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A First Course in

Complex
Analysis
with Applications

Dennis G. Zill
Loyola Marymount University

Patrick D. Shanahan
Loyola Marymount University


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Copyright © 2003 by Jones and Bartlett Publishers, Inc.
Library of Congress Cataloging-in-Publication Data
Zill, Dennis G., 1940A first course in complex analysis with applications / Dennis G. Zill, Patrick D. Shanahan.
p. cm.
Includes indexes.
ISBN 0-7637-1437-2
1. Functions of complex variables. I. Shanahan, Patrick, 1931- II. Title.
QA331.7 .Z55 2003
515’.9—dc21
2002034160
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Printed in the United States of America
06 05 04 03 02
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For Dana, Kasey, and Cody



Contents

7.1

Contents
Preface

ix

Chapter 1.

Complex Numbers and the Complex Plane 1
1.1
Complex Numbers and Their Properties 2
1.2
Complex Plane 10
1.3
Polar Form of Complex Numbers 16

1.4
Powers and Roots 23
1.5
Sets of Points in the Complex Plane 29
1.6
Applications 36
Chapter 1 Review Quiz 45

Chapter 2.

Complex Functions and Mappings 49
2.1
Complex Functions 50
2.2
Complex Functions as Mappings 58
2.3
Linear Mappings 68
2.4
Special Power Functions 80
2.4.1 The Power Function z n 81
2.4.2 The Power Function z 1/n 86
2.5
Reciprocal Function 100
2.6
Limits and Continuity 110
2.6.1 Limits 110
2.6.2 Continuity 119
2.7
Applications 132
Chapter 2 Review Quiz 138


Chapter 3.

Analytic Functions 141
3.1
Differentiability and Analyticity 142
3.2
Cauchy-Riemann Equations 152
3.3
Harmonic Functions 159
3.4
Applications 164
Chapter 3 Review Quiz 172

v


vi

Contents

Chapter 4.

Elementary Functions 175
4.1
Exponential and Logarithmic Functions 176
4.1.1 Complex Exponential Function 176
4.1.2 Complex Logarithmic Function 182
4.2
Complex Powers 194

4.3
Trigonometric and Hyperbolic Functions 200
4.3.1 Complex Trigonometric Functions 200
4.3.2 Complex Hyperbolic Functions 209
4.4
Inverse Trigonometric and Hyperbolic
Functions 214
4.5
Applications 222
Chapter 4 Review Quiz 232

Chapter 5.

Integration in the Complex Plane 235
5.1
Real Integrals 236
5.2
Complex Integrals 245
5.3
Cauchy-Goursat Theorem 256
5.4
Independence of Path 264
5.5
Cauchy’s Integral Formulas and Their
Consequences 272
5.5.1 Cauchy’s Two Integral Formulas 273
5.5.2 Some Consequences of the Integral
Formulas 277
5.6
Applications 284

Chapter 5 Review Quiz 297

Chapter 6.

Series
6.1
6.2
6.3
6.4
6.5
6.6

6.7

and Residues 301
Sequences and Series 302
Taylor Series 313
Laurent Series 324
Zeros and Poles 335
Residues and Residue Theorem 342
Some Consequences of the Residue
Theorem 352
6.6.1 Evaluation of Real Trigonometric
Integrals 352
6.6.2 Evaluation of Real Improper
Integrals 354
6.6.3 Integration along a Branch Cut 361
6.6.4 The Argument Principle and Rouch´e’s
Theorem 363
6.6.5 Summing Infinite Series 367

Applications 374
Chapter 6 Review Quiz 386


Contents

vii

Chapter 7.

Conformal Mappings 389
7.1
Conformal Mapping 390
7.2
Linear Fractional Transformations 399
7.3
Schwarz-Christoffel Transformations 410
7.4
Poisson Integral Formulas 420
7.5
Applications 429
7.5.1 Boundary-Value Problems 429
7.5.2 Fluid Flow 437
Chapter 7 Review Quiz 448

Appendixes: I
II
III

Proof of Theorem 2.1 APP-2

Proof of the Cauchy-Goursat Theorem
Table of Conformal Mappings APP-9

Answers for Selected Odd-Numbered Problems
Index

IND-1

ANS-1

APP-4



Preface

7.2

Preface

Philosophy This text grew out of chapters 17-20 in Advanced Engineering Mathematics, Second Edition (Jones and Bartlett Publishers), by Dennis
G. Zill and the late Michael R. Cullen. This present work represents an expansion and revision of that original material and is intended for use in either
a one-semester or a one-quarter course. Its aim is to introduce the basic principles and applications of complex analysis to undergraduates who have no
prior knowledge of this subject.
The motivation to adapt the material from Advanced Engineering Mathematics into a stand-alone text sprang from our dissatisfaction with the succession of textbooks that we have used over the years in our departmental
undergraduate course offering in complex analysis. It has been our experience
that books claiming to be accessible to undergraduates were often written at a
level that was too advanced for our audience. The “audience” for our juniorlevel course consists of some majors in mathematics, some majors in physics,
but mostly majors from electrical engineering and computer science. At our
institution, a typical student majoring in science or engineering does not take

theory-oriented mathematics courses in methods of proof, linear algebra, abstract algebra, advanced calculus, or introductory real analysis. Moreover,
the only prerequisite for our undergraduate course in complex variables is
the completion of the third semester of the calculus sequence. For the most
part, then, calculus is all that we assume by way of preparation for a student
to use this text, although some working knowledge of differential equations
would be helpful in the sections devoted to applications. We have kept the
theory in this introductory text to what we hope is a manageable level, concentrating only on what we feel is necessary. Many concepts are conveyed
in an informal and conceptual style and driven by examples, rather than the
formal definition/theorem/proof. We think it would be fair to characterize
this text as a continuation of the study of calculus, but also the study of the
calculus of functions of a complex variable. Do not misinterpret the preceding
words; we have not abandoned theory in favor of “cookbook recipes”; proofs
of major results are presented and much of the standard terminology is used.
Indeed, there are many problems in the exercise sets in which a student is
asked to prove something. We freely admit that any student—not just majors in mathematics—can gain some mathematical maturity and insight by
attempting a proof. But we know, too, that most students have no idea how
to start a proof. Thus, in some of our “proof” problems, either the reader
ix


x

Preface

is guided through the starting steps or a strong hint on how to proceed is
provided.
The writing herein is straightforward and reflects the no-nonsense style
of Advanced Engineering Mathematics.

Content We have purposely limited the number of chapters in this text

to seven. This was done for two “reasons”: to provide an appropriate quantity
of material so that most of it can reasonably be covered in a one-term course,
and at the same time to keep the cost of the text within reason.
Here is a brief description of the topics covered in the seven chapters.
• Chapter 1 The complex number system and the complex plane are
examined in detail.
• Chapter 2 Functions of a complex variable, limits, continuity, and
mappings are introduced.
• Chapter 3 The all-important concepts of the derivative of a complex
function and analyticity of a function are presented.
• Chapter 4 The trigonometric, exponential, hyperbolic, and logarithmic functions are covered. The subtle notions of multiple-valued functions and branches are also discussed.
• Chapter 5 The chapter begins with a review of real integrals (including line integrals). The definitions of real line integrals are used to
motivate the definition of the complex integral. The famous CauchyGoursat theorem and the Cauchy integral formulas are introduced in
this chapter. Although we use Green’s theorem to prove Cauchy’s theorem, a sketch of the proof of Goursat’s version of this same theorem is
given in an appendix.
• Chapter 6 This chapter introduces the concepts of complex sequences
and infinite series. The focus of the chapter is on Laurent series, residues,
and the residue theorem. Evaluation of complex as well as real integrals,
summation of infinite series, and calculation of inverse Laplace and inverse Fourier transforms are some of the applications of residue theory
that are covered.
• Chapter 7 Complex mappings that are conformal are defined and
used to solve certain problems involving Laplace’s partial differential
equation.

Features Each chapter begins with its own opening page that includes a
table of contents and a brief introduction describing the material to be covered
in the chapter. Moreover, each section in a chapter starts with introductory comments on the specifics covered in that section. Almost every section
ends with a feature called Remarks in which we talk to the students about
areas where real and complex calculus differ or discuss additional interesting
topics (such as the Riemann sphere and Riemann surfaces) that are related



Preface

xi

to, but not formally covered in, the section. Several of the longer sections,
although unified by subject matter, have been partitioned into subsections;
this was done to facilitate covering the material over several class periods.
The corresponding exercise sets were divided in the same manner in order to
make the assignment of homework easier. Comments, clarifications, and some
warnings are liberally scattered throughout the text by means of annotations
in the left margin marked by the symbol ☞.
There are a lot of examples and we have tried very hard to supply all
pertinent details in the solutions of the examples. Because applications of
complex variables are often compiled into a single chapter placed at the end
of the text, instructors are sometimes hard pressed to cover any applications
in the course. Complex analysis is a powerful tool in applied mathematics. So
to facilitate covering this beautiful aspect of the subject, we have chosen to
end each chapter with a separate section on applications. The exercise sets are
constructed in a pyramidal fashion and each set has at least two parts. The
first part of an exercise set is a generous supply of routine drill-type problems;
the second part consists of conceptual word and geometrical problems. In
many exercise sets, there is a third part devoted to the use of technology.
Since the default operational mode of all computer algebra systems is complex
variables, we have placed an emphasis on that type of software. Although we
have discussed the use of Mathematica in the text proper, the problems are
generic in nature. Answers to selected odd-numbered problems are given in
the back of the text. Since the conceptual problems could also be used as
topics for classroom discussion, we decided not to include their answers. Each

chapter ends with a Chapter Review Quiz. We thought that something more
conceptual would be a bit more interesting than the rehashing of the same
old problems given in the traditional Chapter Review Exercises. Lastly, to
illustrate the subtleties of the action of complex mappings, we have used two
colors.

Acknowledgments We would like to express our appreciation to our
colleague at Loyola Marymount University, Lily Khadjavi, for volunteering to
use a preliminary version of this text. We greatly appreciate her careful reading of the manuscript. We also wish to acknowledge the valuable input from
students who used this book, in particular: Patrick Cahalan, Willa Crosby,
Kellie Dyerly, Sarah Howard, and Matt Kursar. A deeply felt “thank you”
goes to the following reviewers for their words of encouragement, criticisms,
and thoughtful suggestions:
Nicolae H. Pavel, Ohio University
Marcos Jardim, University of Pennsylvania
Ilia A. Binder, Harvard University
Finally, we thank the editorial and production staff at Jones and Bartlett,
especially our production manager, Amy Rose, for their many contributions
and cooperation in the making of this text.






xii

Preface

A Request Although the preliminary versions of this book were class

tested for several semesters, experience has taught us that errors—typos or
just plain mistakes—seem to be an inescapable by-product of the textbookwriting endeavor. We apologize in advance for any errors that you may find
and urge you to bring them to our attention.
Dennis G. Zill
Patrick D. Shanahan
Los Angeles, CA





20
pa


1

Complex Numbers
and the
Complex Plane



π
0
–π
–2π
–3π

1


–1
0

0
1 –1

Riemann surface for arg( z ). See
page 97.

1.1

Complex Numbers and Their Properties

1.2

Complex Plane

1.3

Polar Form of Complex Numbers

1.4

Powers and Roots

1.5

Sets of Points in the Complex Plane


1.6

Applications
Chapter 1 Review Quiz
Introduction In elementary courses you learned
about the existence, and some of the properties,
of complex numbers. But in courses in calculus,
it is most likely that you did not even see a complex number. In this text we study nothing but
complex numbers and the calculus of functions
of a complex variable.
We begin with an in-depth examination of
the arithmetic and algebra of complex numbers.

1


2

Chapter 1 Complex Numbers and the Complex Plane

1.1

Complex Numbers and Their Properties
No one person 1.1
“invented” complex numbers, but controversies surrounding the use of these
numbers existed in the sixteenth century. In their quest to solve polynomial equations by
formulas involving radicals, early dabblers in mathematics were forced to admit that there
were other kinds of numbers besides positive integers. Equations such as x2 + 2x + 2 = 0




and x3 = 6x + 4 that yielded “solutions” 1 + −1 and 3 2 + −2 + 3 2 − −2 caused
particular consternation within the community of √
fledgling mathematical
scholars because

everyone knew that there are no numbers such as −1 and −2, numbers whose square is
negative. Such “numbers” exist only in one’s imagination, or as one philosopher opined, “the
imaginary, (the) bosom child of complex mysticism.” Over time these “imaginary numbers”
did not go away, mainly because mathematicians as a group are tenacious and some are even
practical. A famous mathematician held that even though “they exist in our imagination
. . . nothing prevents us from . . . employing them in calculations.” Mathematicians
also hate to throw anything away. After all, a memory still lingered that negative numbers
at first were branded “fictitious.” The concept of number evolved over centuries; gradually
the set of numbers grew from just positive integers to include rational numbers, negative
numbers, and irrational numbers. But in the eighteenth century the number concept took a
gigantic evolutionary step forward when the German mathematician Carl Friedrich Gauss
put the so-called imaginary numbers—or complex numbers, as they were now beginning to
be called—on a logical and consistent footing by treating them as an extension of the real
number system.
Our goal in this first section is to examine some basic definitions and the arithmetic of
complex numbers.

The Imaginary Unit Even after gaining wide respectability, through
the seminal works of Karl Friedrich Gauss and the French mathematician Augustin Louis Cauchy, the unfortunate name “imaginary” has survived down
the centuries. The
√ symbol i was originally used as a disguise for the embarrassing symbol −1. We now say that i is the imaginary unit and define
it by the property i2 = –1. Using the imaginary unit, we build a general
complex number out of two real numbers.
Definition 1.1


Complex Number

A complex number is any number of the form z = a + ib where a and
b are real numbers and i is the imaginary unit.

Note: The imaginary part of
z = 4 − 9i is −9 not −9i.



Terminology The notations a + ib and a + bi are used interchangeably.
The real number a in z = a+ ib is called the real part of z; the real number b
is called the imaginary part of z. The real and imaginary parts of a complex
number z are abbreviated Re(z) and Im(z), respectively. For example, if
z = 4 − 9i, then Re(z) = 4 and Im(z) = −9. A real constant multiple
of the imaginary unit is called a pure imaginary number. For example,
z = 6i is a pure imaginary number. Two complex numbers are equal if their


1.1 Complex Numbers and Their Properties

3

corresponding real and imaginary parts are equal. Since this simple concept
is sometimes useful, we formalize the last statement in the next definition.

Definition 1.2 Equality
Complex numbers z1 = a1 + ib 1 and z2 = a2 + ib 2 are equal, z1 = z2 , if
a1 = a2 and b1 = b2 .


In terms of the symbols Re(z) and Im(z), Definition 1.2 states that z1 = z2 if
Re(z1 ) = Re(z2 ) and Im(z1 ) = Im(z2 ).
The totality of complex numbers or the set of complex numbers is usually
denoted by the symbol C. Because any real number a can be written as
z = a + 0i, we see that the set R of real numbers is a subset of C.

Arithmetic Operations Complex numbers can be added, subtracted,
multiplied, and divided. If z1 = a1 + ib 1 and z2 = a2 + ib 2 , these operations
are defined as follows.
Addition:

z1 + z2 = (a1 + ib1 ) + (a2 + ib2 ) = (a1 + a2 ) + i(b1 + b2 )

Subtraction:

z1 −z2 = (a1 + ib1 ) − (a2 + ib2 ) = (a1 − a2 ) + i(b1 − b2 )

Multiplication:

z1 · z2 = (a1 + ib1 )(a2 + ib2 )
= a1 a2 − b1 b2 + i(b1 a2 + a1 b2 )

Division:

z1
a1 + ib1
=
, a2 = 0, or b2 = 0
z2

a2 + ib2
b1 a2 − a1 b2
a1 a2 + b1 b2
+i
=
2
2
a2 + b2
a22 + b22

The familiar commutative, associative, and distributive laws hold for complex numbers:

z +z = z +z
1
2
2
1
Commutative laws:
 z z =z z
1 2
2 1

Associative laws:


 z + (z + z ) = (z + z ) + z
1
2
3
1

2
3

z (z z ) = (z z )z
1

Distributive law:

2 3

1 2

3

z1 (z2 + z3 ) = z1 z2 + z1 z3

In view of these laws, there is no need to memorize the definitions of
addition, subtraction, and multiplication.


4

Chapter 1 Complex Numbers and the Complex Plane

Addition, Subtraction, and Multiplication
(i ) To add (subtract ) two complex numbers, simply add (subtract ) the
corresponding real and imaginary parts.
(ii ) To multiply two complex numbers, use the distributive law and the
fact that i2 = −1.
The definition of division deserves further elaboration, and so we will discuss

that operation in more detail shortly.

EXAMPLE 1

Addition and Multiplication

If z1 = 2 + 4i and z2 = −3 + 8i, find (a) z1 + z2 and (b) z1 z2 .
Solution (a) By adding real and imaginary parts, the sum of the two complex
numbers z1 and z2 is
z1 + z2 = (2 + 4i) + (−3 + 8i) = (2 − 3) + (4 + 8)i = −1 + 12i.
(b) By the distributive law and i2 = −1, the product of z1 and z2 is
z1 z2 = (2 + 4i) (−3 + 8i) = (2 + 4i) (−3) + (2 + 4i) (8i)
= −6 − 12i + 16i + 32i2
= (−6 − 32) + (16 − 12)i = −38 + 4i.

Zero and Unity The zero in the complex number system is the number 0 + 0i and the unity is 1 + 0i. The zero and unity are denoted by 0 and
1, respectively. The zero is the additive identity in the complex number
system since, for any complex number z = a + ib, we have z + 0 = z. To see
this, we use the definition of addition:
z + 0 = (a + ib) + (0 + 0i) = a + 0 + i(b + 0) = a + ib = z.
Similarly, the unity is the multiplicative identity of the system since, for
any complex number z, we have z · 1 = z · (1 + 0i) = z.
There is also no need to memorize the definition of division, but before
discussing why this is so, we need to introduce another concept.

Conjugate If z is a complex number, the number obtained by changing
the sign of its imaginary part is called the complex conjugate, or simply
conjugate, of z and is denoted by the symbol z¯. In other words, if z = a + ib,



1.1 Complex Numbers and Their Properties

5

then its conjugate is z¯ = a − ib. For example, if z = 6 + 3i, then z¯ = 6 − 3i;
if z = −5 − i, then z¯ = −5 + i. If z is a real number, say, z = 7, then
z¯ = 7. From the definitions of addition and subtraction of complex numbers,
it is readily shown that the conjugate of a sum and difference of two complex
numbers is the sum and difference of the conjugates:
z1 − z2 = z¯1 − z¯2 .

z1 + z2 = z¯1 + z¯2 ,

(1)

Moreover, we have the following three additional properties:
z1 z2 = z¯1 z¯2 ,

z1
z2

=

z¯1
,
z¯2

z¯ = z.

(2)


Of course, the conjugate of any finite sum (product) of complex numbers is
the sum (product) of the conjugates.
The definitions of addition and multiplication show that the sum and
product of a complex number z with its conjugate z¯ is a real number:
z + z¯ = (a + ib) + (a − ib) = 2a

(3)

z z¯ = (a + ib)(a − ib) = a − i b = a + b .
2

2 2

2

2

(4)

The difference of a complex number z with its conjugate z¯ is a pure imaginary
number:
z − z¯ = (a + ib) − (a − ib) = 2ib.

(5)

Since a = Re(z) and b = Im(z), (3) and (5) yield two useful formulas:
Re(z) =

z + z¯

2

and

Im(z) =

z − z¯
.
2i

(6)

However, (4) is the important relationship in this discussion because it enables
us to approach division in a practical manner.

Division
To divide z 1 by z 2 , multiply the numerator and denominator of z 1 /z2 by
the conjugate of z 2 . That is,
z1 z¯2
z1 z¯2
z1
=
·
=
z2
z2 z¯2
z2 z¯2

(7)


and then use the fact that z2 z¯2 is the sum of the squares of the real and
imaginary parts of z 2 .
The procedure described in (7) is illustrated in the next example.

EXAMPLE 2

Division

If z1 = 2 − 3i and z2 = 4 + 6i, find z1 /z2 .


6

Chapter 1 Complex Numbers and the Complex Plane

Solution We multiply numerator and denominator by the conjugate
z¯2 = 4 − 6i of the denominator z2 = 4 + 6i and then use (4):
2 − 3i
−10 − 24i
2 − 3i 4 − 6i
8 − 12i − 12i + 18i2
z1
=
=
=
=
.
z2
4 + 6i
4 + 6i 4 − 6i

42 + 62
52
Because we want an answer in the form a + bi, we rewrite the last result by
dividing the real and imaginary parts of the numerator −10 − 24i by 52 and
reducing to lowest terms:
5
6
10 24
z1
= − − i = − − i.
z2
52 52
26 13

Inverses In the complex number system, every number z has a unique
additive inverse. As in the real number system, the additive inverse of
z = a + ib is its negative, −z, where −z = −a − ib. For any complex number
z, we have z + (−z) = 0. Similarly, every nonzero complex number z has a
multiplicative inverse. In symbols, for z = 0 there exists one and only one
nonzero complex number z −1 such that zz −1 = 1. The multiplicative inverse
z −1 is the same as the reciprocal 1/z.

EXAMPLE 3

Reciprocal

Find the reciprocal of z = 2 − 3i.
Solution By the definition of division we obtain
1
1

1
2 + 3i
2 + 3i
2 + 3i
=
=
=
=
.
z
2 − 3i
2 − 3i 2 + 3i
4+9
13
Answer should be in the form a + ib.



That is,

3
2
1
= z −1 =
+ i.
z
13 13

You should take a few seconds to verify the multiplication
zz −1 = (2 − 3i)


2
13

+

3
13 i

= 1.

Remarks Comparison with Real Analysis
(i ) Many of the properties of the real number system R hold in the
complex number system C, but there are some truly remarkable
differences as well. For example, the concept of order in the
real number system does not carry over to the complex number
system. In other words, we cannot compare two complex numbers
z1 = a1 + ib1 , b1 = 0, and z2 = a2 + ib2 , b2 = 0, by means of


1.1 Complex Numbers and Their Properties

7

inequalities. Statements such as z1 < z2
meaning in C except in the special case
bers z1 and z2 are real.
See Problem
Therefore, if you see a statement such as
it is implicit from the use of the inequality

bol α represents a real number.

or z2 ≥ z1 have no
when the two num55 in Exercises 1.1.
z1 = αz2 , α > 0,
α > 0 that the sym-

(ii ) Some things that we take for granted as impossible in real analysis,
such as ex = −2 and sin x = 5 when x is a real variable, are perfectly correct and ordinary in complex analysis when the symbol x
is interpreted as a complex variable. See Example 3 in Section 4.1
and Example 2 in Section 4.3.
We will continue to point out other differences between real analysis and
complex analysis throughout the remainder of the text.

EXERCISES 1.1

Answers to selected odd-numbered problems begin on page ANS-2.

1. Evaluate the following powers of i.
(a) i8

(b) i11

(c) i42

(d) i105

2. Write the given number in the form a + ib.
(a) 2i3 − 3i2 + 5i
(c)


(b) 3i5 − i4 + 7i3 − 10i2 − 9

20
2
5
+ 3 − 18
i
i
i

2
−i

(d) 2i6 +

3

+ 5i−5 − 12i

In Problems 3–20, write the given number in the form a + ib.
3. (5 − 9i) + (2 − 4i)

4. 3(4 − i) − 3(5 + 2i)

5. i(5 + 7i)

6. i(4 − i) + 4i(1 + 2i)

7. (2 − 3i)(4 + i)


8.

9. 3i +

1
2−i

1
2

− 14 i

10.

i
1+i

2
3

+ 53 i

11.

2 − 4i
3 + 5i

12.


10 − 5i
6 + 2i

13.

(3 − i)(2 + 3i)
1+i

14.

(1 + i)(1 − 2i)
(2 + i)(4 − 3i)

15.

(5 − 4i) − (3 + 7i)
(4 + 2i) + (2 − 3i)

16.

(4 + 5i) + 2i3
(2 + i)2

17. i(1 − i)(2 − i)(2 + 6i)
19. (3 + 6i) + (4 − i)(3 + 5i) +

18. (1 + i)2 (1 − i)3
1
2−i


20. (2 + 3i)

2−i
1 + 2i

2


8

Chapter 1 Complex Numbers and the Complex Plane

In Problems 21–24, use the binomial theorem∗
n(n − 1) n−2 2
n n−1
B+
B + ···
A
A
1!
2!
n(n − 1)(n − 2) · · · (n − k + 1) n−k k
B + · · · + Bn,
A
+
k!
where n = 1, 2, 3, . . . , to write the given number in the form a + ib.
(A + B)n = An +

1 − 12 i


21. (2 + 3i)2

22.

23. (−2 + 2i)5

24. (1 + i)8

3

In Problems 25 and 26, find Re(z) and Im(z).
1
i
1
26. z =
25. z =
3−i
2 + 3i
(1 + i)(1 − 2i)(1 + 3i)
In Problems 27–30, let z = x + iy. Express the given quantity in terms of x and y.
27. Re(1/z)

28. Re(z 2 )

29. Im(2z + 4¯
z − 4i)

30. Im(¯
z2 + z2 )


In Problems 31–34, let z = x + iy. Express the given quantity in terms of the
symbols Re(z) and Im(z).
31. Re(iz)

32. Im(iz)

33. Im((1 + i)z)

34. Re(z 2 )

In Problems 35 and 36, show that the indicated numbers satisfy the given equation.
In each case explain why additional solutions can be found.


2
2
+
i. Find an additional solution, z2 .
35. z 2 + i = 0, z1 = −
2
2
36. z 4 = −4; z1 = 1 + i, z2 = −1 + i. Find two additional solutions, z3 and z4 .
In Problems 37–42, use Definition 1.2 to solve each equation for z = a + ib.
37. 2z = i(2 + 9i)
2

39. z = i
41. z + 2¯
z=


2−i
1 + 3i

38. z − 2¯
z + 7 − 6i = 0
40. z¯2 = 4z
z
42.
= 3 + 4i
1 + z¯

In Problems 43 and 44, solve the given system of equations for z1 and z2 .
43.

iz1 −

iz2 = 2 + 10i

−z1 + (1 − i)z2 = 3 − 5i

44.

iz1 + (1 + i)z2 = 1 + 2i
(2 − i)z1 +

2iz2 = 4i

Focus on Concepts
45. What can be said about the complex number z if z = z¯? If (z)2 = (¯

z )2 ?
46. Think of an alternative solution to Problem 24. Then without doing any significant work, evaluate (1 + i)5404 .

∗ Recall that the coefficients in the expansions of (A + B)2 , (A + B)3 , and so on, can
also be obtained using Pascal’s triangle.


1.1 Complex Numbers and Their Properties

9

47. For n a nonnegative integer, in can be one of four values: 1, i, −1, and −i. In
each of the following four cases, express the integer exponent n in terms of the
symbol k, where k = 0, 1, 2, . . . .
(a) in = 1

(b) in = i

(c) in = −1

(d) in = −i

48. There is an alternative to the procedure given in (7). For example, the quotient
(5 + 6i)/(1 + i) must be expressible in the form a + ib:
5 + 6i
= a + ib.
1+i
Therefore, 5 + 6i = (1 + i)(a + ib). Use this last result to find the given quotient.
Use this method to find the reciprocal of 3 − 4i.


49. Assume for the moment that 1 + i makes sense in the complex number system.
How would you then demonstrate the validity of the equality


1+i=

1
2

+

1
2



2+i

− 12 +

1
2



2?

50. Suppose z1 and z2 are complex numbers. What can be said about z1 or z2 if
z1 z2 = 0?
51. Suppose the product z1 z2 of two complex numbers is a nonzero real constant.

Show that z2 = k¯
z1 , where k is a real number.
52. Without doing any significant work, explain why it follows immediately from
(2) and (3) that z1 z¯2 + z¯1 z2 = 2Re(z1 z¯2 ).
53. Mathematicians like to prove that certain “things” within a mathematical system are unique. For example, a proof of a proposition such as “The unity in
the complex number system is unique” usually starts out with the assumption
that there exist two different unities, say, 11 and 12 , and then proceeds to show
that this assumption leads to some contradiction. Give one contradiction if it
is assumed that two different unities exist.
54. Follow the procedure outlined in Problem 53 to prove the proposition “The zero
in the complex number system is unique.”
55. A number system is said to be an ordered system provided it contains a
subset P with the following two properties:
First, for any nonzero number x in the system, either x or −x is (but not both)
in P.
Second, if x and y are numbers in P, then both xy and x + y are in P.
In the real number system the set P is the set of positive numbers. In the real
number system we say x is greater than y, written x > y, if and only if x − y
is in P . Discuss why the complex number system has no such subset P . [Hint:
Consider i and −i.]


10

Chapter 1 Complex Numbers and the Complex Plane

1.2

Complex Plane
A complex number

1.2 z = x + iy is uniquely determined by an ordered pair

of real numbers
(x, y). The first and second entries of the ordered pairs correspond, in turn, with the
real and imaginary parts of the complex number. For example, the ordered pair (2, −3)
corresponds to the complex number z = 2 − 3i. Conversely, z = 2 − 3i determines the
ordered pair (2, −3). The numbers 7, i, and −5i are equivalent to (7, 0), (0, 1), (0, −5),
respectively. In this manner we are able to associate a complex number z = x + iy with a
point (x, y) in a coordinate plane.
y-axis
or
imaginary axis
z = x + iy or
(x, y)

y

x

x-axis
or
real axis

Complex Plane
Because of the correspondence between a complex
number z = x + iy and one and only one point (x, y) in a coordinate plane,
we shall use the terms complex number and point interchangeably. The coordinate plane illustrated in Figure 1.1 is called the complex plane or simply
the z -plane. The horizontal or x-axis is called the real axis because each
point on that axis represents a real number. The vertical or y-axis is called
the imaginary axis because a point on that axis represents a pure imaginary

number.

Figure 1.1 z-plane

Vectors
In other courses you have undoubtedly seen that the numbers
in an ordered pair of real numbers can be interpreted as the components of
a vector. Thus, a complex number z = x + iy can also be viewed as a twodimensional position vector, that is, a vector whose initial point is the origin
and whose terminal point is the point (x, y). See Figure 1.2. This vector
interpretation prompts us to define the length of the vector z as the distance
x2 + y 2 from the origin to the point (x, y). This length is given a special
name.

y
z = x + iy

x

Figure 1.2 z as a position vector

Definition 1.3

Modulus

The modulus of a complex number z = x + iy, is the real number
|z| =

x2 + y 2 .

(1)


The modulus |z| of a complex number z is also called the absolute value
of z. We shall use both words modulus and absolute value throughout this
text.

EXAMPLE 1

Modulus of a Complex Number

If z = 2 − 3i, then √from (1) we find the modulus of the number to be
|z| = 22 + (−3)2 = 13. If z = −9i, then (1) gives |−9i| = (−9)2 = 9.


1.2 Complex Plane

11

Properties Recall from (4) of Section 1.1 that for any complex number
z = x + iy the product z z¯ is a real number; specifically, z z¯ is the sum of the
squares of the real and imaginary parts of z: z z¯ = x2 + y 2 . Inspection of (1)
2
then shows that |z| = x2 + y 2 . The relations
2

|z| = z z¯ and |z| =


z z¯

(2)


deserve to be stored in memory. The modulus of a complex number z has the
additional properties.
|z1 z2 | = |z1 | |z2 |

and

|z1 |
z1
.
=
z2
|z2 |

(3)

Note that when z1 = z2 = z, the first property in (3) shows that
2

z 2 = |z| .

(4)

The property |z1 z2 | = |z1 | |z2 | can be proved using (2) and is left as an
exercise. See Problem 49 in Exercises 1.2.

z1 + z2
or
(x1 + x2, y1 + y2)


y
z2

Distance Again The addition of complex numbers z1 = x1 + iy1 and
z2 = x2 + iy2 given in Section 1.1, when stated in terms of ordered pairs:
(x1 , y1 ) + (x2 , y2 ) = (x1 + x2 , y1 + y2 )

z1

is simply the component definition of vector addition. The vector interpretation of the sum z1 + z2 is the vector shown in Figure 1.3(a) as the main
diagonal of a parallelogram whose initial point is the origin and terminal point
is (x1 + x2 , y1 + y2 ). The difference z2 − z1 can be drawn either starting from
the terminal point of z1 and ending at the terminal point of z2 , or as a position
vector whose initial point is the origin and terminal point is (x2 − x1 , y2 − y1 ).
See Figure 1.3(b). In the case z = z2 −z1 , it follows from (1) and Figure 1.3(b)
that the distance between two points z1 = x1 + iy1 and z2 = x2 + iy2
in the complex plane is the same as the distance between the origin and the
point (x2 − x1 , y2 − y1 ); that is, |z| = |z2 − z1 | = |(x2 − x1 ) + i(y2 − y1 )| or

x

(a) Vector sum
y

z 2 – z1
or
(x2 – x1, y2 – y1)

z2


z2

–z

1

z1

|z2 − z1 | =

x

(b) Vector difference
Figure 1.3 Sum and difference
of vectors

(x2 − x1 )2 + (y2 − y1 )2 .

(5)

When z1 = 0, we see again that the modulus |z2 | represents the distance
between the origin and the point z2 .

EXAMPLE 2

Set of Points in the Complex Plane

Describe the set of points z in the complex plane that satisfy |z| = |z − i|.
Solution We can interpret the given equation as equality of distances: The
distance from a point z to the origin equals the distance from z to the point



12

Chapter 1 Complex Numbers and the Complex Plane
y

i. Geometrically, it seems plausible from Figure 1.4 that the set of points z
lie on a horizontal line. To establish this analytically, we use (1) and (5) to
write |z| = |z − i| as:

i
|z – i|
z

x2 + y 2 =

|z|

x2 + (y − 1)2

x2 + y 2 = x2 + (y − 1)2

x

x2 + y 2 = x2 + y 2 − 2y + 1.
Figure 1.4 Horizontal line is the set of
points satisfying |z| = |z − i|.

y


The last equation yields y = 12 . Since the equality is true for arbitrary x,
y = 12 is an equation of the horizontal line shown in color in Figure 1.4.
Complex numbers satisfying |z| = |z − i| can then be written as z = x + 12 i.

Inequalities In the Remarks at the end of the last section we pointed
out that no order relation can be defined on the system of complex numbers.
However, since |z| is a real number, we can compare the absolute values of
two complex
numbers. For

√ example, if z1 = 3 + 4i and z2 = 5 − i, then
|z1 | = 25 = 5 and |z2 | = 26 and, consequently, |z1 | < |z2 |. In view of (1),
a geometric interpretation of the last inequality is simple: The point (3, 4) is
closer to the origin than the point (5, −1).
Now consider the triangle given in Figure 1.5 with vertices at the origin,
z1 , and z1 + z2 . We know from geometry that the length of the side of the
triangle corresponding to the vector z1 + z2 cannot be longer than the sum
of the lengths of the remaining two sides. In symbols we can express this
observation by the inequality

z1 + z2
|z1 + z2|
|z1|

|z2|
z1
x

Figure 1.5 Triangle with vector sides


This inequality can be derived using
the properties of complex numbers
in Section 1.1. See Problem 50 in
Exercises 1.2.



|z1 + z2 | ≤ |z1 | + |z2 |.

(6)

The result in (6) is known as the triangle inequality. Now from the identity
z1 = z1 + z2 + (−z2 ), (6) gives
|z1 | = |z1 + z2 + (−z2 )| ≤ |z1 + z2 | + |−z2 | .
Since |z2 | = |−z2 | (see Problem 47 in Exercises 1.2), solving the last result for
|z1 + z2 | yields another important inequality:
|z1 + z2 | ≥ |z1 | − |z2 |.

(7)

But because z1 + z2 = z2 + z1 , (7) can be written in the alternative form
|z1 + z2 | = |z2 + z1 | ≥ |z2 | − |z1 | = − (|z1 | − |z2 |) and so combined with the
last result implies
|z1 + z2 | ≥ |z1 | − |z2 | .

(8)

It also follows from (6) by replacing z2 by −z2 that |z1 + (−z2 )| ≤ |z1 | +
|(−z2 )| = |z1 | + |z2 |. This result is the same as

|z1 − z2 | ≤ |z1 | + |z2 | .

(9)


1.2 Complex Plane

13

From (8) with z2 replaced by −z2 , we also find
|z1 − z2 | ≥ |z1 | − |z2 | .

(10)

In conclusion, we note that the triangle inequality (6) extends to any finite
sum of complex numbers:
|z1 + z2 + z3 + · · · + zn | ≤ |z1 | + |z2 | + |z3 | + · · · + |zn | .

(11)

The inequalities (6)–(10) will be important when we work with integrals of a
function of a complex variable in Chapters 5 and 6.

EXAMPLE 3

An Upper Bound
−1
if |z| = 2.
Find an upper bound for 4
z − 5z + 1

Solution By the second result in (3), the absolute value of a quotient is the
quotient of the absolute values. Thus with |−1| = 1, we want to find a positive
real number M such that
1
≤ M.
|z 4 − 5z + 1|
To accomplish this task we want the denominator as small as possible. By
(10) we can write
z 4 − 5z + 1 = z 4 − (5z − 1) ≥ z 4 − |5z − 1| .

(12)

But to make the difference in the last expression in (12) as small as possible, we
want to make |5z − 1| as large as possible. From (9), |5z − 1| ≤ |5z| + |−1| =
5|z| + 1. Using |z| = 2, (12) becomes
4

z 4 − 5z + 1 ≥ z 4 − |5z − 1| ≥ |z| − (5 |z| + 1)
4

= |z| − 5 |z| − 1 = |16 − 10 − 1| = 5.
Hence for |z| = 2 we have
1
1
≤ .
| z 4 − 5z + 1 |
5

Remarks
We have seen that the triangle inequality |z1 + z2 | ≤ |z1 | + |z2 | indicates

that the length of the vector z1 + z2 cannot exceed the sum of the lengths
of the individual vectors z1 and z2 . But the results given in (3) are
interesting. The product z1 z2 and quotient z1 /z2 , (z2 = 0), are complex
numbers and so are vectors in the complex plane. The equalities |z1 z2 | =
|z1 | |z2 | and |z1 /z 2 | = |z1 | / |z2 | indicate that the lengths of the vectors
z1 z2 and z1 /z2 are exactly equal to the product of the lengths and to the
quotient of the lengths, respectively, of the individual vectors z1 and z2 .


×