Endre pap auth complex analysis through exambookzz org

Complex Analysis through Examples and Exercises

Kluwer Text in the Mathematical Sciences

VOLUME21

A Graduate-Level Book Series

The titfes published in this series are listed at the end 0/this vofume.

Complex Analysis
through Examples
and Exercises

by

Endre Pap

Institute ofMathematics,
University ofNovi Sad,
Novi Sad, Yugoslavia

SPRINGER-SCIENCE+BUSINESS MEDIA, B.V.

A c.I.P. Catalogue record for this book is available from the Library of Congress.

ISBN 978-90-481-5253-7 ISBN 978-94-017-1106-7 (eBook)
DOI 10.1007/978-94-017-1106-7

Printed an acid-free paper

AII Rights Reserved
© 1999 Springer Science+Business Media Dordrecht
Originally published by Kluwer Academic Publishers in 1999
No part of the material protected by this copyright notice may be reproduced or
utilized in any form or by any means, electronic or mechanical,
inc1uding photocopying, recording or by any information storage and
retrieval system, without written permission from the copyright owner.

Contents

Contents v

Preface IX

1 The Complex Numbers 1

1.1 Algebraic Properties 1

1.1.1 Preliminaries 1

1.1.2 Examples and Exercises 2

1.2 The Topology of the Complex Plane 32

1.2.1 Preliminaries . . . . . . 32

1.2.2 Examples and Exercises 33

2 Sequences and series 37

2.1 Sequences . . . . . 37

2.1.1 Preliminaries 37

2.1.2 Examples and Exercises 38

2.2 Series . . . . . . . . . 44

2.2.1 Preliminaries 44

2.2.2 Examples and Exercises 45

3 Complex functions 53

3.1 General Properties 53

3.1.1 Preliminaries 53

3.1.2 Examples and Exercises 54

vi CONTENTS

3.2 Special Functions .. 64
3.2.1 Preliminaries 64
3.2.2 Examples and Exercises 65
68
3.3 Multi-valued functions 68
3.3.1 Preliminaries 68
3.3.2 Examples and Exercises
73

4 Conformal mappings
73
4.1 Basics . . . . . . . 73
4.1.1 Preliminaries 73
4.1.2 Examples and Exercises 74
74
4.2 Special mappings .. 75

4.2.1 Preliminaries 103
4.2.2 Examples and Exercises
103
5 The Integral 103
104
5.1 Basics ••• 0 ••
129
5.1.1 Preliminaries
129
5.1.2 Examples and Exercises 129
131
6 The Analytic functions 162
6.1 The Power Series Representation
171
6.1.1 Preliminaries . . . . . .
6.1.2 Examples and Exercises 171
6.2 Composite Examples . . . . . . 171
172
7 Isolated Singularities 177
7.1 Singularities . . . . 177
7.1.1 Preliminaries
7.1.2 Examples and Exercises

7.2 Laurent series ....
7.2.1 Preliminaries

CONTENTS vii

7.2.2 Examples and Exercises . . . . . . . . . . . . . . . . . . . . . 179

8 Residues 191
8.1 Residue Theorem · 191
8.1.1 Preliminaries · 191
8.1.2 Examples and Exercises · 193
8.2 Composite Examples . . . . . . .206

9 Analytic continuation 227
9.1 Continuation . . . .227
9.1.1 Preliminaries .227
9.1.2 Examples and Exercises .228
9.2 Composite Examples . . . . . . .233

10 Integral transforms 255
10.1 Analytic Functions Defined by Integrals. .255
10.1.1 Preliminaries . . . . . . .255
10.1.2 Examples and Exercises .256
10.2 Composite Examples . . . . . . .268

11 Miscellaneous Examples 313

Bibliography 333

List of Symbols 335

Index 336

Preface

The book Complex Analysis through Examples and Exercises has come out from the
lectures and exercises that the author held mostly for mathematician and physists .
The book is an attempt to present the rather involved subject of complex analysis
through an active approach by the reader. Thus this book is a complex combination
of theory and examples.

Complex analysis is involved in all branches of mathematics. It often happens
that the complex analysis is the shortest path for solving a problem in real circum-
stances. We are using the (Cauchy) integral approach and the (Weierstrass) power
series approach .

In the theory of complex analysis, on the hand one has an interplay of several
mathematical disciplines, while on the other various methods, tools, and approaches.
In view of that, the exposition of new notions and methods in our book is taken step
by step. A minimal amount of expository theory is included at the beinning of each
section, the Preliminaries, with maximum effort placed on weil selected examples
and exercises capturing the essence of the material. Actually, I have divided the
problems into two classes called Examples and Exercises (some of them often also
contain proofs of the statements from the Preliminaries). The examples contain
complete solutions and serve as a model for solving similar problems given in the
exercises. The readers are left to find the solution in the exercisesj the answers, and,
occasionally, some hints, are still given. Special sections contain so called Composite
Examples which consist of combinations of different types of examples explaining,
altogether, some problems completely and giving to the reader an opportunity to

check his entire previously accepted knowledge.

The necessary prerequisites are a standard undergraduate course on real func-
tions of real variables. I have tried to make the book self-contained as much as
possible. For that reason, I have also included in the Preliminaries and Examples
some of the mathematical tools mentioned.

The book is prepared for undergraduate and graduate students in matheniatics,
physics, technology, economics, and everybody with an interest in complex analysis.

We have used for some calculations and drawings the mathematical software

ix

x PREFACE

packages Mathematica and Scientific Work Place v2.5.

I am grateful to Academician Bogoljub Stankovic for a long period of collabo-
ration on the subject of the book, to Prof. Arpad TakaCi for his numerous remarks
and advice about the text, and to Ivana Stajner for reading some part of the text.
I would like to express my thanks to MarCicev Merima for typing the majority of
the manuscript. It is my pleasure to thank the Institute of Mathematics in Novi
Sad for working conditions and financial support. I would like to thank Kluwer
Academic Publishers, especially Dr. Paul Roos and Ms. Angelique Hempel for their
encouragement and patience.

Novi Sad, June 1998 ENDRE PAP

Chapter 1

The Complex N umbers

1.1 Algebraic Properties

1.1.1 Preliminaries
The field of complex numbers Cis the set of all ordered pairs (a, b) where a and b

are real numbers and where addition and multiplication are defined by:

(a, b) + (c, d) = (a + c, b+ d)
(a, b)(c, d) = (ac - bd, bc + ad).

We will write a for the complex number (a,O). In fact, the mapping a 1-+ (a,O)
defines a field isomorphism of IR into C, hence we may consider IR as a subset of C. If

we put z = (0,1), then (a, b) = a + bz. For z = a + zb we put Re z = a and Imz = b.

Real numbers are associated with points on the x-axis and called the real axis .
Purely imaginary numbers are associated with points on the y-axis and called the
imaginary axis .

Note that z2 = -1, so the equation z2+ 1 = 0 has a root in C. If z = x+zy (x, y E
IR), then we define

Izl = Jx 2 + y2

to be the absolute value of z and z = x - zy is the conjugate of z. We have Izl 2 = zz
and the triangle inequality

Iz + wl s:; Izl + Iwl (z,w E C).

By the definition of complex numbers, each z in C can be identified with a unique
point (Rez,Imz) in the plane IR2•

1

E. Pap, Complex Analysis through Examples and Exercises
© Springer Science+Business Media Dordrecht 1999

2 CHAPTER 1. THE COMPLEX NUMBERS

The argument of z i- 0, denoted by arg z, is the angle 0 (modulo 271") between

the vector from the origin to z and positive x-axis, i.e., which satisfy

Rez . 0 Imz
cosO = Tz! and sm = Tz!.

The point z = x+zy =I- 0 has polar coordinates (r, 0) : x = r cos 0, y = r sin O. Clearly

°r = Izl and 0 is the angle between the positive real axis and the line segment from
to z. Notice that 0 plus any multiple of 271" can be substituted for 0 in the above

equations. The angle 0 is called the argument of z and is denoted by 0 = argz.

Let Zl = rl (cos Ol + zsin Ol and Z2 = r2 (cos O2 + i sin ( 2 ) then

In particular, if z = r(cos 0 + zsin 0), then

zn = rn(cos(nO) + zsin(nO». (1.1 )

As a special case of (1.1) we obtain DeMoivre's formula:

(cos 0 + zsin Ot = cos nO + zsin nO. (1.2)

The n-th root of z = r(cos 0 + zsin 0) are

o+ 2h . 0 + 2h
Zk = y'r(cos + zsm )
n n

for k = 0,1, ... , n - 1.

The complex number z = r( cos O+z sin 0) also has the exponential representation

z = r exp( zO).

For more explanations see the chapter on power series.

1.1.2 Examples and Exercises

Example 1.1 Find the real numbers p and q such that the complex numbers
1

z = p + zq, W = P + z- be equal.

q

1.1. ALGEBRAIC PROPERTIES 3

Solution. We have that z = w is equivalent with Rez = Rew and Imz =

Immw. Therefore p = p and q = !, pEllt, q2 = 1, i.e., ql = 1, q2 = -1 and pEllt.

q

Example 1.2 For z = 1 + z find w such that the real parts of the following numbers

are equal to zero a)z + Wj b) Z· Wj c) fuj d) ~.

Solution. Let z = 1 + z = (1,1) and w = x + zy = (x, y). Then we have

° a)

Re(z +w) = {=} 1 + x = 0.
Hence x = -1 and y E llt is arbitrary.

b) Re (z . w) = °{=} Re (x - y + z(x + y)) = 0. x· z, for arbitrary
by w =
Hence x - y = 0. Therefore w is given x +zx = x(l +z) =
xE llt.
have fu = 0. Hence
c) We

z 1 + z x - zy x + y + z(x - y)
- = - - . - - = ---'--:--'--::--~
W X + zy x - zy x 2 + y2

Therefore Re(-=-) =o{=} ~+Y2 =O:::}x=-y,xi=0.

w x +y

° Finally we have w = x(l - z) for x E llt and x i= 0.
d) From Re (~) = it follows that for every x E llt :

w x + zy 1 - z x + y + z(x - y)
-=--.--=
z l+z 1-z 2

Hence W = x(l - z) for every x E llt.

Example 1.3 Prove that

Hint. It is easy to prove the case n = 2 and then use mathematical induction

to prove the general case.

Example 1.4 Find for z = 1 + 2z the following numbers

a) zn j b) l/zj c) l/zn j d) Z2 + 2z + 5 + z.

4 CHAPTER 1. THE COMPLEX NUMBERS

Solution. a) We have

E zn (1 + 2zt

(~)(2z)k

[.!!.) [n-l)
IJ-l)k(n)22k+Z 't(-l)k( n )22k+1,
k=O 2k k=O 2k + 1

where [xl is the greatest integer part of x.

b) We have

1 1 1 - 2z 1 - 2z 1 2
- = - - . - - = - - = - - - z.
z 1 + 2z 1 - 2z 5 5 5

1 (1 1 e) Sinee ~ = (~)n, we have by b)znZ

2)n n
(1 + 2z)n = 5" - 5" z = 5n (1 - 2z) .

Applying the same proeeclure as in a) we obtain (using that the imaginary part of
z is -2) :

cl) Putting z = 1 + 2z in Z2 + 2z + 5 + z, we obtain 4 + 9z.

Example 1.5 Find the positions of the following points in the complex plane:

a + za, a - za, -a + za, -a - za fOT a E lR?

Solution. Using the trigonometrie representation (p,O) wc obtain

a + za V2 V2
V2aT + zv'2aT

(v'2' a ~) '

'4

a - za V2a . -v'2 - zV2Va2-
2 2
-a + za
(v'2a , -~4)'

(V2a, 3:),

-a - za (V2a, 3:).

1.1. ALGEBRAIC PROPERTIES 5

Hence the four given points are the corners of the square in the circle with the center

at origin and the radius J21a1-

Example 1.6 Which subsets of the complex plain correspond to the complex num-
bers with the following properties:

a) Rez = Imz; b) Rez< 1; c) -1:::;Rez:::;l;

d) Imz 2 0; e) Izl :::; 2; f) 1 < Izl < 3;
g) Izl > 2;
h) -7r < argz < 7r; i) ~ < argz < f?

Solution.

a) Rez = Imz {::::=:} x = y, where z = x + iy.

The desired subset consists of the points of the straight line y = x (Figure 1.1).

x

Figure 1.1 Rez = Imz

b) Rez < 1 {::::=:} x < 1 and y is an arbitrary real number, where z = x +iy. The

desired subset is the half plane left from the straight line x = 1 (without the points

6 CHAPTER 1. THE COMPLEX NUMBERS
of this straight line), Figure 1.2.

x

Figure 1.2 Rez < 1

c) -1 :::; Re z :::; 1 means -1 :::; x :::; 1 and y is an arbitrary real number. The
desired subset is the strip between straight lines x = -1 and x = 1, Figure 1.3.

:..1 x

Figure 1.3 -1 :::; Rez :::; 1

d) Im x ~ 0 means y ~ 0 and x is an arbitrary real number. The desired subset

1.1. ALGEBRAIC PROPERTIES 7

is the upper half plane with respect to the x- axis, Figure 1.4.

Figure 1.4 Imz ~ °

e) The eondition JzJ ~ 2, using the trigonometrie representation of the complex
number z = (p, ()), reduced on JzJ = p ~ 2 and () is an arbitrary angle from [0, 27l'J.

The desired subset is the disc with center at (0,0) and radius r = 2, Figure 1.5.

2 X

Figure 1.5 JzJ ~ 2

f) The ease 1 < JzJ < 3 reduces in a sirnilar way as in e) on 1 < JzJ = P < 3,

where () is an arbitrary angle from the interval [0,27l'J. The desired subset is the

8 CHAPTER 1. THE COMPLEX NUMBERS

annulus between the circles Izl = 1 and Izl = 3 without these circles, Figure 1.6.

Figure 1.6 1 < Izl < 3
g) For the case Izl > 2 we have JzJ = p > 2, () E [0, 27r]. Therefore the desired
subset is the whole complex plane without the disc Jzl ~ 2, Figure 1.7.

Figure 1.7 Izl > 2

h) The condition -7r < arg z < 7r implies that p is arbitrary and -7r < () < 7r.

Therefore the desired subset is the whole complex plane without the negative part

1.1. ALGEBRAIC PROPERTIES 9

of the real axis, Figure 1.8.

x

Figure 1.8 -'Fr < arg z < 'Fr

i) The condition 'Fr /6 < arg z < 'Fr /4 implies that p is arbitrary and e E ('Fr/6, 'Fr/4).

The desired subset are the points between half straight lines y = x . tan ~ and

y = x . tan i without these half straight lines, Figure 1.9.

Figure 1.9 'Fr/6 < arg z < 'Fr /4

Example 1.7 Find the conditions for the real and imaginary part of a complex

number z so that z belongs to the triangle with vertices 0,1 +zY3, 1 +z/ Y3, Figure

1.10.

10 CHAPTER 1. THE COMPLEX NUMBERS

Solution. We denote the vertices as:

Zl = 0, Z2 = 1 + zV3, Z3 = 1 + z/V3.

y

O=z 1 x

Figure 1.10

If we denote z = x + zy, then Rez = x and Imz = y and

-Y = tan (), () E [1-C', -1C'] .
x 6 3

Therefore Im z = Re z . tan (). Since the function tan () is monotone increasing we

have
1
y'3Rez ::::; Imz ::::; V3Rez.

Together with the condition 0 ::::; Re z ::::; 1 we have completely described the points
in the given triangle with vertices Zt, Z2 and Z3.

Example 1.8 Prove that tor every z E C

Izl ::::; IRezl + IImzl ::::; -12 ·14

Solution. The left part of the inequality follows from

Izl = v'Re 2z + Im 2z::::; y'(IRezl + IImzl)2 = IRezl + IImzl

1.1. ALGEBRAIC PROPERTIES 11

The right part of the inequality follows from the following obvious inequality

(IRezl-IImzr ~ O.

Then IRezl 2 + IImzl 2 - 2lRezl·IImzl ~ 0,

Izl2 ~ 2IRezl·IIm z l·

Adding Izl 2 to the both sides of the last inequality we obtain

21z12 ~ 2lRezl'IImzl + Iz12,

21z1 2 ~ (IRezl + IImzlt Hence v'2lzl ~ IRezl + IImzl·

Example 1.9 Find the complex numbers which are the corners 01 the triangle with
equal sides with vertices on unit circle and whose one vertex is on the negative part
the real axis.

Solution. Let us put Zj = pj(cos8j + zsin8j ),j = 1,2,3, for the soughtafter
points.

/ /

Figure 1.11
We have for alt three vertices Pj = 1 and

Zl : 81 = 7r /3, z2: 82 = 7r, Z3: 83 = -7r /3.

Therefore:

Zl = (1, 7r/3) = 1/2 + zV3/2; Z2 = (1, 7r) = -1; Z3 = (1, -7r/3) = 1/2 - zV3/2,

Figure 1.11.