A first course in complex analysis with applications dennis zill (2003, jon

A First Course in with Applications Complex Analysis Dennis G Zill Loyola Marymount University Patrick D Shanahan Loyola Marymount University Copyright © 2003 by Jones and Bartlett Publishers, Inc Lib.

Copyright © 2003 by Jones and Bartlett Publishers, Inc.

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7.1 Contents

Preface ix

1.5 Sets of Points in the Complex Plane 29

Chapter 1 Review Quiz 45

2.4.1 The Power Function z n 81

2.4.2 The Power Function z 1/n 86

Chapter 2 Review Quiz 138

3.1 Differentiability and Analyticity 142

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

4.3 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

Chapter 4 Review Quiz 232

Chapter 5 Integration in the Complex Plane 235

5.5.1 Cauchy’s Two Integral Formulas 273

5.5.2 Some Consequences of the Integral

Chapter 5 Review Quiz 297

6.6 Some Consequences of the Residue

6.6.3 Integration along a Branch Cut 361

6.6.4 The Argument Principle and Rouch´e’s

Chapter 7 Conformal Mappings 389

7.2 Preface

Philosophy This text grew out of chapters 17-20 in Advanced

Engineer-ing Mathematics, Second Edition (Jones and Bartlett Publishers), by Dennis

G Zill and the late Michael R Cullen This present work represents an pansion and revision of that original material and is intended for use in either

ex-a one-semester or ex-a one-quex-arter course Its ex-aim is to introduce the bex-asic ciples and applications of complex analysis to undergraduates who have noprior knowledge of this subject

prin-The motivation to adapt the material from Advanced Engineering

Math-ematics into a stand-alone text sprang from our dissatisfaction with the

suc-cession of textbooks that we have used over the years in our departmentalundergraduate course offering in complex analysis It has been our experiencethat books claiming to be accessible to undergraduates were often written at alevel that was too advanced for our audience The “audience” for our junior-level course consists of some majors in mathematics, some majors in physics,but mostly majors from electrical engineering and computer science At ourinstitution, a typical student majoring in science or engineering does not taketheory-oriented mathematics courses in methods of proof, linear algebra, ab-stract algebra, advanced calculus, or introductory real analysis Moreover,the only prerequisite for our undergraduate course in complex variables isthe completion of the third semester of the calculus sequence For the mostpart, then, calculus is all that we assume by way of preparation for a student

to use this text, although some working knowledge of differential equationswould be helpful in the sections devoted to applications We have kept thetheory in this introductory text to what we hope is a manageable level, con-centrating only on what we feel is necessary Many concepts are conveyed

in an informal and conceptual style and driven by examples, rather than theformal definition/theorem/proof We think it would be fair to characterizethis text as a continuation of the study of calculus, but also the study of thecalculus of functions of a complex variable Do not misinterpret the precedingwords; 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

ma-jors in mathematics—can gain some mathematical maturity and insight byattempting 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

is guided through the starting steps or a strong hint on how to proceed isprovided.

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

logarith-mic functions are covered The subtle notions of multiple-valued tions and branches are also discussed

func-• Chapter 5 The chapter begins with a review of real integrals

(in-cluding line integrals) The definitions of real line integrals are used tomotivate the definition of the complex integral The famous Cauchy-Goursat theorem and the Cauchy integral formulas are introduced inthis chapter Although we use Green’s theorem to prove Cauchy’s the-orem, a sketch of the proof of Goursat’s version of this same theorem isgiven 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 in-verse Fourier transforms are some of the applications of residue theorythat are covered

• Chapter 7 Complex mappings that are conformal are defined and

used to solve certain problems involving Laplace’s partial differentialequation

Features Each chapter begins with its own opening page that includes atable of contents and a brief introduction describing the material to be covered

in the chapter Moreover, each section in a chapter starts with tory comments on the specifics covered in that section Almost every section

introduc-ends with a feature called Remarks in which we talk to the students about

areas where real and complex calculus differ or discuss additional interestingtopics (such as the Riemann sphere and Riemann surfaces) that are related

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 tomake the assignment of homework easier Comments, clarifications, and somewarnings 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 allpertinent details in the solutions of the examples Because applications ofcomplex 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 toend each chapter with a separate section on applications The exercise sets areconstructed in a pyramidal fashion and each set has at least two parts Thefirst part of an exercise set is a generous supply of routine drill-type problems;the second part consists of conceptual word and geometrical problems Inmany exercise sets, there is a third part devoted to the use of technology.Since the default operational mode of all computer algebra systems is complexvariables, 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 inthe back of the text Since the conceptual problems could also be used astopics for classroom discussion, we decided not to include their answers Eachchapter ends with a Chapter Review Quiz We thought that something moreconceptual would be a bit more interesting than the rehashing of the sameold problems given in the traditional Chapter Review Exercises Lastly, toillustrate the subtleties of the action of complex mappings, we have used twocolors

Acknowledgments We would like to express our appreciation to ourcolleague at Loyola Marymount University, Lily Khadjavi, for volunteering touse a preliminary version of this text We greatly appreciate her careful read-ing of the manuscript We also wish to acknowledge the valuable input fromstudents 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 contributionsand cooperation in the making of this text

✐ ✐

A Request Although the preliminary versions of this book were classtested for several semesters, experience has taught us that errors—typos orjust plain mistakes—seem to be an inescapable by-product of the textbook-writing endeavor We apologize in advance for any errors that you may findand urge you to bring them to our attention

Dennis G ZillPatrick D Shanahan

Los Angeles, CA

Complex Numbers and the

1

1 0

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 plex number In this text we study nothing but complex numbers and the calculus of functions

com-of a complex variable.

We begin with an in-depth examination of the arithmetic and algebra of complex numbers.

1

1.1 Complex Numbers and Their Properties

1.1

No one person “invented” complex numbers, but controversies surrounding the use of thesenumbers existed in the sixteenth century.In their quest to solve polynomial equations byformulas 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, “theimaginary, (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 evenpractical.A famous mathematician held that even though “they exist in our imagination nothing prevents us from employing them in calculations.” Mathematiciansalso 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, negativenumbers, and irrational numbers.But in the eighteenth century the number concept took agigantic 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 realnumber system

Our goal in this first section is to examine some basic definitions and the arithmetic ofcomplex numbers

The Imaginary Unit Even after gaining wide respectability, throughthe seminal works of Karl Friedrich Gauss and the French mathematician Au-gustin Louis Cauchy, the unfortunate name “imaginary” has survived down

the centuries.The symbol i was originally used as a disguise for the

embar-rassing 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 generalcomplex number out of two real numbers

Definition 1.1 Complex Number

A complex number is any number of the formz = 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

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 + ib1 and z2= a2+ ib2are equal, z1= z2, if

a1= a2andb1= b2

In terms of the symbols Re(z) and Im(z), Definition 1.2 states that z1= z2if

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 + ib1and z2= a2 + ib2, these operationsare defined as follows

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 discussthat operation in more detail shortly

EXAMPLE 1 Addition and Multiplication

If z1= 2 + 4i and z2=−3 + 8i, find (a) z1+ z2 and (b) z1z2

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 z2is

z1z2= (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

num-ber 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 beforediscussing 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,

then its conjugate isz = 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 complexnumbers is the sum and difference of the conjugates:

z1+ z2= ¯z1+ ¯z2, z1− z2= ¯z1− ¯z2. (1)Moreover, we have the following three additional properties:

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) = a2− i2b2= a2+ b2. (4)

The difference of a complex number z with its conjugate ¯ z is a pure imaginary

number:

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

Re(z) = z + ¯ z

z − ¯z

However, (4) is the important relationship in this discussion because it enables

us to approach division in a practical manner

Solution We multiply numerator and denominator by the conjugate

¯2= 4− 6i of the denominator z2= 4 + 6i and then use (4):

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:

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

☞

Answer should be in the form a + ib.

Find the reciprocal of z = 2 − 3i.

Solution By the definition of division we obtain

(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 numbersystem.In other words, we cannot compare two complex numbers

z = a + ib , b = 0, and z = a + ib , b = 0, by means of

inequalities.Statements such as z1 < z2 or z2 ≥ z1 have no

meaning in C except in the special case when the two

Therefore, if you see a statement such as z1 = αz2, α > 0,

it is implicit from the use of the inequality α > 0 that the bol α represents a real number.

sym-(ii ) Some things that we take for granted as impossible in real analysis, such as e x =−2 and sin x = 5 when x is a real variable, are per-

fectly correct and ordinary in complex analysis when the symbol x

is interpreted as a complex variable.See Example 3 in Section 4.1and Example 2 in Section 4.3

We will continue to point out other differences between real analysis andcomplex 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.

−i

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)

17. i(1 − i)(2 − i)(2 + 6i) 18. (1 + i)2(1− i)3

19. (3 + 6i) + (4 − i)(3 + 5i) + 1

In Problems 21–24, use the binomial theorem∗

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.

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).

33. Im((1 + i)z) 34. Re(z2)

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 i Find an additional solution, z2.

36. z4=−4; z1= 1 + i, z2=−1 + i Find two additional solutions, z3and z4.

In Problems 37–42, use Definition 1.2 to solve each equation for z = a + ib.

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

sig-nificant work, evaluate (1 + i)5404

∗ Recall that the coefficients in the expansions of (A + B)2, (A + B)3, and so on, can

47. For n a nonnegative integer, i n 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,

(c) i n=−1 (d) i n=−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

1 + i = a + ib.

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

51. Suppose the product z1z2 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 z1z¯2+ ¯z1z2= 2Re(z1z¯2)

53. Mathematicians like to prove that certain “things” within a mathematical tem are unique For example, a proof of a proposition such as “The unity inthe complex number system is unique” usually starts out with the assumption

sys-that there exist two different unities, say, 11and 12, and then proceeds to showthat 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.]

1.2 Complex Plane

1.2

A complex number 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 numberz = x + iy with a

point (x, y)in a coordinate plane

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

coor-dinate 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

y-axis

x-axis

or real axis

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

two-dimensional 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

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

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

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+ y2.Inspection of (1)then shows that|z|2

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

z2 =|z|2

The property |z1z2| = |z1| |z2| can be proved using (2) and is left as an

exercise.See Problem 49 in Exercises 1.2

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)

is simply the component definition of vector addition.The vector

interpre-tation of the sum z1+ z2 is the vector shown in Figure 1.3(a) as the maindiagonal of a parallelogram whose initial point is the origin and terminal point

is (x1+ x2, y1+ y2).The difference z2− z1can be drawn either starting from

the terminal point of z1and 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

|z2− z1| =(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

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) towrite|z| = |z − i| as:

Figure 1.4Horizontal line is the set of

points satisfying|z| = |z − i|. The last equation yields y = 1

2.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 +12i.

Inequalities In the Remarks at the end of the last section we pointedout 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).

Figure 1.5Triangle with vector sides

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 thisobservation by the inequality

|z1+ z2| ≤ |z1| + |z2| (6)

☞

This inequality can be derived using

the properties of complex numbers

in Section 1.1 See Problem 50 in

Exercises 1.2. 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

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 finitesum of complex numbers:

|z1+ z2+ z3+· · · + z n | ≤ |z1| + |z2| + |z3| + · · · + |z n | (11)The inequalities (6)–(10) will be important when we work with integrals of afunction of a complex variable in Chapters 5 and 6

EXAMPLE 3 An Upper Bound

Find an upper bound for

z4− 5z + 1 −1

if |z| = 2.

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

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

that the length of the vector z1+ z2cannot exceed the sum of the lengths

of the individual vectors z1 and z2.But the results given in (3) are

interesting.The product z1z2 and quotient z1/z2, (z2 = 0), are complex

numbers and so are vectors in the complex plane.The equalities|z1z2| =

|z1| |z2| and |z1/z2| = |z1| / |z2| indicate that the lengths of the vectors

z1z2and z1/z2 are exactly equal to the product of the lengths and to the

quotient of the lengths, respectively, of the individual vectors z and z

EXERCISES 1.2 Answers to selected odd-numbered problems begin on page ANS-2.

In Problems 1–4, interpret z1 and z2 as vectors Graph z1, z2, and the indicatedsum and difference as vectors

1. z1= 4 + 2i, z2=−2 + 5i; z1+ z2, z1− z2

2. z1= 1− i, z2= 1 + i; z1+ z2, z1− z2

3. z1= 5 + 4i, z2=−3i; 3z1+ 5z2, z1− 2z2

4. z1= 4− 3i, z2=−2 + 3i; 2z1+ 4z2, z1− z2

5. Given that z1= 5− 2i and z2=−1 − i, find a vector z3 in the same direction

as z1+ z2 but four times as long

6. (a) Plot the points z1=−2 − 8i, z2= 3i, z3=−6 − 5i.

(b) The points in part (a) determine a triangle with vertices at z1, z2, and z3,respectively Express each side of the triangle as a difference of vectors

7. In Problem 6, determine whether the points z1, z2, and z3are the vertices of aright triangle

8. The three points z1 = 1 + 5i, z2=−4 − i, z3= 3 + i are vertices of a triangle Find the length of the median from z1 to the side z3− z2

In Problems 9–12, find the modulus of the given complex number

In Problems 15 and 16, determine which of the given two complex numbers is closest

to the origin Which is closest to 1 + i?

4i, 23+16i

In Problems 17–26, describe the set of points z in the complex plane that satisfy

the given equation

17 Re((1 + i)z − 1) = 0 18 [Im(i¯ z)]2= 2

30. Find an upper bound for the reciprocal of the modulus of z4−5z2

(b) In general, how would you describe geometrically the relationship between

a complex number z = a + ib and its conjugate ¯ z = a − ib?

(c) Describe geometrically the relationship between z = a+ib and z1=−a+ib.

34. How would you describe geometrically the relationship between a nonzero

com-plex number z = a + ib and its

(a) negative,−z?

(b) inverse, z −1 ? [Hint : Reread Problem 33 and then recall z −1= ¯z/|z|2

.]

35. Consider the complex numbers z1 = 4 + i, z2=−2+i, z3 =−2−2i, z4= 3−5i.

(a) Use four different sketches to plot the four pairs of points z1, iz1; z2, iz2; z3, iz3;

and z4, iz4

(b) In general, how would you describe geometrically the effect of multiplying

a complex number z = x + iy by i? By −i?

36. What is the only complex number with modulus 0?

37. Under what circumstances does|z1+ z2| = |z1| + |z2|?

38. Let z = x + iy Using complex notation, find an equation of a circle of radius

5 centered at (3, −6).

39. Describe the set of points z in the complex plane that satisfy z = cos θ + i sin θ, where θ is measured in radians from the positive x-axis.

40. Let z = x + iy Using complex notation, find an equation of an ellipse with foci

(−2, 1), (2, 1) whose major axis is 8 units long.

41. Suppose z = x + iy In (6) of Section 1.1 we saw that x and y could be pressed in terms of z and ¯ z Use these results to express the following Cartesian

ex-equations in complex form

42. Using complex notation, find a parametric equation of the line segment between

any two distinct complex numbers z1 and z2 in the complex plane

43. Suppose z1, z2, and z3 are three distinct points in the complex plane and k is

a real number Interpret z3− z2= k(z2− z1) geometrically

44. Suppose z1 = z2 Interpret Re(z1z¯2) = 0 geometrically in terms of vectors z1

and z2

45. Suppose w = ¯ z/z Without doing any calculations, explain why |w| = 1.

46. Without doing any calculations, explain why the inequalities|Re(z)| ≤ |z| and

|Im(z)| ≤ |z| hold for all complex numbers z.

49. In this problem we will start you out in the proof of the first property|z1z2| =

|z1| |z2| in (3) By the first result in (2) we can write |z1z2|2

= (z1z2)(z1z2).Now use the first property in (2) of Section 1.1 to continue the proof

50. In this problem we guide you through an analytical proof of the triangle equality (6)

in-Since|z1+ z2| and |z1| + |z2| are positive real numbers, we have

(c) Use parts (a) and (b) along with the results in Problem 46 to derive (6).

1.3 Polar Form of Complex Numbers

1.3

Recall from calculus that a point P in the plane whose rectangular coordinates are

(x, y) can also be described in terms of polar coordinates.The polar coordinate system, invented by Isaac Newton, consists of point O called the pole and the horizontal half-line emanating from the pole called the polar axis.If r is a directed distance from the pole to

P and θ is an angle of inclination (in radians) measured from the polar axis to the line OP,

then the point can be described by the ordered pair (r, θ), called the polar coordinates of

Figure 1.6Polar coordinates

We say that (1) is the polar form or polar representation of the complex

number z.Again, from Figure 1 7 we see that the coordinate r can be preted as the distance from the origin to the point (x, y).In other words, we shall adopt the convention that r is never negative † so that we can take r to

inter-† In general, in the polar description (r, θ) of a point P in the Cartesian plane, we can

have r ≥ 0 or r < 0.

be the modulus of z, that is, r = |z| The angle θ of inclination of the vector

z, which will always be measured in radians from the positive real axis, is

positive when measured counterclockwise and negative when measured

clock-wise.The angle θ is called an argument of z and is denoted by θ = arg(z)

An argument θ of a complex number must satisfy the equations cos θ = x/r and sin θ = y/r.An argument of a complex number z is not unique since cos θ and sin θ are 2π-periodic; in other words, if θ0 is an argument of z, then necessarily the angles θ0± 2π, θ0± 4π, are also arguments of z.In

practice we use tan θ = y/x to find θ.However, because tan θ is π-periodic,

some care must be exercised in using the last equation.A calculator will giveonly angles satisfying−π/2 < tan −1 (y/x) < π/2, that is, angles in the first and fourth quadrants.We have to choose θ consistent with the quadrant in which z is located; this may require adding or subtracting π to tan −1 (y/x)

when appropriate.The following example illustrates how this is done

Be careful using tan−1 (y/x)

EXAMPLE 1 A Complex Number in Polar Form

Express− √3− i in polar form.

Solution With x = − √ 3 and y = −1 we obtain r = |z| = − √3 2

= π/6, which is an angle whose terminal side is in the first quadrant.But

since the point (− √ 3, −1) lies in the third quadrant, we take the solution of

tan θ = −1/(− √ 3) = 1/ √

3 to be θ = arg(z) = π/6 + π = 7π/6.See Figure

1.8 It follows from (1) that a polar form of the number is

z = 2

cos7π

For example, if z = i, we see in Figure 1.9 that some values of arg(i) are

π/2, 5π/2, −3π/2, and so on, but Arg(i) = π/2.Similarly, we see from

Figure 1.10 that the argument of− √3− i that lies in the interval (−π, π),

the principal argument of z, is Arg(z) = π/6 − π = −5π/6.Using Arg(z) we

can express the complex number in (2) in the alternative polar form:

z = 2

cos

x

i

π 3 2

π 5 2

π 2 –

Figure 1.9Some arguments of i

For example, arg(i) = π

2 + 2nπ For the choices n = 0 and n = −1, (3) gives

arg(i) = Arg(i) = π/2 and arg(i) = −3π/2, respectively.

Multiplication and Division The polar form of a complex number

is especially convenient when multiplying or dividing two complex numbers.Suppose

z1= r1(cos θ1+ i sin θ1) and z2= r2(cos θ2+ i sin θ2), where θ1 and θ2 are any arguments of z1 and z2, respectively.Then

z1z2= r1r2[cos θ1cos θ2− sin θ1sin θ2+ i (sin θ1cos θ2+ cos θ1sin θ2)] (4)

Inspection of the expressions in (6) and (7) and Figure 1.11 shows that the

lengths of the two vectors z1z2 and z1/z2are the product of the lengths of z1

and z2 and the quotient of the lengths of z1 and z2, respectively.See (3) of

Section 1.2 Moreover, the arguments of z1z2 and z1/z2 are given by

arg(z1z2) = arg(z1) + arg(z2) and arg

EXAMPLE 2 Argument of a Product and of a Quotient

We have just seen that for z1= i and z2=− √3− i that Arg(z1) = π/2 and Arg(z2) =−5π/6, respectively.Thus arguments for the product and quotient

Integer Powers of z We can find integer powers of a complex number

z from the results in (6) and (7).For example, if z = r(cos θ + i sin θ), then

with z1= z2= z, (6) gives

z2= r2[cos (θ + θ) + i sin (θ + θ)] = r2(cos 2θ + i sin 2θ)

Since z3= z2z, it then follows that

When n = 0, we get the familiar result z0= 1

EXAMPLE 3 Power of a Complex Number

Compute z3 for z = − √3− i.

Solution In (2) of Example 1 we saw that a polar form of the given number

is z = 2[cos (7π/6) + i sin (7π/6)].Using (9) with r = 2, θ = 7π/6, and n = 3

we get

− √3− i 3= 23

cos



37π6



+ i sin



37π6



= 8

cos7π

Note in Example 3, if we also want the value of z −3, then we could proceed

in two ways: either find the reciprocal of z3=−8i or use (9) with n = −3.

de Moivre’s Formula When z = cos θ + i sin θ, we have |z| = r = 1,

and so (9) yields

This last result is known as de Moivre’s formula and is useful in deriving

certain trigonometric identities involving cos nθ and sin nθ.See Problems 33

and 34 in Exercises 1.3

EXAMPLE 4 de Moivre’s Formula

From (10), with θ = π/6, cos θ = √

(i ) Observe in Example 2 that even though we used the principal ments of z1 and z2that arg(z1/z2) = 4π/3 = Arg(z1/z2).Although

argu-(8) is true for any arguments of z1 and z2, it is not true, in eral, that Arg(z1z2) =Arg(z1)+Arg(z2) and Arg(z1/z2) = Arg(z1)−

gen-Arg(z2).See Problems 37 and 38 in Exercises 1.3

(ii ) An argument can be assigned to any nonzero complex number z However, for z = 0, arg(z) cannot be defined in any way that is

meaningful

(iii ) If we take arg(z) from the interval ( −π, π), the relationship between

a complex number z and its argument is single-valued; that is, every nonzero complex number has precisely one angle in ( −π, π).But

there is nothing special about the interval (−π, π); we also establish

a single-valued relationship by using the interval (0, 2π) to define the principal value of the argument of z.For the interval ( −π, π),

the negative real axis is analogous to a barrier that we agree not to

cross; the technical name for this barrier is a branch cut.If we use

(0, 2π), the branch cut is the positive real axis.The concept of a

branch cut is important and will be examined in greater detail when

we study functions in Chapters 2 and 4

(iv ) The “cosine i sine” part of the polar form of a complex number is

sometimes abbreviated cis.That is,

z = r (cos θ + i sin θ) = r cis θ.

This notation, used mainly in engineering, will not be used in thistext

EXERCISES 1.3 Answers to selected odd-numbered problems begin on page ANS-2.

In Problems 1–10, write the given complex number in polar form first using an

argument θ = Arg(z) and then using θ = Arg(z).

In Problems 11 and 12, use a calculator to write the given complex number in polar

form first using an argument θ = Arg(z) and then using θ = Arg(z).

In Problems 15–18, write the complex number whose polar form is given in the form

a + ib Use a calculator if necessary.

5 + i sin

π

5

In Problems 19 and 20, use (6) and (7) to find z1z2 and z1/z2 Write the number

in the form a + ib.

12+ i sin

π

12

In Problems 21–24, write each complex number in polar form Then use either (6)

or (7) to obtain the polar form of the given number Finally, write the polar form

in the form a + ib.

21 (3− 3i)(5 + 5 √ 3i) 22 (4 + 4i)( −1 + i)

In Problems 25–30, use (9) to compute the indicated powers.

12

30. √

3

cos2π

9 + i sin

2π

9

6

In Problems 31 and 32, write the given complex number in polar form and in then

in the form a + ib.

31.

cosπ

9+ i sin

π

9

122

cosπ

cos3π

cos π

37. For the complex numbers z1=−1 and z2= 5i, verify that:

(a) Arg(z1z2)= Arg(z1) + Arg(z2)

(b) Arg(z1/z2)= Arg(z1)− Arg(z2)

38. For the complex numbers given in Problem 37, verify that:

(a) arg(z1z2) = arg(z1) + arg(z2)

(b) arg(z1/z2) = arg(z1)− arg(z2)

Focus on Concepts

39. Suppose that z = r(cos θ + i sin θ) Describe geometrically the effect of plying z by a complex number of the form z1= cos α + i sin α, when α > 0 and when α < 0.

multi-40. Suppose z = cos θ + i sin θ If n is an integer, evaluate z n+ ¯z n and z n − ¯z n

.

41. Write an equation that relates arg(z) to arg(1/z), z = 0.

42. Are there any special cases in which Arg(z1z2) = Arg(z1) + Arg(z2)? Proveyour assertions

43. How are the complex numbers z1 and z2 related if arg(z1) = arg(z2)?

44. Describe the set of points z in the complex plane that satisfy arg(z) = π/4.

45. Student A states that, even though she can’t find it in the text, she thinks thatarg(¯z) = − arg(z) For example, she says, if z = 1 + i, then ¯z = 1 − i and

arg(z) = π/4 and arg(¯ z) = −π/4 Student B disagrees because he feels that

he has a counterexample: If z = i, then ¯ z = −i; we can take arg(i) = π/2

and arg(−i) = 3π/2 and so arg(i) = − arg(−i) Take sides and defend your

position

46. Suppose z1, z2, and z1z2 are complex numbers in the first quadrant and that

the points z = 0, z = 1, z1, z2, and z1z2 are labeled O, A, B, C, and D, respectively Study the formula in (6) and then discuss how the triangles OAB and OCD are related.

47. Suppose z1 = r1(cos θ1+ i sin θ1) and z2 = r2(cos θ2+ i sin θ2) If z1 = z2,

then how are r1 and r2 related? How are θ1 and θ2 related?

48. Suppose z1is in the first quadrant For each z2, discuss the quadrant in which

(b) Use part (a) and appropriate results from this section to establish that

1 + cos θ + cos 2θ + · · · + cos nθ =1

2+sin

n +12

θ

sin12θ

for 0 < θ < 2π The foregoing result is known as Lagrange’s identity

and is useful in the theory of Fourier series

50. Suppose z1, z2, z3, and z4 are four distinct complex numbers Interpret metrically:

Recall from algebra that –2 and 2 are said to be square roots of the number 4 because

(−2)2= 4 and (2)2= 4.In other words, the two square roots of 4 are distinct solutions of

the equation w2= 4.In like manner we say w = 3 is a cube root of 27 since w3= 33= 27.This last equation points us again in the direction of complex variables since any real number

has only one real cube root and two complex roots.In general, we say that a number wis

an n th root of a nonzero complex number z if w n = z , where n is a positive integer.For example, you are urged to verify that w1 = 12√

√

2i are the two square roots of the complex number z = i because w2= i and w2= i.See Problem 39

in Exercises 1.1

We will now demonstrate that there are exactly n solutions of the equation w n = z.

Roots Suppose z = r(cos θ + i sin θ) and w = ρ(cos φ + i sin φ) are polar forms of the complex numbers z and w Then, in view of (9) of Section 1.3, the equation w n = z becomes

ρ n (cos nφ + i sin nφ) = r(cos θ + i sin θ). (1)From (1), we can conclude that

See Problem 47 in Exercises 1.3

From (2), we define ρ = √ n

r to be the unique positive nth root of the

positive real number r.From (3), the definition of equality of two complex

numbers implies that

cos nφ = cos θ and sin nφ = sin θ.

These equalities, in turn, indicate that the arguments θ and φ are related by

nφ = θ + 2kπ, where k is an integer.Thus,

φ = θ + 2kπ

As k takes on the successive integer values k = 0, 1, 2, , n − 1 we obtain

n distinct nth roots of z; these roots have the same modulus √ n

r but different

arguments.Notice that for k ≥ n we obtain the same roots because the sine

and cosine are 2π-periodic.To see why this is so, suppose k = n + m, where



, cos φ = cos



θ + 2mπ n



.

We summarize this result.The n nth roots of a nonzero complex number

z = r(cos θ + i sin θ) are given by

w k = √ n

r

cos



θ + 2kπ n



+ i sin



θ + 2kπ n



where k = 0, 1, 2, , n − 1.

EXAMPLE 1 Cube Roots of a Complex Number

Find the three cube roots of z = i.

Solution Keep in mind that we are basically solving the equation w3 = i Now with r = 1, θ = arg(i) = π/2, a polar form of the given number is given

by z = cos(π/2) + i sin(π/2).From (4), with n = 3, we then obtain

w k=√3

1

cos

Hence the three roots are,

Principal n th Root On page 17 we pointed out that the symbol

arg(z) really stands for a set of arguments for a complex number z.Stated another way, for a given complex number z = 0, arg(z) is infinite-valued.In

like manner, z 1/n is n-valued; that is, the symbol z 1/n represents the set of

n nth roots w k of z.The unique root of a complex number z (obtained by using the principal value of arg(z) with k = 0) is naturally referred to as the

principal n th root of w.In Example 1, since Arg(i) = π/2, we see that

w0= 1 2

√

3 +1

2i is the principal cube root of i.The choice of Arg(z) and k = 0

guarantees us that when z is a positive real number r, the principal nth root

is √ n r.

☞

√

4 = 2 and √3

27 = 3 are the

cipal square root of 4 and the

prin-cipal cube root of 27, respectively.

y

x

w2

Figure 1.12Three cube roots of i

Since the roots given by (4) have the same modulus, the n nth roots of a nonzero complex number z lie on a circle of radius √ n

r centered at the origin

in the complex plane.Moreover, since the difference between the arguments

of any two successive roots wk and wk+1 is 2π/n, the n nth roots of z are equally spaced on this circle, beginning with the root whose argument is θ/n Figure 1.12 shows the three cube roots of i obtained in Example 1 spaced at equal angular intervals of 2π/3 on the circumference of a unit circle beginning with the root w0 whose argument is π/6.

As the next example shows, the roots of a complex number do not have

to be “nice” numbers as in Example 1

EXAMPLE 2 Fourth Roots of a Complex Number

Find the four fourth roots of z = 1 + i.

Solution In this case, r = √

2 and θ = arg(z) = π/4.From (4) with n = 4,

we obtain

w k=√4

2

cos

Figure 1.13Four fourth roots of 1 + i

As shown in Figure 1.13, the four roots lie on a circle centered at the origin of

radius r = √4

2≈ 1.19 and are spaced at equal angular intervals of 2π/4 = π/2

radians, beginning with the root whose argument is π/16.

(i ) As a consequence of (4), we can say that the complex number system

is closed under the operation of extracting roots.This means that

for any z in C, z 1/n is also in C.The real number system does not

possess a similar closure property since, if x is in R, x 1/n is not

necessarily in R.

(ii ) Geometrically, the n nth roots of a complex number z can also be interpreted as the vertices of a regular polygon with n sides that is

inscribed within a circle of radius √ n

r centered at the origin.You

can see the plausibility of this fact by reinspecting Figures 1.12 and1.13 See Problem 19 in Exercises 1.4

(iii ) When m and n are positive integers with no common factors, then

(4) enables us to define a rational power of z, that is, z m/n.It

can be shown that the set of values (z 1/n)mis the same as the set of

values (z m)1/n This set of n common values is defined to be z m/n.See Problems 25 and 26 in Exercises 1.4

EXERCISES 1.4 Answers to selected odd-numbered problems begin on page ANS-3.

In Problems 1–14, use (4) to compute all roots Give the principal nth root in each case Sketch the roots w0, w1, , w n −1 on an appropriate circle centered at theorigin

15. (a) Verify that (4 + 3i)2= 7 + 24i.

(b) Use part (a) to find the two values of (7 + 24i) 1/2

16. Rework Problem 15 using (4).

17. Find all solutions of the equation z4+ 1 = 0

18. Use the fact that 8i = (2 + 2i)2 to find all solutions of the equation

(b) Find n nth roots of unity for n = 3, n = 4, and n = 5.

(c) Carefully plot the roots of unity found in part (b) Sketch the regular

polygons formed with the roots as vertices [Hint : See (ii ) in the Remarks.]

20. Suppose w is a cube root of unity corresponding to k = 1 See Problem 19(a).

(a) How are w and w2 related?

(b) Verify by direct computation that

1 + w + w2 = 0.

(c) Explain how the result in part (b) follows from the basic definition that w

is a cube root of 1, that is, w3= 1 [Hint : Factor.]

21. For a fixed n, if we take k = 1 in Problem 19(a), we obtain the root

23. Consider the equation in Problem 22

(a) In the complex plane, determine the location of all solutions z when n = 6.

[Hint : Write the equation in the form [(z + 2)/ ( −z)]6

= 1 and use part(a) of Problem 19.]

(b) Reexamine the solutions of the equation in Problem 22 for n = 1 and n = 2.

Form a conjecture as to the location of all solutions of (z + 2) n + z n = 0.

24. For the n nth roots of unity given in Problem 21, show that

25. (a) First compute the set of values i 1/2 using (4) Then compute (i 1/2)3using

(9) of Section 1.3

(b) Now compute i3 Then compute (i3)1/2 using (4) Compare these valueswith the results of part (b)

(c) Lastly, compute i 3/2using formula (5)

26. Use (5) to find all solutions of the equation w2= (−1 + i)5

Figure 1.14Figure for Problem 27

28. Suppose n denotes a nonnegative integer Determine the values of n such that z n = 1 possesses only real solutions Defend your answer with soundmathematics

29 (a) Proceed as in Example 2 to find the approximate values of the two square

30. Discuss: What geometric significance does the result in Problem 24 have?

31. Discuss: A real number can have a complex nth root Can a nonreal complex number have a real nth root?

32. Suppose w is located in the first quadrant and is a cube root of a complex number z Can there exist a second cube root of z located in the first quadrant?

Defend your answer with sound mathematics

33. Suppose z is a complex number that possesses a fourth root w that is neither

real nor pure imaginary Explain why the remaining fourth roots are neitherreal nor pure imaginary

34. Suppose z = r(cos θ + i sin θ) is complex number such that 1 < r < 2 and

0 < θ ≤ π/4 Suppose further that w0 is a cube root of z corresponding to

k = 0 Carefully sketch w0, w2, and w3 in the complex plane

Computer Lab Assignments

In Problems 35–40, use a CAS§ to first find z n = w for the given complex number and the indicated value of n Then, using the output and the same value of n, determine whether w 1/n = (z n)1/n = z If not, explain why not.

37. 1 + 3i; n = 8 38. 2 + 2i; n = 12

§

has only one real cube root and two complex roots .In general,... 1.19 and are spaced at equal angular intervals of 2π/4 = π/2

radians, beginning with the root whose argument is π/16.

(i ) As a consequence of (4), we can say that the... greater detail when

we study functions in Chapters and