
n=l
n
for lz -zol < R,
then : F2(z) = h(z), and so His a primitive of /2. For Ji, we have
z
00
-
00
� -Cm nWn+l,
_ � _Cm-nn+l _
-�
f1 (z) � (z zo)
where w
1
z -zo
= - -, and this is valid for R > lz -zol > r. So
1
converges for lwl < -. If we set
r
then
!
G(w)
=-
f
1
for lwl < -,
r
n l
Cm-n w - . Hence if define Fi by
n=l
f
1
Cm-n
(z -zo)-n for z EA,
F1 (z) = G (--) =
z -zo
n
n=l
then
00
= (z -zo)-2 LCm n (z -zo)-n+l = fi(z).
n=l
□
140
A Friendly Approach to Complex Analysis
Note that the uniqueness of coefficients is valid only if we consider a par­
ticular fixed annulus. It can happen that the same function has different
Laurent expansions, but valid in different annuli, as shown in the following
example.
Example 4.12. Consider
Fig. 4.3
f defined
IC\ {O, 1}, A:= {z E IC: 0
by
f(z)
< lzl <
=
1
zz-l
)
(
1} and A:= {z E IC: 1
f is holomorphic in the annulus A:= {z EC : 0
<
Laurent expansion there? Since lzl < 1, we have
1
f(z) =
=- 1 +z + z2 + z3 + · · · ) = -
!(
z(z-l)
z
and so the Laurent series coefficients are given by
C-2 =C_ 3
C-1 =Co
=
···
= C1
, z EC\ {0, 1}.
=
lzl
<
< lzl < +oo}.
1}. What is its
! - 1-z- z -z
2
z
3
-••• ,
0,
= · · · = -1.
But f is also holomorphic in the annulus A:=
{z EC: 1 <
lzl}, with inner
radius 1, and outer radius infinite. So f has a Laurent series expansion in
A too!
Since lzl
> 1,
we have
l
=
!
l
= 2_
1 + + 2_
z z2
z2 (
z2 (1-1/z)
1
1
= 2 + 3 +
z
z
....
Thus now the Laurent series coefficients are given by
f (z)
=
z(z-l)
+ 2_3 + · · ·)
z
c_z = c_ 3 = ... = 1,
C-1 =Co
= C1 =
· · · = 0.
So we notice that the coefficients are different, but this is not surprising,
since the annuli we considered for the Laurent expansions were different
too.
◊
Exercise 4.29. Reconsider Example 4.12, and now find Laurent expansions for f
also in the annuli A1 := {z E IC: 0 < lz-11 < 1} and A1 := {z E IC: 1 < lz-11}.
141
Taylor and Laurent series
4.8
Classification of singularities
If we look at the three functions
sinz
z
1
z3
'
1
exp-,
z
then we notice that each of them is not defined at 0, and refer to O as a "sin­
gularity" of these functions, because the function is not defined there. But
we will see that each of these functions behave very differently near their
common singularity. In other words, the "nature of the singularity" differs
in each case. We will explain precisely how the behaviour is different in
each case, and this is what we mean by classification of singularities. More­
over, we will learn two results, which will allow us to find out the type of
singularity at hand. Of these two characterization results for singularities,
one result will be in terms of limits, while the other will be in terms of what
happens with Laurent coefficients. We first give the following definition.
Definition 4.4. Let f be a complex valued function which is not defined
at a point z0, and suppose that it is holomorphic in some punctured disc
{z E <C: 0 < lz - zol < R} centered at zo with some radius R > 0. Then
we call zo an isolated singularity of f.
Example 4.13. For example, each of the functions
sinz
z
1
'
z3
1
exp-,
z
has an isolated singularity at 0. On the other hand, f given by
f(z) :=
1
sin (l/z)
has a singularity at 0, but it is not an isolated singularity. (At z =l/mr,
n E Z, the function f is not defined.)
◊
Definition 4.5. An isolated singularity z0 off is called
(1) a removable singularity off if there is a function F, holomorphic in
the disc {z E <C: lz - zol < R} such that F =fin the punctured disc
{z E <C: 0 < lz - zol < R}.
( )I = +oo, that is,
2
( ) a pole offif lim lfz
z➔zo
for all M > 0 there is a 8 > 0 such that whenever O < lz - zol
( )I > M.
lfz
3
( ) an essential singularity off if z0 is neither removable nor a pole.
<
8,
142
A Friendly Approach to Complex Analysis
Example 4.14.
(1) The function f given by f(z)
since for z =/=- 0, we have
=
sin z
has a removable singularity at 0,
z
l
sin z
z3 z 5
-- =; ( z - ! +
3
z
5! - + · · ·
) =];
00
t
( -1
2n
z
2n
'
( + l )!
and the right hand side, being a power series with an infinite radius
of convergence (why?) defines an entire function F. Since this entire
function F coincides with the given function f in the punctured plane
(C \ {0}, it follows that f has a removable singularity at 0.
1
(2) The function \ has a pole at 0, since lim - 3 = +oo.
z--+0 Z1 1
Z
!
(3) The function exp has an essential singularity at 0. Indeed,
z
(a) 0 is not a removable singularity, because for example lime½ = +oo.
X',,0
1
x
(b) 0 is also not a pole, since lime = 0, and so it can't be that
x/'O
lim lf(z)I = +oo.
◊
z--+0
We will now learn our first characterization result for singularities, in terms
of limiting behaviour.
Theorem 4.8. ( Classification of singularities via limits) Suppose z0
is an isolated singularity of f. Then
z0 is removable
{cc}
zo is a pole
{cc}
lim (z - zo)f(z) = 0.
z➔zo
(a) ,( lim (z - zo)f(z) = 0) and
z➔zo
(b) =in EN such that lim(z
-zot + 1 f(z) = 0.
a
z--+z
zo is essential
Proof.
I
{cc}
( The smallest such n is called
the order of the pole zo of f.)
VnEN,( lim(z- zorf(z)=O).
(1) z0 removable==>
z➔zo
z
��
a
(z - zo)f(z) = 0
I·
Let z0 be removable, and let F be holomorphic in
D(zo, R) := {z EC: lz - zol < R}
Taylor and Laurent series
143
such that F = f for O < lz - zol < R. Then using the fact that F is
continuous at z 0, we obtain
lim (z-zo)F(z) = lim (z - zo)F(z) = 0· F(zo) = 0.
z➔ zo
z➔zo
Hence for everyE > 0there exists a 8 > 0such that for O < lz - zol < 8,
there holds that
and so
l(z -zo)f(z) - OI =l(z -zo)F(z)-OI <E,
lim (z-zo)f(z) = 0.
z➔zo
Next we will show that lim (z - zo)f (z) = 0 =} zo is removable .
z➔zo
Suppose that
lim (z-zo)f(z)
z--+zo
= 0,
and that f is holomorphic in the punctured disc {z EC: 0< lz-zol< R}.
Then f has a Laurent expansion there:
f(z) =
L C (z-zo) ,
n
n
nEZ
0< lz -zol< R,
where the coefficients Cn 's are given by contour integrals along any circle
Cr with center z 0 in the punctured disc and r < R:
1
f(z)
dz,
n
Cr (z-zo) +l
We show that C-n = 0for n EN. Given E > 0, take r small enough so that
l(z - zo)f(z)I <Eon Cr. Then we have for n ENthat
Cn
1
I C- n I = 121ri
= 2_l_
1ri
1
1-1
1
(z - zo)f(z) d
f(z)
d
zl
n+l zl = 21ri
(z
- zo)- n +2
(z
-z
o
)Cr
Cr
1
E
n 1
n
< ---2
1r r = Er - < 1:R -l_
21r r-n +2
Since the choice ofE> 0was arbitrary, it follows that c_n = 0for all n EN.
Consequently, with
(X)
n=O
we see that Fis holomorphic in {z E C: lz - zol < R}, and F=fin the
punctured disc {z EC: 0< lz - zol< R}.
144
A Friendly Approach to Complex Analysis
(2) zo is a pole ::::} {
,( lim (z - zo)f(z)
z--+zo
= 0)
and
:3n EN such that lim (z - zor+l f(z)
z➔za
= 0.
Suppose that z0 is a pole of.f. Then z0 is not removable (why?), and so by
the first part, it follows that
,( lim (z - zo)f(z)
z➔zo
= o).
(Here ---, is the symbol for negation, to be read as "it is not the case that".)
There is some R > 0 such that lf(z)I > 1 in the punctured disc
D := {z E (C: 0
< lz - zol < R}.
Define g in this punctured disc D by
g(z) =
1
f(z) ·
Since z0 is a pole off, it follows that lim g(z)
z➔zo
lim (z - zo)g(z)
z➔�o
= 0.
In particular, also
= 0,
and so by the first part above, g has a holomorphic extension G to
{z E (C: lz - zol < R}.
Also,
G(zo) = lim G(z) = lim g(z) = 0.
z--+zo
z--+zo
So z0 is a zero ofG, and since G is not identically zero in a neighbourhood
ofz0, it follows from the result on the classification ofzeros that z0 has some
order n EN, and there is a holomorphic function H defined for lz- zol < R
such that H(zo) =/- 0 and
G(z) = (z - zo) n H(z).
In particular, for 0
< lz - zol < R, we have
1
1
1
f(z) = g(z) = G(z) = (z - zo) n H(z) ·
Hence
0 - 1·Im (z - zo) 1Im
. (z - z0) n +lj (z) 0
.
z--+zo H( Z)
H(zo)
z--+zo
Done!
145
Taylor and Laurent series
o)
• lim (z - zo)f(z) =
and
z--+zo .
(
3n EN such that lim (z - z0 )n+l f(z) = 0.
z--+zo
==> z0 is a pole.
}
Choose the smallest such n, call it n*: thus
lim (z - zo)n .+l f(z) = 0,
z➔zo
o).
· lim (z - zo)n * f(z) =
z--+zo
(
So (z - zor• f(z) has a removable singularity at z = zo. Thus there exists
an F, holomorphic in {z EC: lz - zol < R} such that for 0 < lz - zol < R,
(z - zo)n • f(z) = F(z).
Note that
F(zo) = lim F(z) = lim (z - zor• f(z)-=/- 0
z➔zo
z➔zo
(owing to our choice of n*.) From
F(z)
fz ( ) (z - zo)n•
for 0 < lz - zol < R, we obtain
lim IJ(z)I = lim
z--+zo
z--+zo
So zo is a pole off.
IF(z)I
1
= +oo.
= IF(zo)I · lim
n
z--+zo
z
ZoI n .
lz - zol · '--.,.--'
I
,60
(3) This follows easily from the first two parts. Indeed, if z0 is an essential
singularity of f, then it is not removable, and so
o),
• lim (z - zo)f(z) =
z--+zo
(
and so we get the desired claim at least for n = 1. But since z0 isn't a pole,
by (2) we also get the desired conclusion for all other n's as well.
Conversely, if
o) ,
for all n EN, · lim (z - zorf(z) =
z--+zo
(
then in particular, from the n = I case, we get by (1) that zo can't be
a removable singularity. And from the cases for all other n ?: 2, we can
conclude, in light of (2), that zo can't be a pole either. This zo must be an
essential singularity of f.
□
146
A Friendly Approach to Complex Analysis
Let us reconsider Example 4.14.
Example 4.15.
sin z
= z--+0
lim sin z = sin 0 = 0, and so 0 is a removable
z--+0
z
sinz
.
.
smgulanty of -- .
z
1
and lim z 4 • \ = lim z = 0, 0 is a pole of 3, and
lim z · \ =
(2) As , ( z--+0
Z
z--+0
Z
z--+0
Z
1
moreover the order is 3 since , ( lim z2 • 3 =
and , ( lim z3 • \ =
z--+0
z--+0
z
z
1
1
1
1
1
+
+
-+
-For
x
>
0
ex
=
1
+
•
•
•
>
-Thus
for all
(3)
3!x3
n!xn
'
l!x 2!x 2
1
1
n EN, xn e;;; > 1 and
n.
(1) We have lim z •
o)
o)
o).
So for all n EN,
, ( lim Z n exp!
Z
z--+0
= o).
1
Thus 0 is an essential singularity of exp -.
z
◊
We will now learn our second characterization result for singularities, in
terms of the Laurent series coefficients.
Theorem 4.9. (Classification via Laurent coefficients) Let
I: c (z - zot for 0 < lz -zol < R, for some R > 0.
(1) zo be an isolated singularity of f, and
(2) f(z)
Then
=
nEZ
n
zo is removable
zo is a pole
z0 is essential
{=}
{=}
{=}
For all n < 0, Cn = 0
There exists an m EN such that
(a) C-m i- 0 and
(b) for all n < -m, Cn = 0
Then the order of the pole zo ism.
There are infinitely many
negative indices n such that Cn i- 0.
147
Taylor and Laurent series
removable
pole
-m
essential
Proof.
0
infinitely many nonzero
I
(1) zo is removable =? (for all n
< 0,
Cn
= 0)
I-
Suppose that z0 is a removable singularity. Then f has a holomorphic
extension F defined for lz - zol < R. But then this holomorphic F has a
Taylor series expansion
00
F(z)
In particular, for O
= Lcn(Z - zo) n
n =O
< lz - zol < R,
for lz-zol
< R.
we obtain
L Cn(z-zo) = L C (z-zo) ,
00
f(z) =
n
n =O
nEZ
n
n
and by the uniqueness of the Laurent series expansion in an annulus, it
follows that Cn = Cn for n � 0, and Cn = 0 for all n < 0.
I
(For all n
< 0,
Cn
Suppose that Cn
= 0)
=? z0 is removable
= 0 for all n < 0.
1·
Then for O
<
lz - zol
< R,
00
nEZ
and so defining F for I z - zo I
F(z)
n =O
< R by
=
L C (z - zo) ,
00
n =O
n
n
we see that Fis holomorphic in {z E C : lz - zol < R} (because it is a
power series!) and moreover F = f for O < lz - zol < R.
(2) We will show that if z0 is a pole of order m, then
f(z )
=
C-m
(Z - Zo ) m
00
C-1
�
(
)n
+···+--+co+
�en z-zo ,
Z - Zo
n =O
148
A Friendly Approach to Complex Analysis
with C- m =/- 0. Suppose zo is a pole of order m. Using Theorem 4.8,
+
lim (z - Zo)((z - zo)m f(z)) = lim (z - zor l f(z) = 0,
z-+zo
z-+zo
the function (z - zorf(z) has a removable singularity at zo. We have
+
(z - zorf(z) = (z - zo)m
Cn(z - zo)n =
Cn(Z - zo)n m ,
L
L
nEZ
nEZ
and by the previous part, this last series has all coefficients of negative
powers of z - zo equal to 0, that is, 0 = c-(m+ ) = c-(m+2) = · · ·. Hence
l
for O < I z - zo I < R,
(z - zo)m f(z) = C- m + C-m+1(z - zo) + C-m +2(z - zo)2 +···
(4.6)
and so
c 1
C- m
+co +c1(z - zo) +c2(z - zo)2 +···.
+···+ _
(z - zo)
(z - zo)m
Moreover, the function C-m + C-m+1(z - zo) + C- m +2(z - zo)2 + ··· is
holomorphic, and so it has the limit C- m as z-+ zo. Thus
f(z)=
lim(z-zorf(z) = lim(c_ m +C-m +1(z-zo)+c_ m +2(z-zo)2 +- ··) = C- m ·
z-tzo
z-+zo
On the other hand, since z0 is a pole of order m, we know that
•( lim (z - zo)m f(z) =
z-+zo
o).
Consequently, (4.6) implies that C- m =/- 0.
Suppose now that
f(z) =
L
c m
( Z -_ Zo
+···+�+co +
Cn(z - zo)n,
Z - Zo
n=O
with c_ m =/- 0. We will show that z0 pole of order m.
Suppose that there is some m E N such that C- m =/- 0 and Cn = 0 for all
n < -m. Then (z-zorf(z) = C- m +C- m +1(z-zo)+c_ m +2(z-zo)2 +···,
and since the right hand side defines a holomorphic function, say h, in
{z E <C: lz - zol < R}, it follows that
)m
lim (z - zo)m f(z) = lim h(z) = h(zo) = C- m =/- 0,
z-+zo
m+
lim (z - zo) l f(z) = 0 · C- = 0.
z-+zo
z--+zo
and
m
Thus zo is a pole of order m off, by Theorem 4.8.
(3) This is immediate from the previous two parts and the fact that an
D
essential singularity is neither a removable singularity nor a pole.
149
Taylor and Laurent series
Let us reconsider Example 4.14.
Example 4.16.
(1) We have for z
i- 0,
si z
: = � (z _ ;� + �: _ ;� + _ ...) = 1 _ ;� + �: _ ;� + _ ... .
Since there are no negative powers of z appearing in the Laurent series
sinz
.
.
.
.
.
expansion, 1t follows that 0 1s a removable smgulanty of --.
(2) For z
z
i- 0,
1
1
- =···+0+-+0+··· .
z3
z3
Thus c_3 = 1
order 3.
(3) For z 0,
i-
i- 0, while 0 = c_4 = c_5 =··· . So 0 is a pole of z\ of
1
1
1
1
exp-;; = 1 + l! + 2! 2 + ! +··· ·
z
z
3 z3
For infinitely many negative indices n (in fact for all of them),
C-n
1
=I-/- 0.
n.
Thus 0 is an essential singularity of exp .!_.
z
◊
Exercise 4.30. Let D be a domain and f be holomorphic in D such that f has
only one zero zo in D, and the order of zo is m > 1. Show that the function
z f---t 1/f(z) has a pole of order matzo.
Exercise 4.31. Let D be a disc with center zo. Suppose that f is nonzero and
holomorphic in D \ {zo}, and that f has a pole of order mat zo. Show that the
function z f---t 1/ f (z) has a holomorphic extension g to D, and that g has a zero
of order m at zo.
Exercise 4.32. Let D be a domain and zo ED. Suppose that f has a pole of order
m at zo and that f has the Laurent series expansion
f(z) =
L:Cn(z
nEZ
- zof
for O
< lz - zol < R,
where R is some positive number. Show that
C-1
1
dm-1
lim - _1 ((z - zo) m f(z)) .
= (m- l '. z--+zo
z d
)
m
A F'riendly Approach to Complex Analysis
150
Exercise 4.33. True or false?
(1) If f has a Laurent expansion z-1 +eo+c1z+- ••,convergent in some punctured
disc about the origin, then f has a pole at 0.
(2) A function may have different Laurent series centered at zo, depending on
the annulus of convergence selected.
(3) If f has an isolated singularity at zo, then it has a Laurent series centered at
zo and is convergent in some punctured disc O < [ z - zo[ < R.
(4) If a Laurent series for f convergent in some annulus R1 < [z - zo[ < R2
is actually a Taylor series (no negative powers of z - zo), then this series
actually converges in the full disc given by [z - zo[ < R2 (at least).
(5) If the last conclusion holds, then f has at worst removable singularities in
the full disc given by [z - zo[ < R2 and may be considered holomorphic
throughout this disc.
Exercise 4.34. Decide the nature of the singularity, if any, at O for the following
functions. If the function is holomorphic or the singularity is isolated, expand
the function in appropriate powers of z convergent in a punctured disc given by
0 < [z[ < R.
. 1
sinz, sm ;;,
1
sin l'
z
sinz
z
. 1
z sm­
z
Exercise 4.35. True or false?
I
!I=
(1) lim exp
+oo.
z➔O
Z
(2) If f has a pole of order m at zo, then there exists a polynomial p such that
fhas a holomorphic extension to a disc around zo.
(z -pzo)m
(3) If f is holomorphic in a neighbourhood of 0, then there is an integer m such
that 1._ has a pole at 0 whenever n
zn
> m.
(4) If f, g have poles of order m f,mg respectively at zo, then their pointwise
product f · g has a pole of order m1 + mg at zo.
Exercise 4.36. Give an example of a function holomorphic in all of (C except for
essential singularities at the two points 0 and 1.
Exercise 4.37. The function f given by f(z) = l/(z - 1) clearly does not have a
singularity at 0. As it has the Laurent series z-1 + z-2 + z-3 + • • • for [z[ > 1,
one might then say that this series has infinitely many negative powers of z, and
fallaciously conclude that the point 0 is an essential singularity of f. Point out
the flaw in this argument.
Exercise 4.38. Prove or disprove: If f and g have a pole and an essential singu­
larity respectively at the point zo, then fg has an essential singularity at zo.
Taylor and Laurent series
151
Wild behaviour near essential singularities
4.8.1
We now show a result which illustrates the "wild" behaviour of a function
f at its essential singularity
any E
>
zo.
It says that given any complex number
w,
z 0, there
w. So the
0, and any arbitrary small punctured disc� with center
is a point
z
in � such that
lies within a distance E from
f(z)
image of any punctured disc centered at the essential singularity is dense
in C. Or in even more descriptive terms, f comes arbitrarily close to any
complex value in every neighbourhood of
zo.
Theorem 4.10. ("Casorati-Weierstrass" 3) Suppose z 0 is an essential
singularity of f. Then
(1) for every complex number w,
(2) for every 8 > 0, and
(3)
for every
E
> 0,
there exists a z EC such that lz - zol
8 \
.
zo <>----------;
2'.,/
Proof.
and 8
8
and lf(z) - wl
•
f
�
<
E.
f(z)
wa----�
E
'
Suppose the statement is false. Then there exist
w E
C, E
> 0 such that whenever z ED:= {z EC: 0 < lz - zol < 8},
lf(z) - wl
for
<
z ED
2: E. Then
g
defined by
g(z)
1
=
f(z) -
w
is holomorphic there, and lim (z z➔zo
l z
O:; g( )I =
1
lf(z) - wl
1
zo)g(z)
:; �)
>
0
we have
= 0, since
zED.
3This result was published by Weierstrass in 1876 (in German) and by the Sokhotski
in 1873 (in Russian), So it was called Sokhotski's theorem in the Russian literature and
Weierstrass's theorem in the Western literature. The same theorem was published by
Casorati in 1868, and by Briot and Bouquet in the first edition of their book (1859),
called Theorie des fonctions doublement periodiques, et en particulier, des fonctions
elliptiques. However, Briot and Bouquet removed this theorem from the second edition
(1875).
152
A Friendly Approach to Complex Analysis
g has a removable singularity at zo. Let its extension be denoted again
g. Let m be the order of the zero of g at zo. (Set m = 0 if g(zo) =/- 0.)
Then g(z) = (z - z0) m h(z), for some function h holomorphic in D and such
that h(zo) =/- 0. Then for 0 < lz - zol < 8
So
by
(z - zo)m + 1 (J(z) - w + w)
(z - zo) m +l f(z)
1
(z - zo)m +l __ + (z - zo) m +l · w
g(z)
z - zo
+ (z - zo)m +l . w
h(z)
z-tzo
0
+ 0 . w = 0.
--t h(zo)
zo (when m = 0) or a pole
z0 ( when m E N). Hence zo can't be an essential singularity of f, a
Thus either f has a removable singularity at
at
□
contradiction.
Example 4.17. The function exp(l/ z) has an essential singularity at 0.
We show that it takes on any given nonzero w (= p exp(i0) E <C,
p,0 E JR),
in any arbitrarily small neighbourhood of 0. Setting z = rexp(it), we need
to solve
1
cost
sint
exp - = exp (-- - i--) =
r
z
r
-
pe'0 .
By equating the absolute values, we obtain
cost
-- = 1ogp.
r
On the other hand, by looking at arguments, we see that a solution is given
when
_
Using (cost)2
+ (sint)2 = 1,
sint
r
= 0_
we have
1
r=--;=====�
J(logp)2
+ 02
Moreover
tant = ---.
logp
But we are allowed to increase 0 by integral multiplies of 21r, without chang­
ing w. Bearing this in mind, it is clear from the above expression for r that
we can make r as small as we please.
◊
153
Taylor and Laurent series
The above example illustrates a much stronger theorem than the Casorati­
Weierstrass Theorem, due to Picard, which says that the image of any
punctured disc centered at an essential singularity misses at most one point
of IC! In our example above, the exceptional value is w = 0. A proof of
Picard's Theorem is beyond the scope of these notes, but can be found in
the book [Conway (1978)].
Exercise 4.39. Prove, using the Casorati-Weierstrass Theorem, that if f has an
essential singularity at zo, and if w is any complex value whatever, then there
exists a sequence z1, z2, Z3, · · · such that
lim f (zn) = w.
lim Zn = zo and
n--+cx:i
n--tcx:i
4.9
Residue Theorem
Suppose that D is a domain and that a holomorphic f : D \ { z0} ---+ IC has
an isolated singularity at zo. Let
J(z) =
L Cn(z - zo)
nEZ
n
for O < lz - zol < R.
Then we call the coefficient C- 1 the residue off at zo, and denote it by
res(!, zo)-
Why the name? In everyday language, "residue" means something which
is "left over". In the above too, one can think of c_ 1 as something which
is left over in the following manner. We know that
(
f(z)dz =
21ric_ 1 =
Cn (Z - zo)n )dz,
� O�� 1 + 1 dz=
(z
Cr nEZ
Cr
lcr
where Cr is given by Cr(t)= z0 +rexp(it), E [0,21r], and r < R. Note that
we have
0 for n-=/=- -1,
Cn Z-Zo n )dz= {
2.
1ric_ 1 £or n _
- - 1,
Cr
1
1 (L
r
1(
(
)
and so if we imagine integrating termwise formally in the Laurent series
expansion off, then 21ric_ 1 is what is left over.
What is the big fuss about this residue? The equation
21ric_ 1
=
1
f(z)dz
Cr
gives a way of computing contour integrals via calculating the residue of
f at z0 (which amounts to finding the value of the coefficient c_1 in the
154
A Friendly Approach to Complex Analysis
Laurent expansion off). So if there is way of calculating c_1 easily, then
we can compute
1
'Y
f(z)dz ( = f f(z)dz) = 21ric_1
lcr
(4.7)
for any closed path I which is D\ {zo}-homotopic to Cr. And we have seen
in Exercise 4.32 that there is a way of calculating c_l via the formula
dm-1
1
c_l =
' lim - m _1 ((z - zo)m f(z)),
(m-1.) z ➔ zo dz
when z0 is a pole of order m off. It turns out that some awkward real
integrals can be computed by first relating them to a contour integral
for an appropriate holomorphic f and some path 1, and then using this
route via the residue to evaluate the contour integral, and eventually also
the awkward real integral. Here is an example.
1
Example 4.18. Consider the real integral
1
3cos
5
0
0
+
We view this as the contour integral along a circular path as follows. First
we write
1
z+ �
exp(i0) + exp(-i0)
z2 + 1
cosB =
=
=
2
2z '
2
271"
---d0.
where z := exp(i0). So if, is the circular path with radius 1 and center at
0 defined by
1(0) = exp(i0),
then ,'(0)d0 = iexp(i0)d0
f 21r
}0
1
d0 =
5+3cos0
1
'Y
0 E [O, 21r],
= izd0, and so
l
- z2
5+3 (
1
:1) iz
dz =
2
Let f be the function defined by
2i
f(z) = - (3z + l)(z + 3 ·
)
1'Y
2i
(3z+l)(z+3)
dz
·
155
Taylor and Laurent series
Then f has two poles, one at -1/3 and the other at -3, both of order 1.
Of these, only the one at -1/3 lies inside '"Y· Thus,
1271"
0
1
d0 =
5+
3 cos 0
1
'Y
f(z)dz = 21ri · res
(1, _!3 ) •
r
What is c_1 ? We have in a punctured disc with center -1/3 that
f ( Z)
=
C l
� !
Z
+
Co +
C1
(Z
2
+
�) +
C
3
(Z
+
�
... ,
+
and so (z +
co (z +
.... Thus
�) f(z) = c_ 1 +
�) +
C -1
Consequently,
.
1 Im
1271"
=
z➔-1/3
o
1)
- !( Z) =
(Z +
3
1
1·Im
z➔-1/3
7r
d0 = -.
5 +cos
3
2
0
2i
i
= --.
3(z + 3)
4
----
◊
More generally, if f has a finite number of poles in D, a result similar to
(4.7) holds, and this is the content of the following.
Theorem 4.11. (Residue Theorem) Let
(1) D be a domain,
(2)
(3)
(4)
(5)
f be holomorphic in D\ {P1, ... ,PK},
f have poles at pl,···,PK oforder m1,••·,mK, respectively,
'"Y be a closed path in D\ {p1, ... ,PK} and
"( be such that for each k = l, ..., K, "( is D \ {Pk}-homotopic to a
circle Ck centered at Pk such that the interior ofCk is contained in D
and contains only the pole Pk.
Then
1
'Y
K
f(z)dz = 21riLres( f,pk)k=l
156
Proof.
A Friendly Approach to Complex Analysis
For each k = l, ..., K, we can write
f(z) =
=
L
C-n,k(z - Pk)-n + L Cn,k(z - Pkt= fk(z) + hk(z),
n=l
n=O
mk
0 < lz - Pkl < Rk, for some Rk > 0, where the sum with negative powers
of z - Pk is denoted by fk, and the sum with nonnegative powers of z - Pk
is denoted by hk. Note that hk is holomorphic for lz - Pk! < Rk, and that
fk is a rational function, defined in <C \ {pk}, having only one singularity,
namely a pole at Pk· Thus f - !k is holomorphic in a small disc around Pk·
Set g := f - (!1 +···+ fK)- Fix a k E {1, ..., K}. Write
g = (f - fk) k
L
j#
Observe that both f - fk and each Ji for j -/- k is holomorphic in a small
disc around Pk· Thus g is holomorphic in a small disc around Pk· This
happens with each k E {1, ..., K}. Thus, g is holomorphic in D. We note
that as 'Y is D \ {pi}-homotopic to a circle 01 centered at p1, by the Cauchy
Integral Theorem,
that is,
So
j
"I
i
i
g(z)dz = 0,
(!-(Ji+···+ fK))dz = 0. By the Cauchy Integral Theorem,
1
fk(z)dz =
"/
f(z)dz =
1
f(z)dz = 2nic-1,k,
k = l, · · · , K.
j fk(z)dz = Lk=l 2nic_1,k = 2ni Lk=l res(f,Pk)L
k=l
Ck
K
K
K
"I
o
Exercise 4.40. Evaluate/ L g(z) dz along the path I shown below.
"/ 1 + expz
-5+10i
10+10
v
5+5i
-5+5i
-5-2i
,
,
,
I
-5-5i
5-5i
�
10-2i
□
157
Taylor and Laurent series
Integral of a real rational function using the Residue Theorem.
As we mentioned earlier, the Residue Theorem can be used to calculate
contour integrals, and sometimes gives an easy way to calculate some real
integrals. Let us see how it can be used to calculate the improper integrals
of rational functions. Consider a real integral of the type
1_:
f(x)dx.
Such an integral, for which the interval of integration is not finite, is called
an improper integral, and it is defined as
100
-oo
1
a--+-oo l
f(x)dx = lim
°
f(x)dx + lim
b--++oo
a
b
l J(x)dx,
lo
(4.8)
when both the limits on the right hand side exist. In this case, also
1_:
f(x)dx = rE/_f!- 1:r f(x)dx.
00
(4.9)
We call the right hand side in (4.9), if it exists, the Cauchy principal value
of the integral. (However, it can happen that the Cauchy principal value
exists, for example,
lim
r--++oo
J
r
-r
r
xdx = lim (
r--++oo
2
-
2
r2 )
2
= 0,
2
b
but the improper integral doesn't exist: l xdx = b .)
2
lo
We assume that the function fin (4.8) is a real rational function whose
denominator is different from O for all real x, and the denominator has
degree at least two more than the degree of the numerator. Then it can
be seen that the limits on the right hand side in (4.8) exist, and so we can
start from (4.9).
Consider
l
f(z)dz, around a path, shown in Figure 4.4.
(j
-r
0
r
Fig. 4.4 'Y consists of the semicircular arc u and the straight line joining -r tor.
158
A Friendly Approach to Complex Analysis
Since
f
is rational,
1
f(z)dz =
H
z
f (z)
has finitely many poles in the upper half
plane, and if we choose r large enough, 'Y encloses all of these poles in its
1
interior. By the Residue Theorem, we then obtain
'Y
CT
f(z)dz +
r
J
-r
f(x)dx = 21ri
L
k: lm(pk)>0
res(f,Pk),
where the sum consists of terms for all the poles that lie in the upper
half-plane. From this, we obtain
r
J
f(x)dx = 21ri
-r
L
k: Im(pk)>O
res(f,Pk) -
1
CT
f(z)dz.
We show that as r increases, the value of the integral over the corresponding
semicircular arc u approaches 0. Indeed, from the fact that the degree of
the denominator off is at least two more than the degree of the numerator,
it follows that there are
Hence for r
>
Consequently,
ro,
lim
1=
r-++=
1
-=
a
M, ro
large enough such that
M
lf(z)I <
ll
(lzl > ro).
W
f(z)dzl � � 1rr = � .
1r
f(z)dz = 0,
and so
f(x)dx = 21ri
L
k: lm(pk)>O
res(f,pk)-
Let us see an example of this method in action.
Example 4.19. We will show that
The function
i
p1 = exp (7r ) ,
4
f
given by
f (z)
1
1
-- dx
O (X) 1 + X 4
=�
l+z
= 7r1n·
2y2
has four poles of order 1:
31ri
51ri
P2 = exp ( ) , p3 = exp ( ) ,
4
4
1
p4
= exp (
71ri
4) .
159
Taylor and Laurent series
The first two of these poles lie in the upper half plane. We have
res(f,Pl )
=
= z➔
r1mp1 z1+- Pi
Z
4
1
res(f,p2 ) = �
z �2
=4
z-p2
= z��2
1+ z4
4
1:
As
1
f is
21ri (
4
even
-i
(f(x) = f(-x)
/
00
1
exp (
i
:)+
for all
x
1
E
pf
41
= - exp
(1+ z4 )
4�
=
!(1 +z t=p2
: x dx =
z -pi
1
1
-
(7ri)
4
i
iexp ( -
exp ( - : ))
JR),
00
�
=
i
we have
'
(1+ p�)
Z -p2
___1_�_
Thus
1
(1+ Z 4) - (1+ P4i )
r
1m
z➔
p1
i
: ).
= 7r sin
i = �.
7r
1
dx = 1
2 _ 00 1 + x4 dx = 2v12·
lo 1 + x4
◊
Here is an example of the computation of an exotic integral where the
integrand is not a rational function.
Example 4.20. (Fresnel Integrals4 ) We will show that
We consider
1
f
lo
00
cos(x 2 )dx =
exp(iz2 )dz,
f
lo
00
sin(x )dx
2
= �2v2
where, is shown below.
lzl =R
4These integrals arise in optics, in the description of diffraction phenomena.
160
A Friendly Approach to Complex Analysis
i
Since exp(iz 2 ) is entire, by the Cauchy Integral Theorem we have
0=
=
1
exp(iz 2 )dz
R
exp(ix 2 )dx +
-1
R
14
1!:
exp(iR2 exp(2i0))iR exp(i0)d0
exp (it2 exp (ii)) exp (i�)dt.
We will show that the middle integral goes to 0 as R -+ oo. First note that
I
jexp(iR2 exp(2i0))iRexp(i0) = jRexp(R2 (i cos(20) - sin(20)))
= Re -R
2 sin(20)
I
.
But Figure 4.5 shows that whenever the angle t is such that 0 < t < i,
2
sin t
-<-.
1f t
Indeed, the length of the arc PAQ, which is 2t, is clearly less than the
length of the semicircular arc PBQ, which is 1r sin t, and so the inequality
above follows. (Alternately, we could note that t H sin tis concave in [0, 1r],
because its second derivative is - sin t, which is nonpositive in [0, 1r], and so
the graph of sin t is lies above that of the straight line graph of 2t/1r joining
the two points (0, 0) and (1r/2, 1).)
p
B
Q
Fig. 4.5 Here P is any point on the circle with center O and radius 1 such that OP
makes an angle t with the positive real axis. We reflect P in the real axis to get the
point Q, and let M be the intersection of PQ with the real axis. With Mas center and
radius PM, we draw a circle, meeting the real axis on the right of A at the point B.
Taylor and Laurent series
= 20 yields
exp(2i0) )iR exp(i0) I = Re-R
Applying this inequality with t
!exp(iR
2
and so
I 1 ,,.
4
I
2
exp(iR exp(2i0))iRexp(i0)d0 s;R
which tends to O as
lim
R-H oo
f
R
lo
R➔
1 ..
4
2
161
2
sin( 2 6) s;Re-R �,
20
7r
2
e-4R nd0=
(l-e-R ),
4R
+oo. Hence we obtain
exp(ix )dx
2
=
f
lim
R
lo
R-H oo
i
= 1+ f
v2 lo
oo
2
exp (it exp (i�)) exp (i�)dt
2
4
e-t
Here we have used the known 5 fact that
f
e_"'
oo
lo
dx
2
=
2
dt
= (1 + i)-Jrr_
2v2
J1r.
2
So by equating real and imaginary parts,
f
00
lo
Exercise 4.41. Evaluate
1
=
2
cos(x )dx
1
o
f
00
lo
sin(x )dx
2
= �2v2
◊
2
"
cos0
5 + 4 cos0
d0.
Exercise 4.42. Evaluate the following integrals:
(1)
(2)
()
3
(4)
00
o
1
- -2 dx.
1 +x
roo
roo (1
1
1
(a2 + x2 )(b2 + x2) dx,
lo
lo
00
o
1
+ x2)2 dx
where a> b > 0.
.
1 +x 2
.
1 +x 4 dx
Exercise 4.43. If n E N and C is the path C(t) = exp(i0),
2
{ e:�: dz. Deduce that f " ecose cos(n0 - sin0)d0 = 2�.
kz
5With J
h
:= o
l
[
00
2
e-x dx, 1 2
=
=
(
f 00
l/l
e-x dx
121
0
2
00
0
2
) (
�
f 00
lo
2
e-Y dy
7l'
e-r rdrd0 = -.
4
)
0
E [O, 271'], find
2
f 00
2
= [o lo e-(x +Y ldxdy
l
00
162
A Friendly Approach to Complex Analysis
Exercise 4.44. Let f have a zero of order 1 at zo, so that 1/f has a pole of order
1 at zo. Prove that res (y, zo)
=
'� .
f ( o)
1
Exercise 4.45. Prove that res (-.- -,k1r)
smz
=
(-l)k .
Exercise 4.46. The nth Fibonacci number fn, where n � 0, is defined by the
following recurrence relation: Jo = 1, Ji = 1, fn = fn-1 + fn- 2 for n � 2.
Let F(z) :=
L fnz .
(X)
n
n=O
(1) Prove by induction that fn S:: 2n for all n EN.
( 2) Using the estimate fn S:: 2n , deduce that the radius of convergence of Fis at
least 1/2.
_: _ 2•
1
z
Hint: Write down the Taylor series for zF(z) and z F(z) and add.
(3) Show that the recurrence among the fn implies that F(z) =
2
(4) Verify that res ( n+l l � _ 2 , 0)
z
(
z z )
(5) Using the Residue Theorem, prove
fn
=
Js ( (1
= fn •
+2�r
+l
- c �)
2
n +l
) ·
1
around a circle with center O and radius R
zn+l(l-z-z2 )
and show that this integral vanishes as R-+ +oo.
Hint: Integrate
4.10
Notes
§4.8 follows closely [Beck, Marchesi, Pixton, Sabalka (2008)]. Exercise 4.11,
4.24, 4.33, 4.35, 4.36, 4.37, 4.38, 4.39 are taken from [Flanigan (1972)].
Exercise 4.14 is taken from [Volkovyski'i'., Lunts, Aramanovich (1991)]. Ex­
ercise 4.15, 4.25 are taken from [Rudin (1987)]. Exercise 4.28 is taken from
[Beck, Marchesi, Pixton, Sabalka (2008)]. Exercise 4.40 is taken from [Ash
and Novinger (2007)].
Chapter 5
Harmonic functions
In this last chapter, we study:
(1) harmonic functions, which are real-valued functions that solve a certain
partial differential equation, called the Laplace equation,
(2) that real and imaginary parts of holomorphic functions are harmonic,
and that the converse holds locally and globally on simply connected
domains,
(3) some consequences of the above interplay between harmonic and holo­
morphic functions, in particular in a certain "boundary value problem",
called the Dirichlet problem.
5.1
What is a harmonic function?
Definition 5.1. Let Ube an open subset of JR.2 . A function u: U-+ JR. is
called harmonic if uhas continuous partial derivatives of order 2 ( abbrevi­
ated by writing uE C2 ), and it satisfies the Laplace equation:
{J2u
{J2u
(�u)(x, y) := a 2 (x, y) + 2 (x, y) = 0 for all (x, y) EU.
x
ay
Example 5.1. Let U = JR.2 • Consider the function u : U -+ JR. given by
u(x, y) = x2 - y2 for (x, y) E JR.2 • Then
a2 u
a
au
2x
ax =
au
2
ay = - y
a2 u
2,
ax2 =
a2 u
ay2 = -2.
2u
y
u E C 2 and �u = O in JR.2 ,
x
xy
2 2
+
)
,
(
ax2
ay2 ( , ) = - = 0. Since
u is harmonic in JR.2 .
◊
Thus
163
164
A Friendly Approach to Complex Analysis
Of course, not all functions are harmonic.
Example 5.2. Consider the function u given by u(x,y)
(x,y) E JR2. Then for all (x,y) E JR2,
cP u
a2 u
ox 2 (x,y) + oy2 (x,y) = 2 + 2 = 4 # 0.
Since
x2
+ y2 for
6..u is never O in JR2 , u is not harmonic in any open subset of JR2 •
◊
Exercise 5.1. Show that the following functions u are harmonic in the correspond­
ing open set U.
2
2
2
(1) u(x,y) = log(x + y ), U = ffi. \ {(O, O)}.
(2) u(x,y)=e°' siny,U=ffi.2•
Exercise 5.2. Show that the set Har(U) of all harmonic functions on an open set
U forms a real vector space with pointwise operations.
Exercise 5.3. Is the pointwise product of two harmonic functions also necessarily
harmonic?
Why bother about harmonic functions? Harmonic functions are
important because they satisfy the Laplace equation, which is important
among other things for two primary reasons:
(1) The Laplace equation is the prototype of an important class of PDEs,
namely "elliptic equations", which is one of the three main classes of
PDEs.
I Class of PDE I Main example
Parabolic
Hyperbolic
Diffusion equation
Wave equation
82 u 8 2 u
=0
8x 2 + 8y2
82 u 82 u
---=0
8x 2
at
2
8 u 82 u
---=0
8t 2
8x 2
(2) The Laplace equation arises in many applications, for example in
physics in the following scenarios. In hydrodynamics, the "velocity
potential" of the fluid flow satisfies the Laplace equation, while in elec­
trostatics, the electrostatic potential satisfies the Laplace equation. The
Laplace equation also has an important link with stochastic processes.
We describe this very roughly below.
Harmonic functions
Imagine the open unit disc lIJ) :=
{z
165
E <C: lzl
< 1}.
Consider a particle
starting at a point z ED and undergoing Brownian motion (for exam­
ple think of a pollen grain in water, bombarded by many tiny water
molecules producing "random" motion). Intuitively one feels that since
the motion is random, eventually the particle will leave the boundary
:= {z
1['
E <C : lzl
=
1} of the disc lill. Let us denote by (z the point
on 1[' where the particle first exits the unit circle 1[', having started at
z.
So we get a random variable (z which lives on the unit circle. See
Figure 5.1.
lIJ)
f lives on 1['
Fig. 5.1 Brownian motion and the Dirichlet problem.
Now let f : 1[' -+ ffi. be a given continuous function. Then we can think
of f((z) as being a real-valued random variable on T. Let us denote its
expectation by IE(f((z )). This depends on where one starts initially,
that is, it depends on z. Let u: lIJ)-+ ffi. be given by u(z) = IE(f((z )),
z E lill. It turns out that u is then harmonic, and in fact it is a solution to
the "Dirichlet problem", which is the boundary value problem, where
given f : 1[' -+ ffi. on the boundary 1[', we have to find a function u,
solving the Laplace equation in the interior lIJ) of 1[', such that it has a
continuous extension to 1[', matching with the given data
f:
{Au= 0 in ID),
uhr =
5.2
f.
What is the link between harmonic functions and holo­
morphic functions?
It might appear that harmonic functions should belong just to the realm of
Real Analysis. In this section, we will now learn about two results, which
will amply justify their study in Complex Analysis. Roughly a function is
166
A Friendly Approach to Complex Analysis
harmonic in an open set if and only if locally, it is the real part of some
holomorphic function.
Theorem 5.1.
Let
(l) U be an open subset of C and
(2) f: U-+ C be holomorphic in U.
Then : :: ;::���,} are harmonic functions in U.
Vice versa, we will also learn the following converse to this.
Theorem 5.2.
Let
(l) U be simply connected and
(2) u: U-+ ffi. be harmonic in U.
Then there exists a function v : U -+
f := u + iv is holomorphic in U.
ffi.,
such that v is harmonic in U and
Note that for the fin the conclusion of the above result, we have Re(!)=
and Im(!) =
v.
u,
So what we are saying is that every harmonic function in
a simply connected domain is the real part of some holomorphic function
defined there. Since a unit disc is simply connected, in particular it follows
that every harmonic function is
locally the
real part of a holomorphic func­
tion defined (at least in that disc). We will see later on in Exercise 5.5 that
the assumption of simply connectedness is not superfluous, and that given
a harmonic function in a non-simply connected domain, there may fail to
exist a
globally
defined holomorphic function in the whole domain whose
real part is the given harmonic function. But we are jumping ahead. Let us
first focus on the first result mentioned above. Before proving Theorem 5.1,
let us revisit Example 5.1.
U
Example 5.3. In our previous example, when
we have u = Re(z 2 )= Re(x2 -y2
= ffi.2 and
+ 2xyi) and z 2 is entire.
u = x2
-
y2 ,
So Theorem 5.1
delivers again our old observation that u is harmonic in ffi.2 • In fact from our
calculation, Theorem 5.1 also gives us that
v
(We can of course check this by brute force:
o
and so
o2u
ox2
+
2u
OU
= 2Y'
=
ox2
ox
o 2v
= 0 + 0 = 0.)
oy2
O'
:=
ov = 2x
ox
2xy =
'
o2 v
oy2
Im(z 2 ) is harmonic.
= 0,
167
Harmonic functions
On the other hand, it follows from Example 5.2 and Theorem 5.1 that
for any open subset
U
of C, u
holomorphic function defined in
:=
U.
x2
+ y2
is not the real part of any
◊
Proof. (of Theorem 5.1.) We have f(x + iy) = u(x, y) + iv(x, y) for
(x, y) E U. Since f is infinitely many times differentiable, we know that
u, v have partial derivatives of all orders, and so by the Cauchy-Riemann
equations, we have
{J2 u =� au
( )
ax ax
ax2
and so
av
( )
ax ay
(C,;,R) �
2
(u� )
u is harmonic. Similarly,
a v =� av
( )
ax ax
ax2
2
(C,;,R) � (- au
ax
ay
)
2
(v� )
�
ay
av
( )
ax
(C,;,R) � (- au
ay
ay
2
au
- ay2 '
�ay (- axau)
)·
(C,;,R) � (- av
ay
a2 v
ay
)
ay2 '
and so v is harmonic as well. (Alternately, we could have also noted that
v
□
= Re(-if).)
Now we show Theorem 5.2, which gives a converse to the above result
when the open set in question is a simply connected domain. As mentioned
earlier, for more general domains, it can happen that there are harmonic
functions which aren't globally the real part of a holomorphic function; see
Exercise 5.5, where we take
(1)
(2)
U := JR.2 \ {(O, 0)} (which is not simply connected) and
u := log(x 2 + y 2 ) (which is harmonic in U).
Then there is no holomorphic f in (C \
{O}
such that
u = Re(!) in U.
Proof. (of Theorem 5.2.) We will construct a holomorphic f with real
part
u, and then v
First set
:= Im(!) will serve as the required harmonic function.
au .au
g=--i.
ax
ay
We will prove that g is holomorphic, and then to construct a primitive of
g, necessarily holomorphic, which will be the f we seek. To show that g
168
A Friendly Approach to Complex Analysis
is holomorphic, we will use Theorem 2.2. First we note that since u is
harmonic, the functions
Re(g)
OU
= ox
and Im(g)
= - OU
oy
have continuous partial derivatives. Moreover, using the fact that u satisfies
the Laplace equation, we can now see that the real and imaginary parts of
g satisfy the Cauchy-Riemann equations:
2
2
=
(Re(g)) = O
=
=
=
(Im(g)),
OX
x
X
x2
y
y2
y
y
o (ou) o u o u o ( ou) o
o
o o -o
o
ou
ou
ou
= !_ ( ) = !_ ( ) = _!_ (- ) = _!_(Im(g)).
o!_(Re(g))
oy ox ox oy
ox oy
ox
y
o
o
Hence g is holomorphic in U, and by Theorem 3.5, since U is simply con­
nected, it has a primitive G in U. Decompose G = + into its real and
imaginary parts
Then
u iv
u, v.
i ov
ox + ox'
OU - i OU =
' OU
g=G =
OX
y
o
and so
(5.1)
o(u-u) =O,
OX
and so it follows from the Fundamental Theorem of Integral Calculus, that
u - u is locally constant along horizontal lines. Also, (5.1) gives
OU
OV (C_:_R)
oy ox
OU
oy '
where have used the Cauchy-Riemann equations to obtain the last equality.
Hence
o(u-u) =0
oy
showing that u - u is locally constant along vertical lines as well. Since U
is a domain, any two points in U can be joined by a stepwise path, and
so it follows from here that u - u must be a constant, say C ( E ffi.), in U.
Consequently,
f
== a - c = (u - c) + iv= u + iv,
and so we see that f is holomorphic in U, u = Re(!), v := v = Im(G) is
□
harmonic in U and f = u + iv.
169
Harmonic functions
Definition 5.2. Let U be an open set and u : U-+ IR be harmonic in U.
v : U -+ IR such
harmonic conjugate of u.
Any harmonic function
U
is called a
Example 5.4.
u
v := 2xy
that f
:= u + iv is holomorphic in
is a harmonic conjugate of the harmonic function
:= x2 -y2 considered in Example 5.1 because
f
where
:= u +iv= x2 -y2 + 2xyi = (x + iy)2 = z2,
z = x + iy,
◊
is entire.
Harmonic conjugates are obviously not unique since we can just add a con­
stant to a harmonic conjugate and get a new harmonic conjugate. In a
simply connected domain, for each harmonic function, Theorem 5.2 guar­
antees the existence of a harmonic conjugate, but it is not particularly
useful for finding a harmonic conjugate (from the above proof, we see that
it relies on the construction of the primitive G). A more direct way is to
look at the Cauchy-Riemann equations, as shown in the following example.
Example 5.5. Let
u
= -(sinx)(sinhy).
Then
�:� = +(sinx)(sinh y),
�:� =
-(sinx)(sinh y),
and so u is harmonic in IR2 . As IR2 is simply connected, we know that there
is a harmonic conjugate. Can we find one? If
then we know that
u
v is
a harmonic conjugate,
+ iv will be holomorphic, and so the pair
u, v
satisfy the Cauchy-Riemann equations. So we need av such that
av
OU
= - = +(smx)(coshy),
ay
ax
av au
.
= -(cosx)(smh
y).
=
ax
ay
Integrating (5.2) with respect to
v=
for some constant
C(y) is
C(y)
(
)( . h
sm
(5.3)
while keeping y fixed, we see that
depending on
y
(5.2)
-(cosx)(coshy) + C(y),
differentiable, we have
- cos X
x
) (5.3)
=
must
y.
(5.4)
So assuming for the moment that
ov (5.4)
(
)( . h )
dC ( )
= - cos X sm y + d y .
ay
y
170
A Friendly Approach to Complex Analysis
Consequently
dC
(y) = 0,
dy
and so C(y) = C for all y. So we could take any constant C, and in
particular we may choose C = 0. So based on the above rough reasoning,
we take
v = -(cosx)(coshy)
as a candidate for a harmonic conjugate of u. In order to verify this guess,
we can quickly note that the Cauchy-Riemann equations are satisfied by
the pair (u, v), so that f := u + iv is holomorphic. But instead, let us
use an old calculation we performed in Exercise 1.37 to check directly that
f := u + iv is holomorphic by actually finding the f:
f = u +iv= -(sinx)(sinhy) - i(cosx)(coshy)
= -i ((cosx)(cosh y) - i(sinx)(sinh y)) = -i cos z,
◊
which is entire.
Exercise 5.4. Find harmonic conjugates for the following harmonic functions in
ll�.2 :
2
3
x
e siny, x - 3xy - 2y, x(l + 2y).
Exercise 5.5. Show that there is no holomorphic function f defined in C \ {O}
whose real part is the harmonic function u defined by u(x,y) = log(x2 + y 2),
(x, y) E JE. 2 \ {(O, O)}.
Hint: If v is a harmonic conjugate of u, then also h(z) := z 2 exp( -(u + iv)) is
holomorphic. Find lhl, and conclude that h' = 0. Show that h' = 0 implies that
2/z has a primitive in C \ {O}, which is impossible.
Exercise 5.6. Is it possible to find a v : JE.2 -+ lE. so that f defined by
J(x + iy)
= x3 +y 3 +iv(x,y),
(x,y) E JE.
2
,
is holomorphic in C?
5.3
Consequences of the two way traffic: holomorphic ++
harmonic
In the previous section, the two results given in Theorems 5.1 and 5.2 show
that there is a two way traffic between the real analysis world of harmonic
functions and the complex analysis world ofholomorphic functions, allowing
a fruitful interaction between the two worlds.
171
Harmonic functions
Holomorphic functions
In the previous three chapters we have learnt many pleasant properties
possessed by holomorphic functions. Let us now use some of these to derive
some important properties of harmonic functions. In particular, we will
show the following results in this section:
(1)
If u is harmonic, then it is
c= .
(2) The Mean Value Property for harmonic functions.
(3) The Maximum Principle for harmonic functions.
(4) Uniqueness of solutions to the Dirichlet Problem.
5.3.1
Harmonic functions are smooth
Corollary 5.1. Harmonic functions are infinitely many times differen­
tiable.
(Note that the definition of a harmonic function demands only
twice con­
tinuous differentiability. The remarkable result here say s that thanks to the
fact that the Laplace equation is satisfied, in fact the function has got to
be infinitely differentiable. A result of this ty pe is called a
regularity result
in PDE theory.)
Proof.
zo
=
U. Let
r > 0 such that the disc D with center
Suppose that u is a harmonic function in an open set
(xo, Yo) E U. Then there is a
zo and radius r is contained in U. But ulv is harmonic in D and D is
simply connected. So there is a holomorphic function f defined in
that Re(!)
in
= u in D.
D, such
But f is infinitely many times complex differentiable
D. Consequently, u is infinitely many times differentiable in D, and in
particular at zo E D. As the choice of zo E D was arbitrary, the result
follows.
□
Exercise 5.7. Show that all partial derivatives of a harmonic function are har­
monic.
172
5.3.2
A Friendly Approach to Complex Analysis
Mean value property
Using the Cauchy Integral Formula, we immediately obtain the following
"mean value property" of harmonic functions, which says that the value of
a harmonic function is the average (or mean) of the values on a circle with
that point as the center.
Theorem 5.3. (Mean-value property of harmonic functions) Let
(1) U be an open set,
(2) u: U-+ ffi. be harmonic in U,
(3) Zo EU,
(4) R > 0 be such that the disc {z EC: lz - zol < R} CU.
271"
1
Then u(zo) = - 1 u(zo + rexp(it))dt for all r such that O < r < R.
21r 0
Proof. The disc D := {z EC : lz - zol < R} is simply connected, and
so there is a holomorphic function f defined in D, whose real part is u.
But now by the Cauchy Integral Formula, if C is the circular path given by
C(t) = zo + rexp(it) (t E [O, 21r]), then
2 f(zo + rexp(it)) .
.
f(z) =1
1
71"
-dz
1 ------irexp(it)dt
f(zo ) =r exp(it)
21ri c z - zo
21ri O
271"
= _!_
f(zo+rexp(it))dt.
21r lo
1
f
Equating real parts, the claim is proved.
5.3.3
D
Maximum Principle
From the Maximum Modulus Theorem (see page 129), we also obtain the
following.
Theorem 5.4. (Maximum Principle) Suppose that
(1) U is a simply connected domain,
(2) u: U-+ ffi. is harmonic in U,
(3) z0 EU is a point such that u(z0) 2: u(z) for all z ED.
Then u is constant in U.
There is a holomorphic function f defined in U whose real part
equals u. But then the function g : U -+ C defined by g(z) = exp(f(z))
Proof.
173
Harmonic functions
(z EU) is holomorphic too. We have
lg(zo)I
=
I exp(f(zo))I
= e Re(zo) = eu(zo)
� e u(z)
= lg(z)I
for all z EU.
By the Maximum Modulus Theorem applied tog, it follows that g must be
constant in U. Thus lgl is also constant in U, that is, /gl = e Re(f) = e u is
a constant in U. Taking the (real) logarithm, it follows that u is constant
in U.
□
5.4
Uniqueness of solution for the Dirichlet problem
The Maximum Principle has an important consequence about the unique­
ness of solutions to the Dirichlet problem, as explained below. Let
:=
11' :=
][J)
{z EC: /z/ < 1},
{z EC: lzl = l}.
r.p lives here
Then the Dirichlet problem is the following:
Given r.p : 11' -+ IR, continuous,
find a u : ][J) U 11' -+ JR such that
( 1) u is continuous on ][J) U 11',
(2) uhr = r.p,
(3) u has continuous partial derivatives of order
(4) -6.u = 0 in ][J).
2 in
][J),
The given function r.p is called the boundary data.
The reason one is
interested in solving the Dirichlet problem is that the need arises in ap­
plications, for example in heat conduction, electrostatics, and fluid flow.
Using the Maximum Principle, we can show the following.
Proposition 5.1. The solution to the Dirichlet problem is unique.
174
A Friendly Approach to Complex Analysis
Proof.
Indeed, let u1, u2 be two distinct solutions corresponding to the
boundary data cp. In particular, u1 = u2 on 1l'. So u1 must differ from
u2 somewhere inside lDl. Without loss of generality suppose that there is
a point w E lDl where u1(w)
>
Then u := u1 - u2 is such that
u2 (w). (Otherwise exchange their labels.)
(1) u=0on'll',
(2)
u(w)
> 0,
(3) u is harmonic in lDl.
Let zo E [)) U 1l' be the maximizer for the real-valued continuous function
u on the compact set [)) U 1l'. From (1),(2), zo e/. 1l'. So zo E IDl. We have
u(z0) 2: u(z) for all z E IDl, and so by the Maximum Principle, u must be
constant in IDl. But u is continuous on[)) U 1l', and u is 0on 1l'. So it follows
that the constant value of u must be 0everywhere in IDlU'll'. Hence u1 = u2,
a contradiction.
D
Remark 5.1. It can be shown that the following expression, called the
Poisson Integral Formula,
boundary data cp:
u(rexp(it))= -
1
21r
1
o
2
-rr
gives the solution to the Dirichlet problem with
1
1- r2
- 2rcos (O - t) + r2
cp(exp(i0))d0
(( E 1l').
This can be derived using the Cauchy Integral Formula, but there are some
technical subtleties, and so we will not prove this here.
Exercise 5.8. (Some half-plane Dirichlet problems.) Given the "boundary data"
b : JR ➔ JR, we consider the problem of finding a continuous, real-valued function
h defined in the closed upper half-plane y 2: 0, such that h is harmonic in the
open upper half-plane y > 0 and moreover, h(x, 0) = b(x).
(1)
If b is just a polynomial p, then show that we can simply take h given by
h(x,y) = Re(p(x+iy)).
(2) Prove that if
b(x) =
1
-1+x
-2,
then (x,y) I-+ Re(b(x + iy)) is not a solution (because of the pole at z
Show that
1 2
y
= 2
h(x,y) := Re
t
z+i
X + y+ 1)
(--i .)
gives a solution to the Dirichlet problem.
Exercise 5.9. Let
u:
JR2 ➔ JR be a harmonic function such that
(x,y) E JR2 . Prove that u is constant.
u(x, y)
= i).
> 0 for all
Hint: Let f be an entire function whose real part is u. Consider exp(-!).
Harmonic functions
175
Exercise 5.10. The regularity of functions satisfying the Laplace equation is not
completely for free. Here is an example to show that a discontinuous function
may satisfy the Laplace equation! Consider the function u : IB.2 ➔ ffi. defined to
4
be the real part of e-l/z when z #- 0 and O at the origin.
(1) Verify that u is discontinuous at 0.
(2) Check that u(x, 0)
= e-l/x
4
, u(0, y) = e-l/y .
4
(3) Being the real part of a holomorphic function in C\ {O}, we know already
that u satisfies the Laplace equation everywhere in IB.2 \ { (0, O)}. Show that
also
2
{J u
8x2
exist, and that
82 u
ax 2 (0, 0) +
(0, 0) and
82 u
8y2
(0, 0)
82 u
ay2 (0, 0) = 0.
Exercise 5.11. Let D1, D2 be domains in C. Let rp : D1 ➔ D2 be holomorphic.
Show that if h : D2 ➔ ffi. is harmonic, then h o rp: D1 ➔ ffi. is harmonic as well.
Now suppose that rp : D1 ➔ D2 is holomorphic, a bijection, and also that
rp- 1 : D2 ➔ D 1 is holomorphic. We call such a map rp a biholomorphism. Con­
clude that a function h : D2 ➔ ffi. is harmonic if and only if h o rp : D1 ➔ ffi. is
harmonic.
Thus the existence of a biholomorphism between two domains allows one to
transplant harmonic (or even holomorphic) functions from one domain to the
other. This mobility has the advantage that if D1 is "nice" (like a half plane or a
disc), while D2 is complicated, then problems (like the Dirichlet Problem) in D2
can be solved by first moving over to D1, solving it there, and then transplanting
the solution to D2.
A first natural question is then the following: Given two domains D1 and D2,
is there a biholomorphism between them? An answer is provided by the Riemann
Mapping Theorem, a proof of which is beyond the scope of this book, but can be
found for example in [Conway (1978)].
Theorem 5.5. (Riemann Mapping Theorem) Let D be a proper (that is,
D #- q simply connected domain in C. Then there exists a biholomorphism
rp: D ➔ lill := {z EC: lzl < l}.
Thus the above result guarantees a biholomorphism between any two proper
simply connected domains (by a passage through llll). Unfortunately, the proof
does not give a practical algorithm for finding the biholomorphism. Show that
the "Mobius transformation" rp: lHI ➔ llll, where lHI := {s EC: Re (s) > O}, given
by
s-1
, s E lHI,
rp(s) =
s+ l
is a biholomorphism between the right half plane lHI and the disc llll.
176
5.5
A Friendly Approach to Complex Analysis
Notes
Exercises 5.5, 5.6 and the proof of Theorem 5.2 is taken from [Beck, March­
esi, Pixton, Sabalka (2008)]. Exercise 5.8 is based on [Flanigan (1973)].
Solutions
Solutions to the exercises from the Introduction
Solution to Exercise 0.1
Suppose the derivative of f' at O exists, and is equal to L. Then taking
E := 1 >0, there exists a 8 >0 such that whenever0 < Ix -0I < 8, we have
I
f'(x) - f'(O)
x-0
-
LI <
E.
In particular, with x := 8/2, we have0 < Ix -0I = 8/2 < 8, and so there
must hold that
f'(x) - f'(O)
2(8/2)-0
= 1
- LI = 12 - LI < E.
(5.5)
x-0
(8/2) -0
I
LI
On the other hand, with x := -8/2, we have0 < Ix -0I = 8/2 < 8, and so
again there must hold that
I
f'(x) - f'(O)
x-0
LI = ,-2(-(-8/82/2)-)-00 - LI = 12+ LI < E.
(5.6)
From (5.5) and (5.6) it follows, using the triangle inequality for the real
absolute value, that 4 = l2+L+2-LI::::; l2+LI + 12- LI < E+ E = 2E = 2,
a contradiction. Hence f' cannot be differentiable at0.
177
178
A Friendly Approach to Complex Analysis
Solutions to the exercises from Chapter 1
Solution to Exercise 1.1
Since (x,y) =/- 0, at least one among x,y is nonzero, and so x2+ y2 =/- 0.
Thus
(
(
X
-y ) E JR2
x2+y2 ' x2+y2
Moreover,
(x,Y) ·
-
=
X
(
x2+y2
-y
' x2+y2 )
)
(
X
X
-y -) X• (--y -)+y -X•--- y• -· x2 yx2+y2
x2+y2 '
x2+y2
+ 2
x2+y2 -xy+xy
(x2 y2 ' x2 y2 ) = (1,0).
+
+
Hence for (x,y) =/- (0,0), we have (x,y)-1 = (
x
X2+y2
,
-y
X2+y2 )
in C.
Solution to Exercise 1.2
Since 0 E
(-i, i), tan0 ER We have
__1__ --,------,-1---,--,tan 0
+i ( 2
= 2
2)
2
+(tan0)
1
1- itan0
1 +(tan0)
sin 0
_ _ _ . (cos0)2
2
(cos0)
c
0
=----)
+i ( _ �os�---(cos0)2+(sin0)2
(cos0)2+(sin0)2
=
. (sin0)(cos0)
(cos0)2
•
+i
= (cos 0 )2+i"(sm 0 )(cos 0 ).
1
1
Hence
..
1+itan0
.
= (l+itan0)((cos0)2+i(sm0)(cos0))
.
1- itan 0
sin0 .
= (cos0)2 - - · (sm0)(cos0)
cos 0
+i ((sin0)(cos0)+ :�:: • (cos0)2 )
= (cos0)2 - (sin0)2+i2 (sin0)(cos0) = cos(20)+isin(20).
179
Solutions
Solution to Exercise 1.3
Let P C C be a set of positive elements of C. Then since i -/=- 0, by (P3),
either i E P or (if/. P and -i E P). By (P2), we have
-1
= i · i = ( -i) · ( -i) E P.
(5.7)
1 = (-1) · (-1) E P.
(5.8)
Again by (P2),
But 1-/=- 0, and (5.7), (5.8) contradict (P3) for x
= l.
Solution to Exercise 1.4-
See Figure 5.2.
cos j
i
0
+ isin j
1
-v'2i
Fig. 5.2 Location of the complex numbers 0, 1, -3/2, i, -v'2i, cos�+ isin�-
3
3
180
A Friendly Approach to Complex Analysis
Solution to Exercise 1.5
We have for 0 E � that (cos 0+ isin 0) 3
(cos0+ isin0)3
= cos(30)+ isin(30). But
= (cos0+ isin0)((cos0)2 - (sin0)2 + i2(cos0)(sin0)
= (cos0)((cos0)2 - (sin0)2 ) - (sin0)2(cos0)(sin0)
+i(· .. ) .
Hence equating the real parts on both sides, we obtain
cos(30)
= Re((cos0+ isin0)3 )
= (cos0)((cos0)2
-
(sin0)2 )
- 2(cos0)(sin0)
2
= (cos0)((cos0)2 -1+ (cos0)2 )-2(cos0)(1-(cos0)2 )
= (cos0)3 - cos0 + (cos0)3 - 2 cos0 + 2(cos0)3
= 4(cos0)3 - 3cos0.
Alternatively, the Binomial Formula (a + b) n
=
to hold for a, b E C, n E N and so
t (�)
k=O
n k
ak b -
continues
cos(30) = Re((cos0+ isin0)3)
= Re((cos0)3 + 3 (cos0)2 (isin0)+ 3 (cos0)(isin0)2 + (isin0)3 )
= (cos0)3 - 3 (cos0)(sin0)2
= 4(cos0)3 - 3cos0.
Solution to Exercise 1. 6
We have 1+ i = J2 ( � + i�)
(1+ i) 10
= J2 (cos�+ isin�). Hence
10
= (0) 10 (cos�+ isin�) = 25 (cos (10 • �)+ isin (10 • �))
= 32 (cos (21r+ i) + isin (21r+ i))
= 32 (cos (i) + isin (i)) = 32(0+ i -1) = 32i.
Solution to Exercise 1. 7
The angle made by 2+ i with the positive real axis is tan- 1 (1/2), and the
angle made by 3+i with the positive real axis is tan- 1 (1/ 3 ). Thus the angle
Solutions
181
made by (2+i)(3+i) with the positive real axis is tan-1(1/2)+tan-1(1/3).
On the other hand, since
(2 + i)(3 + i) = 6 - 1 + i(2 + 3) = 5 + 5i,
the angle made by (2 + i)(3 + i) with the positive real axis is
tan- 1 (5/5) = tan- 1 1 = n:/4.
Consequently,
7r:
11
11
4 = tan- 2 + tan- 3
Solution to Exercise 1. 8
Suppose that the vertices A, B, C of the equilateral triangle are at the
complex numbers ZA, ZB, zc, and that they are labelled in the anticlockwise
fashion. Since t'(AC) = t'(AB) and L.CAB = n:/3, we have
zc - ZA = (cos
i + i sin i) (zB- ZA)-
(5.9)
We argue by contradiction and let p, q, m, n E Z be such that
zc - zA = p + iq, and
ZB - ZA = m + in.
Then (5.9) becomes p + iq = ( � + � i) (m + in), that is,
p=
q
=
m J3
n, and
2 2
m-v'3 n
-2-+2•
(5.10)
(5.11)
Thus (by multiplying (5.10) by -n and (5.11) by m and adding), we obtain
qm - pn =
v'3 (m2 + n2 ).
2
But m 2 + n2 =/. 0 (since ZB =/. ZA!), and so we obtain
V3 =
a contradiction.
2(qm - pn)
E (Ql
m2 +n2
182
A Friendly Approach to Complex Analysis
Solution to Exercise 1. 9
We write -1 = 1 • (cos1r + i sin1r),and we seek w = p(coso: + i sino:) such
that w4 = p4 (cos(4o:) + i sin(4o:)) = 1 • (cos1r + i sin1r). Hence p4 = 1 and
sop= l. Also,we have 4o: E {1r,1r ± 21r,1r ± 41r,•••},and so
o:E{i, �±i,
±1r,
�
···}·
So we get w = p(cos a + i sin a) = 1 (cos a + i sin a) belongs to
31r
. . 31r
51r
. . 51r
71r
. . 71r
cos +ism 4, COS
4
4 +iSlll 4, COS 4 +iSlll 4, COS 4 +iSlll 4}
{ 7r • • 7r
-
l+i -l+i -1-i 1-i
{
}.
- J2' J2' J2'J2
The four fourth roots of unity are depicted in the complex plane in Fig­
ure 5.3.
1-i
v'2
Fig. 5.3
Location of the complex numbers w that satisfy w 4
= - 1.
Solution to Exercise 1.10
We have
0 = z 6 - z 3 - 2 = (z 3 ) 2
-
2 z 3 + z 3 - 2 = (z 3
-
2) (z3 + 1),
183
Solutions
and so z 3
= 2 or z 3 = -1. The equation z 3 = 2 holds if and only if
. . 21r
. . 41r
3tn (cos 21r +ism
3tn (cos 41r +ism
3tn} ,
z E { v2
) , v2
) , v2
3
3
3
3
that is,
J3)
2 ,
3tn (- 1 +i.
zE { v2
2
3tn (- 1 - i.
v2
2
J3)
2 ,
3tn} .
v2
On the other hand, z3 = -1 holds if and only if
51r . . 51r
. .
1r . . 1r
+ism }
z E { cos +ism , cos 1r +ism 1r, cos
that is,
3
3
3
vf3
3 ,
vf3
zE{!+i
-1 !_i
}·
2
2' ' 2
2
So z6 - z 3 - 2 = 0 if and only if [(z 3
if
that is,
= 2) or (z 3 = -1)], that is, if and only
J3)
J3)
1 . vf3
. vf3
.
.
3tn (--1 +i3tn (--1 -i3tn v2
+i- -1 -1 -i- } .
v2
z E { v2
2
2 '
2
2 ' '2
2' '2
2
Solution to Exercise 1.11
Suppose w E (C \JR is such that w 3 = 1. Then (w- l)(w 2 + w +1) = 0, and
since w I- 1, we have w 2 + w + l = 0. Hence
((b - a)w+(b - c))((b - a)w 2
+ b - c)
= (b - a) w + (b - a)(b - c)(-1) + (b - c)2
= (b- a)2 · 1 + (b - a)(b - c)(-1) + (b- c)2
= (b - a)(b - a - b + c) + (b - c)2
= (b - a)(c - a)+ (b - c)2
= (be - ca - ab+ a2 + b2 - 2bc + c2
2
= a2
3
+ b2 + c2 - ab - be - ca = 0.
184
A Friendly Approach to Complex Analysis
Hence either (b - a)w = c - b or (b - a)w2 = c - b. But the latter case is
the same as (b- a)w3 =(c-b)w, that is, (c-b)w =b- a. So we have that
lb- al =le-bl, and the angle between the line segments joining a to b and
a to c is 7f /3; see Figure 5.4.
a
'
J_,
'
�
a
or
2,r
'
3
a, b, c
Fig. 5.4
'
2,r
3
form an equilateral triangle.
In either case, we obtain that the triangle formed by a, b, c is equilateral.
If a, b, c are all real, then the equilateral triangle must degenerate to a
point r E IR, and so a =b =c ( = r). Thus we recover the real case result.
Solution to Exercise 1.12
Let w E C\IR be such that w3 = 1. Since (1-w)(l+w+w2) = ,0 and w =/- 1,
we have l+w+w2 = 0 . Also, l+w2 +w4 = l+w2 +w-w3 =l+w2 +w = .0
We have
3
n
(l + l) n + (l + w) n + (l + w ) n =�
3
2 3
3
( )
But
1 +1 +1
if k
(1 + wk + w2k ) = { 1 + w + w2 if k
1 + w 2 + w 4 if k
_
3 if k
{ 0 if k
0 if k
=0
=1
=2
Thus
(l + l)3n + (1 + w)3n + (1 + w2 )3n = 3 . (
3
; (l + wk + w2k ).
=0
=1
=2
mod 3,
mod 3,
mod 3
mod ,3
mod ,3
mod .3
(3;) (3;)
+
+ ...+
G:)).
Solutions
185
But also
(1 + 1)3n + (1 + w)3n + (1 + w2)3n =
=
23n + (-w2)3n + (-w)3n
23n + (-lt + (-l)n
= 23n + 2(-l)n,
and so the claim follows.
Solution to Exercise 1.13
Q
' P
B'
Fig. 5.5 RP and SQ have equal lengths and meet at right angles.
See Figure 5.5. Let the points A, B, C, D in the plane correspond to the
complex numbers a, b, c, d, respectively. Since AB' is obtained from AB by
rotating AB about A in an anticlockwise fashion by 90 ° , we have that B'
corresponds to the complex number a - i(b - a). Since Pis the midpoint
of BB', it follows that P corresponds to
a+ b - i(b - a)
2
Similarly, Q, R, S correspond to
c + d - i(d - c)
d + a - i(a - d)
b + c - i(c - b)
2
2
2
186
A Friendly Approach to Complex Analysis
respectively . If we denote the complex numbers corresponding to P, Q, R, S,
by p, q, r, s, respectively, then
· b+c-i(c-b) - d+ a-i(a-d)
.
iq-s
(
) =i (
)
2
2
-b+c- a+d+ib
( + c-d- a)
2
b-----'a) + c+d-i(
d-c) =
-------'= -a-b+i(
---'------'-p + r.
2
2
Hence lq - sl =IP - rl, showing that f(QS) =f(PR). Also, since mul­
tiplication by i produces a rotation about the origin by 90 °, we see that
PR J_ QS.
Solution to Exercise 1.14
Let z1 =x1 + iy1, z2 =x2 + iy2, where x1, x2, Y1, Y2 belong to JR. Then
z1z2 =X1X2 - Y1Y2 +i(X1Y2 + Y1X2 ), and
lz1z2 1 2 =(x1x2 - Y1Y2 )2 + (x1Y2 + Y1X2 )2
=XiX�- �+ YiY�+ XiY�+ �+ YiX�
=Xi (X�+ y�)+ Yi (Y�+ x�)=(xi + yi )(x�+ y�)
=lz1 l 2 lz2 1 2 Since lz1I, lz2I, lz1z2I are all nonnegative, it follows that lz1z2I =lz1I lz2ISolution to Exercise 1.15
Let z =x+iy, where x, y E R Then
(z)=X-iy=X-i(- y)=X+iy=Z.
Also,
zz=(x+iy)(x-iy)=x2 + y2 +i(-xy+ xy)=x2 + y2 =lzl 2Finally,
z+ z x+ jef+ x- jef 2 x
= =x=Re (z) and
=
2
2
2
z- z ;i+iy-;i+iy 2iy
=y=Im (z).
=
=
2i
2i
2i
187
Solutions
Solution to Exercise 1.16
Let z = x+iy, where x, y E R Then
lzl =Ix+iyl = Jx 2 +y2 = Jx 2 +(-y) 2 =Ix- iyl = lzl,
IRe(z)I = lxl = H::; Jx 2 +y2 =Ix+iyl = lzl,
Jx 2 +y2 = lx+iyl =lzl.
IIm(z)I =IYI =
#::;
Since z is obtained by reflection of z in the real axis and since O E �, the
distance ofz to O is equal to the distance ofz to 0. In other words, lzl = lzl.
The inequalities IRe(z)I ::; lzl and IIm(z)I ::; lzl just say that the length of
any side in a right angled triangle is at most the length of the hypotenuse.
z
Solution to Exercise 1.17
First we note that lazl = lal lzl = lal lzl < 1-1 = 1, and so az-=/- 1. We have
z-a z-a
z-a ( z -a )
zz-az-az + aa
.
.
=
=
1-az 1-az 1-az-az + aazz
1-az
1-az
lzl2 -az-az + lal2
1-az-az + lal2 lzl2
1-az-az + lal2 1zl2 + lzl2 + lal2 -1-lal2 lzl2
1-az-az + lal2 lzl2
2
2
2
lzl +lal -1-lal2 lzl
= 1 + -------,---,,.,.,-cc:--1-az-az+lal2 lzl2
2
lzl + lal2 -1-lal2 lzl2
=1+
ll-azl2
_ l _ (1-lzl2 )(1-lal2 )
·
11-azl2
188
A Friendly Approach to Complex Analysis
Thus I
2
(1- lzl2 )(1- lal2 ) <
z-a
1 = 1- 1- 0 = 1.
1- az
11- azl2
2:0 as [z[::;1, [a[<l
Solution to Exercise 1.18
Let w E C be such that p(w)
=
0, that is, eo + c1 w + • • • + cd wd
Co+ C1 W + • · · + Cd W d = 0 = 0,
and so, using the fact that the
Ck
=
0. Hence
are all real for O � k � d, we obtain
0 =Co+ C1 w + ... + Cd W d = Co+ C1 w + ... + Cd W d
=Co+ ciw + · · · + Cd W d = Co+ C1W + · · · + cd (w)d
where the last equality follows from the observation that
w )k
w k = -- = - - = (1 � k � d.
� �
k times
k times
Consequently, 0 = c0 + c1w + · · · + cd (w)d =p(w).
Solution to Exercise 1.19
Let a= lal(cosa+ i sina), and b = lbl(cos,B + i sin,B), where a, ,B E [O, 21r).
Then
ab= lal(cosa+ i sina) · lbl(cos,B - i sin,B)
= !al lbl(cosa + i sina)(cos,B - i sin,B),
and so Im(ab) = !al lbl(-(cosa)(sin,B) + (sina)(cos,B)) = !al lbl sin(a -,B).
lal
A
a
Bb
Fig. 5.6
The area of .6.OAB formed by the triangle with vertices at 0, a, b.
189
Solutions
=
The area of the triangle OAB formed by 0, a, b (where O
B b) is
1
1
.
.
2 f(OA)-f(OB)-smL'.AOB = 21al·lbl·lsm(a-,B
)I
But for a square matrix M
.
l
For z1, z2, Z3 EC, we have
=
I I
1
Im(ab)
= 1Im(ab)I = 2
2
Solution to Exercise 1.20
1 Z1 Z1
w := i · det [1 z2 z2]
1 Z3 Z3
= 0, A= a and
1 Z1 Z1
-i · det [1 z2 z2 .
1 Z3 Z3
= [mi1],
detM =
L (sign o-) · mia(i),
aESn
where Sn is the set of all permutations on the set { 1, • • • , n}. Hence
detM =
L (sign o-) · mia(i) = detM,
aESn
where M is the matrix obtained from M by taking entry wise complex con­
jugates. Hence
1 Z1 Z1
det [ 1 z2 z2
1 Z3 Z3
l
1 Z1 Z1
= det [1 z2 z2
1 Z3 Z3
l
[ l
1 Z1 Z1
= - det 1 z2 z2 ,
1 Z3 Z3
where the last equality is obtained by interchanging the second and the
third columns. Consequently,
1 Z1 Z1
i · det [ 1 z2 z2
1 Z3 Z3
l
[ l
1 Z1 Z1
1 z1 z1
1 Z3 Z3
1 Z3 Z3
= -i · det 1 z2 z2 = -i · ( - det [1 z2 z2
1 Z1 Z1
= i · det [1 z2 z2]
1 Z3 Z3
Thus w is its own complex conjugate, and hence it is real.
l)
190
A Friendly Approach to Complex Analysis
Solution to Exercise 1.21
We have
2
2
lz1 + z21 + lz1 - z21
= (z1 + z2)(z1 + z2) + (z1 - z2)(z1 - z2)
= Zl • Zl + Zl ·
Z2 + Z2 •
Z1 + Z2 •
Z2
+z1 · z1 + z1 · (-z2) + (-z2) · z1 + (-z2) · (-z2)
2
2
2
2
= lz1l + � +� + lz2l + lz1l - � - � + lz2l
2
2
= 2(lz1l + lz2l ).
Consider the paralellogram P with vertices at 0, z1, z2, z1 +z2 in the complex
plane. Then lz1 +z2I denotes the length of one diagonal of P, while lz1 -z21
is the length of the other diagonal of P. Also, lz1I, lz21 are the lengths of
the sides on P. So the above equality says "In a parallelogram, the sum
of the squares of the lengths of the diagonals equals twice the sum of the
squares of the sides."
Solution to Exercise 1.22
For z1, z2 EC, we have lz1I
= lz1 - z2 + z2I � lz1 - z2I + lz2I, and so
lz1I - hi� lz1 - z2ISince this holds for all z1, z2 EC, by swapping z1 and
(5.12)
z2
in (5.12),
lz2I - lz1I � lz2 - z1I = I - (z1 - z2)I = I - 11 lz1 - z2I = lz1 - z2I- (5.13)
(5.12) and (5.13) give llz1I - lz2II � lz1 - z2ISolution to Exercise 1.23
(1),(2),(3):
Fig. 5.7 Left to right: The set of points described by lz - (1- i)I = 2, lz - (1- i)I < 2
and 1 < lz - (1- i)I < 2, respectively.
Solutions
191
(4) Let z = x + iy, where x, y ER Then Re(z - (1- i)) = 3 if and only if
x - 1 = 3, that is x = 4.
4
0
(5) Let z = x + iy, where x, y ER Then IIm(z - (1- i))I < 3 if and only
if IY + 11 < 3, that is -4 < y < 2.
(6) {z EC: lz - (1- i)I = lz - (1 + i)I} is the set of all complex numbers z
whose distance to 1-i is equal to its distance to 1 + i. So it is the set of all
points which lies on the perpendicular bisector of the line segment joining
1- i to 1 + i. So the set is the real line �l+i
1-i
Fig. 5.8
The set of points z satisfying [z - (1 - i)[
(7) The equation lz - (1 - i)I + lz - (1 + i)I
= [z - (1 + i)[ is R
= 2 says that the sum of the
192
A Friendly Approach to Complex Analysis
distances of z to 1 + i and to 1-i is 2. But the distance between 1-i and
1 + i is 2. So z lies on the line segment joining 1-i to 1-i.
This conclusion can also be arrived at analytically. If z = x + iy, where
x,y E �, then
2 = J(x -1) 2 + (y + 1) 2 + J(x -1) 2 + (y -1) 2
?:': IY + 11 + IY-11 ?:': 1 +JI+ 1-JI= 2.
Thus IY + 11
+ IY-11 = 2 and x = 1.
l+i
1-i
Fig. 5.9 The set of points
joining 1 - i to 1 + i.
z satisfying lz - (1 - i) I+ iz - (1 + i)I
= 2 is the line segment
(8) The set of z such that lz -(1-i)I + lz -(1 + i)I = 3 lies on an ellipse E
with foci at 1 + i and 1-i, and so {z E (C: lz - (1-i)I + lz - (1 + i)I < 3}
is the region in the interior of the ellipse E.
Fig. 5.10 The set of points
the ellipse E.
z satisfying lz - (1- i)I + lz - (1 + i)I < 3 is the interior of
193
Solutions
Solution to Exercise 1.24-
Co s·
C1
···+ zd-l + zd) . mce
lcd-1
1 +...+ lc �I + �)= O,
lim
n ➔ oo ( n
nd 1
nd
there exists an N large enough so that
lcoI
l iI
lcdl
lcd- 1 I
+···+ ed-1
<
2·
+Nd
N
�
Hence for lzl> N =: R, we have
c
cd- 1
d
lp(z)I= l z lled +
+ od I
+...+ �
z
zd-1
z
cd l
d
� lzl (lcdl- ; +···+ �� 1 +
:�
z
l
ed- i i + +
+�
� l z ld ( lcdl- (
· ··
zd-1
z
zd ))
For z r 0, we have p()z = z
_j_
C d- 1
d( Cd + -+
z
I
hl
I)
� lzld (1 cdl- c � +...+ )�� 1 +
c
il
lc I
lc I •
� lzld (lcdl- ; ) = ; lz ld
�J))
.._.,...,
=:M
Solution to Exercise 1.25
("If' part) Suppose that the two real sequences (Re(Zn)) nE fil and
(Im(zn)) nEN are convergent respectively to Re(L) and Im(L). Then given
E > 0,
lzn - LI = J(Re(zn) - Re(L)) 2 +(Im(zn) - m
I (L))2
2
2 =E
<
(J2) + (J2)
for n > N, where N is large enough to ensure that for all n > N,
and IIm(zn) - m
I (L)I<
IRe(zn) - Re(L)I <
J2
So (zn) nEN converges to L.
J2"
("Only if' part) Suppose (zn) nEN converges to L. Given E > 0, let N be
such that for all n > N, l z - LI< E. But then for n > N,
IRe(zn) - Re(L)I= IRe(zn - L)I� lzn - LI< E, and
I (zn - L)I� lzn - LI< E.
I (L)I= Im
IIm(zn) - m
Hence the two real sequences (Re(zn)) nEN and (Im(zn)) nEN are convergent
respectively to Re(L) and Im(L).
194
A Friendly Approach to Complex Analysis
Solution to Exercise 1.26
("Only if' part) Suppose (zn )nE N converges to L. Then (Re(zn ))nE N and
(Im(zn ))nE N converge to Re(L) and Im(L), respectively. Hence (Re(zn ))nE N
and (-Im(zn ))nE N converge to Re(L) and -Im(L), respectively, that is,
(Re(zn ))nE N and (Im(zn))nE N converge to Re(£) and Im(£), respectively.
Consequently (zn)nE N converges to L.
("If' part) Suppose (zn)nE N converges to L. By the previous part, ((zn ))nE N
converges to (£), that is, (zn )nE N converges to L.
Solution to Exercise 1.27
Let (zn )nE N be a Cauchy sequence in C. The inequalities
IRe(zn ) - Re(zm)I = IRe(zn - Zm)I S lzn - Zml and
IIm(zn ) - Im(zm)I = IIm(zn - Zm)I S lzn - Zml
show that the two real sequences (Re(zn ))nE N and (Im(zn ))nE N are then
also Cauchy sequences. Since � is complete, they are convergent to, say,
a, bE�, respectively. But then (zn )nE N converges in (C to a+ ib. Hence (C
is complete.
Solution to Exercise 1.28
Let zoEC and
E
> 0. Set 8 = E > 0. Then whenever lz - zol < 8, we have
IRe(z) - Re(zo)I = IRe(z - zo)I S lz - zol < 8 = E.
So z f---t Re(z) is continuous at zo. Since the choice of zoEC was arbitrary,
it follows that z f---t Re(z) is continuous (in C).
Solution to Exercise 1.29
Let U:= {zEC: Re(z) • Im(z) > 1}, and set F:= CU (the complement of
U). If (zn )nE N is a sequence in F such that (zn )nE N converges to L in C,
then we have
Re(zn ) · Im(zn ) S 1 for all nEN,
(5.14)
and (Re(zn ))nE N, (Im(zn ))nE N converge respectively to Re(L) and Im(L).
Thus (Re(zn ) · Im(zn ))nE N is also convergent, with limit Re(L) · Im(L).
(5.14) then gives Re(L) • Im(L) s 1. So LE F. Hence Fis closed, and so
CF= U is open.
Solutions
195
Next we show that U is not a domain. Suppose that it is. Then there
is a (stepwise) path 1: [a, b] ➔ U that joins 1(a)=2+2i EU to the point
1(b) = -2 - 2i E U. Since the map z f-t Re(z) : <C ➔ JR is continuous, it
follows that t 8 Re(,(t)) : [a, b] ➔ IR is continuous too. We have
<p(a)= Re(,(a))= Re(2+2i)= 2, and
<p(b)= Re(,(b))= Re(-2 - 2i)=-2.
Since <p(a) = 2 > 0 > -2 = <p(b), it follows by the Intermediate Value
Theorem that there exists a t* E [a, b] such that O = <p(t*) = Re(,(t*)).
But then Re(,(t*)) • Im(,(t*)) = 0 • Im(,(t*)) = 0 -:f 1, showing that
1(t*) (/. U, a contradiction. So U is not path-connected, and hence not a
domain.
Solution to Exercise 1.30
Since D is open, it is clear that its reflection in the real axis, D*, is also
open.
Let w1, w2 E D*. Then w1, w2 E D. As D is a domain, there exists
a stepwise path 1 : [a, b] ➔ <C such that 1(a) = w 1, 1(b) = w2 and for
all t E [a, bl, 1(t) E D. Now define
[a, b] ➔ <C by ,*(t) = 1(t),
t E [a, bj. Then ,*(a) = w1 = w1, 1*(b) = w2 = w2, and for all t E [a, bl,
1*(t)=1(t) E D*. As 1* is the composition of the continuous functions 1
and z f-t z, 1* is continuous. Also, since I is a stepwise path, there exist
points
to =a < t1 < · · · < tn < tn+l =b
such that for each k = 0, 1, · · · , n, the restriction ,l[t k ,t k +il has either con­
stant real part or constant imaginary part. Consequently 1*l[t k ,t k +il (which
has the same real part as ,l[t k .f k +il and has imaginary part which is mi­
nus the imaginary part of ,l[t k ,t k +il) also has either a constant real part
of a constant imaginary part. So 1* is a stepwise path too. Hence D* is
path-connected.
As D* is open and path-connected, it is a domain.
,* :
Solution to Exercise 1.31
We have
9
exp(i ;) =exp(i(41r+i))=e 0 (cosi+i sini)=l(O+i-l)=i,
and exp(3 +1ri)=e3 (cos1r+i sin1r)=e3 (-1+i • 0)=-e3 .
196
A Friendly Approach to Complex Analysis
Solution to Exercise 1.32
Let z = x + iy, where x,y E IR. Then ex (cosy + i siny) = 1ri. Taking
absolute values on both sides, we obtain ex = 1r, and so x = log1r. Hence
cosy + i siny = i, which means that siny = 1 and cosy = 0. Thus y =
� +21rk, k E Z. See Figure 5.11.
1
Fig. 5.11
Possible values of y when cosy+ i siny = i are given by y =
Hence if expz = 1ri, then
z E { log1r +i
Ci
� + 21l'k, k E Z.
+21rk) , k E Z} .
Vice versa, if z E log1r +i ( � +21rk) for some k E Z, then
expz =
e
log-,r
(cos
(i +21rk) +i sin (i +21rk)) =1r(0 +i · 1) = 1ri.
Consequently, exp z = 1ri if and only if z E { log1r +i (
i +21rk) , k E Z} .
Solution to Exercise 1.33
Let ,(t) :=exp( it), t E [0, 21r]. Then
,(t) =exp( it) = e0 ( cos t +i sint) =cost+ i sin t.
The point ( cost, sint) lies on a circle of radius 1 and center (0,0), and with
increasing t, this point moves anticlockwise. Hence the curve t r-+ 1(t) is
the circle traversed in the anticlockwise direction, as shown in Figure 5.12 .
197
Solutions
1(t) = cost+isint
1(0) = 1( 21r) = 1
---+------+----!---
Fig. 5.12 The curve t >-+ ')'(t) := exp(it), t E [0,27r].
Solution to Exercise 1.34
We have exp(t+it) = e t(cost+isint). So the image curve is given by
ti---+ (et cost,et sint). We have sketched this (not to scale!) in Figure 5.13.
Thus the curve is a spiral , and as t \. - oo, et(cost+isint) converges to
0, while the curve spirals outwards as t / +oo.
--
t/'+oo
t=O
Fig. 5.13
The image of the line y
= x under the map z = x + iy >-+ expz.
Solution to Exercise 1.35
We have
2
exp(z 2 ) =exp((x+iy)2 ) =exp(x2 -y2+2xyi) =ex -Y\cos(2xy)+isin(2xy)),
198
A Friendly Approach to Complex Analysis
and so I exp(z 2 )1 = ex -Y , Re( exp(z 2 )) = ex -Y cos(2xy),
2
2
Im( exp(z 2 )) = ex -Y sin(2xy). Also, we have for z-=/- 0 that
2
1
- iy
)
exp ! = exp (- -.-) = exp ( :
2
Z
X
= e '"
2
�Y
2
X
+iy
+y
) + i sin ( 2
cos ( 2
x �\2
x
(
and so it follows from here that
lexp
2
2
2
�I= e
7
y2
and
)) ,
:c2�y2 ,
Re ( exp �) = e :c
2
2
2
2
Im ( exp �) = e :c
�Y
�Y
7
7
cos ( 2
x
sin ( 2
x
y2
y2
),
).
Solution to Exercise 1.36
For z1, z2 EC, we have
(sin z1)(cos z2 )+ (cos z1)(sin z2 )
exp(iz1) � exp(-iz1)
exp(iz2 ) + exp(-iz2 )
)
) (
=(
i
2
exp(iz1) + exp(-iz2 )
exp(iz2 ) � exp(-iz2 )
) (
)
+ (
i
2
2 exp(i(z1 + z2 )) - 2 exp(-i(z1 + z2 ))
. ( + )
=��
=
�.
�
Solution to Exercise 1.37
Let z = x + iy, where x,y ER Then
cos z = cos(x + iy) = (cos x)(cos( iy)) - (sin x)(sin(iy))
= (cosx) (
e- Y
+ eY
2
e- Y _ eY
)
) - (sinx) (
2i
sinhy
= (cosx)(coshy) - (sinx) (-- -)
i
= (cosx)(cos hy) - i(sinx)(sinhy).
199
Solutions
Thus
I coszl 2 = (cosx)2 (coshy)2 +(sinx)2 (sinhy)2
= (1- (sinx)2 )(coshy)2 +(sinx)2
=
(
e2Y -
2 + e-2Y
·
)
4
e2Y +2+e- 2Y
(coshy)2 - (sinx)2 (coshy)2 +(sinx)2 (
- 1)
4
= (coshy)2 - (sinx)2 (coshy)2 +(sinx)2 ((coshy)2 - 1)
= (coshy)2 -�+�- (sinx)2
= (coshy)2 - (sinx)2 .
Solution to Exercise 1.38
Let z
= x+iy, where x, y E JR. Then cos z = 3 gives
= 3,
(sinx)(sinhy) = 0.
(cosx)(cosh y)
(5.15)
(5.16)
We note that sinhy = 0 if and only if y = 0. But y = 0 is impossible,
since this would mean z (= x + iy = x) would be real, and there are no
real solutions x to cosx = 3! Thus we must have sinhy cf. 0, and so (5.16)
implies that sinx = 0. Hence x E {mr: n E Z}. But then cosx = ±1. As
eY + e- Y
> 0 for all y E IR,
2
we see from (5.15) that cosx can't be -1. So x E {21rn : n E Z} and
cos x = l. Then cosh y = 3 yields
coshy =
eY + e- Y
---=3 '
that is,
(eY )2 - 6eY +1
eY
= 0. Thus
2
6
"9=1
= ± � = 3 ± y�
-l = 3 ± 2V2 '
2
and soy= log(3+2J2°) or
y = log(3- 2V2) = log
9�
3+2 2
= log
J2 = - log(3+2V2).
1
3+2 2
200
A Friendly Approach to Complex Analysis
Consequently, z E {21rn ± i log(3+2J2), n E Z}.
Vice versa, if z =21rn ± i log(3+2J2) for some n E Z, then
cosz = (cos(21rn))(cosh(±(3 ± 2V2))) - i(sin(21rn))(sinh• • • ))
"-v---'
'-..--'
=1
=0
=cosh(±(3 ± 2V2)) =
log(3+2v'2)
e
+ e- log(3+2v'2)
2
3+2J2+3 - 2J2
3+2J2+(3+2J2)-l
=3_
=
=
2
2
Consequently, cos z =3 if and only if z E {27rn ± i log(3+2J2), n E Z}
Solution to Exercise 1.39
0
Fig. 5.14
The set
{z E IC: z =f. 0,
¾ < IArg(z)I <
n-
Solution to Exercise 1.40
We have
Log(l+i) =Log ( V2 ( � +i �)) =Log(V2 (cos�+i sin�))
=Log(V2 exp(i�)) =log V2+ii
Solution to Exercise 1.41
We have
Log(-1) =Log(l · exp(i1r)) =log l+i1r =0+i1r =i1r,
Log(l) =Log(l · exp(iO)) =log 1+iO =0+iO =0.
Solutions
201
With z = -1, we have Log(z 2 ) = Log((-1) 2 ) = Log(l) = 0, while 2 •
Log(z) = 2 · Log(-1) = 2 • i1r. So we see that when z = -1,
Log(z 2 ) = 0 =/- 2 · i1r = 2 · Log(z).
Solution to Exercise 1.42
Let A:= {z EC: 1 < z < e}. Then z EA if and only if z = rexp(iArg(z)),
where 1 < r < e and Arg(z) E (-1r,1r]. For such a z,
Log(z) = Log(rexp(iArg(z))) = log r +iArg(z)
and O = log 1 < log r < log e = 1. So the image lies in the rectangle
]I:= {x +iy: 0 < X < 1, -7r < y < 1r}.
Log
�
Vice versa, if x + iy E ll, then z := exp(x + iy) = ex exp(iy) E A since
lzl = ex E (1, e), and log(z) = Log(e x exp(iy)) = log ex + iy = x + iy.
Hence the image of A under Log is precisely ll.
Solution to Exercise 1.43
The principal value of (1 +i) 1 -i is exp((l - i)Log(l +i)). We have
Log(l +i) = Log ( v'2 exp (i�)) = log v'2 +ii.
Thus the principal value of (1 +i) 1 -i is
exp( (1 - i)Log(l +i))
= exp ( (1 - i) (log v'2 +ii))
= e10g v'2+¾ exp (i (i - log h))
In ... (1 +i)
exp(-i log vIn2)
v'2
= v � e4
= e l (1 +i) ( cos(log v'2) - i sin(log v'2)).
202
A Friendly Approach to Complex Analysis
Solutions to the exercises from Chapter 2
Solution to Exercise 2.1
For z -=/- 0, we have
f(z) - f( 0) _ = lzl 2 - 0 = ff .
0
z
z- 0
z- 0
Thus given€> 0, we setb =€> 0, and then whenever 0 < lz-01 = lzl <b,
f(z)-f(O) 01 = lffl = ff = lzl < b =
z-o
z
lz l
Hence f is complex differentiable at 0, and f'(0) = 0.
l
€.
Solution to Exercise 2.2
Let w0 E D*. Then wo E D. Since f is holomorphic in D, given € > 0,
there exists ab> 0 such that whenever 0 < lz- w0 1 <b, z ED and
I
f(z) - !J_Wo) J'
- (wo)I <
z-wo
€.
Now let w be such that 0 < lw-wol < b. But then
0 < lw-wol
= lw-wol = lw-wol <b,
1j(W)-f(Wo)
and so w ED. Thus w E D*. Moreover,
l
f*(w) -f*(wo)
-f'(wo)I =
w-wo
(5.17)
w-wo
-f'(wo)I
= I f(wj_ - !J_Wo) - f'(wo)I
w-wo
= I f(w_}_ !J_Wo) - J'(wo)I < € (using (5.17)).
W-Wo
So f* is complex differentiable at w0 and (f*)'(wo) = f'(wo)- As wo ED*
was arbitrary, J* is holomorphic in D*.
Solution to Exercise 2.3
Since f is complex differentiable at z0, there exists an r > 0 and a function
h: D(zo,r) ➔ <C, where D(zo,r) := {z E (C: lz-zol < r} CD such that
f(z) = f(zo) + (f'(zo) + h(z))(z - zo) for lz-zol < r,
203
Solutions
and
lim h(z) =0.
We can choose an r' < r so that in D*(zo,r'):= {z E <C: 0 < lz - zol < r'}
(c D(zo,r) CD), we have lh(z)I < 1. Now given E > 0, set
z➔zo
, r' .
{ lf'(zoE)I + l }
Then whenever 0 < lz - zol < 8, we have z ED(zo,r') and so
8 =min
IJ(z) - f(zo)I =lf'(zo) + h(z)llz - zol :::; (IJ'(zo)I + lh(z)I)
lf'(zoE)I + 1
< (lf'(zo)I + 1)
=.
lf'(zo)E I + 1
E
Hence f is continuous at zo.
Solution to Exercise 2.4
Using the fact that f, g : U ➔ <C are complex differentiable functions at
z0 EU, it follows from Lemma 2.1 that there exists an r > 0, and functions
hf , hg : D(zo,r) ➔ <C, where D(zo,r) := {z E <C: lz - zol < r}, such that
for lz - zol < r
f(z) =f(zo) + (f'(zo) + h 1 (z))(z - zo),
g(z) = g(zo) + (g'(zo) + hg (z))(z - zo),
and lim h 1 (z) =0 = lim hg (z).
z➔zo
(5.18)
(5.19)
z➔zo
(1) Adding (5.18) and (5.19), we obtain for lz - zol < r that
(f + g)(z) =(f + g)(zo) + (J'(zo) + g'(zo) + hf + g (z))(z - zo),
where hf + g (z):=h 1 (z) + hg (z) in D(zo,r). Moreover,
lim h J + g (z) = lim (h1 (z)+hg (z)) = lim h 1 (z)+ lim hg (z) =0+0 =0.
So by Lemma 2.1, it follows that f + g is complex differentiable and
(f + g)'(zo) =f'(zo) + g'(zo)(2) Multiplying (5.18) by a, we obtain for lz - zol < r that
z➔zo
z➔zo
z➔zo
z➔zo
(a· f)(z) =(a· f)(zo) + (a· J'(zo) + h0 - J (z))(z - zo),
where h0 . 1 (z):=a· h 1 (z) in D(zo,r). Moreover,
lim h0 . 1 (z) = lim (a· h 1 (z)) =a· lim h 1 (z) =a· 0 =0.
z➔zo
z➔zo
So by Lemma 2.1, it follows that a · f is complex differentiable and
(a· f)'(zo) =a· f'(zo)z➔zo
204
A Friendly Approach to Complex Analysis
(3) Multiplying (5.18) and (5.19), we obtain for lz - zol < r that
(fg)(z) = (fg)(zo) + (J'(zo)g(zo) + f(zo)g'(zo) + h19 (z)) (z - zo),
where
h19 (z):= f(zo)hg (z)+g(zo)h1(z)+(z-zo)(J'(zo)+h1(z))(g'(zo)+h9 (z))
in D(z0,r). Moreover,
lim h19 (z)
z➔zo
= J(zo) · 0 + g(zo) · 0 + 0 · (J'(zo) + 0) · (g'(zo) + 0) = 0.
So by Lemma 2.1, it follows that fg is complex differentiable and
(fg)'(zo) = J'(zo)g(zo) + J(zo)g'(zo)Solution to Exercise 2. 5
Suppose that Hol(l[])) is finite dimensional with dimension d. Then the d+ 1
vectors 1, z, z2, • • • , zd E Hol(l[])) must be linearly dependent. So there exist
scalars ao, · · · ,ad, not all zeros, such that
ao · 1 + a1 · z +···+ad · zd = 0 (z E ][])).
Let k E {0, 1, · · · , d} be the smallest index such that ak -/- 0. Then by
differentiating k times and evaluating at 0 E ][]), we get
0 + ak · k! + 0 = 0,
and so ak
= 0, a contradiction.
Solution to Exercise 2. 6
Let z0 E U. Since f is holomorphic at z0, there exists a r > 0 and a
complex-valued h defined on D(zo,r) := {z EC: lz - zol < r} CU such
that
f(z) = J(zo) + (J'(zo) + h(z))(z - zo), z E D(zo,r),
and
lim h(z)
z➔zo
= 0.
Thus with g := 1/ f, we have
-1
g( z)
=
-(1
g zo)
+ (J'(zo) + h(z))(z - zo),
(5.20)
205
Solutions
and so g(zo) = g(z) + (f'(zo) + h(z))g(zo)g(z) · (z - zo)- Rearranging, we
obtain
g(z) = g(zo) + (-f'(zo)g(zo)g(z) - h(z)g(zo)g(z)) · (z - zo)
h(z)
f'(zo)
f'(zo)
f'(zo)
) (z zo)
= g(zo) + ( - (f(zo))2 + (f(zo))2 f(zo)f(z) f(zo)f(z)
f'(zo)
= g(zo) + ( - (f(zo))2 + <p(z)) · (z - z0),
where
h�)
f'(�)
f'�o)
<p(z) ·- ---2 - --z E D(zo,r) .
- -----,--------,-------,--,(
f(zo))
f(zo)f(z)
f(zo)f(z)'
.
Using the continuity off at zo and (5.20), we obtain
Hence g is differentiable at z0 and
f'(zo)
'
g (zo) = - (f(zo))2.
Solution to Exercise 2. 7
We have already shown this for m 2: 0. Suppose now that m = -n for
some n E N. Then consider the map
z f---7 zm = z -n =
1
zn
=
1
f(z)'
where f(z) := zn , z E C \ {O}. Since f is holomorphic and pointwise
nonzero in C \ {O}, it follows that 1/ f is holomorphic too, and
1 '
1
1
m-1
f'(z)
nzn -l
( ) (z) = = -n. n +l = m. z-m+l = mz
=
2
(f(z))
f
z
(zn )2
This completes the proof.
Solution to Exercise 2.8
Let f : lIJ) -t C be defined by
z
f(z)=_l+ ,
1-z
ZE
lIJ) '
206
A Friendly Approach to Complex Analysis
and
g: C--+ C be define d by g(z) = expz, z EC.
go f is holomorp hic in ][}), and more ove r,
and so
(go f)'(z)
= g'(f(z) ) •
1+ z
1+ z • .!}_
)
)
= exp ((1-z
dz
1-z
f'(z)
= exp (-��;)· (-(l+z):
= exp (=-
2
z
C� )- � d�(l+z))
l z
z
l+z
(l+z)
) (-z ·
( 1 -z)2
1
(1-z)2
exp
z))
d
1+
Thus - (exp (--1- z
dz
The n f(][})) c C = D9,
(- 1 + z)
1-z
2
=---( 1- z)2
exp
1
1
-z
)
z) z
1+
(--1 -z
,
E][})_
Solution to Exercise 2. 9
Le t
z = x + iy, whe re x, y E R We have lzl 2 = x2 + y2 . So if u, v de note
the re al p art and the imaginary p art of lzl 2 , the n we have u = x2 + y2 ,
v = 0. So
Since
au
2x,
ax=
au
= 2y,
ay
z
f=. 0, it follows that
x
av
=0,
ay
av
O
ax=
or y is nonze ro.
Cauchy-Rie mann equations is violate d, since e ithe r
So at le ast one of the
So lzl 2 is not diffe re ntiable at any nonze ro comp lex numb e r.
Solution to Exercise 2.10
Le t
z = x + iy,
z3
whe re
x,y ER The n
2
3
3
= (x +iy) = x + 3x (iy) + 3x(iy) + (iy)
2
= x3 - 3xy2 + i(3x2 y - y
3
).
3
207
Solutions
So if
u, v
denote the real and imaginary parts of z3, then we have
= x3 - 3xy2 and
v(x, y) = 3x2 y - y3.
u(x, y)
We have
u, v
are continuously differentiable, and
av
au 3 2 3 2
= x - y
ax
ay'
av
au
-=-6xy= --.
ax
ay
and
So the Cauchy-Riemann equations are satisfied everywhere in IR 2 • Hence
z r-+ z3
is entire.
Solution to Exercise 2.11
Let
z = x+iy,
where
x, y
Re(z) = Re(x+iy) = x.
Re(z), then we have
ER We have
denote the real and imaginary parts of
So if
u, v
U=X,
V
So we have
= 0.
au
av
=I-:/- 0 =
for all (x, y) E
ax
ay
2
IR .
Thus the Cauchy-Riemann equations are satisfied nowhere in IR2 • Hence
Re(z)
is not complex differentiable at any point in C.
Solution to Exercise 2.12
Let
u, v
be the real and imaginary parts of
au
ax
=
av
ay
= 0 and
Hence
u(x, Yo) - u(xo, Yo) =
au
ay
1.
x
xo
f.
=
-
Then
_ av
ax
v=
0. Thus
= 0.
au
(�, Yo)d�
aX
= 0
whenever the straight line segment joining ( x, Yo) E D to ( xo,
in D. Similarly,
u(xo, y) - u(xo, Yo) =
1,
Y au
Yo 8
y
(xo, 17)d17 =
0
Yo)
E D lies
208
A Friendly Approach to Complex Analysis
whenever the straight line segment joining (xo, Yo) E D to (x0, y) E D lies
in D. So the value of u along horizontal and vertical line segments lying
inside D is constant. As D is path-connected, it now follows that u is
constant in D (because any two points in D can be joined by a stepwise
path). Hence f = u + iO = u is constant in D.
Solution to Exercise 2.13
OU .av
+i
Let u, v be the real and imaginary parts off. Then f'(z) =
ox
ox
gives
=0
8u_ov_ .
0 D
ox-ox- m '
and using the Cauchy-Riemann equations, also that
Hence
u(x, Yo) - u(xo, Yo)=
1.
ou
!l(t, Yo)dt = 0
XO UX
x
whenever the straight line segment joining (x, Yo) E D to (xo, Yo) E D lies
in D. Similarly,
u(xo, y) - u(xo, Yo) =
1
y ou
(xo, 'TJ)d'TJ = 0
Yo 8
y
whenever the straight line segment joining (xo, Yo) ED to (xo, y) E D lies
in D. So the value of u along horizontal and vertical line segments lying
inside D is constant. As D is path-connected , it now follows that u is
constant in D (because any two points in D can be joined by a stepwise
path). Similarly v is also constant in D. Consequently, f = u + iv is
constant in D.
Solution to Exercise 2.14
We have using the chain rule that
av
OU
(x, y) = hI (v(x, y)) (x, y),
ox
ox
ov
OU
(x, y) = hI (v(x, y)) (x, y).
oy
oy
209
Solutions
Using these and the Gauch-Riemann equations, we obtain
�:(x,y) = h'(v(x,y)) · �:(x,y) = h'(v(x,y)) · �:(x,y)
= h'(v(x,y)) • (h'(v(x,y))�:(x,y)) = (h'(v(x,y)))2 • ;:(x,y)
= -(h'(v(x,y)))2
• �: (x,y),
au
(x,y) = 0. As (1 + (h'(v(x,y)))2 ) � 1
ay
au
(x,y) = 0.
ay
Using the Cauchy-Riemann equations, we also obtain
av
au
= .
(x,y) =
- ay(x,y) 0
ax
From this, it follows that
and so (1 + (h'(v(x,y)))2 )
> 0,
�:(x,y) = h'(v(x,y));: (x,y) = h'(v(x,y)) • 0 = 0,
and using the Cauchy-Riemann equations again, we have
av
au
(x,y) = 0.
(x,y) =
ay
ax
But now it follows that u is constant locally along horizontal and along
vertical lines. Since D is a domain, it is path-connected, and so we know
that any two points in D can be connected by a stepwise path. This implies
that u is constant in D. Similarly v is constant in D. So f = u + iv is
constant in D as well.
Solution to Exercise 2.15
("If'' part) Suppose k = 2. Then
f(z) = x 2 -y2 + 2xyi = x 2 + (iy) 2 + 2x( iy) = (x + iy) 2 = z 2,
which we have seen is entire in Example 2.1.
("Only if"' part) Now suppose that f is entire. Then the Cauchy-Riemann
equations must be satisfied everywhere in IR2 • So in particular,
av
au
for all x,y E IR.
= 2x = kx =
ax
ay
If we take x = 1, then the above yields that k = 2.
A Friendly Approach to Complex Analysis
210
Solution to Exercise 2.16
z
0
0
We have that the length of z-zo is lzol tan(d0) � lzold0, while the length
of zn-z0 is \z0 \ n tan(nd0) � \zol n nd0, and so the magnification produced
locally by z c-+ zn is given by
lzn-zlJ'I
lz-zol
�
\zo\ n nd0
lzo\d0
= n lzol n -1.
Also, from the picture, we see that the anticlockwise rotation produced
locally is equal to (n-1)0. Hence
f'(zo)
Consequently,
d
= n \zol n -I(cos((n-1)0) + isin((n-1)0) )
= n (lzol (cos0 + isin0n n -1 = nz;-1.
zn
dz
= nzn -I, z EC.
Solution to Exercise 2.17
i( y
+ 8)1---------<>z
Fig. 5.15
t5
iy1----------<>Zo
expz
exp
�
expzo
Calculation of the amount of local magnification produced by exp.
From Figure 5.15, we see that the magnification produced is
ex . t5
-8
= ex .
211
Solutions
expz
exp
Zo
Z
iy1-------<)--�o-<>
y
expzo
�
y
From the picture, we see that the anticlockwise rotation produced is y.
Thus the complex derivative at z0 must be e "' (cosy+isiny) = exp(x+iy).
Consequently, exp' z
= exp z.
Solution to Exercise 2.18
Let
zo E C. Move the point z0 along the line with slope 1 through a
distance of o to get the point z. Similarly move z0 horizontally to the left
by a distance
o to get the point z.
Re(•) is complex differentiable at
Suppose that the real part mapping
z0•
In Figure 5.16, by looking at the
images of z, z under the mapping Re(·), we see get conflicting values of the
local rotation produced as being
45 °
0 ° , which cannot happen.
at z0• As the choice of zo E
and
map cannot be complex differentiable
arbitrary, the map is complex differentiable nowhere.
z
0
0
Zo
z
Re(·)
�
C was
Clockwise
45 °
rotation
No rotation
�
Re(z)
Fig. 5.16
So the
Non complex differentiability of Re(·).
,)
Re(zo) Re(z)
212
A Friendly Approach to Complex Analysis
Solution to Exercise 2.19
For f = u + iv with smooth u, v E C2, that is, twice continuously differen­
tiable, we have
213
Solutions
Solutions to the exercises from Chapter 3
Solution to Exercise 3.1
We have ,-1 =cos t+ i sin t, ,-2 =cos(2t)+ i sin(2t), and ')'3 =cos t - i sin t,
and in each case, (Re(1'k(t)))2 + (In1(1'k))2 = 1, k = 1,2,3, and so the
image of ')'k is contained in the circle 11' with center O and radius 1. Also if
z =exp(i0) with 0 E [O, 21r), then z = ,-1(0) = ,-2(0/2) = ')'3(21r - 0), and
so every point on 11' belongs to the image of each of the curves ,-1, ,-2, 1'3 ·
We have
1
1
1
1
1
1'i
-dz=
z
1'2
!dz = f
z
lo
1'
!dz= f
z
lo
3
O
2
1r
--1 .- • i exp(it)dt =21ri,
exp(it)
21r
. • 2i exp(2it)dt =41ri,
exp 2it)
!
�
21r
. • (-i) exp(-it)dt =-21ri.
exp -it)
Solution to Exercise 3.2
Let ,-(t) = x(t) + iy(t), t E [O, 1], where x, y are real-valued. Also, let u, v
be the real and imaginary parts of f, respectively. Then
f' (,'(t)) · ,-' (t) = ( �: (x(t), y(t)) + i :: (x(t), y(t))) (x'(t)+ iy'(t))
au
av
= a (x(t), y(t)). x'(t) - a (x(t), y(t)) · y'(t)
x
x
+i ( �: (x(t), y(t)) · y'(t)+ �: (x(t), y(t)) · x'(t))
au
au
= o (x(t), y(t)) · x'(t) + a (x(t), y(t)). y'(t)
y
x
+i ( �� (x(t), y(t)) · y'(t)+ :: (x(t), y(t)) · x'(t))
(using Cauchy-Riemann equations)
d
d
= u(x(t), y(t)) + i v(x(t), y(t)) (using the Chain Rule)
dt
dt
= !(u(x(t), y(t)) + iv(x(t), y(t))) = !f(,-(t)).
214
A Friendly Approach to Complex Analysis
Solution to Exercise 3.3
Let, be the c ir cular path ,(t)=2exp(it), t E [0,27r].
I,
(1) We have
(z+z)dz=
121r
1
(2exp(it)+2exp(-it)) • 2i • exp(it)dt
=4i
I,
(2) We have
(z 2
-
2
O 7'
121r
121r
2z+3)dz=
=
(exp(2it)+l)dt=4i • 0+4i • 271"=87ri.
(4exp(2it) - 4exp(it) +3) • 2i • exp(it)dt
i(8exp(3it)- 8exp(2it)+6exp(it))dt
=0+0+0=0.
I,
(3) We have
xydz=
121'
1
1-'Ir.._,._.,
1'
2 cos t-2 sint-2i·(cost+i s int)dt
=4i
2
O 7'
(sin(2t))(cos t+i s int)dt
(s in(2t)) cos t dt- 2
=4i
odd function
121' (
0
cos t- cos(3t))dt
=0- 2(0- 0)=0.
Solution to Exercise 3.,4.
(1) ,(t)= (1+i)t, t E [0,1), and so
1
'Y
1
Re(z)dz=
O
1
l+i
t(l+i)dt= -.
[ 7r/2,0], and so
(2) ,(t)=i+exp(it), t E o
o
Re(z)dz=l (cos t)iexp(it)dt =l i(cost)2
1
'Y
/
o 2
-7r
=l
-1r/
2
(i
cos(2t)+1
2
/
2
_ s in(2t)
)dt
-7r
2
. 1 7r 1 1 .71"
=0+i · - · - + - = - +i-.
2 2 2 2
4
2
-
(cost)(sint)dt
215
Solutions
(3) 1(t) = t + it2, t E [O, 1], and so
1
7
Re(z)dz =
1
o1
1
t · (1 + 2it)dt = 2
1
1
2
+ 2i • -3 = -2 + -3 i.
Solution to Exercise 3. 5
By the Binomial Theorem,
Thus for O :S k :S n,
+ zr = � (n) l-k-1
'
� f, Z
zk+l
l=O
(1
and so
21ri
_1
1
C
(1 + zr dz= _1
z k+l
21ri
1t
C
(n)zl-k-ldz
l=O £
Solution to Exercise 3.6
Let u,, VJ, U9, Vy : [a, b] --+ IB. be such that f('Y(t))'Y'(t)
g('Y(t))'Y'(t) = U9(t) + iVg(t), t E [a, b]. Then
1
(f + g)(z)dz =
= U1 (t) + iV1 (t),
b
l (f + g)('Y(t)) • "f'(t)dt
b
= l (!b(t)) • "f'(t) + g('Y(t)) • "f'(t))dt
b
b
= l (U,(t) + Ug(t))dt + i l (V,(t) + Vg(t))dt
b
b
b
b
= l u1 (t)dt + l U9(t)dt + i l v1 (t)dt + l Vg(t)dt
=
1
f(z)dz +
1
g(z)dz.
216
A Friendly Approach to Complex Analysis
(2) Let a = p + iq where p, q E IB., and let U, V : [a, b] ➔ IB. be such that
f('y(t)) · 'Y'(t) = U(t) + iV(t), t E [a, b]. Then
1
b
(a· f)(z)dz = 1 (p + iq)(U(t) + iV(t))dt
b
b
= 1 (pU(t) - qV(t))dt + i 1 (pV(t) + qU(t))dt
b
b
b
b
= p(1 U(t)dt + i1 V(t)dt) + iq(1 U(t)dt + i1 V(t)dt)
b
b
= (p+iq)(1 U(t)dt+i 1 V(t)dt) =a·
1
f(z)dz.
Solution to Exercise 3. 7
-(-'Y): [a, b] ➔ <C is given by
(-(-'Y))(t) = (-'Y)(a + b - t)
= 'Y(a + b - (a+ b - t)) = 'Y(t), t E [a, bl,
and so -(-'Y) = 'Y· (This is obvious visually.)
Solution to Exercise 3.8
We do have that "f(b) = (-'Y)(a), and so the two paths 'Y and -'Y can be
concatenated, and we have
1
-r+(--y)
f(z)dz =
1
')'
f(z)dz + f f(z)dz =
J_')'
1')'
f(z)dz
-1')'
f(z)dz = 0.
Solution to Exercise 3.9
Let 'Y: [O, 1] ➔ <C be defined by 'Y(t) = (1 + i)t, t E [O, 1]. By Pythagoras's
Theorem, the length of 'Y is Jl 2 + 1 2 =
v'2.
l+i
1
0
1
Also, l('y(t))2 1
We have
1
and so
7
217
Solutions
= It+itl2 = 2t2 , and so max l('y(t))2 1 = 2 · 1 2 = 2. Thus
z2dzl
11"I
tE[0,1]
:S ( max j('y(t)) 1) · (length of 'Y) = 2V2.
2
tE[0,1]
1
1
( +i)3
,
z2 dz = ( (t+it)2 · (1+i)dt = ( (1 +i)3 t2 dt = l
lo
lo
2v12
)3
J2z2 dz = = - -.
3
3
117
I
(
3
Solution to Exercise 3.10
We have
(2:) I (2:)1 I la
=
=
2�i
(
n
l ;:t
dzl
+ l n
( +_ )_ _n I) .
<- 2_
271" . 1 = max l_I _ _z__
(max - l_ z
1
271" izl=l 1 z
izl=l
:S (1+1) n = 2 n = 4 _
2
2
n+l
2
2
n
Solution to Exercise 3.11
Suppose that F = U+iV is a primitive of z in C. Then
au
. av
-+iax
ax
. . lT])2
ln JN.. •
= - - i- = F' = Z = X - iy
av
. au
ay
ay
_
Fix x0 ER Then for (x,y) E JR2, we have
V(x,y) - V(xo,y) =
1x 8V (�,y)d� = 1x -yd�= -xy + xoy.
a
xo
X
xo
So V(x,y) = -xy+cp(y), where cp(y) := V(xo,y) +xoy. Hence
X
=
aV
ay
= -X+'{)'(y),
that is, cp'(y) = 2x for all x E JR, which is clearly impossible, since in
particular, we would obtain 2 · 1 = 2 = cp'(y) = 2 • 0 = O!
218
A Friendly Approach to Complex Analysis
Solution to Exercise 3.12
The map ( i---+ f(()g'(() + f'(()g(() has a primitive since (Jg)'= f g' + f'g.
So by the Fundamental Theorem of Contour Integration,
1
(!(()g'(() + f'(()g(()) d( = f(z)g(z) - f(w)g(w),
and a rearrangement proves the claim.
Solution to Exercise 3.13
We have sin ' z = cos z, and so cos z has a primitive in IC. Thus by the
Fundamental Theorem of Contour Integration
· i") e -1 -e1
exp (.i·i") - exp (-i·
..
=
COS Z dZ ...
= Sln i- Sln (-i.) = 2Sln
i= 2
.
.
2i
i
1'Y
=(e-¾)i.
Solution to Exercise 3.14
Since exp' z = exp z in C, we have
1
1
exp zdz = exp(a+ib)-exp O = ea (cos b+i sin b)-1 = ea cos b-1+iea sin b,
for a path 'Y joining O to a + ib.If 'Y(x) =(a + ib)x, x E [O, 1], then
exp zdz = fo\xp(a+ib)·(a+ib)dx =
Hence (a -ib)
(a2 + b2)
1
exp zdz=
fo 1 e x(cos(bx)+i sin(bx))(a+ib)dx.
a
fo 1 e x(cos(bx) + i sin(bx))(a + b )dx.Thus
a
fo 1 e x cos(bx)dx = Re((a -ib) 1 exp zdz)
2
2
a
= Re((a -ib)(ea cos b - 1 + iea sin b))
Hence
11
=a(ea cos b - 1) + bea sin b.
a
a
_ a(e cos b - 1) + be sin b ·
e a x cos(bx)dx 2
2
a +b
o
219
Solutions
Solution to Exercise 3.15
Consider the closed circular path C with center at O and radius r > 0
traversed in the anticlockwise direction: C(0) = rexp(i0), 0 E [0,21r]. By
the Fundamental Theorem of Contour Integration, we have
0
=
=
la
exp zdz =
121r
121r e
r cos 0+ir sin 0
. ri exp(i0)d0
e r cosO • r • i • exp(i(rsin0 + 0))d0.
So we obtain in particular (by equating real parts) that
{21'
e r cosO cos(rsin0 + 0)d0 = 0.
Jo
Solution to Exercise 3.16
Suppose Fis holomorphic in C\ {0} and that F' = 1/z in C\ {0}. Consider
a circular path C with a positive radius centered at O traversed in the
anticlockwise direction. Then by the Fundamental Theorem of Contour
Integration, since C is closed, we have
la
1
F'(z)dz = 0.
On the other hand, we know that
C
F'(z)dz =
1
�dz= 21ri,
cZ
a contradiction.
Solution to Exercise 3.17
(ERl) Let 'Y : [0, 1] -+ D be a path which is closed. Define the map
H: [0, 1] x [0, 1]-+ D by H(t,s) = 'Y(t), t, s E [0, l]. Then His continuous,
and
H(t, 0) = 'Y(t), for all t E [0, 1),
H(t, 1) = 'Y(t), for all t E [0, 1),
H(0,s) = "f(0) = "f(l) = H(l,s), for alls E [0, l].
Hence 'Y is D-homotopic to itself. So the relation is reflexive.
220
A Friendly Approach to Complex Analysis
(ER2 ) Let 'Yo,'Yi: 0
[ ,1] ➔ D be closed paths such that 'Yo is D-homotopic
to'Yl· Then there exists a continuous H: 0
[ ,1] x 0
[ ,1] ➔ D such that
H(t,0)='Yo(t), for all t E 0
[ ,1],
H(t,1) ='Y1(t), for all t E 0
[ ,1],
H(0,s) = H(l,s), for alls E 0
[ ,l].
Let H: 0
[ ,1] x 0
[ ,1] ➔ D be defined by H(t,s) = H(t,1 -s) for all
t, s E 0
[ ,l]. Then ii is continuous and
[ ,1],
H(t,0) = H(t,1) = '}'1 (t), for all t E 0
H(t,1) = H(t,0) ='Yo(t), for all t E [0,1],
H(O,s) = H(0,1- s) = H(l,1- s) = H(l,s), for alls E 0
[ ,1].
Thus'Yl is D-homotopic to'YO· Hence the relation is symmetric.
(ER3) Let 'Yo,'Y1, '}'2 be closed paths such that 'Yo is D-homotopic to '}'1
and 'Yi is D-homotopic to '}'2. So there exist two continuous functions
H,K:[0,1] x 0
[ ,1] ➔ D such that
H(t,0) = 'Yo(t), for all t E 0
[ ,1],
K(t,0)='Y1(t), for all t E [0,1],
K(t,1) ='Y2(t), for all t E 0
[ ,1],
H(t,1) = 'Yi (t), for all t E 0
[ ,1],
H(0,s) = H(l,s), for alls E [0,1], K(0,s) = K(l,s), for alls E [0,1].
Let L: [0,1] x 0
[ ,1] ➔ D be defined by
sE O
H(t,2s)
[ , ½L
L(t,s) = {
K(t,2(s- 21 )) s E (21 ,1]
Then
L(t,0) = H(t,0) = 'Yo(t), for all t E 0
[ ,1],
[ ,1].
L(t,1) =K(t,1) = ','2 (t), for all t E 0
Also, for 0 ::; s ::; ½, L(0,s) = H(0,2s) = H(l,2s) = L(l,s), and for
½ < s ::; 1,
L(O,s)=K(0,2(s-�)) =K(l,2(s-�)) =L(l,s).
Moreover, if ((tn ,sn ))nE N is a sequence in 0
[ ,1] x
(to,½), then
(½,1] that converges to
lim L(tn , S n ) = lim K(tn ,2(sn - � )) = K(to,0) = '}'1 (to)
n-too
2
n-too
= H(to,1) = L (to, �) =L ( }�� (tn , S n )) .
Hence it follows that L is continuous. Consequently 'Yo is D-homotopic to
,.,,2. So the relation is transitive.
As the relation of D-homotopy is reflexive, symmetric and transitive, it
is an equivalence relation.
221
Solutions
Solution to Exercise 3.18
From the picture , we see that C is C \ {O}-homotopic to S.
-1
+i
"/2
s
"/3
"/1
-1-i
Let
for
t
"/4
1-i
'Y1 ( t ) := (1- t )(l-i)+ t (l+i) = 1+i(2 t - 1),
'Y2( t ) := (1- t )(l+i)+ t (-1+i) = (1- 2 t )+i,
"f3( t ) := (1- t )(-1+i)+ t (-1-i) = -1 +i(l- 2 t ),
"f4( t ) := (1- t )(-1-i)+ t (l-i) = 2 t - 1-i,
E [O, l]. Then
We have
1
1
1
1
f !dz =
!dz.
!dz+
!dz+
!dz+
�z
�z
nz
�z
kz
1
2i
_ 1 2i(l-i(2 t - 1))
/
dt /
dt
}0 l+i(2 t - 1) - }0 1+ (2 t - 1)2
1
1
2 t- 1
1
/
2
dt
+
2i /
dt
2
}0 l+ (2 t - 1)
}0 l+ (2 t - 1)2
1
U
d
1 d
(u=2t-1) i· jl -t+ 1 --2 t
=
2
-1
1
+u
-1
1
+u
i(tan-11- tan- 1 (-1))+ 0 = i(�- ( - �)) =ii.
Similarly, using the facts that
1
1
/
1r
2
2 t- 1
/
dt =
and
dt = O,
2
2
}0 l+ (2 t - 1)2
lo l+ (2 t - 1)
222
A Friendly Approach to Complex Analysis
we obtain
1
1
1
-2 dt= f -2·(-(2t- 1) - i)dt
! dz = (
1+(2t- 1)2
lo 1 - 2t+i
lo
z
72
1
73
.
.
=0+(-1)(-i)
2 =i2,
7f
7f
1
1
-2i
-2i-(-l+i(2t- l))dt
! dz = (
dt= f
lo
1+(2t- 1)2
lo -1+i(l - 2t)
z
= -i•(-1)-�+0=i� and
2'
2
l
1
2
2·((2t- l)+i)dt
! dz = f
dt= f
lo 2t- 1 - i
lo 1+(2t- 1)2
74Z
1
.
.
= 0 +i=i-.
2
2
7f
Thus
Is�
dz
7f
=4 · (ii) =21ri, as expected.
Solution to Exercise 3.19
For a circular path C with center O and a positive radius, traversed in the
anticlockwise direction, we have
1
- dz =21ri.
cZ1
But the elliptic path Eis IC\ {O}-homotopic to C, as shown in Figure 5.17.
C
E
Fig. 5.17 E, C are lC \ {O}-homotopic.
223
Solutions
So by the Cauchy Integral Theorem,
/ 2"
f
! dz= f ! dz= 21ri, and so
lE z
le z
I
21ri= f !dz=
. . -(-asin0+ib cos0)d0
+i
0
b sm 0
z
cos
a
lE
lo
2
/ " (-a sin0 +ib cos0)(a cos0 - ibsin0) B
d
=
a 2 (cos0)2 +b2 (sin0)2
lo
/ 2" (b2 - a 2 )(cos0)(sin0) +iab((cos0)2 +(sin0)2 ) B
d
=
a 2 (cos0)2 +b2 (sin0)2
lo
/ 2" (b2 - a 2 )(cos0)(sin0) +iab · 1 B
_
d ·
a 2 (cos0)2 +b2 (sin0)2
lo
/2,r
21r
I
d0 Equating the imaginary parts,
2 (cos0)2 +b2 (sin0)2
o
ab·
a
l
-
Solution to Exercise 3.20
See Figure 5.18.
Fig. 5.18
(1) z H Log(z - 4i) is holomorphic in IC\ {r + 4i : r :::; O}. So by the
Cauchy Integral Theorem
i
Log(z - 4i)dz= 0.
(2) If C denotes the circle with center 1 and any positive radius, say, r,
then we know that
( _]_ dz= 21ri.
I
le z-
A Friendly Approach to Complex Analysis
224
Since 1 /(•-1) is holomorphic in re\ { 1}, and since the two circular paths
C and C are re \ {1 }-homotopic, it follows from the Cauchy Integral
Theorem that
1
1
f --dz = f --dz = 27ri.
1
l
z
z
le e -1
(3) We have
(
i z - 3 = exp((z -3)Logi)= exp ( (z - 3) log 1 +ii))
= exp ( (z -3)( 0 +ii))
= exp (ii· (z -3)),
i
and so z t-+ i z - 3 is entire. By the Cauchy Integral Theorem,
i z - 3 dz = 0.
Solution to Exercise 3.21
(1) We have
<p (t)= exp
I
s)
( rt ,'(s) ds) . d ( lot ,'(
)
'Y ds = <p(t)
lo
,(s)
1
and so cp'1 - <p')' = 0. Hence
dt
�i�j �gj.
,2
( s)
!__('!'_' ) = cp',- cp,' = _Q_ = 0
dt
and so
=
,2
,'(t)
,
. ,(t)
,
But ')'(0)= 1(1) (since I is closed). Hence
cp(l) = cp(0)= exp ( lo
Consequently, w(1) E Z.
(2) We have
o
�(�j ds) = exp(0)= 1.
1 r 1 r�(t)
dt
w(I'i) =
27ri lo r 1 (t)
_1_ [ 1 27ri exp(27rit)
dt
=
27ri lo
exp(27rit)
1
= - · 27ri = 1.
27ri
225
Solutions
1
(3) We have (11· ,2)'(t) =,Ht)· ,2(t) + ,1(t),�(t), t E [0, 1], and so
1
1
(,1 . ,2)'(t) d
w(,1· ,2) = - .
t
2 7r2 o (11· 12)(t)
1
1
(t) + ,1(t),�(t) d
, )
t
= _ _ r W · ,2 .
21ri lo
,1(t) ,2(t)
1 / yi{th�(t)
1 / 'Y� (t)·yA()
dt
dt + _
=_
2ni}0 �-,2(t)
2ni}0 ,1(t)·yA(J
1
1
= w(,1) + w(,2).
(4) We have rm =r1····· r1 (m times), and so
w(rm) = w(r1) + · · · + w(r1) (m times)
= m· w(r1) =m· 1 =m.
(5) Consider the map <p: [0, 1] --+JR.given by <p(t) = l,o(t)/, t E [0, 1], which
measures the distance of 1(t) from 0. Being a continuous function, it
has a minimum value d0, and do > 0 since 10 does not pass through 0.
Take i5 =do/2 > 0. Let I be a smooth closed path such that
max l,(t) - ,ol < 8.
II, - ,olloo :=tE[0,1]
We will show that I is (C \ {0}-homotopic to ,o. Define the function
H: [0, 1] x [0, 1]--+ C\ {0} by H(t, s) =(1-s),o(t) +s,(t), t, s E [0, l].
Then H is continuous,
H(t, 0) =,o(t) for all t E [0, 1],
H(t, 1) =1(t) for all t E [0, 1], and
H(O, s) =(1 - s),o(0) +s1(0)
= (1 -s),0 (1) +s1(1) = H(l, s) for alls E [0, l].
Also we note that H(t, s) is never 0, because it is a convex combination
of ,o(t) and 1(t), and if (1 - s),o(t) + s1(t) =0 for some t, s, then we
arrive at a contradiction. See Figure 5.19.
1(t)
-----<>--------<>:
'
o--:
'
Fig. 5.19
0
< do/2
'
,o(t)
'
That H(t, s) is never 0.
226
A Friendly Approach to Complex Analysis
Indeed,
1·
do
2
=
> sl,,o(t) - 'Y(t)!
= l'Yo(t)I
bo(t) -
((1 - s)'Yo(t) + S')'(t))I =
bo(t) - OI
� do.
Thus by the Cauchy Integral Theorem,
w('Y)
l
= -.
21ri
J
J
l
1
l
-dz= -.
-dz= w('Yo)21ri 7 z
7z
Solution to Exercise 3.22
(1)
1r-nt
t
0
Q
nr
The length of the arc on the big circle subtended by LQ'0Q is t · nr. As
the small coin rolls without slipping, the angle made by 0' P with 00' is
(t • nr)/r = n • t. From the picture, we see that 0' (nr + r) exp(it), and
P
= (nr
=
+ r) exp(it) + exp(-i(1r - nt)) · rexp(it)
-------- '--v--'
produces a
O'P'
clockwise rotation
= (n + l)rexp(it) + (-1) • r • exp((n + l)it).
227
Solutions
(2) The area enclosed by the epicycloid 'Y is
1 { 27r
1
dz=
}o r((n+l) exp(it) - exp((n+l)it)).
1
2i
2i /
=
;
fo
i
r((n+l)i exp(it) - (n+l)i exp((n+l)it))dt
2 1r
r2 ((n+1) exp(-it) - exp(-(n+l)it)) •
(n+l)i ( exp(it) - exp((n+l)it))dt
2 1r
(n+l)r
= --1 ((n+1) - (n+1) exp(int) - exp(-int)+ l)dt
2
2
0
2
= (n+l)r ((n+l) · 27r+0+0+27r)
2
= (n+l)r2 ?T(n+2) = ?Tr2 (n+l)(n+2).
Solution to Exercise 3.23
Take for example f given by f(z) = 1/z for z E D := (C \ {O}. Then we
have seen that f does not have a primitive in D. (See Example 3.7 and
Exercise 3.16.)
Solution to Exercise 3.24
Let 'Y(t)
1
7
= exp(it), t E
.
i
(z - a)(az - 1)
[O, 1] and so
dz = 1
=
27r
0
2 1r
0
2,r
0
2 ,r
1
i
(exp(it) - a)(a exp(it) -1)
.
. ) dt
iexp
(it
- exp(it)
----,----,-----.....,..,.......,......,..dt
(exp(it) - a)(a - exp(-it)) exp(it)
1
-...,..dt
= 1 -----,--,------,--,------,-----,-
=1
0
2 ,r
(exp(it) - a)(exp(-it) - a)
1
--�dt
I exp(it) - al2
1
= 1 ------2
2 dt
0
2 ,r
((cos t) - a) +(sin t)
1
= 1 ----dt •
1 - 2a cost + a2
0
228
A Friendly Approach to Complex Analysis
Since the mapping z f-+ i/ (az -l) is holomorphic in a disc containing the
unit circle 'Y (because 0 < a < l), by the Cauchy Integral Formula,
1
_l_
So
1
21ri 'Y
2
0
7r
azi_ l
z
-a
1
------2 dt=
+
1- 2a cost
a
dz =
1
-I
_i_·
a z -l z=a
=
_i_· -.
a2 - 1
i.
2 7f
.
az-1 d
z=2 1r i---=--.
-2
a -l
1- a2
'Y z i a
Solution to Exercise 3.25
-1
-1
-1
( 2)
(3)
( 4)
( 5)
(2)
j
1
1
1
1
'Y
expzd
-l
Z
z2
+
: l dz
z
'Y
1
z= 21riexpzl
+ 1d =
z
'Y z - 1
2
z -l
'Y
3
(5)
(1)
(1)
1
1
(3)
(4)
= 21riexp l= 21r ie.
2
z::l d z = 21riz +
z
= 0.
z=l
1
-l
z
+
12 + 1
11
= 21ri.
= 21r i
l z=l
1+ 1
2
(-1) 2 + 1
z2 + 1d
� d z=2 1ri z + 11
=-21r i.
=21ri
z=j
2
'Yz - 1
'Yz-(-1)
z-l z=-1
-1- 1
2
z2 + 1
d z = z + 1(-1- - _1_) d z
2
z -l z + l
'Y
'Yz2 - 1
1
1
1
z +1
z +1
z2 + 1
z2 + 1
2 d
2
d z = 21ri-1
- z-21ri-- 1
'Y z-l
'Yz-(-1)
2 z=l
2 z=-1
= 21ri(l) - 21r i(l) =0.
=
2
2
Solutions
229
Solution to Exercise 3.26
Suppose F is a primitive. Consider the closed path I given by lz - =I0 ½
traversed anticlockwise. By the Fundamental Theorem of Contour Integra­
tion,
1
1
z(z}_ l)
dz=
1
F'(z)dz =
0,
.
I
since I is closed. On the other hand, by the Cauchy Integral Formula,
1
1
d
z
2
Z ( Z - l)
1
1
.
z -1 d
-z = 211'i�
1
= 211'i2 -0 l
0
Z - 1 z=O
So we arrive at a contradiction. Hence
1
z(z2 - 1)
does not p ossess a primitive in {z EC: <0 lzl < 1}.
'Y
=
2
'Y Z -
= -211'i.
Solution to Exercise 3.27
iR
T
s
i
-R
-i
R
Fig. 5.20 The path u = S + T.
(1) By the Cauchy Integral Formula, we have
1
1
1
exp(iz)
ex iz
exp (.i�)
z+i
;
dz=
. dz= 211'i
z +
z+i
z-i
(. 1)
1
e
ex
i)
(i
p
_ .
• _ .
-�
2
2
- 11'i
i +i - 11'i 2i - e ·
(2) Let z = z + iy, with x, y real and y 2:
Then
0.
O'
F(z)dz =
O'
O'
I .
z=i
I exp(iz)I = I exp(i(x + iy))I = I exp(-y + ix)I = e-Y � 1
(since y 2:
)0 Thus
.
IF(z)I =
1_
I exp(iz)I
< _
2
2
lz +ll
1z +11·
230
A Friendly Approach to Complex Analysis
But lz2 I- I - ll::; lz2 -
(-1)1 =
<
IF(z)I - 2
lz
1
1
lz
2
+ 11, and so if lzl
<
+ 11 -
1
lzl - 1
2
<
1
l
---3_
\!'2.
F(z)dzl::; 21rR • 1_!1 � IF(z)I ::; 21rR• �2
E
\!'2,
- lzl 2
since lzl 2::; 2lzl 2 - 2 for lz2 :2: 2, that is, lzl 2:
(3) We have
2:
(for R 2:
v'2)
R
_ 47f --=:_:f O .
- R
/ F(z)dz
So lim
R➔ oolT
= 0.
Since
f F(z)dz = f F(z)dz - f F(z)dz = '!!_ - f F(z)dz,
ls
la
it now follows that lim
(4) Let S(x)
1
= x, x E
e
lr
f F(z)dz = '!!__
e
R➔ ools
[-R,R]. Then
lim
lr
f F(z)dz = '!!__O = !!__
R ➔ oolr
e
e
R
· R sinx
exp(ix) l
R cosx
• dx= J -- dx+i J -dx
F(z)dz= J -- R x2 + 1
- R x2 + 1
- R x2 + 1
s
R
=J
cosx
--x
d +O
-R x2
+1
where we have used the fact that
last equality. Hence
R
lim J
R➔ oo
cosx
- R x2
+1
sin x
x2 +l
is an odd function to get the
r
F(z)dz = !!__
dx = lim
R➔ ools
e
Solution to Exercise 3.28
Consider the entire function exp z and let C : [O, 21r] -+ (C be the circular
path with center O and radius 1 given by C(0) = exp(i0), 0 E [O, 21r]. By
the Cauchy Integral Formula,
1
I
exp z
l
-- dz = expz
-.
21ri CZ - 0
z=O
= expO = 1,
231
Solutions
1
But
211"
211"
exp z
exp(exp(i0)) . .
· )d0 =i· 1 exp(expi
dz=1
iexpi
(0
( ·0))d0
.
( 0)
expi
cz- 0
0
0
2 11"
=i Jo{ exp(cos0+i sin0)d0
=i Jo{
Hence {o
J
211"
=- {
}0
211"
e cos 0 (cos(sin0)+i sin(sin0))d0
2 11"
ec089 sin(sin0)d0+i {
}0
211"
ecos0 cos(sin0)d0
ecos e cos(sin0)d0 = 21r.
Solution to Exercise 3.29
If f is holomorphic, then f(n) is holomorphic too, and so is its derivative
j(n+l). But j(n+l), being complex differentiable, is in particular continu­
ous. So f(n) has a continuous complex derivative.
Solution to Exercise 3. 30
Since for all z EC, lf(z)I � 8 > 0, in particular f(z) =/- 0 for all z EC, and
so 1/ f is entire. But
for all z EC
I I :::; �'
ftz)
and so by Liouville's Theorem, 1/ f is constant. Thus f is constant as well.
Solution to Exercise 3.31
Let g be defined by g(z) := f(z) - wo for z E C. Then g is entire and
lg(z)I � r for all z E C, and so g is bounded away from 0. Hence by
Exercise 3.30, g is constant, and so f = g + w0 must be constant too.
Solution to Exercise 3.32
Consider the compact set K := {(x, y) : 0 :::; x:::; T1 , 0 :::; y :::; T2}. The
continuous function (x, y) 1----r lf(x +iy)I assumes a maximum value Mon
232
A Friendly Approach to Complex Analysis
K. But for
mEZ
nEZ
there are integers n, m such that x + iy =xo + nT1 + i (yo + mT2) for some
xo E [O, T1) and Yo E [O, T2). Owing to the periodicity off,
f(x + iy)=f(xo + nT1 + i (yo + mT2))=f(xo + iyo) Ef(K),
and so for all x, y E JR, lf(xo + iyo)I SM. Hence f is bounded on C, and
by Liouville's Theorem, must be constant.
Solution to Exercise 3.33
(1) Let g be defined by g(z) = exp(-z) • f(z) for z EC. Then g is entire.
Moreover, since lf(z)I SI expzl , it follows by rearranging that
lg(z)j = I exp(-z) · J(z)I S 1
for all z EC. So by Liouville's Theorem, g is constant with value say,
c. Since jg(z)j S 1, we obtain jcj S1. Consequently,
g(z) = exp(-z) · f(z) =c,
and so f(z) = c · expz for all z E C, where c is a constant such that
leis 1.
(2) We know that if p is a polynomial of degree d � 1, then there exist
M, R > 0 such that
jp(z)I � Mjzj d
for jzj > R. Thus with z = x < -R < 0, we have jzj > Rand so
Mlxl d Slp(z)I Sle x l =ex S 1 ( because x < 0). Hence lxl d S 1/M for
all x < -R, a contradiction. Sop is a constant, say equal to co. But
then lp(z)I S I expzl again gives with z = x < 0 that leol S lex l = e x
for all x < 0, and so leol = 0. Consequently, p=co= 0.
Solution to Exercise 3.3,4.
(1) We have for z E C that
lz - ail � lzl - la1I= R - la1I,
lz - a2I � lzl - la2I= R - la2I-
233
Solutions
Thus
11
lf(z)I
f(z)
< max
dzl • (length of C)
zEC
lz
1
z
a2)
- a2I
a1llz
c (z a )(
M
<
· 21rR,
- -,------,----,,.....,---,--...,,..
(R-la1l)(R - la21)
where M := maxlf(z)lzEC
(2) Since a1 =/. a2, we have
and so
1
z - a1
1
z - a2
(3) We have
1
z - a2 - (z - ai)
(z - a1)(z - a2)
-1
1
f(z) _ f(z) dz
f(z)
dz
(
c (z - a1)(z - a2) - c a1 - a2 z - a1 z - a2 )
1
f(z) dz)
f(z) dz
a1 - a2
c z - a2
c z - a1
If we consider a small disc .6.1 with center a1 and radius r1 > 0 with
boundary C1, then we see that C and C1 are C\ {a1}-homotopic, and
the function
(z)
, z EC\ {ai}
g(z) := f
z - a1
is holomorphic. Thus by the Cauchy Integral Theorem,
f(z) dz.
f(z) dz=
c z - a1
c1 z - a1
f(z) dz= (a1). Thus
But by the Cauchy Integral Formula, 1.
f
21ri c1 Z - a1
f(z) dz= 21rif(a1).
c z-ai
Similarly,
f(z) dz= 21rif(a2)c z-a2
21ri(f(a1) - f(a2))
f(z)
-----.
dz = ------Consequent1y,
a1 - a2
c (z - a1)(z - a2)
1
1
1
1
-1
(1
1
1
234
A Friendly Approach to Complex Analysis
(4) Suppose that f is a bounded entire function with a bound Mon lfl,
that is, lf(z)I ::::; M for all z E C. Suppose that a1, a2 are any two
distinct points in C. Let C be a circular path with radius R > 0 and
center 0, traversed once in the counterclockwise direction that contains
a1, a2 in its interior. Then using the results in parts (i) and (iii),
I
la1- a2I · 21ri(f(a1)- f(a2))
lf(ai) _ f(a2)I =
21r
a1- a2
f(z)
= la1- a2I
dz
21r
· c (z- a1)(z- a2) I
21rRM
< la1- a2I .
21r
(R- la1l)(R - la2I)
As R can be made as large as we please and since
21rRM
--------+ 0
(R- la1l)(R - la2I)
11
I
as R --+ oo, it follows that lf(a1)- f(a2)I = 0. This implies that
J(a1) = J(a2). Hence f is constant.
235
Solutions
Solutions to the exercises from Chapter 4
Solution to Exercise 4.1
=
=
=
L
L
lm(an )Re(an ) and
Since Lan converges, so do the real series
n= l
n= l
n= l
Hence lim Re(an ) = 0 and lim lm(an ) = 0. Thus lim an = 0 too.
n➔oo
n➔oo
n➔oo
Solution to Exercise 4.2
As
=
L la l converges, and for all n EN, Re(a ) :S la l, lm(a ) :S la l,
n= l
n
n
=
=
n= l
n= l
n
n
n
=
converge by the Comparison Test. Hence Lan converges.
n= l
Solution to Exercise 4.3
With S n := l+z+• · ·+zn - 1 +zn , we have that ZS n = z+z2 +• • •+zn +zn+l,
and so (1-z)s n = 1-zn+l_ Since lzl < 1, z-/:- 1 and so 1-z-/:- 0. So
1-zn+l
S n = 1 + Z + · · · + Zn -l + Zn = --1-z
(5.21)
Thus
=
lim S
n➔= n
=
lim
n➔ =
1-zn+l
1-z
=
1-0
1-z
1
1- z'
1
zn = lim S n = --.
n➔
L..,
L..,
1
Z
n=O
n=O
(Note that we have used the fact that since lzl < 1, one has
and so
'°'
zn converges, with
'°'
=
=
lim zn+l = 0,
n➔
and this can be justified as follows: lzn+l -0I
r:= lzl < 1.)
lzJ n+l
n
..::..f 0, since
236
A Friendly Approach to Complex Analysis
Solution to Exercise 4.4
For n E N, let S n := 1 + 2z + 3z2 + • • · + (n - l)zn -2 + nzn -l_ Then
zsn = z + 2z2 + · · · + (n - l)zn -l + nzn. Hence
l-Z n
(1 - z)Sn = l + z + z2 + · · · + zn-1 - nzn = -- nzn.
l-z
Thus
l-zn
nzn
---.
Sn =
2
1-z
(1-z)
(We could have also obtained this expression from (5.21) by differentiating
with respect to z.) If we set r := lzl, then O::; r < l and so
1
r=-l+h
1
where h := - - 1 > 0. We have
r
n. ( -l) . 2 .
n
h
(1 + h) = l + (;) h + (;) h2 + ... + (:) hn � (;) h2 =
�
Hence
n
2
n (n-1) h
2
(n-1) h
=----2 '
0<nrn=---<n·----n • 2
•
•
(l + h)
and so by the Sandwich Theorem, lim nrn
n➔oo
lim S n
n➔oo
= 0. Consequently,
n
n
_
1
= lim ( 1 - Z 2 - nz ) = 1- 0 2 -_0 =
1-z (1-z)2 "
n➔oo
1-z
(1-z)
(1-z)
Solution to Exercise 4.5
I I I
We have
:s
I I
I
=
exp(s · �og(n))
exp(s�log n)
1
1
1
1
- eRe(s·logn) - e(logn)·(Re(s)) - (elogn)Re(s) - nRe(s)·
=
Recall that � _!._ converges if p > l. Hence if Re(s) > 1, then
L., nP
n=
l
1
Re(s)
n
n=l
converges. Thus
1
L
00
00
Lns
n=l
converges absolutely for Re(s) > 1, and in particular , it converges for
Re(s) > 1.
237
Solutions
Solution to Exercise 4. 6
Let L =/=- 0. We have that for all z such that lzl < 1/L that there exists a
lzl :S: q < l for
q < l and an N large enough such that ,Vlcn zn l =
n
all n > N. This is because � lzl � Llzl < l. (For example take
q = (Llzl + 1)/2 < 1.) So by the Root Test, the power series converges
absolutely for such z.
If L = 0, then for any z E C, we can guarantee that there exists a
q < l such that v'lcn zn l =
lzl :S: q < l for all n > N. This is
n
because � lzl � 0lzl = 0 < 1. (So we may arrange for example
that q = 1/2 < 1.) So again by the Root Test, the power series converges
absolutely for such z.
On the other hand, if L =/=- 0 and lzl > 1/L, then there exists an N
large enough such that v'lcn zn l =
lzl > 1 for all n > N. This is
n
because � lzl � Llzl > l. So again by the Root Test, the power
series diverges.
vfcJ
vfcJ
vfcJ
Solution to Exercise 4. 7
When z = 0, the series clearly converges with sum 0. Suppose z =/=- 0.
Then lzl =/=- 0. Choose N E N such that N > 1/lzl. Then for n > N,
lnzl > Nlzl > 1, and so lnn zn - 0I = lnzln > ln = 1, showing that
Ln z
00
Thus if z =/=- 0, then
n n
diverges.
n=l
Solution to Exercise 4. 8
V�
We have lim nf"l = lim � = 0. So the radius of convergence of
n-+oo
n-+oo n
oo
n
I::n
n=l
is infinite, and the power series converges for all z E C.
238
A Friendly Approach to Complex Analysis
Solution to Exercise ,4.9
(1) We have
(-1r+1
n
.
l
lim n +
= 1lffi --=1 '
n--+oo
n--+oo n + 1
(-l)n
n
(-1r n
and so the radius of convergence of
- - z is 1.
n
n=l
(2) We have
.!. 2012 _
(n +1)2012
.
.
hm l
hm (1 + )
- 1,
I
2012
n--+oo
n--+oo
n
n
L
00
and so the radius of convergence of
(3) We have
lim
n--+oo
1
(n + 1)!
1
n!
Ln2012
00
z
n
n=O
is 1.
1
= lim _ _ = 0 '
n--+oo n
and so the radius of convergence of
f�
n.
n=O
z
+1
n
is infinite.
Solution to Exercise 4.10
For l z l < 1, we know that
and so
f (z ):=1 +2z +3z2 +4z3
z f (z )=g(z ):= z
+· · ·=
(l
1
_
+2z2 +3z3 +4z4 +•··=
z )2
,
z
2
(1- z )
for l z l < 1. So g(z ):= z +2z2 +3z3 +4z4 +··· converges for l z l < 1, g is
holomorphic in the disc l z l < 1, with g' (z ) = 1 +22z +32z 2 +42z3 + ...
for l z l < 1. On the other hand,
g(z)= zf (z )=
for l z l < 1,
(1- z )2
z
and so
_
z
1
,
_ d
2
_ 1- z +2z _ 1 + z
g (z )- (
)-1
+z
2
2
3
dz (1- z )
•(l- z )
(1- z )3 -(1- z )3"
(l- z )
0
239
Solutions
Solution to Exercise 4.11
L :2 converges}= {z EC: lzl � 1}.
n
oo
(1) False. For example, { z EC:
n=l
(2) True.
(- 1 r
L --z
n
00
(3) False. For example
z = -1.
n
n=l
converges for z = 1, but diverges for
(4) False. See the example in (3).
(5) True. Same example as in (3).
L zn2
n
oo
(6) True. For example, consider
.
n=l
(7) True. The radius of convergence is � 1 and 1
1 +ii=
v'2 > 1.
Solution to Exercise 4.12
Since
d2n
d 2n+l
sinz = (-ltsinz and d 2n+l sinz = (-ltcosz,
2n
dz
z
and sinO = 0 and cosO = 1, we have
1 dn
z3 z5 + •••
+- zn = z- sinz = '"' - ( .
sinz)
n
z=O
3! 5!
L.., n! dz
n =O
I
00
z
z
- - •· •. Alternately,
+
1
2. 41. +
oo
oo
1 ·
1
exp(iz) + exp(-iz) 1
cosz = -----'-----'-----'-� = - ( L -i nz n + L - ( -1) ni·nz n ) .
2
2 n =O n!
n!
n =O
2
n
Since i = (-lt, we have
2
4
Similarly, cos z = 1-
z2 iz 3 z 4 + iz 5 z6
1 + iz +
+ 4!
(
6! · · ·
2!
2
4
2
5
6
3
iz
z
z
z
iz
...
1 - iz + 4! 2! +
6! + )
1 4 1
1
= 1- -z 2 + -z - -z 6 + - · · · .
2!
4!
6!
cos z =
1
3!
5! -
3!
5! -
240
A Friendly Approach to Complex Analysis
Solution to Exercise ,4.13
Let p(z) = z6 - z4 + z2 - 1, z E C. Then
p'(z)=6z5 - 4z3 + 2z,
p"(z) = 30z4 - 12z2 + 2,
p"' (z)=120z3 - 24z,
p (4l (z)=360z2 - 24,
p (5l (z)= 720z,
p (6l (z) = 720,
p (7l (z) = p (8l (z)= · · · = 0.
Hence
p(l)=1- 1 + 1- 1 = 0,
p'(l)
= 6 - 4 + 2 = 4,
1!
p"(l) 30- 12+2
=
= 10 '
2!
2
p"' (l) 120- 24
--=---=16 '
3!
6
4
(
p ) (1) = 360- 24
= 14,
�
24
p (5 l (l) 720
=6 '
=
5!
120
p (6)(1) = 720 =
1.
-6!- 720
Thus for all z E <C,
z6 -z4 + z2 - 1
p'(l)
p (6)(1)
= p(l) + �(z
- 1) + · · · + - -(z- 1)6 +0
61
= 4(z -1) + lO(z-1) 2 + 16(z-1)3 + 14(z-1)4 +6(z-1)5 +(z-1)6 .
Solution to Exercise 4.14
(1) The function z H exp(z2 ) possesses a primitive, say g, in the simply
connected domain <C. Thus
f(z)=
1
'YDz
exp((2 )d( =
1
'YOz
g'(()d( = g(z) - g(0).
241
Solutions
So f'(z)= g'(z)= exp(z 2 )=
I
L 1n1
00
2n
z
n=O .
. Consequently,
I
d2n+l
1 d2n
1
1
f'(z)
=
and
(z)
= 0.
2
(2n)! dz n
(2n+1)! dz2n +J' z=O
z=O n!
(2n)'
Hence j( 2n+ll(O) = -,-· and J( 2n+ 2 l(O) = 0. Also, f(O) = 0. Thus
n.
oo j(n)(0)
oo j( 2n+l)(O)
oo
l
---z 2n+l =
--Zn =
z 2n+l.
f(z)=
n!
+
1)!
+
l)(n!)
(2n
(2n
n =O
n =O
n =O
(2) For lzl < 1,
1
-- = l -z+z2 -z3 +z4 -+···
z+l
and since power series are holomorphic in the region of convergence
with complex derivative given by termwise differentiation, we obtain
for lzl < 1 that
l
d l
---2 = - --= -1+2z - 3z2 +4z3 -+ · · • .
(z+1)
dzz+l
Multiplying both sides by -z 2 gives
L
z
L
L
L
2
oo
= z 2 - 2z3 - 3z4 +- ...=
(-1 r . ( n - l) .Zn
(z + 1) 2
n=2
for lzl < 1. So we have co = c1 = 0 and en = (-lt · (n - 1) for n � 2.
Solution to Exercise 4.15
Let z EC. Let R > lzl. Then
(n+l)!
· max lf(z)I
Rn +l lzl:":'.R
. max M-lzln= (n+l)! -M-Rn = (n+l)!M_
< (n+l)!
Rn +l lzl:','.R
Rn +l
R
But the choice of lzl > R was arbitrary, and so f(n +l)(z)= 0. Since z EC
was arbitrary, we obtain that j(n +l) 0 in C. By Taylor's Theorem, for
all z EC,
n
oo j( k )( )
j( k )( )
f(z) =
�zk ,
�(z - ol =
k =O
k =O
since j(n+l)(0) = f(n + 2 )(0) = f(n +3)(0) = · · · = 0. So we see that f is a
polynomial of degree at most n.
If n = 0, then f is a bounded entire function, and our conclusion ob­
tained above says that f is constant. So the special case when n = 0 is
Liouville's Theorem.
lln+l)(z)I �
=
L
L
A Friendly Approach to Complex Analysis
242
Solution to Exercise 4.16
1
By the Cauchy Integral Formula,
2013!
-27ft .
Hence
1
c
d 2012 .
sin z
dz =
s nz I
d Z 2012 r
Z2013
z=0
= (-1) 2012/2 sinzl
z=O
=0.
sinz
dz
2013
=0.
cZ
Solution to Exercise 4.17
z0, there exists a 8 > 0, which can be chosen
R, such that g(z) =/-0 for lz -zol < 8. We have f(zo) =0, but
f(z) = (z-zorg(z) (lz-zol < R) shows that f(z) =/-0 for O < lz-zol < 8.
By the continuity of g at
smaller than
So by the Theorem on Classification of Zeros, there exists an m E N, which
is the order of zo as a zero off and a gholomorphic in
Then for lz - zol
that this implies
0 =/-
while if m
0 =/-
< R,
D such that g(z0) =/-0.
(z - zo)m g(z) = (z - zorg(z). We show
m = m. For if m > m, then we get the contradiction that
g(zo) =
> m,
we have
lim
z➔zo
g(z) =
lim
z➔zo
(z - zo)m - m g(z) =0 · g(zo) =0,
then we get the contradiction that
lim g(z) = lim (z - zo)m - mg(z) =0 · g(zo) =0.
g(zo) = z➔zo
z➔zo
Consequently, m
= m,
and so Zo is a zero of order m
= m.
Solution to Exercise 4 .18
(1) We have
f(z) =
(1
+ z2 ) 4 = ((z - i)(z +i)) 4 = (z - i)4 (z +i) 4 ,
g(z) := (z +i) 4 , g is entire, g(i) =
f(z) = (z - i)4g(z). Soi is a zero off of order 4.
and so with
(2) We have
f(2n1ri) =
1 -1
J'(2mri)= expzl
So 2n1ri is a zero of f of order 1.
=0,
and
(2i) 4 = 16 =/- 0 and
. = 1 =f. O.
z=2n1ri
Solutions
243
1
(3) f(O) = 1 - 1 + (0)2 = 0, and we have
2
1 (1-cos(2z))
1 .
f(z) = cosz -1 + (smz)2 = cosz-1 + ·
2
2
2
3 1
= cosz ---- cos(2z)
4 4
= (1- �� + :; - :� +-...) - �
_ ! (l _ 4z2 + 16z4 _ 26 z6 + _ ...
)
4
2!
4!
6!
1
+ ! . )z 2 + ( .!_ _ ! . 6 )z4 + ...
= (1 _ � _ ! ) +
4 4
2! 4 2!
4! 4 4!
(-.!.
..___.....,
0
±
________,
0
'-----v------"
#0
and so zo is a zero of order 4.
Solution to Exercise ,4..19
First we note that if z is a point that is distinct from z0 in the disc, then
f(z)-=/- 0. By the result on the classification of zeros, f(z) = (z-zo)g(z),
where g is holomorphic in the disc, and g(zo)-=/- 0. Thus
_1_
21ri
1
zf'(z) d
z
"I f(z)
l
+ 1
= _ _ z( · g(z) (z zo) · g'(z)) dz
21ri "I
(z-zo)g(z)
z(g(z) + (z - zo) · g'(z))
1
g�)
�
=
21ri "I
(z -zo)
+ = z(g(z) (z zo) · g'(z)) J
(Cauchy Integral Formula)
z=zo
g(z)
zo(g(zo) + 0 · g'(zo))
g(zo)
= zo.
1
1
A Friendly Approach to Complex Analysis
244
Solution to Exercise 4-.20
By the result on the classification of zeros, f(z) = (z - z0)mg(z), where g
is holomorphic in D and g(z0) =/=- 0. Thus
(f(z))2 = (z - zo)2m ( g(z))2 .
....__.,
=:G z
(
)
Clearly (f(z0)) 2 = 0, and G is holomorphic in D with G(zo) = (g(z0)) 2 =/=- 0.
Hence z0 is a zero ofz r-+ (f(z))2 of order 2m. Also,
J'(z) = m(z - zor- g(z) + (z - zor9'(z)
m 1
= (z - zo) - (mg(z) + (z - zo)g'(z)),
1
and so f'(zo) = (zo-zo)m- 1g1(zo) (m� ) O· g1(zo) = 0. As 91 is holomorphic
and
l
g1 (zo)
= mg(zo) + 0 · g'(zo) = mg(zo) + 0 = mg(zo) =/=- 0,
it follows that zo is a zero off' oforder m - l.
Solution to Exercise 4-.21
Consider the function f : <C ➔ <C given by
f(x,y) = x sin ! ifx =/=- 0,
X
and f(0, *) = 0. Then f is obviously continuous at any point (x o, Yo) where
x0 =/=- 0. Moreover, since
lf(x, Yo) - f(O, Yo)I = Ix sin� - 01 = lxl I sin� I ::':'. lxl · 1 = Ix - 0I
for all x =/=- 0. Thus f is continuous also at the points (0, *). Hence f is
continuous everywhere in <C. By the definition off, 0 is a zero off. 0 is
clearly not an isolated zero off, since
f (_!_,
nn
sin(mr) = 0,
o) = _!_
nn
n E N.
Also, f is not identically zero in any disc centered at 0 because for n E N,
f
C2n
!
O = 2
l)n' )
( n
!
l)n
sin ((2n + l)i) =
(2n
!
( l =/=l) n - r
o.
245
Solutions
Solution to Exercise 4.22
We know that for all x1, x2 EJR,
cos(x1 + x2) = (cosx1)(cosx2)- (sinx1)(sinx2).
Fix x EJR and consider the entire function f given by
f(z)
:= cos(z+x)- ((cosz)(cosx)- (sinz)(sinx)),
(5.22)
z EC.
We have f(y) = 0 for ally E JR, thanks to (5.22), and so by the Identity
Theorem, f(z) = 0 for all z EC, that is,
cos(z + x) = (cosz)(cosx)- (sinz)(sinx), z EC.
(5.23)
But the choice of x E JR was arbitrary, and so (5.23) holds for all x E R
Next, fix a z EC. Consider the entire function g given by
g(w)
:= cos(z + w)- ((cosz)(cosw)- (sinz)(sinw)),
w EC.
Then we have g(x) = 0 for all x EJR by (5.23). Another application of the
Identity Theorem yields g(w) = 0 for all w EC. Hence
cos(z + w) = (cosz)(cosw)- (sinz)(sin w), w EC.
(5.24)
But the choice of z E C was arbitrary. Consequently, (5.24) holds for all
z EC (and for all w EC).
Solution to Exercise 4.23
Suppose f, g EHol(D) are such that
(5.25)
(f · g)(z) = f(z) · g(z) = 0, z ED.
Suppose there exists zo E D such that f(zo) -:/- 0. By the continuity off,
there exists a 8 > 0 such that f(z)-:/- 0 whenever lz-zol < 8. (5.25) then
implies g(z) = 0 for lz-zol < 8. By the Identity Theorem, g 0 in D. So
Hol(D) has no zero divisors.
On the other hand, C(D) is not an integral domain, and we show this
below. Let zo ED and let 8 > 0 be such that the disc
Li:= {z ED: lz-zol < 8} CD.
Consider the continuous function r.p: JR---+ JR defined by
0 if t :S 0,
r.p(t) = {
t if t > 0.
Define f, g by
f(z) := r.p(Re(z-zo)),
=
for z ED.
g(z)
:= r.p(-Re(z-zo)),
246
A Friendly Approach to Comp lex Analysis
g > 0 here
Fig. 5.21
f
> 0 here
Construction off, g E C(D) using cp.
Being the composition of continuous functions, f, g E C(D). Also f(z) > 0
for all z in the right half of�, and so f-:/- 0 in C(D). Similarly g(z) > 0 for
all z belonging to the left half of�, and so g -:/- 0 in C(D). Nevertheless,
f ·g=O.
Solution to Exercise 4.24
(1) False. Take D = C, f = exp, g = l. Then for all n E N, we have
f(21rin) = exp(21rin) = 1 = g(21rin), but f -:/- g (for example, because
f(i1r) = -1-:/- 1 = g(i1r)).
(2) 'Irue.
(3) 'Irue. Let 'Y(t) = x(t) + iy(t), t E [a, b]. Consider a point to where
either x'(to) or y'(to) is nonzero. (If they are both always 0, then
a = b, a contradiction.) Suppose x'(to) > 0 (the other cases are handled
similarly). Then x(t) > 0 in a neighbourhood of t0. So x is increasing
there. Take tn = t0 +
n ?: N, where N is large enough so that
to+ if E [a, b]. Set Zn = 'Y(tn)- Then (zn)n2: N is a sequence of distinct
points (since their real parts ate distinct), which converges to 'Y(t0). By
the Identity Theorem, f = g in D.
(4) 'Irue. By the Taylor expansion around w, f =gin a disc with center
w, and so by the Identity Theorem, f =gin D.
¼,
Solution to Exercise 4.25
Let K = {z E C : lzl :::; l}. For each z E K, there exists a smallest
n(z) E {0, 1, 2, 3, · · •} such that for all w near z,
f(w) =
L C (z)(w - zr
00
n=O
n
and Cn(z)(z) = 0. Hence u< n(z))(z))/((n(z))!) = 0, and so f( n(z))(z) = 0.
Let rp : K-+ NU {0} be defined by cp(z) = n(z). Since K is uncountable,
247
Solutions
while NU {0} is countable, there exists an N such that rp-1(N) is infinite.
Let (zn )nE N be a sequence of distinct points in rp-1(N). As K is compact,
this has a convergent subsequence (znk hE N with limit, say z* E K. As
j(N)(znk ) = 0 for all k, by the Identity Theorem (applied to j(N)), we have
j(N) = 0 in K, and so also in C. By Taylor's Theorem,
oo j(n)(O) n N-1 j(n)(0) n
_
f(z) z _
z ,
I:-,
n.1n.- n=O
for all z E <C, and so f is a polynomial.
L-
n=O
Solution to Exercise 4.26
Let zo E D be such that Jf(zo)I 2: Jf(z)J for all z E D. By the Maximum
Modulus Theorem, f is constant in D, a contradiction.
Solution to Exercise 4.27
Let J(zo) =/- 0. Then Jf(zo)I > 0 and so for all z ED, Jf(z)J 2: Jf(zo)I > 0,
implying that for all z E D, f(z) =/- 0. Now consider the holomorphic
function g := 1/ f in D. We have
1
1
Jg(zo)I = IJ
= Jg(z)J, z ED,
(zo)I 2: IJ(z)J
and so by the Maximum Modulus Theorem, g is constant. But then f is
constant too.
Solution to Exercise 4.28
Let zo be a maximizer, which exists since z f-+ Jf(z)J is continuous and
K := {z E (C : JzJ :::; 1} is compact. But z0 can't be in the interior of L:
indeed, if JzoJ < 1, then by the Maximum Modulus Theorem (applied to f
on lill := { z E (C: JzJ < 1}), f would be constant in lill, which it clearly isn't.
Hence z E 'll' := {z E (C: JzJ = 1}. So
max Jf(z)J = max Jf(z)J = max I exp(2it) - 2J = 1- 1 - 2J = 3.
zEK
tE[0,21r)
izl=l
Similarly, if z1 is a minimizer, z1 can't be in the interior of K. Indeed,
zf 2 =/- 0, and so by the Minimum Modulus Theorem, f would be a
constant, a contradiction. So z1 E 'll'. Hence
min Jf(z)J = min Jf(z)J = min I exp(2it) - 2J = JI - 2J = 1.
zEK
tE[0,21r)
izl=l
See Figure 5.22.
-
248
A Friendly Approach to Complex Analysis
']['
exp(2it)
minimizer
/
maximizer
+-----.-----+=====at
-1
2
1
0
Maximizer and minimizer for Jz 2 - 21 in the unit disc.
Fig. 5.22
Solution to Exercise 4.29
For z E A1 := {z E (C: 0 < lz-11 < 1}, we have
1
z(z-l)
1
(z-l+l)(z-1)
=
=
z
z
� (1 - (z-1) + (z-1) 2
- (z -1) 3 +- · • ·)
1
� -1 (z-1) - (z-1) 2 + (z-1) 3
+
1
-
+ .. • .
On the other hand, for z E A1 := {z E (C: 1 < lz -11}
1
z(z-1)
1
(z-1+l)(z -1)
=
=
1
(z-1) 2
1
(z _ l) 2 (l +
1
1
(1- z-l + (z -1) 2
1
1
(z-1) 2 - (z -1) 3
+
1
-
_1
_)
z-l
1
(z-1) 3
1
(z-1)4 - (z-1) 5
-
+ .. ·)
+- .. · ·
Solution to Exercise 4.30
By the result on classification of zeros, f(z) = (z-zorg(z), z ED, where
g is holomorphic in D and g(zo) =I- 0. Also, since zo is the only zero off
in D, g( z) =f=. 0 for all z E D. So 1/g is holomorphic, and it has a Taylor
expansion in a disc with center zo: there exists an R > 0 such that
1
g (z)
=
L en(z - zot for lz-zol < R,
00
n=O
249
Solutions
and co-/- 0 (because g( zo)-/- 0). Thus for 0
1
J( z )
=
=
1
( z- zo)mg( z )
Co
+
( z- zO )m
1
< lz - zol < R,
00
cn ( z- zot
( z - zo)m �
=
00
Cm-1
�
n
+ ···+ -- + � Cm+n (Z-Zo) •
m -1
( z - zo)
z - zo
n =O
C1
Hence 1/ f has a pole of order m at zo.
Solution to Exercise 4.31
z
t---+ ( z - z0 )m f( z ) has a holomorphic extension, say h, to
·( lim ( z - zor f( z )
z-tzo
= o),
D. Also,
and so h( zo)-/- 0. Moreover, since f( z )-/- 0 for zED, also h( z )-/- 0 for all
zED. Hence
1
f( z )
=
( z - zor
h( z )
for all zED\ {zo}
and g defined by
g( z )
is holomorphic in
D.
Since
=
:
( z - zor
1
h( zo)
h( z )
-/- 0,
z0
'
zED
is a zero of g of order m.
Solution to Exercise 4.32
We must have Cn
= 0 for all n <
-m. Thus
L
00
C-m
C-m
r1
+ ...+ � +
Cn (Z - zot+
m
z - zo
( z - zo m-1
( z - zo)
n =O
-1
So ( z- zorf( z ) = C-m +c-m+1(z-zo)+·· •+c-1(z-zor +···. Hence
m
( z - z0 ) f( z ) has a holomorphic extension, say g, to
f( z ) =
�:={zEC: lz-zol <R}.
So for lz-zol
< R, g( z) = c_m +c-m+1(z-zo)+···+c-1(z-zo)m-l +···.
By Taylor's Theorem,
c_ 1
=
dm-l g
1
(zo).
m
!
( - 1) dzm-l
250
A Friendly Approach to Complex Analysis
But gCm-l) is holomorphic in b. and in particular, continuous at
z0.
So
Also, for 0 < lz- zol < R, g( z ) = ( z - z or f( z ), and so for z -/- Zo in b.,
m 1
d g(m-l)( z ) = --((
z - z orf( z )).
dz m- 1
Hence
c_1 =
1
1
im-l)(zo) = --- lim g(m-ll(z)
(m - l)!
(m -1)! z-tzo
m -1
1
d
lim m _ 1 (( z - z orf( z )).
( m- 1)1. z ➔zo dz
Solution to Exercise 4.33
(1)
(2)
(3)
(4)
(5)
True, since c_1 = 1-/- 0, and c_2 = c_3 = · · · = 0.
True.
True.
True.
True.
Solution to Exercise 4.34
(1) sin z does not have a singularity at 0, and for
sin z
z
E (C,
5
z3
z
= z- +- +• • • .
3!
5!
(2) sin ! has an essential singularity at 0, since for z-/- 0,
z
. 1
sin-=••• z
1
1
1
+-- --3 + -.
5
5!z
3!z
z
sinz
(3) -- has a removable singularity at 0, since we have that
z
sinz
lim z • -- = lim sinz = 0.
sinz
Also, -z
z➔O
z
z➔O
1
1
= 1- 1
, 0.
-z2 + z 4 - z 6 + - · · · £or z 4
3!
5!
7!
251
Solutions
sinz
(4) -2- has a pole of order 1 at 0,since for z =f. 0,
z
sinz
1
z z3 z 5
= +
+ - ... .
7 � 3! 5!
7!
(5) 1/( sin(l/z)) does not have an isolated singularity at 0 because with
Zn = 1/(mr),n EN,we have
sin _!_ =sin(mr) =0
1
Zn
n--+oo
=- --+ 0. ( We h ad also observed this
. m
. Example 4.13.)
n1r
( 6) z sin - has an essential singularity at 0 since for z =f. 0
z
1
1
. 1
z sm - = · · · - +-4 - -2 + 1.
3!z
5!z
z
and
Zn
Solution to Exercise 4.35
( 1) False. limje ½j =lime ½ =O,and so,(limlexp !l=+oo).
x,?'O
x,?'O
z--+0
z
(2) '!rue. There exists an R > 0 such that
C-m
C-1
C-m +l
"""'
f( Z ) =
+ · · · + -- + � Cn ( Z - zo)n,
+
(z - z0 )m (z - zo)m -l
z - zo n =O
for 0 < !z - zol < R, and so with
P :=C-m + C-m +1(z - zo) + · · · + C-1(z - zo)m -l,
we have for 0 < !z - zo[ < R,
f(z) -
(Z
p(z)
-
Zo )m
=�
� cn (z - zo)n .
n =O
( 3) 'Irue. Let the order of0 as a zero off be m. (Take m =0 iff(0) =/- 0.)
Then there exists a holomorphic function g such that f(z) =zm g (z)
and g(0) =/- 0. Hence for n > m,and z =f. 0,
m
f(z)
g (z)
-- z g(z) --zn - m
Thus
[g(z)[
f(z)
1
=lim
lim
=+oo
=[g(zo)[ · lim - -z--+0 I zn I z--+0 [z[n-m
z--+0 [z[n m
since g (zo) =f. 0 and n > m.
252
A Friendly Approach to Complex Analysis
(4) True. In some punctured disc D= {z E C : 0 < [z - zo[ < R}, f,g
are nonzero, and there exist holomorphic functions h f , h9 such that
h 1 (zo ) =/- 0,h 9 (zo ) =/- 0,and for all zED,
™
t =(z - zo )™ 1 h 1 (z ) , tz =(z - z0 ) 9 hg(z ) .
g )
f z)
Thus h 1 (zo ) h 9 (zo ) =/- 0 and for all zED
l
=(z - zo )™ 1 +m 9 h 1 (z ) h 9 (z) .
f(z ) g(z )
Consequently,fg has a pole of order m1 +m9 at z0•
Solution to Exercise 4.36
Consider f given by
f(z) = (exp�)+exp( � )' zEC\{0,1 }.
l z
Then f is holomorphic in C\{0,1 }. The function exp( 1 /( 1 - z)) is holo­
morphic in a neighbourhood of z= 0, while the function exp(l/z ) has an
essential singularity at 0. Thus,their sum,namely f, has an essential sin­
gularity at 0. (Why?) On the other hand exp(l/z) is holomorphic in a
neighbourhood of 1 ,while exp(l/( 1 - z)) has an essential singularity there.
So f has an essential singularity at z=l.
Solution to Exercise 4.37
We have seen that if zo is an isolated singularity of a function g with the
Laurent series expansion
g(z )=
Cn(z - zot
L
nEZ
for O < [z - z0[ < R and for some R > 0, and there are infinitely many
indices n < 0 such that Cn =/- 0,then z0 is an essential singularity of g.
However,for the given f, the annulus for the Laurent expansion
z-1 +z-2 +z-3 +...
is given by [z[ > l. The correct annulus to consider for deciding the nature
of the singularity at z = 0 is of the form O < [z[ < R for some R > 0. In
fact,for [zl < 1 we have
1
f(z )=- _ =-(l+z+z2 +z3 +···),
1
z
showing that f is holomorphic for lz[ < 1 ,and f does not have a singularity
at z=0.
253
Solutions
Solution to Exercise 4.38
It is clear that oz
is an isolated singularity of Jg . Indeed, since J and
g
both have an isolated singularity atzo, we have that J is holomorphic in a
punctured disc 0 < lz-zo
<l Rf for some Rf > 0, and g is holomorphic in a
punctured disc 0 < -z
lz
o<
l R9 for some R9 > 0. ThusJg is holomorphic
in the punctured disc 0 < zI zo I < min{ RJ, R9}.
Suppose thatJg has a removable singularity or a pole atz 0• Then there
exists an m � 1 such that
=
lim
(z-zo)
J(z)g(z)
= 0.
z➔zo
SinceJ
has a pole atoz , say of order m1, J
C-m1
(z-zo )mf
is nonzero nearzo
and
C-m1+1
C-1
n
Z
Zo ) ,
+ + z -+�
+ -z
L..., Cn (o
(z o )m _ 1 · · · -z
n =O
for 0 < lz
zo<
l Rand for some R > 0. Here C-m-:/, 0. So forz-:/- zo,
J (Z ) =
f
but nearoz , we have
=
(z-zo) g(z
)
=
1
Z
Zo ) m1J()Z
(-
1 -zo) =J(z)g(z)
(z
• -zo) =(z
...______...., __________
-tO
-tO
1
0. 0- 0.
--+--·
C-m1
So g must have a pole at z0 or a removable singularity at z0, a contradiction.
Consequently Jg has an essential singularity at z 0.
z-tzo
Solution to Exercise 4.39
Set E := 1/n =: o (> 0). By the Casorati-Weierstrass Theorem, there exists
a Z n in the punctured disc aroundzo with radius o such that (
JI z )-wl
< E.
n
That is, zl o
z
<
l
l/n
and
(
J
I
z
)
wl
<
E.
Hence
(z
N
converges
to
)
n
n
n nE
oz , and
(J(z )n) nE N converges tow.
Solution to Exercise 4.40
1 + exp z = 0 if and only if z E {'1ri + 21rni: n E Z}. So
Log( z )
J(z) :=
1 + expz
is holomorphic in C
( \(
oo,0]) \ {1ri + 21rni: n E Z}. J has poles of order
1 at the points {1ri + 21rni: n E Z}, of which two lie inside the given path
--y:-1r and 31ri. See Figure 5.23 .
A Friendly Approach to Complex Analysis
254
3ni
�
✓
,1
�
::c
' -?ri
,
,,
✓
,2
Fig. 5.23
1
We have
J(z)dz =
'Y
1
J(z)dz +
1
The curves ,1 and 12.
J(z)dz = 2ni (res(J, 3ni) - res(!, -ni)).
12
11
So we need to calculate res(!, 3ni) and res(!, -ni). We can write
Log(z) = C-1,31ri h
--�
. + 31ri,
1 +expz z - 3ni
h
w here 31ri is holornorphic in a neighbour hood of 3ni. Thus
z - 3ni
(z - 3ni)Log(z) _ .
_ .
- 1llll
. · L Og (Z )
C-1 ' 3,ri. - 1llll
z--+31ri
z--+37ri expz - exp(3ni)
1 +expz
1
· Log(3ni) = -1 ( logl3nil + i '!!..2)
=
expz 1 z=31ri
= -log3 - logn - i'i.
We can write
Log(z)
C-1,-1ri
" + h-,ri,
1 +expz z - (-ni)
w here h-1ri is holornorphic in a neighbour hood of -ni. Thus
z - (-ni)
(z - (-ni))Log(z)
.
.
· Log(Z )
= 1llll
C-1 ' -1ri· = 1llll
z--+-1ri expz -exp(-ni)
z--+-1ri
1 +expz
1
-Log(-ni) = -l ( logl - nil + i( - '!!..))
=
2
expzlz=-1ri
1
. 7f
= -logn+ i .
2
Log(z)
dz=2ni(-log3-logn- i'!!..+logn- i'!!..) = 2n2 - (2nlog3) i.
So
2
2
l+expz
1
255
Solutions
Solution to Exercise 4.41
Let 'Y be the circular path given by "!(0)
{ 2n
}0
cos 0
d0
5+4cos 0
=
=
1
1
=
exp(i0) for 0 E [O,27r). Then
1
z+ !.
z
__
z2 +l
2
. _!_dz =
dz
2 +5z+2)
!.
"'
"'2iz(2z
'5+4z+
'
--z iz
2
z2 +l
dz.
2iz(2z+l)(z+2)
7
Let f be defined by
z 2 +1
f (z) : =
2iz(2z+l)(z+2)·
Then f has three poles, at 0, -1/2, -2, and each is of order 1. Of these,
the poles at O and -1/2 lie inside 'Y· So by the Residue Theorem,
1
2n
0
cos 0
dB
5+4cos 0
= 27ri(res(f, 0) +res(!, -1/2))
z·(z +1)
·(i·Im ----'-------'-+
2
= 27ri
(z+l/2)·(z2 +1)
1.Im
)
z ➔ -I/ 2 2iz(2z+l)(z+2)
z ➔ O 2iz(2z+l)(z+2)
1 . .§.4
1
= 27fi· (--- +
)
2i·1·2 2i.•(-½)·2·!
7f
3
=
1
5
27ri· (- - -)
4i 12i
Solution to Exercise 4.42
(1) Let
Ji be defined by
1
fi(z) = l+z2·
Then Ji has poles at i and -i, both of order 1. Hence
1
1
.
.
. . z-i
-- dx = - •2m •res(JI,i) = 7ri· hm-2
z ➔ i 1+Z 2
2
1+ X
0
1
1
7r
= 7ri· lim -- = 7ri· - = -.
2i
2
z ➔ i Z +i
(2) Leth be defined by
1
h(z) = (a2 + z2 (b2 + z2 ) .
)
1
00
256
A Friendly Approach to Complex Analysis
Then h has poles atai, -ai, bi, -bi, all of order 1. As h is even,
rlo = (a2 +x2)(b1 2 +x2)dx
=
1
2·
7f
= i
=
27ri (res(h,ai) +res(h,bi))
1__ + __1__ )
(-a
(b
(a
)2a
b )2b
2 -
2(a2�b2)
(3) Leth be defined by
2
i
(i-¾)
2 -
=
2
i
2ab(:+b)"
1
h(z) = ( 1+z2 )2
Then h has poles at i and -i, both of order 2 . We have
ro = ( 1+x1 ) dx = !2 · 27r · res(h, )
l
i
2 2
i
1
7ri m
· li ..!!:_ ((z- i )2 •
. )
. 2
1! z---+idz
(z-i) (z+i)2
-2
. -2
. .
7r
= 7rz • hm --- = 7rz • - = -.
z---+i(z+i)3
-8i 4
=
(4) Let f4 be defined by
Then f4 has poles at
i
Pl = exp (: ), P2 = exp
1=
Ct),
p3
=
exp
c;i ), p4 = exp c;i ),
all of order 1. Hence
o
1+x 2
--4 dx
1+x
1
= 2 · 27ri (res(f4,p1)+res(f4,p 2 ))
+p12
l+p 22
7ri ( l pf +
)
4
4P�
+
1
+ _ _ + � + _l_) = 7ri (-Pl P2 + _l_ + _l_)
= 7ri
4pf 4p1 4p� 4p2
4
4p1 4p2
i
i
e
e
(i
)
e
((
)-e
xp 7r/4)- xp 7r /4 + xp 7r/4
_ .
xp( i7r/4)
-ITT ()
4
4
-i sin(7r/4)
i sin(7r/4)
= 7ri (+
= �) = 7ri · (-i ) . _2.._
2
y'2
y'2
2
(_E!_
=
Solutions
257
Solution to Exercise 4.43
1
By the Residue Theorem,
expz
expz
z
21ri 1· dn
exp
--·
d
( z n+l ·-1m-z1ri-res
-_
2
.
n
( z n +l,
c zn +l
n! z--+0 dz
zn +l)
n
21ri . d
21ri .
21ri
= - · hm
- expz = - · hm expz = - · expO
n! z--+0 dz n
n! z--+0
n!
21ri .
21ri
=
l=
.
n!
n!
Hence
exp(cos0+isin0)
21ri
.
- =
· i.(cos0+ism
. O)dO
.
n!
cos((n+1)0)+ism((n+1)0)
0
o) _
121r
121r
121r
1
=i
exp(cos0+isin0) • ( cos(n0) - isin(n0))d0
=i
exp(cos0)( cos(n0 - sin0) - isin(n0 - sin0))d0 .
Equating the imaginary parts, we obtain
exp(cos0) • cos(n0 - sin0)d0
ok
Solution
�
= 1.
n.
to Exercise 4.44
For z in a small punctured disc D centered at zo, f(z) =I= 0 and
J(z) = (z - zo)h(z)
(5.26)
for some holomorphic function h such that h(zo) =I= 0. From (5.26) we have
f'(z) = h(z)+(z - zo)h'(z) and in particular, f'(zo) = h(zo)- Now
1
1
for zED\ {zo},
J(z) (z - zo)h(z)
¾
and since
where do
=
is holomorphic in D,
1
h(zo)
1
h(z)
=
=
do+d1(z-zo)+··· for zED,
1
f'(zo)"
Hence for zED\ {z0},
1
1
do
-·(do+d1(z - zo)+·· ·) = --+d1+d2(z - zo)+···
=
f(z) z - zo
z - zo
and so res
(y,
zo)
=
do
=
f'(�o)
.
258
A Friendly Approach to Complex Analysis
Solution to Exercise 4.45
Let J be given by J(z) = sinz. J has zeros of order 1 at kn, k E Z. So by
the previous exercise,
1
1
1
res (--, kn) =
, 1
= -l k = (-1) k
k'Tf
sinz
sin zlz=k1r
( ) ( )
Solution to Exercise 4.46
(1) We have Jo = l ::=; 2° = 1, Ji = 1 ::=; 21 = 2, and if Jm ::=; 2m for all
m ::=; n (for some n � l) , then
m
m
m
m
m
Jm+I = Jm + Jm- 1 ::=; 2 + 2 -l = 2 -l · 3 < 2 -l · 4 = 2 +l.
(2) If lzl < 1/2, then v'lcn zn l =
lzl ::=; ffn · lzl = 2lzl < 1 for all
n EN. So by the Root Test,
vfcJ ·
00
n
Llenz l
n=O
converges when lzl < 1/2. Hence the radius of convergence of F is
� 1/2.
(3) We have for lzl < 1/2:
zF(z) = Joz+ fiz2 + hz3 + • • • ,
Joz2 + fiz3 + · · · .
z2 F(z) =
Adding these,
zF(z)+ z2 F(z) = 1 · z +(Ji+ Jo)z2 + (h + fi )z3 + · · ·
= fiz+ hz2 + hz3 + · · ·
= (Jo+ fiz+ hz2 + hz3 + · · ·) - Jo
= F(z) - l.
Hence
1 = F(z) - zF(z) - z2 F(z) = (1 - z - z2 )F(z).
1
1
So for lzl < - , we have F(z) =
2
l-z-z2
(4) We have
F(z)
1
Jo+···+ Jn -IZ n -I + Jn Z n + Jn+1zn+I + · · ·
zn+I
Jn
Ji
Jo
= n+I + n + · · ·+ + Jn+I + Jn+2 z+ · · · ,
--;
z
z
(5.27)
259
Solutions
and so
1
,
res ( n+1
(l- z - z2 )
z
o) = coefficient of! in (5.27) =f n·
z
(5) We have for Jzl = R > 2:
11- z - z2 1 2::: Jz2 + z l - 1 =l z l · lz+ 11- 1 =R · lz+ ll - 1
2: R· (lz l - 1)- 1 = R · (R- 1)- 1 = R2 - R- l
> 0 (since R > 2).
Hence if CR : [0,27r] -+ (C is the circular path given by
CR(t) = Rexp(it),
t E [0,271"], then
I
}{0R
I
l
l
< _l_ .
· 271" R
dz n
n
2
2
+
+
R l R - R- l
z
l(l- z - z )
1
1
R ➔ oo
.
Rn R2 - R- l -+ O
1
Define G by G(z) := n+
. Then G has
z
l(l- z - z 2)
(a) a pole at 0 of order n + l,
-1+v'5
of order 1,
2
-1- v'5
(c) a pole at
of order 1.
2
Thus
(b) a pole at
1
-l+v'5
-1-v'5
1
res(G,0)+res (G,--- ) +res (G,--- ) =-.
G(z )dz
2
2
271"Z CR
for all R > 2. Hence
res(G,0)+res
( -l+v'5
,a, 2 ) +res (G, -1-v'5
2 )
\
= Rlim � { G(z )dz =0,
➔ O 27ri JcR
.
that 1s, fn
= -res ( G,
-l+v'5
-1- v'5
) - res ( G,
).
2
2
260
A Friendly Approach to Complex Analysis
We have
res
1 + v'5) =
(G , ---2
1·
1m
z➔ -1!v's
( z- -1 + v'5)
2
1
Also,
Hence
)
f =-1 . ( l+v'5 )n+l _ _!__ ( 1-v'5 n+l
n
y'5
y'5
2
2
. ( ( 1 +2
- �
Js) •+' - ( Js) •+') .
1-2
1
z n+l (l - z - z 2 )
261
Solutions
Solutions to the exercises from Chapter 5
Solution to Exercise 5.1
(1) We have for (x, y) E JR2 \ {(O, 0)} that
2x
1
--2x=--,
2
2
2
+
x +y 2
x
y
2x
2
2y 2 + 2x 2 -4x 2
(2x)
- --,-,--....,........,
(x 2 +y 2)2
x 2 +y 2 (x 2 +y 2)2 ·
2(y 2 -x 2)
(x 2 +y 2)2 ·
Similarly noticing the symmetry in the roles of x and y, we have
2y
a 2u 2(x 2 -y 2)
and 2 = 2 + 2 2 .
y )
(x
xz + yz'
ay
au
ay
Consequently
a
a
2 u 2(y 2 -x 2) 2(x 2 -y 2)
2u
-+= ---'---------'---+---=0 .
2
a y2 (x 2 +y 2)2 (x 2 +y 2)2
ax
Since u is C2 and 6-u=0 in JR2 \ {(0, 0)}, u is harmonic there.
(2) We have for (x, y) E JR2 that
aU
a zu
ax
ax 2
aU
a 2u
X
= e cos y,
ay
a y2
•
= e smy,
So
X
x
= e sm y, and,
.
.
= ex( -smy
).
a2 u + a2 u= e x siny + ex(-siny)=0 . As u is C2 and 6-u=0
ax 2
a yz
IR , u is harmonic in IR2 .
in
2
Solution to Exercise 5.2
Consider the vector space V of all real-valued functions defined on U, with
pointwise operations. Then we know that V is a real vector space. We will
show that Har(U) is a subspace of this vector space V, and hence a vector
space with pointwise operations. We have
(81) The constant function O assuming value 0 everywhere on U belongs to
Har(U). Indeed
262
A Friendly Approach to Complex Analysis
(S2) Let u , v E Har(U). Then
82 (u +v) 82 (u +v) [J2 u 82 v 8 2 u 8 2 v
----+----=-+-+-+8x2
8y 2
8x2 8x2 8y 2 8y 2
_ 82 u 02 u
82 v 02 v
) + (-+)
- (-+2
2
8x
8y
8x2 8y 2
=o+o=o.
(S3) Let a: E JR and u E Har(U). Then
82 u
82 u
c.92 (o: · u) c.92 (o: · u)
=
+
+
0:
0:
8x2
8y 2
• 8x 2
• 8y 2
= a: ( ::� + ::�) =a: · 0=0.
Hence Har(U) is a real vector space with pointwise operations.
Solution to Exercise 5.3
u :=x=Re(z) and u :=x +y=Re(z - iz) are harmonic in JR2, and their
pointwise product is u • u=x • (x +y)=x2 +xy. We have
a
a
a2 (u . u) a2 (u . u)
8x2 + oy 2 = 8x (2x +y)+ oy (x)=2+0=2#0.
So the pointwise product of two harmonic functions need not be harmonic.
Solution to Exercise 5.4
(1) Let u =ex sin y. We seek a v such that u + iv is holomorphic. So the
Cauchy-Riemann equations must be satisfied. Hence
av
OU
- = -- = -eX cosy.
ax
&y
So ifwe keep y fixed, we obtain by integrating that v = -e x cosy+C(y),
for a constant C(y), which depends on y. Thus
:: =ex siny+ C'(y)= :: =ex siny,
and so C'(y)=0, giving C(y)=K. So we try v :=-e x cosy. Then
u + iv = ex siny+ i(-ex cosy)= ex (siny - icosy)
= -iex (cosy+isiny)= -iexp(x+iy)= -iexp(z),
where z=x+iy. Hence u+iv = -iexp z, which is indeed holomorphic.
Hence v = -ex cosy is a harmonic conjugate for u :=ex sin y.
263
Solutions
(2) Let u = x 3 - 3xy2 - 2y. We seek av such that u + iv is holomorphic.
So the Cauchy-Riemann equations must be satisfied. Hence
av
au
-=--= 6xy+2.
ax
ay
So if we keep y fixed, we obtain by integrating that
x2
2
v = 6 y + 2x+C(y) = 3x y + 2x+C(y),
2
for a constant C(y), which depends on y. Thus
av
au
- = 3x2 + C' (y) = -= 3x2 - 3y 2
ay
ax
and so C'(y) = -3y2 , which gives
3
C(y) =-3 y + C = -y3 + C,
3
and so we try v = 3x 2 y + 2x - y3• With this v, we have
u +iv= x 3 - 3xy2 - 2y + i(3x 2 y+2x - y3)
= x 3 +3x(iy) 2 +3x2 (iy)+(iy) 3 - 2y + i2x
= (x + iy)3+2i(x+iy) = z2 + 2iz
for z = x + iy. Thus u +iv= z2 + 2iz, which is indeed holomorphic.
So v := 3x 2 y+2x -y3 is a harmonic conjugate of u := x 3 - 3xy2 - 2y.
(3) Let u := x(l + 2y). We seek a v such that u + iv is holomorphic. So
the Cauchy-Riemann equations must be satisfied. Hence
av
au
-=--=-2x.
ax
ay
So if we keep y fixed, we obtain by integrating that
x2
v =-2 + C(y) = -x2 + C(y),
2
for a constant C(y), which depends on y. Thus
OV
OU
- = CI (y) = -= 1 +2y.
ay
ax
Thus
y2
+ C = y + y2 + C.
2
So we try v :=-x2 + y + y2 . With this v, we have
u +iv= x(l + 2y)+i(-x 2 +y+y 2 ) = x+iy + 2xy+i(y2
C(y) = y+2 ·
-
x2 )
= x + iy - i((x2 - y2 ) + i2xy) = x +iy - i(x + iy) 2
= z - iz2
for z = x + iy. Hence u + iv = z - iz2 is indeed holomorphic, and
v :=-x2 +y+y2 is a harmonic conjugate of u := x(l + 2y).
264
A Friendly Approach to Complex Analysis
Solution to Exercise 5. 5
Let v be a harmonic conjugate of u. Then f := u + iv is holomorphic in
C \ {0}. Hence h := z 2 exp(-f(z)) is holomorphic in C \ {0}. We have
2
lhl = lzl 2 1exp(-f(z))I = lzl 2 e-Re(f(z)) = lzl 2 e-u = lzl 2 e-loglzl
w
2
1 = 1.
= lzl ·
But then h must be constant in each disc contained in C\ {0}. Consequently
h must be constant in C \ {0}. Thus h' = 0. But
2
h' = 2z exp(-f(z)) + z exp(-f(z)) · (-J'(z)),
and so
f'(z) = �So 1/ z would have a primitive in C\ {0}. Now if, is the path 'Y(t) = exp(it),
0 ::; t ::; 21r, we have the contradiction that
1 1
�dz= f'(z)dz = 0.
z
'Y
Hence u has no harmonic conjugate in C \ {0}.
2 · 21ri =
'Y
Solution to Exercise 5.6
Set u := x 3 + y 3• If f were holomorphic, then u would be harmonic. But
fP u
fP u
&
&
= (3x2 ) + (3y2 )
&y
&xZ &yZ
&x
for x =/- -y. Hence the answer is "no".
+
= 6x + 6y = 6(x + y) =/- 0
Solution to Exercise 5. 7
It suffices to show that if u is harmonic, then so are �: and �:.
We know that u is infinitely many times differentiable. We have
265
Solutions
Similarly
Solution to Exercise 5.8
(1) Let b(x) = p(x) =co+ c1x +
• · · + cdxd . Then
p(z) := p(x + iy) =Co+ C1Z +
· · · + Cd Z d
is entire, and so h := Re(p(x+iy)) is harmonic. Moreover for all x E �,
h(x, 0) = Re(p(x + iO)) = Re(p(x)) = Re(b(x)) = b(x).
(2) We have b(z)
:= b(x + iy) :=
1
is not defined at z = i. But
1 + z2
i
z+i
is holomorphic in the upper half-plane, and so its real part is harmonic
there. Moreover,
0+1
h(x, 0) = x2 + (0 + 1)2
1
= x2 + 1 = b(x)
for all x ER
Solution to Exercise 5. 9
Let f be an entire function whose real part is u. (Note that C is simply
connected.) Then exp(-f) is entire too. We have
Iexp(- !)I= e-Re(f) = e-u:::; 1,
since u(x, y) > 0 for all x, y E �- By Liouville's Theorem, exp(-f) is
a constant. Hence I exp(-f) I is constant too, that is, e-u is constant.
Consequently the real logarithm log(e-u) = -u is constant, and so u is
constant as well.
A Friendly Approach to Complex Analysis
266
Solution to Exercise 5.10
(1) For z =r exp(i0), we have
exp (- :4
) = exp (-r14 exp(-i40)) .
l/n and 40 = -1r, that is, with
1
= : Xn + iyn ,
Zn : = :;:; exp ( -i.
4
4
we have u(xn ,Yn ) = exp(-n4 exp(i1r)) = exp(-n4 • (-1)) = en • So we
have that (xn ,Yn ) -+ (0,0), but it is not the case that u(xn ,Yn ) -+ 0,
showing that u is not continuous at (0,0).
So if we taker
(2) We have
u(x, 0) = exp
=
(-
7r)
\ )
(x + i)4
=
u(0,y) exp (- (0 +\i)4 )
(3) We have
,::iu
_u (0
ax '
0) = lim
x---+0
u(x
=
=
4
e-l/x , and
exp (- i4�4 )
' 0) - u(0 ' 0)
X-0
=
=
and so 0 ::;
I
e- x
�
4
I
4
4
4
. e-l/x - 0
hm --x---+0
X
(The last equality follows from the fact that
e 1 /x
(- 1 \ ) = e- /y .
exp
=
4
. e- 1 /x
hm -x---+0
X
1
4
=
0.
1
1
1 1
l+-+(-) +··•>2! x4
x4
x4
2
=
::; lxl3 .) Similarly, :: (0,0) = 0. Thus
4
au
u
d
-(x, 0) - a-e- i/x - 0
(0, 0)
a2u
ax
x ____
-(0
0) = lim a x
= lim -=d=
x---+0
x---+0
X
a x2 '
x-0
4
-4
4
1 . e - 1/ x . �
-4e- 1 /x
= 1m ---- = lim ---- = 0.
x---+0
x---+0
x6
x
4
2
1
1
1 1
e-l/x
4
_
+
·
·
·
>
-,
__
::; 2lxl2 -)
0::;
(As e 1/x = 1 +-+
)
(
2x8
x6
2! x4
x4
Similarly,
a2u
(0,0) = 0.
ay 2
a2u
a2u
Hence
(0,0) + 2 (0,0) = 0 + 0 = 0.
2
ay
ax
I I
267
Solutions
Solution to Exercise 5.11
(1) Let zo E D1. Then cp(zo) E D2. Let A be a disc with center cp(z0)
and radius E > 0 small enough so that A C D 2 and cp- 1 (A) C D 1.
Since A is simply connected, there is a holomorphic function G defined
in A such that g = Re(G) in A. The composition of the holomorphic
maps cpl,r•(�) : cp- 1(A) ---+ A and G : A ---+ <C is holomorphic, and
so Re(G o cpl,r•(�)) is harmonic in cp- 1 (A). But for z E cp- 1(A),
'Pl cp -•(�)(z) = cp(z) EA, and so
(Go 'Pl,r•(�))(z) = G(cp(z)) = g(cp(z)) =(go cp)(z).
So (gocp)l cp -•(�) is harmonic in cp- 1 (A). As z0 Ecp- 1 (A), in particular,
we have A(g o cp)(zo) = 0. Since the choice of z0 E D1 was arbitrary,
g o <p is harmonic in D1.
(2) If h : D2 ---+ JR is harmonic, then by the first part, it follows that
h o <p : D1 ---+ JR is harmonic.
Now suppose hocp: D1 ---+ JR is harmonic. Then since cp-1 : D 2 ---+ D1 is
holomorphic, by the first part, (hocp) ocp- 1 : D 2 ---+ JR is harmonic. But
(ho cp) o cp- 1 =ho (cp o cp- 1) =ho (idvJ = h, where idv2 : D2 ---+ D2
is the identity map z Hz (z E D2). So h : D 2 ---+ JR is harmonic.
p (= s)
A
(= -1)
Fig. 5.24
0
B
(= 1)
Triangle inequality in t:..PO' B
(3) By the triangle inequality in APO'B shown in Figure 5.24, for s (= P)
in IHI, we have
Is+ 11 = £(PA) =£(PO')+ £(0'A) =£(PO')+ £(0'B)
> f(PB) =
Is- 11,
where we have used the fact that O' is on the perpendicular bisector of
AB to get the third equality. So cp(s) ElI)) for all s EIHI. The function
<p is clearly holomorphic: for s EIHI,
1
s
=
cp'(s) = l· ! +(s-l)· (- : 2 =
)
�s\:;);
(s 1 )
(s: 1 ) 2"
s l
268
A Friendly Approach to Complex Analysis
Now consider 'lj; : lI)) ---+ IHI given by
'lj;(s)=��;,
zElDl.
(This expression for 'lj;, which is a candidate for cp- 1, is obtained by
8 -1
.)
solving for s in the equation z = cp(s)=
s+l
I
-l
N' '
'
I
A
0
1
+3_ 1
B
Triangle inequality in b.PO' B
Fig. 5.25
=
In the Figure 5.25, we know that the angle subtended by the diameter
AB at any point of the circle is go o, and so for any point P ( z) in
lDl, LAPE> go 0 • So
Re('lj;(z))=Re(��;)
=
l'I/J(z)I cos0 = l'I/J(z)I cos(?T-LAPE)> 0.
Thus 'lj;(z) E IHI for all z E lDl. The map 'lj; is holomorphic in lDl: for
z ElDl,
,
1
1
l-z+l+z =
2
+ (l + z) (
) =
'lj; (z) = l·
·
2
2
(1 z)
l-z
(l-z)2·
(1 z)
Finally, for s EIHI, we have
s-1
1 + -+ +
;
s +1 = s 1 s - 1 = 2s = ,
('lj; 0 cp)(s)=
s
1
s + l-s+l
2
1_
s+l
and for z ElDl, we have
l+z
---1 1 + z-l + z 2z
--------z
(cpo'lj;)(z) =
l+z+l-z
- 2 --+1
l-z
So cp is a bijection and cp- 1 ='lj;.
�+:
Bibliography
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Complex Analysis, http://math. sf su. edu/beck/papers/complex.pdf
Conway, J. (1978). Functions of One Complex Variable I, 2nd Edition. (Springer).
Fisher, S. (1999). Complex Variables, 2nd Edition. (Dover).
Flanigan, F. (1972) Complex Variables (Dover).
Flanigan, F. (1973). Classroom Notes: Some Half-Plane Dirichlet Problems: A
Bare Hands Approach. American Mathematical Monthly 80, 1, pp. 59-61.
Gelbaum, B. and Olmsted, J. (1964). Counterexamples in Analysis (Dover).
Gilman, J., Kra, I. and Rodriguez, R. (2007). Complex Analysis. In the Spirit of
Lipman Bers (Springer).
Howie, J. (2003). Complex Analysis (Springer).
Needham, T. (1997). Visual Complex Analysis (Oxford University Press).
Ash, R. and Novinger, W. (2007). Complex Analysis, 2nd Edition. (Dover).
Remmert, R. (1991) Theory of Complex Functions (Springer).
Rudin, W. (1987). Real and Complex Analysis, 3rd edition. (McGraw-Hill).
Shastri, A. (2000). An Introduction to Complex Analysis (Macmillan Publishers
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Shaw, W. (2006). Complex Analysis with MATHEMATICA (Cambridge Univer­
sity Press).
Shurman, J. (2012). Course Materials for Mathematics 311: Complex Analysis,
http://people.reed.edu/-jerry/311/mats.html.
Tall, D. (1970) Functions of a Complex Variable (Dover).
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on Complex Analysis (Dover).
269
Index
concatenation of paths, 71
conjugate, 11
contour integral, 62
convergent series, 106
curve, 15
fl-contractible curve, 82
C, 1
absolute value, 10
absolutely convergent series, 106
Argand plane, 5
d-bar operator, 57
de Moivre's formula, 8
degree of a polynomial, 100
Dirichlet problem, 173
divergent series, 106
domain, 15
biholomorphism, 175
binomial coefficients, 10
Binomial Formula, 180
Binomial Theorem, 10
Bombelli, 4
boundary data, 173
elliptic PDE, 164
entire function, 30
epicycloid, 87
essential singularity, 141
Euler Product Formula, 107
exponential, 18
Cardano, 3
Casorati-Weierstrass Theorem, 151
Cauchy Integral Formula, 91
Cauchy Integral formula, 120
Cauchy Integral Theorem, 77
Cauchy principal value, 157
Cauchy sequence, 14
Cauchy's inequality, 122
Cauchy-Riemann equations, 37
chain rule, 34
classification of zeros, 124
closed path, 76
commutative ring, 128
complete, 14
complex differentiable function, 30
complex exponential, 18
complex number, 1
complex plane, 5
Fibonacci number, 162
field, 2
Fresnel integrals, 159
Fundamental Theorem of Algebra, 99
Fundamental Theorem of Contour
Integration, 74
Gaussian integer, 8
harmonic conjugate, 169
harmonic function, 163
holomorphic function, 30
271
272
A Friendly Approach to Complex Analysis
homotopic paths, 78
Identity Theorem, 127
imaginary part, 1
improper integral, 157
integral domain, 128
Integration by Parts Formula, 76
isolated singularity, 141
Laplace equation, 163
Laurent series, 131
Liouville's Theorem, 99
Log, 24
logarithm function, 24
Prime Number Theorem, x
primitive, 88
principal argument, 24
principal value, 26
radius of convergence, 108
real analytic function, 115
real part, 1
removable singularity, 141
Residue Theorem, 155
Riemann zeta function, 107
Riemann Hypothesis, x, 107
Riemann Mapping Theorem, 175
Riemann zeta function, x
root of a polynomial, 100
roots of a complex number, 8
Mobius transformation, 175
Maximum Modulus Theorem, 129
Mean Value Property, 172
Minimum Modulus Theorem, 130
Morera's Theorem, 101
simply connected domain, 82
smooth path, 61
stepwise path, 16
order of a pole, 142
order of a zero of a function, 123
ordered field, 4
Taylor series, 120
triangle inequality, 12
trigonometric functions, 22
partial sums, 106
path, 15
path-connected, 15
Picard's Theorem, 153
Poisson Integral Formula, 174
pole, 141
power series, 107
winding number, 85
zero of a function, 123
zero divisor, 128
zero of a polynomial, 100
zero of a certain order of a function,
123
A FRIENDLY APPROACH TO
COMPLEX ANALYSIS
The book constitutes a basic, concise, yet
rigorous course in complex analysis, for
students who have studied calculus in one
and several variables, but have not
previously been exposed to complex
analysis. The textbook should be particularly
useful and relevant for undergraduate
students in joint programmes with
mathematics, as well as engineering
students. The aim of the book is to cover
the bare bones of the subject with minimal
prerequisites. The core content of the book
is the three main pillars of complex
analysis: the Cauchy-Riemann equations,
the Cauchy Integral Theorem, and Taylor
and Laurent series expansions.
Each section contains several problems,
which are not purely drill exercises, but are
rather meant to reinforce the fundamental
concepts. Detailed solutions to all the
exercises appear at the end of the book,
making the book ideal also for self-study.
There are many figures illustrating the text.
World Scientific
www.worldscientific.com
9047 SC
1 1111 1 11 1 1 111 1 1111 1 1
ISBN 978-981-4578-99-B(pbk)
9 789814 578998