Student Solutions Manual for use with Complex Variables and Applications [7 ed.]

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Student Solutions Manual for use with

Complex Variables and Applications Seventh Edition

Selected Solutions to Exercises in Chapters 1-7

by

James Ward Brown Professor of Mathematics

The University of Michigan- Dearborn

Ruel

V. Churchill Late Professor of Mathematics

The University of Michigan

Mc Graw Higher Education Hill

Boston

Burr Ridge, IL

Bangkok Milan

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Montreal

Dubuque, IA

Caracas

New Delhi

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Seoul

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San Francisco St. Louis London Madrid Mexico City Singapore Sydney Taipei Toronto

Madison, Wl

Kuala Lumpur

Lisbon

Table of Contents

Chapter

1

1

Chapter 2

22

Chapter 3

35

Chapter 4

53

Chapter 5

75

Chapter 6

94

Chapter 7

118

;

COMPLEX VARIABLES AND APPLICATIONS" (7/e) by Brown and Churchill

Chapter 1 SECTION 2

1.

2.

3.

(a)

(V2-/)-/(l-V2i) = V2-/-/-V2=-2i;

(b)

(2 -3)(-2,l)

(0

(3 >

(a)

Re(iz)

= Re[i(x + iy)] = Re(-y + ix) = -y = - Im z

Im(i'z)

= Im[i*(;t + iy)] = Im(-y + a) = x = Re z.

(l

= (-4 + 3,6 + 2) = (-1,8);

D(3,-l)(^) = (10,0)(i, L) = (2,l). 1

+ z) 2 =(l + z)(l + z) = (l + zH + (l + z)z = l-(l + z) + z(l + z) = l+z+z + z 2 = l + 2z + z 2

z=4±i,

4.

If

5.

To prove

then z

2

.

2 -2z + 2 = (l±i) -2(l±t) + 2 = ±2i-2T2i' + 2 = 0.

that multiplication is commutative, write

Z1Z2

= (*i.yi)(*2.y2 ) = ~ Wa» y^2 + *iy2 ) = (*2*i -yzVp^ H-x^) = (x2 ,y2 )(x ,y ) = z2 z l

6.

(a)

To

verify the associative

(Zj

+ z2 ) + z3 = l(x

l

»

,y l ) + (x2 , y2 )]

yi )

(y,

+ (*2 + ^3

= z +(z2 +z3 ). 1

.

law for addition, write

= ((*i + x2 ) + jcj, = (^1

l

l

.

+ (x 3 ,y 3 ) = (x, + x2

+ y2 ) + y3 ) = y%

+ v3 ) = (*i

,

vt

+ y2 ) + (*

+ (x2 + x3 ), .

yi )

+ [( x2

>

y* )

y,

3

,y3 )

+ (y2 + y3 ))

+ (*3

.

^

)]

2

(b)

To

verify the distributive law, write

z(zl

+ z2 ) = (*,30K*i.yi) + (x2 ,y2 )] = (jc,30(*i + x2 = (ja, + xx2 - yy, - yy2 = (xx ~yy + xx2 -yy2 l

l

=

,

,

+ y2 )

,

yjc,

+ yx2 + xy + xy2 )

yx,

+xy +yx2 +xy2 )

x

l

~ yyv yx + xy,) + (xx2 - yy2 yx2 + xy2 ) = (x, y)(x ,y ) + (x, y)(x2 ,y2 ) = zz +zz2 l

l

10.

The problem here

is to

,

l

.

x

solve the equation z

2

+z+1=

(x,y)(x,y) + (x,y)

for z

= {x,y) by writing

+ (1,0) = (0,0).

Since (x

2

- y 2 + x + 1, 2xy + y) = (0,0),

follows that

it

x

2

-y 2 +;e + l =

and

2xy + y = 0.

By writing the second of these equations as (2x + l)y = 0, we see that either 2x + 1 = or 2 If y = 0, the first equation becomes x +x + l = 0, which has no real roots y = 0. (according to the quadratic formula). Hence 2x + 1 = 0, or x = -1/2. In that case, the first 2 equation reveals that y = 3/4, or y = ±V3/2. Thus

SECTION 3 + 2/

1

!

(a )

2-/

|

3-4/

5/

(1

- 0(2 - 0(3 4

(c)

(l-/)

(a)

(-l)z

(b)

77- =

1

1/z

~

(1

=[(l-0d-0]

2

=(-2/)

— £=7= 1

z

z

z

1

-

- Q(-5Q _ -5 + lOj

2

_ 5/ _1 ~ " ~ ^10/ 2

5

=-4.

= z[l + (-1)] = z

z fe*0).

25

(5/)(-5/)

~ 3/)(3 -

since z + (-l)z

= -z

z

(2 |

5/

5/

>

2.

+ 2Q(3 + 4Q (3-4/)(3 + 4/) (1

=

•

= 0;

+

-5-10/ 25

=

2. 5'

3

3.

(z^Xzj^) = ^[ZaCz^)] = z^iztzjzj = zJC^K)] = z^z^zj] = (z&XZiZj.

1

6.

'

Z\Z2

= z.

—

£2

Z3 Z4

ZyZ 7.

V

_

Z2 Z

\ Z 2j

SECTION 4

w

^=(-73,1),

z2 =(V3,o)

Z\

^

|=

1

Z K l)

c2

/

=

(z 3

(z2

*0,z4 * 0).

*0,z*0).

4

(c)

^=(-3,1),

z2 =(l,4)

Z.

(d)

2.

z,

=

+ i>i

,

z2

= *i -

%

Inequalities (3), Sec. 4, are

Rez IRezI + llmzl, we rewrite it in the following V2~V*

2(*

and

as

x'i

-y2 )

- Vi )~(x2 - iy2 ) = z -z2

(*i

l

and ZyZ 2

+ iy2 ) = {x x2 - yj2 ) + i(y^2 + xj2 ) = (X& - y2 ) - i{y x2 + x y2 ) = (x - fy,)^ - iy2 ) = z,z2 = (x + iy

l

l

)(x2

x

x

4.

6.

(a)

zfaZs

W

z

4

1

2

z

2

= z 2 z 2 = zzzz = (z

z)(z z)

.

;

= zzzz = z 4

.

fa) ZyZ2^3

J

Z2 Z3

__lz l_ i

(b)

z2 z3

8.

t

x

= (ziZ2 )z3 = zfo z3 = (*i £2)^3 = *i 22 z3

=z

lz2Z3

In this problem,

l

Z2 Z3

= Jzi l_ lz2 llz3

l

we shall use the inequalities IRezl|lz

Iz

observe that

when

Izl

=2,

= |lzl 2 -l| = l4-ll=3

2

l-lll|

and

Thus,

when

= |lzl2 -3| = l4-3l=l.

2

2

-3l>|lz M3l|

lz

lzl=2, 4

lz

Consequently,

when

z

2 2 2 -4z +3l=lz -lHz -3l>3-l = 3.

on the

lies

circle lzl= 2,

1 4

-4z +3

z

11.

(a)

Prove that z () Suppose that z

Prove that z

that z

only

x=

either

2

-4 as

>2

exterior to the circle with center at

to see that the set in question consists of all points

-3/2 and radius

2. It is

a domain.

19

(c)

Write

Imz >

1

as y

>1

to see that this is the half plane consisting of all points lying

above the horizontal line y

= 1.

It is

a domain.

y

y =i

X

(d)

The set Imz =

1 is

simply the horizontal line y

= 1.

It is

not a domain.

y

y=

\

X

(e)

The set

(f)

The

< argz < —

(z

* 0)

is

indicated below.

It is

not a domain.

4

set Iz

- 4l£l zl can be

written in the form (x

- 4) 2 + y 2 1 x 2 + y\ which

reduces to

indicated below, is not a domain. The set is also geometrically evident since it consists of all points z such that the distance between z and 4 is greater than or equal to the distance between z and the origin.

*^2.

This

set,

which

is

— 20 4.

(a)

The closure of the

set -it

< argz < n

(z

# 0)

is

the entire plane.

1

(b)

We

first

write the set IRezklzl

inequality is the

same

as

2

y >

as

2

\x\ 0.

+y

Hence

2 ,

or

x

2

0. Finally, by completing the square, we arrive at the inequality 2 2 2 which describes the circle, together with its exterior, that is centered (x - 1) + y > l (x

,

at

z=1

with radius

1.

The closure of this

set

is itself.

21

(d)

Since z

2

2 2 = (x + iyf =x -y + ilxy,

The closure of shown below.

\y\

,

- 2I< 1,

as

shown below.

Since every polygonal line joining zt and z2 must contain at least one point that is clear that S is not connected.

8.

2

can be written as y < x or set consists of the lines y = ±x together with the shaded the set Re(z

not in S,

is

it

We are given that a set S contains each of its accumulation points. The problem here is to show that S must be closed. We do this by contradiction. We let Zq be a boundary point of S and suppose

that

it is

not a point in S.

The

fact that z

is

a boundary point

every neighborhood of zQ contains at least one point in S; and, since z that every deleted

neighborhood of S must contain

accumulation point of S, and that Zq is not in S. is,

S is

closed.

it

follows that z

is

at least

is

one point in

means

not in

S.

we

see

Zq is

an

S,

Thus

that

a point in S. But this contradicts the fact

We may conclude, then, that each boundary point

Zq

must be

in S.

That

22

Chapter 2 SECTION 1.

(a)

11 *

The function f(z) =

+1

z points

(fc)

z

(d)

z

is

z

The function /(z) =

is

z+z

imaginary axis. This

is

The function /(z) =

-

circle

]

defined throughout the entire finite plane except for the

= 0.

Izl

= 1, where

and y =

Using x =

f(z)

defined everywhere in the finite plane except for the

because the equation z + z

—

1

3.

+ 1 = 0.

The function /(z) = Arg( point z

(c)

2

= ±i, where

defined everywhere in the finite plane except at the

is

2

is

the

same

as

x = 0.

defined everywhere in the finite plane except on the

2

1 -Izl

^—-,

= 0.

write

= x*-y 2 -2y+ i(2x - 2xy)

_2

-2

_

2

2

+

+

_ fet Sfe - o

-2

2

= ir + 4r + 2iz-^- + ^- = z 2

5.

is

— Izl

. fa±tf + ftdg, +

SECTION

=

2

+2/z.

2

17

Consider the function '

/w-if where z = x + iy. Observe

that if z

-

x + iy Y

l

= (jc,0),

then

x + iO N2 and

if

z

= (0,y),

(z*0),

—

23 But

if

z

=

(*,*),

/fe)

= l,I^

This shows that /(z) has value value -1 at

cannot

10. (a)

all

nonzero points on the line y

4z

= x. Thus

the limit of /(z) as z tends to

To show that lim

2

= 4, we use statement

-r

- 1)

-P

4 lim

To establish the limit lim

kZ ,

——

V-»i

(z-l)

lira

-7—^

2

To

verify that lim

z

+l

= lim

\-,

j J

(2), Sec. 16,

and write

2

——

-?

= cosy = -e x cosy => 2e* cosy = => cosy = 0. Thus

/(z)

«x

y= uy

^ + nn

= -v* =* -e'siny = e'siny => 2e*siny =

v

= -e x siny.

(n

=> siny =

0.

= 0,±l,±2,...).

Hence

y-nit (« = 0,±1,±2,...). Since these are two different sets of values of y, the Cauchy-Riemann equations cannot be satisfied anywhere.

26 3.

(aj

/(z ) =

- = -.- =

—

z

Izl

z

z

=

—

_+

+y

x

u

= —2 :

2

x

x

x +y

2

+y

a and

t 2

So

2

v

-y

=

, 2

x +y

, 2

.

Since

/'(z) exists

when

z

»

Moreover, when z

0.

y y

.

,

2

2

(*

(*-»)

_

(x

(fc)

/(z)

2

+y

-* 2 2xy 2 2+ +v ) V+/) 2 " ,

2

.

= vy

v

So /'(z) exists only when y =

x,

2

+y

2 2 )

2-

2

z

(z) (z)

= y 2 Now .

= 2y => y = jc

=> 2x

(x

2

\_

2

(zz)

)

2

x -i'2jcy-y

ay

2

= jc 2 + /y 2 Hence u = x 2 and ux

.

(gr _

=

2

* 0,

and

and

M

= 0.

= -v, =>

y

we find that

f\x + ix) = ux (x,x) + ivx (x,x) = 2x + 1O = 2x.

W

f(z)

= z Im z = (x + iy)y = xy + iy 2 Here .

wx

Hence /'(z)

= vy => y = 2y => y =

exists only

when

/'(0)

4.

(flj

/(z)

z

= 0.

u

= xy and v = y 2

and In

«y

= -vx

=

—

4

= Mje (0, 0) + iv, (0, 0) =

cos 40

= vg

and

We observe that

* = 0.

fact,

+ iO = 0.

= -^ = ^cos4^ + ^~sin40j (z*0).

rur

.

ue

=

Since

—

4

-sin 40

= -rvr

/is analytic in

its

domain of definition. Furthermore,

f'(z)

= e- w (ur + ivr ) = e- w ^~cos40 + i^-sin4aj = ~e- ie (cos46-isin4d) = --^"'V' 48

r

r

-4 3

r e

(b)

f(z)

'2

= -Jre'

,e

(re

rur

its

e-

w

(ur

— 2^

«" w

2^e

ien

M

'

V

U

rur

r

= ~

rT--cos| + irVsiiif

= A=e- W e wn 2V^

1

2f(z)'

y

(r>0,0< 0

is

.

entire since

,

,

V

ux

(d)

f(z)

= e~ y cosx - vy

= (z 2 - 2)e' x e"y = z 2 -2

g(z)

and

is

and

entire since

h(z)

uy

it is

= -e~ y sinx = -vx

.

the product of the entire functions

= e~ x e~ = e~ x (cosy - isiny) = e~ x cosy + i(-e~ x siny). ty

'

>

'

v

v

U

The function g

is entire

ux

2.

(a)

/(z)

= xy + iy U

is

= -e~

since x

it is

cos y

nowhere

a polynomial, and h

= vy

and

uy

= -e~

x

which means

sin y

= - vx

.

analytic since

that the

= vy =>y = l

and

uy

=-vx =>x = 0,

Cauchy-Riemann equations hold only

lx

f(z)

entire since

V

ux

(c)

is

'

V

= e y e = e y (cosx + /sin x) = g y cosj: + Vsinjc i

U

Mx

= vy =* -e y sinx = e y sinx

is

at the point

z

= (0,1) =

i.

nowhere analytic since

V

2e y sinx

=

=> sinx

=

and

= -v, => e y cosx = -e y cosx => 2^ cosx = More

precisely, the roots of the equation sinx

cos /iff

= (-1)" * 0. Consequently,

the

=

=> cosx are nit

= 0. (n

= 0,±l,±2,...), and

Cauchy-Riemann equations are not

satisfied

anywhere.

7.

(a)

Suppose

= u(x,y) + iv(x,y)

is analytic and real- valued in a domain D. Since f(z) is real- valued, it has the form f(z) = u(x,y) + i0. The Cauchy-Riemann equations ux =vy ,uy = -vx thus become ux =0,u = 0; and this means that u{x,y) = a, y where a is a (real) constant. (See the proof of the theorem in Sec. 23.) Evidently, then, f(z) = a. That is, / is constant in D.

that

a function f(z)

32 (b)

Suppose that a function Write

constant there.

f(z)

-

/

|/(z)|

throughout D.

If,

analytic in a

is

= c, where

Example 3

(a)

=

C

•

24 then

tells

its

a (real) constant.

* 0,

modulus

is

= 0, we

see that

If c

write f(z)7(z)

=c

2 ,

or

"

m

analytic and never zero in

in Sec.

and that

D,

the conjugate f(z)

us that f(z) must be constant in

must be D.

analytic in

D.

(Yl

SECTION 25 1.

is

is

on the other hand, c /(*)

Since f(z)

c

D

domain

straightforward to

It is

show

+uyy =0 when

we start with

harmonic conjugate v(x,y),

ux

that u xx

u(x,y)

= 2x(l-y). To

find a

= 2-2y. Now

ux (x,y)

=vy =>vy =2-2y=> v(x,y) = 2y-y 2 + (x).

Then uy

= -vx

=>

= * 2 + c.

-2x- -vy =2-3x 2 + 3y 2 =»

w^Oc.y)

v(*,y)

u(x,y)

= 2x - x 3 + 3xy 2 To find a .

= 2-3* 2 + 3y 2 Now .

= 2y- 3x 2 y + y 3 + 6xy = 6xy - 0O) = c.

Consequendy, v(jc,y)

fcj

It is

straightforward to

show

harmonic conjugate v(x,y),

= vy

vy

= 2y - 3x 2 y + y 3 + c.

that u xx

+uyy =0 when

«(jc,y)

= sinhjcsiny. To

we start with ux (x,y) = cosh x sin y. Now

= coshxsiny => v(x,y) = -coshxcosy + sinhxcosy = sinhxcosy - 0(*) =

c.

find a

(d)

It is

show

straightforward to

m„+w=0

that

when

u{x,y)

we start with ux (x,y) =

harmonic conjugate v(x,y),

(x

= v,

2

=~

=> vy

^

+

(jc2

2)2

=

-

,

.

To

find a

Now

r~^~TT+ y2 ) 2 2

=> v(*,y) = -^£-5.+

,

^(jc).

Then 2

2

"y

= "Vx

^ (x

2

+y

2

2

2

=

2

(x

)

2

^ *'

+ y2 ) 2 ~

(jc)

= 0=

*

= °-

Consequendy, v(x,y)

=

— x

Suppose

that

v and

w = v - V,

= vy

uy

,

= -vx

and

23).

w(jc,y)

= c, where

that

uy

,

= -Vx

.

and

wy =vy -Vy =ux -ux = 0.

= c.

u and v are harmonic conjugates of each other in a domain D. Then ux

It

= Vy

c is a (real) constant (compare the proof of the theorem in Sec.

That is, v(x,y)-V(x,y)

Suppose

ux

This means that

then,

wx =vx -Vx =-uy +Uy=0 Hence

T + c.

V are harmonic conjugates of u in a domain D. ux

If

+y

= vy

Uy

,

= -vx

and

vx

= uy

vx

= 0,

v

,

y

= -ux

.

follows readily from these equations that

ux

— 0,

uy

=

and

vy

= 0.

Consequently, w(*,;y) and v(x,y) must be constant throughout/) (compare the proof of the theorem in Sec. 23).

The Cauchy-Riemann equations

in polar coordinates are

rur

=ve

and

ue - -rvT

Now ru r

= ve =>

run

+ ur = ver

.

K

and

Thus

ru n + ru + u gg =rv6r - rvr6 r

and, since v^.

= vr6 we have ,

r

which

is

;

the polar

2

un

+ rur + Ugg = 0,

form of Laplace's equation. To show

that v satisfies the

same

equation,

we observe that "e

= -rvr =>

vr

=

— 1

1

u g => v„

= —u g

r

r

— 1

fr

r

and rur Since u Br

= urg

,

w(r,0)

= lnr,

.

then,

r

If

=vg =*v gg = rur0

2

vn

+ rvr + vee = u e - ru^ -ue + rur8 = 0.

then

rV + m

r

+ u gg = r 2

[

~

+ rfi + 1

= 0.

J

This

tells

us that the function w

= In r

is

harmonic

follows from the Cauchy-Riemann equation rur thus v(r,6)

=

+ (r), where

0(r)

is at

in the

domain r>O,O