COMPLEX VARIABLES AND APPLICATIONS SOL 7ED J W BROWN R V CHURCHILL

Download 2 May 2012 ... The University ofMichigan. Mc. Graw. Hill. Higher Education. Boston ... COMPLEX VARIABLES AND APPLICATIONS" (7/e) by Brown a...

0 downloads 443 Views 3MB Size
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

Bogota

Montreal

Dubuque, IA

Caracas

New Delhi

Santiago

Seoul

New York

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
if

we write them

In order to verify the inequality

Imz< llmzl< Izl.

and

y
2

2

+y 2

>l*l

2 .

2

-2l;tllyl

ways:

+ lyl,

+ y 2 )>l;cl 2 + 2|jtllyl + lyl 2 lxl

,

+ lyl 2 >0,

(l*l-lyl)

2

This last form of the inequality to be verified perfect square.

2

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

2(*

and

as

x<\x\<^x 2 + y 2

3.

y

+ Z,

>0. is

obviously true since the left-hand side

is

a

+ 1

1

;

5

4.

(a)

Rewrite It is

\z

- 1 + il= 1

as

- (1 - 1)| = 1.

|z

Write z I

- 4il +

z+

4/1

points z such that the

curve (b)

the circle centered at 1

-i

with radius

1.

IX\ \-i

I

(a)

is

shown below.

(

5.

This

Write

is

\z

= 10

as

I

J

- 4il

z

sum of the

- (- 4i)l = 10

\z

to see that this is the locus of all

distances from z to 4i and

-4i

is

a constant. Such a

an ellipse with foci ±4i.

- \\=\z + i\

as \z

- ll=lz - (-i)l

to see that this is the locus

of

all

points z such

always the same as the distance to -i. The curve then, the perpendicular bisector of the line segment from 1 to -i.

that the distance

from z

to 1 is

is,

SECTION 5 1.

2.

+ 3i' = z + 3i=z-3i;

(a)

z

(b)

iz

= iz=-iz;

(c)

(2

+ if = (2 + if = (2 - if = 4 - 4i +

M

I(2z

(a)

Rewrite Re(z-/)

2

= 4 - 4/ - 1 = 3 - 4/

+ 5)(V2-0I=I2z + 5IIV2-jI=I2TT5IV2+T = V3I2z + 5I.

through the point

= 2 as Re[* + i(-y - 1)] = 2, z = 2, shown below.

or

3>

2

x

x=

2.

This

is

the vertical line

y

6

(b)

Rewrite I2z - i\= 4 as 2 z

--

= 4,

= 2.

or

This

is

the circle centered at

2

2

shown below.

radius 2,

i/2, j

3.

Write

= x + iy and

z,

l

l

Zr

z2

= x2 + ry2

Then

.

+%) = (*i -^ +

~z2 = (*, + (y )-(x2

2)

1

= (*i - *2 ) -

- ft ) =

I

'(>'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
Specifically,

when

|Re(2

lzl<

and

\zx

(see Sec. 4)

+z2 + z3 |^|£i| + |z2 + M|

1,

+ z + z 3 < 12 + z + z 3 < 2+lzl )|



l

+lz

3 l

= 2+lzl+lzl 3 < 2 + 1 + 1 = 4.

with

7

10.

First write z

4

- 4z 2 + 3 = (z 2 - l)(z 2 - 3). Then

2

-ll>|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 (<=)

is real <=>

Suppose

that z

= x,

Thus z = * + iO

(=>) Suppose that z

Prove that z

that z

only

x=

either

2

-
2 -4z +3l

3

- iy = x + iy. This means

so that x

that i'2y

= 0,

or y

= 0.

is real.

is real,

is either real

lz

= z.

= z,

or z

(^) Suppose if

z

1 4

2

so that z

= x + iO. Then

or pure imaginary <=» z

=z

2

or y

.

Then

= 0,

2

(jr-i'y)

or possibly

2

z

= z2

= x - iO = jc + i'O = z. .

2

=(x + iy) or i4xy = 0. But this can be jc = y = 0. Thus z is either real or pure ,

imaginary.

(^) Suppose that z is either real or pure imaginary. If z is real, so that z-x, 2 2 2 2 2 2 2 If z is pure imaginary, so that z = iy, then z = (-iy) = (iy) = z z =x =z .

.

12. fa)

then

We shall use mathematical induction to show that Zi+z2 +-"+z„ =z +z2 +---+zn

(n

1

This

is

known when n = 2

(Sec. 5).

Assuming now

that

it is

write

Zt

+Z2 +*"+Zm +Zm+1 - (Z +Z2 + -,- +£m ) + Zm+l t

= (z +z2 +-+zm ) + zm+1 = Z + Z2 H l"Z ) + Z m m+1 1

t

= Zj + Z2 H

l"Z

m

+ Zm+1

.

true

= 2,3,...).

when n = m, we may

1

(b)

In the

same way, we can show

that

^z1 ---zn =z This

is

true

when n = 2 ^l Z 2

'

' '

Assuming

(Sec. 5).

(n

z1 --zn

l

that

it is

true

~ (^2 " = (^1^2 " 2m )^m+l = ^2 ~

£m Zm+l

"'Zm ) Zm+l

( Z1 Z 2

'

*

'

'

14.

The identities

2

= -^-^

and Rez

(Sec. 5) zz =\z\

'

= 2,3,...).

when n = m, we write

Zm ) ^nAr

enable us to write

lz-z

l=

R

as

2

(z-Z )(z-Z zz-(z5, 2

15.

Since x

- -^-^ and y =

= i? 2

,

+ z!Q+Zo5>

=/?2

2

= /? 2

-2Re(zz

lzl

)

)

+ lz

x

the hyperbola

l

2

'

.

-y 2 = 1

can be written in the following

ways: .

_n2

_\2

/ fz-z

z+z\ 2 2

+ 2zz + z 2

z

= 1,

1 2i

;

2 .

z

2

2z + 2z

2

z

-2zz+z

2

2

= 1,

_

2 +z =2.

SECTION 7 1.

(a)

Since arg l

*'

one value of

argj

—j

_2 -2i

is

= 3X8 '

~ afg(

y ~f~~£~J'

or

~2

~ 2/)

'

Consequently, the principal value

is

9

(b)

Since 6

arg(V3-i) =6arg(V3-0,

one value of arg(V3

4.

The

evident

we recall

if

the points e

5.

d = n of

solution

ie

and

1.

6

-i*)

is

6f™\ or e

the equation

that e'

e

lies

\e'

on the

-it.

-11= 2

So the

principal value

and

-it + 2n, or

it.

< < In is geometrically - II is the distance between

in the interval

circle \z\= 1

is

that le'*

See the figure below.

We know from de Moivre's formula that + isin 0) 3 = cos30 + isin30,

(cos or

cos

+ 3cos 2 0(/'sin 0) + 3cos 0(isin 0) 2 + (/sin 0) 3 = cos30 + i sin 30.

3

That is, (cos

By equating

- 3cos 0sin z

3

real parts

z

0) + i'(3cos 0sin

- sin 3 0) = cos30 + isin30.

and then imaginary parts here,

we

arrive at the desired trigonometric

identities:

(a)

8.

3 cos30 = cos 0-3cos0sin

Here z = Also,

9

re'

is

2

0;

(b)

2 3 sin30 = 3cos 0sin0-sin

0.

any nonzero complex number and n a negative integer (n =

m = -n = l,2

-l,-2,...).

By writing

and m

l

(z-

we

see that m

z"=(z

T

l

-

m (z

)

=(z

)

6)

=[l^~ m )

.

Thus

J

=[-Je

K - mB)

the definition

=\e z"

i(-~

= {z

me

\

m )

can also be written as

10

9.

two nonzero complex numbers zi and z 2 suppose that there are complex numbers c and c 2 such that Zj = qc2 and z^ = c c2 Since First of all, given

,

x

.

t

and

IzJHqllcJ it

follows that IzJ^ZjI.

Suppose, on the other hand, that

Zi

If

IzjIHcJIqIMcJIql,

we

=r

{

we know only that 1^1=^1.

exp(i 6l )

and

We may write

= r, exp(/02 ).

z2

introduce the numbers

q =r

x

expl

2

and

i

= expl

c2

we find

*

i

J

2

J'

that

c,c2

= r exp^i-^- Jexpl

= r exp(i^) = z,

t

t

J

and

qc2 =

rt exp|

i

i+il v i ,.0i-0 Jexp^-i

-^y*

|

= n exp

2

= z2

.

That is,

Zi

,

10. If 5

= l + z+z 2 +"-+z",

= qc2

l-z" —

z2

= c,c2

.

then

5 - zS = (1 + z + z 2 +•

Hence S =

and



.+z" )

- ( Z + z 2 + z 3 +.

. .

)

= i _ £"+1

+1

.provided z*l. That is,

1-z

H-z + z 2 +-+z"=

— 7" +1

— 1

(z*l).

1-z

Putting z

= e' e

(0

<

< 2tt)

l

in this identity,

we have

+ e m +e i20 +-+e ine =

1

l

~ eJ(n+l)6 \-e ie

——



11

Now the real part of the left-hand side here is evidendy l

+ cos0 + cos20+-+cosn0;

we write that side in the form

and, to find the real part of the right-hand side,

l-exp|7(n + l)0]



\(2n + l)0"

-exp

exp^-i

exp

i-

f)

l-exp(/0)

exp

exp

(-!)

2

H) -exp ®

which becomes

— — 9

cos

(2n + l)0

.

.

i

sin

2

.

2

.

ism

cos

(2n + l)0 •

1

2

2 -2i'sin—

2 or

r I

(2n + l)0l

sm — + sin .

,

^

J

+1

1

cos

— - cos -(2n + l)0l

j

^

j

2 sin—

The

real part of this is clearly

.

sin

(2/i

+ l)0

i+

-ir-

2

2sin^ 2

and

we arrive at

Lagrange's trigonometric identity:

1

1

+ cos + cos20+- -+cosn0 = — + •

2

sm

(2n + l)0

^ 2sin—

(0< 0<2;r).

12

SECTION 9 1.

(a)

Since 2/

= 2exp i\^ + 2kn

1/2

(2/)

(k

= 0,±1,±2,...),

the desired roots are

=V2exp

(*

=
That is,

= i+i and

c being the principal root These are sketched below.

(b)

Observe that 1-V3i

= 2exp

(l-V3/)

The

1/2

i\

— j + 2Jbr

=V2exp

i'I



= 0,±1,±2,...). Hence

-^ + kn

(A:

principal root is

c

= V2V''*' 6 = V2[ cos| - /sin | ] = V2

and the other root

V3-i

,2

is

V2 These roots are shown below.

'

2J

V2

= 0,1).

13

2.

(a)

Since

-16 = 16exp[i(;r + 2*7r)]

(-16)

1/4

(k

= 0,±1,±2,...),

= 2exp

kit —+— it

,

i

4 The

the needed roots are

(*

2

= 0,1,2,3).

principal root is

c

=2,- = 2(cosf +

The other three

/S

i„l] =

2(-^ + -^j = V2(l +i ).

roots are

cx

= {ij^y*11 = cQ i = V2 (i + /)/ = - V2 (i - o, c2

=(2O*'* = -c =-V2(l +

i),

and c3

The four roots

(b)

First write

= (2e'*/4 )e»*12 = c are

= V2 (1 + OH) = V2(l - /).

shown below.

-8 -8V3/ = 16exp

(-8-8V3i)

The

(-i)

^-~ + 2kn^

l/4

=2exp

,

it

(*

= 0,+l,±2,...). Then

kit

(*

principal root is

c

=2e-

,Vr/6

=2 cos^-/sin- =2

= V3-i. I 2

2

= 0,1,2,3).

The

others are

c2

c3

These roots are

(a)

all

=(2,-/6 )e = -c =-(V3-0, i'r

wy

= (2e-'

3 *' 2

=c

shown below.

By writing -1 = lexp[i(?r + 2kn)]

(-1)

1/3

= exp

(k

.,

i

= 0,±1,±2,.

.

.),

we see that

—% + 2kit 3

The

= -(1 + V3/).

(-/)

3

principal root is

c

The other two

=e

Jt jc 1 + V3i = cos— + jsin— = .

.

roots are

c,

= e'* = -l

and „ c 2

K _ = ejsxii _ = cos„ =e nx-ixn e

All three roots are

shown below.

7t 1-V3i isin— = .

.

,

15

(b)

Since 8

= 8exp[i(0 + 2for)]

8

the principal

(k

1/6

= 0,±1,±2,...),

=V2exp(/^

(*

= 0,1,2,3,4,5),

one being c

The

the desired roots of 8 are

=V2.

others are

= V2V- = V2(cos| + /sin|] = Vlfi + v2

:

^/V^. V2 2

3 (V^"*' )** = V^cosy - isin-0-1) = - V2

c3

1

V3

2"T

N

1-V3/ *

V2

=j2e'*=-j2,

=(j2e M3 )e b, = -c

c4

,

=-

1

+ V3/ '

i

V2

and c5

All six roots are

The

three

=(V2V 2 *V* = -c2

=i^. V2

shown below.

cube roots of the number z

(zo

= -A^2 + A-Jli = 8exp^/-^j

r=2exp[

I

+ 2|E)]

In particular,

= 2exp^-|j = V2(l + i).

are evidently

(*

= 0,U).

16

1 H"

o =

With the aid of the number

"^3/

2

v

c2

5.

(a)

=c ©3 =(c

Q)3 )fi) 3

•(V3

=

we obtain the

,

3

two

other

roots:

V2

J

+ 1) + (V3-1)/

-1

V2

+ VIA _(V3-1)- (V3 + l)i 2

V2

J

Let a denote any fixed real number. In order to find the two square roots of a + i in exponential form, we write

A = \a +

i\

=Var+T

a = Arg(a + i).

and

Since

a+i

= Aexp[i(a + 2kn)]

(*

=

0,±1,±2,...),

we see that V2

(a + i)

That

is,

=VAexp

Since a +

tells

(*

= 0,1).

the desired square roots are

VaV " 2

(b)

y + Jbr

if

i

lies

above the

us that cosf

^

]

>

and

real axis,

and

VaV^V* = -4Ae ia, \

sinf

)

>

0.

\2

\2

a

/1

< a<

we know that

+ cosor

Since cos a

1

[.

n.

=—

A

,

Thus

it

<

and

.

a

1

- cos a

1

|^__a

sin

2

V

2

-7211

A

_

VA-a 42-fA'

Consequently,

VA ±VAVa/2 =±VA^cos| + isin|j = ±VA^

+a

^Va

= ±-^(VA + a+iVA-a).

.

+i

cc

it

2

2

follows that

^fA + a

~a

—<—

VA-fl

V2VA

^|

J

,

and

this

c

17

6.

The four

roots of the equation z

4

+4=

are the four fourth roots of the

we write -4 = 4expD'(;r + 2ibr)]

find those roots,

(-4)

1/4

=V2exp

(f

number -4. To

(& = 0,±1,±2,...). Then

+ Y)l = V2^/4 e^ 2

(*

= 0.1,2,3).

To be specific,

c

= V2V™ = V2f cos^ + /sin^l = V2 f-L + i * = l + i, 4 4j V V2j VV2 c/2

q =c

e"

c2

=c

e"

c3

= Co e i3 */2 = (1 + /)(-/) = 1-/.

r

=(l + i)i = -l + i,

= (l + 0(-l) = -l-i,

This enables us to write 4

z

+4 = (z-

)(z

- q)(z - c2 )(z - c3 )

= [(z - q)(z - c2 )]



[(z

-c

= Kz + 1) - i][(z + 1) + 1]



)(z

[(z

- c3 )]

- 1) - i][(z - 1) + 1]

= [(z + l) 2 + l][(z-l) 2 + l] = (z 2 + 2z + 2)(z 2 -2z + 2).

7.

Let c be any nth root of unity other than unity Exercise 10, Sec. 7),

l+z + z.2z +-+z",

,

I

itself.

With the aid of the

identity (see

l-Z" =-^-

(z*l),

l-z we find that

l

9.

Observe

+ c + c 2 + ... +c «-i = l_£l =: J__L = 1-c 1-c

first that

(z-r^^exp^^)

1

W

i(-0 -2tor) i(-2kn) = -7=exp i(-0)-expeXP_ m 1

"ilr

m

m

18

and



i(-e + 2kn) 1 i(-8) ,.-wm_Ji i(2kn) = - =rexp- -exp-T-exp -, )

(z

where

Jfe

= 0, 1, 2,

7

. . . ,

m-L

Since the set

i{-2kn) L exp-i

(*

m is

the

same

as the set

i{2kn)

exp ^

but in reverse order,

SECTION 1.

(a)

= 0,l,2,...,ro-l)

we find that

1/m

(z

1

)

L

(*

m = (z_1 ) 1/m

= 0,l,2,...,m-l),

.

10

Write z - 2 I

+ £1 il

as

I

z

- (2 - i)l £ 1 to see that this 2-i with radius 1. It is

circle centered at the point

is

the set of points inside and

on the

not a domain.

O

(b)

Write I2z + 3I>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\<^Jx

0, or \y\> 0.

+y

Hence

2 ,

or

x

2


.

But

this last

the closure of the set IRezklzl

is

the

entire plane.

(c)

Since

-= z

2

— = 7-7 = -5x 2

zz

2

\z\

-rr,

+y

2

the set

Re - < - can be VzJ 2

2

written as —.2

x

+y

= < -, or 2 2

-2x) + y >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\<\x\.

region

5.

The

set

this

2

2

S consists of all points z such

that lzl< 1 or lz

)

>

,

- 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

——

-?

= <», we refer to



*-»il/(z-l)

(c)

nonzero points on the real and imaginary axes but

1 at all

exist.

(z

(b)

=137.1

r 3

= lim «-»

(z

= 4.

statement (1), Sec. 16, and write

- 1) 3 = 0.

= °°, we apply statement (3),

Sec. 16, and write

z-1

i-i = lim

lim —^-j

«^fly

11. In this problem,

j

=-

^» l+z 2

= 0.

we consider the function 7-( z )

= f£±A CZ +

(ad- be *0).

fi?

fa)

Suppose

that c

= 0.

Statement

(3), Sec. 16, tells

= hm

lim «-»°

T(l/z)

»-»o

us that lim 7(z)

= — = 0. a + bz

a

= °°

since

24 (b)

Suppose

that c

* 0.

Statement

(2), Sec. 16, reveals that

lim T(z)

=—

*-»-

limT|'il = lim

\zJ

Also,

we know from statement

(1),

a + fe

a

+ dz

c

c

lim T(z)

Sec. 16, that

since

c

= 00

since

l-*-d/c

Um -i-=

SECTION 1.

(a^

If

£+4,0. az + &

19

/(z)

= 3z 2 -2z + 4,then

/'(z)

= ^(3z 2 -2z + 4) = 3^-z 2 -2-^-2 + 4-4 = 3(2z) - 2(1) + = 6z - 2. az

fife

If

lim z -»-rf/c

7(2)

/(z)

az

az

= (l-4z 2 ) 3 ,then

/'(z)

= 3(1 - 4zT T-d - 4 *

>

= 3 (! " 4z 2 f(-8z) = -24z(l - 4z 2 )

dz

(2z +

l)f (z-D-(z-l)f (2z + l)

rL) JK)

+ l)

(2z

(d)

If

/(z)

=

(2,

2

+ 1)(1) . (z _ 1)2 _ 2 (2z + l)

3 (2z

+ l) 2

)4

(1+

(z^0),then

f z

^An^^_ ~ a + Zz 2yd + dz nz) = ^ '

Z

<1

(l

)

2

(z

_

2z(l + z

2 3 )

2

2

)

g

2 2

=

2 3 )

(2z)-(lH-z 2

(z

)

[4z -(l

z 4(lH-z

+ z 2 )] _

2(l

+ z 2 ) 3 (3z 2 - 1)

2

)

2 4 )

2z

25 3.

If

/(z)

= l/z (z*0),

then 1

Aw = f(z + Az)-f(z) = z

Hence /'(z)

=

lim A*->o

We

are given that

According

f(z

)

= g(z

)



=

=

+ Az

that

z

(z

+ Az)z

—=—

lim *z-*0( z

and

-Az

1

+ Az)z

f\z

)

\r.

z

and g'(z

)

exist,

where g'(z )*0.

to the definition of derivative,

lira rfe) = «-*«o

Similarly,

g-(z

)

=

lim

/Mzfe) = lim /w. Z - Zq

Z-Zq

^ -^ )

)

= lim-^-.

Thus

=

lim *-*zog(z)

Urn

fWb-**) s(z)/(z-z

-_

\^

Z)I{Z

-^

lim^(z)/(z-z

)

)

_ f(Zo) ^'(z,,)'

SECTION 22 1.

(a)

= z= x — iy. So u = x, v = -y. = Vj, => 1 = -1, the Cauchy-Riemann Inasmuch as /(z)

equations are not satisfied

anywhere. (b)

= z-z=(x + iy)-(x-iy) = + i2y. So u = 0, v = 2y. Since mx = vy => = 2, the Cauchy-Riemann equations are not satisfied anywhere. f(z)

2

( c)

Here u = 2x, v f( z ) = 2x + ixy ux = vy => 2 = 2xy => xy = 1.

w

= -vT =>

Substituting

= xy 2

.

= -y 2 => y = 0. y = into Ay = 1, we have

= 1. Thus

the

Cauchy-Riemann equations do

not hold anywhere.

(d)

= e x e~'y - e*(cosy - isiny) = e x cosy - ie x siny. So u = e x cosy, = vy => 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<2tt).

Since

1

=-e~ e sin(ln r) = ve its

and

=-e~ e cos(ln r) = -rvr

ug

domain of definition. Also,

ie w f'(z) = e- (ur +ivr ) = e-

a

e~ sin(lnr)

g

cos(lnr)

^. |

r

r

= -^f e~ e cos(ln r) + ie~ e sin(lnr)l = re

/(z)

ux

,

V

/is analytic in

When

Since

.

ue

fcos- + /sin-) 2 2J I

= e~ e cos(lnr) + ie- e sin(lnr) „

a + In).

=-—sm- = -rv

,

and

e

+ ivr ) =

1

V

<

domain of definition. Moreover,

-

f(z)

> 0, a <

(r

2*

^ cos- = v =—

f\z) =

(c)

f

= Vr cos — + n/r sin— 2*

/is analytic in

4

4

,5B

L

= * 3 + i(l - yf, we have k = x 3 and

J

v

z

= (1 - y)\ Observe that

2 = vy => 3x = -3(1- yf=*x 2 + (1- yf=0 and wy

=- =>0 Vje

Evidently, then, the

That

is,

Cauchy-Riemann equations are satisfied only when x = when z = i. Hence the expression

and y =

1.

they hold only

= ux + ivx = 3x 2 + iO = 3x 2 in which case we see that /'(/) = 0. f'(z)

is

valid only

when z =

i,

Here u and v denote the

and imaginary components of the function / defined by means

real

of the equations

t —

Az) =

when z*0,

z

when

z

= 0.

Now x

when z * 0, and

l

+y 2

the following calculations

",(0,0)

2

x +y

show

= v, (0,0) and

2

that

«y (0,0)

= -v,(0,0):

w(0 + Ajc,0)--«(0,0) AjC Ax

Ax-»0

w(0,0

+ Ay)-

A*->0 AjC

«(0,0)

Ay

Ay-»0

v(0 + Ajc,0)- v(0,0)

Ay-»0

Ay

A"->0

Ax

Ay->0

£y

_

Ax v(0,0 + Ay)- v(0,0)

Ay

Ay-»0

Equations

(2),

Sec. 22, are

ux cos

+ uy sin

-w,rsin0 + iycos0

= "r>

v

Solving these simultaneous linear equations for ux and u

ux

a = ur cosO-u g

6

sin

y

,

we find that

uy

=ur smdn + u e

v

= vr sm0 + vg

Cauchy-Riemann equations

in polar form,

,

.

and

cos0 r

Likewise,

sin0

= vr cos a -

Assume now

that the

and

e

=vg

rur

y

ue

,

=-rv

r

cos0 .

,

are satisfied at z„. It follows that

wx

=

i/

r

a cos0-w e

sin

6

= ve

cos 6 r

r

a+u = «r sin d e .

My

(a)

,

Write f(z)

cos0

= ve

_ + vsin0 = vr smd + vg cos0 = vv y .

.

,

,

r

sin0

_

vr cos0

.

= -lf vr cos0-ve

sin0\ \

= ~ vx-

= u(r, 0) + iv(r, 0) Then recall the polar form .

r«r

=ve

M„=-rvr

,

of the Cauchy-Riemann equations, which enables us to rewrite the expression (Sec. 22)

f'(z

for the derivative of /at a point z

f'(Zo )

)

= e- ie (ur +ivr )

= (r

= e~ i0 (-vg --u (A = \r

r

J

O)

,



lg

re

in the following

(u e

way:

+ ivg ) = —(u e + ivg ). zn

30 (b)

Consider

now the function 1 _ f {^\ _ = j\Z)--

z

_ —ig=-e 1

1

re

-is

1 _ /„ „ 6 Q -ism Q \ = = -(cos 6) .

r



cos

sin

6

1

r

r

.

r

With ,.(*. Q\ u{r, d)

At

C0S &

=

and

as

v(r, 0)

S"VS\B

=

,

r

the final expression for /'(z

£K(z)\ =

~ ( —

sin

z \

r

l

/

)

\

r

in part (a) tells us that

#

.cos0

1

1

z\

r

zKre'l

r J

cos

f

z

- /sin 6 r

2

when z*0.

10.

(a)

We consider a function

F(x,y), where

z

X=

+z

= y '

and

2

z-z 2i

Formal application of the chain rule for multivariable functions yields

dxdz

dz

(b)

dy dz

+l dx \2 J

dy \

2\dx

dy j

2i)~ 2{dx

dy

Now define the operator dz suggested by part

(a),

and formally apply

2{dx

dz

=

dy)

2dx

fa + *J + fa +

it

to

a function f(z)

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

2dy

iv

y )=

\[{ux - v

y)

If the

Cauchy-Riemann equations uz = vv u = -v*r y y

df/dz

= 0.

,

+ i(vx + uy )].

are satisfied, this tells us that

31

SECTION 24 1.

(a)

= 3x + y + i(3y - x)

f(z)

U

is

V

=3 = vy

ux

(b)

f(z)

= sin x cosh y + icos x sinh y v

'



>

uy

=l = -vx

.

entire since

V

= cosx cosh y = vy

ux

f(z)

is

and

,

v

U

(c)

entire since

and

uy

= sinxsinhy = -vx

= e~ y sinx - ie~ y cos x = e~ y s'mx + i(-e~ y cosx) '

v

'

>

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- - 0'(jc) = 2*

Consequently,

= 2y-y 2 + (jc 2 + c) = x 2 - y 2 + 2y + c.

v(;c,y)

(b)

It is

straightforward to

show

harmonic conjugate v(x,y),

ux

that

uxx

+ uyy =0 when

we start with

= Vy ^>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 +
Then uy

= -vx => 6xy = 6xy -


= 0=> 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 +
Then «j,

= -vx => sinhxcosy = sinhxcosy -

Consequently, v(*,y)

= - cosh x cosy + c.

0'(jc)

=

=> 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<0<27T.

= vg and

the derivative u

r~~

Now it

mat ve ~ ^

present an arbitrary differentiable function of

r.

The other Cauchy-Riemann equation u g =-rvr then becomes O = -r0'(r). That is, Hence \r) = 0; and we see that (r) = c, where c is an arbitrary (real) constant. v(r, 6)

= 8+c

is

a harmonic conjugate of u(r, 6)

= In r.

35

Chapter 3 SECTION 28 2

1.

exj>(2±3m) = e ex$(±3m) = -e

(b)

( r ( cos- + isinexp—^— = exp- exp— = V«l

(c,)

exp(z

2+

3.

2

(a)

IV

m

m\

II

1

since exp(±3;ri)

,

n

.

.

n\

1

+ 7ri) = (expz)(exp7ri) = -expz,

= -l.

I

since exp7ri

= -l.

First write

exp(z )

where z

= x + iy.

This

= exp(* - iy) = e x e~ = e x cos y - ie x sin y, !y

tells

us that exp(z)

u{x,y)

- e x cosy

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

and

v(x,y)

= -e x smy.

Suppose that the Cauchy-Riemann equations ux = vy and uy = -vx are satisfied point z

= x + iy.

cos y =

conclude

« and v here, these equations no value of y satisfying this pair of equations. We may since the Cauchy-Riemann equations fail to be satisfied anywhere, the It is

and sin y that,

some become

at

easy

to see that, for the functions

= 0. But there

is

function exp(z) is not analytic anywhere.

4.

The function exp(z

2 ) is

entire since

and the chain rule for derivatives

-|exp(z

2

it is

tells

2

a composition of the entire functions z and expz;

us that

2 2 2 =exp(z )-|z = 2zexp(z ). )

Alternatively,

one can show

exp(z

that exp(z

2 ) is

entire

by writing

2

2 )

= exp[(* + iy) j = exp(* 2 - y 2 )exp(i2*y) = exp(x 2 -y 2 )cos(2ry) + iexp(jc 2 - y 2 )sin(2;ty) >

w

<

»

U

and using the Cauchy-Riemann equations. ux



/

V

To be

specific,

=2xexp[x 2 -y 2 )cos(2;ty)-2yexp(;t 2 -y 2 )sin(2xy) = vy

and uy

= - 2y exp(x 2 - y 2 ) cos(2 xy ) - 2x exp(x 2 - y 2 )sin(2.xy ) = -vx

.

y

y

,

36

Furthermore,

-y-exp(z

2 )

= ux + ivx = 2(x + iy)[exp(* 2 -

2

)cos(2xy) + i exp(x

2

-

2

)sin(2xy)]

= 2zexp(z 2 ).

5.

We first write |exp(2z + i)\

= \exp[2x + i(2y + 1)]| = e 2x

and 2

|exp(iz

)|

= |exp[-2xy + i(x 2 - y 2 )]| = e

2xy .

Then, since |exp(2z

it

+ 1) + exp(iz 2 < |exp(2z + 1)| + |exp(iz 2 )|

follows that

|exp(2z

6.

)|

+ 1) + exp(/z 2 < e 2x + e~2xy )|

.

First write

|exp(z

2 )|

= |exp[(x + ry) 2 = |exp(j: 2 -/) + i2xy| = expU 2 -y 2 ) ]|

and exp(lzl

Since the

2 2 2 2 x -y
,

it is

2 )

= exp(jc 2 + y 2 ).

clear that exp(*

2

- y 2 ) < exp(* 2 +y 2 ). Hence

2

|exp(z )|
7.

To prove that

|exp(-2z)|

< 1 <=> Re z > 0,

|exp(-2z)|

It is

it

follows from

above that 2 ).

write

= |exp(-2;c - /2y)| = exp(-2*).

then clear that the statement to be proved

is

the

same

as

exp(-2*) <

obvious from the graph of the exponential function in calculus.

1 <=>

x>

0,

which

is

37

(a)

Write e

That

l

= -2

eV = 2*'*.

as

This

tells

us that

n + 2nn

e*=2

and

y=

* = ln2

and

y = (2n + l);r

(n

= 0,±l,±2,...).

(n

= 0,±1,±2,...).

(n

= 0,±l,±2,...).

(n

= 0,±1,±2,...).

(n

= 0,±1,±2,...).



= 0,±1,±2,...).

(n

= 0,±l,+2,...).

(n

= 0,±1,±2,...);

(n

= 0,±l,±2,...).

is,

Hence z

(b)

Write *

That

l

= 1 + V3

1

as

= ln2 + (2n + l)«f


,(

* /3) ,

from which

'

we see that

e*=2

and

y=

* = ln2

and

y = ^2» + |j«

j + 2nn

is,

Consequently,

z

(cj

Write exp(2z

- 1) = 1

as e

2x_l
= \xi2 + {ln + ^jti

2x ~ l i2y

e

=l

= le i0 and note how

and

2y =

+ 2nx

it

follows that

Evidendy, then,

x=\

and

y

= n^

2

and

this

means

that

z

This problem

is

= -^ + nm

actually to find all roots of the equation

exp(iz)

= exp(/z).

38

To do

this, set

z

= jc +

iy

and rewrite the equation as y

e

Now, according

a

e' =e y e".

to the statement in italics at the beginning of Sec.8 in the text, y

e~

= ey

-x = x + Inn,

and

= 0,±l,±2,.„.

where n may have any one of the values n

y=

The

Thus

and x = nn

Suppose

10. (a)

that e

z

that

Since e

is real.

Moreover, since e

=

x

never zero,

is

0,±l,±2,...).

= e x cosy + ie x sin y, this means that e x siny = 0. siny = 0. Consequently, y = nn(n = 0,±1,±2,...);

Imz = n^(« = 0,±l,±2,...).

is,

On the

other hand, suppose that e

z

is

pure imaginary.

follows that cosy

It

= 0,

or that

— + /i7r(n = 0,±l,±2,...). 2

21

We start by writing 1

_

z

Because Re(e z )

Re(
)

= exp

Since

is

ensures that Re(e

If

f(z)

) is

e

Sec.

25

is

that

fit)

2

+y

2

x 2

-y

.

x +y

2

+ i-

'

x

2

+y

2

follows that

2

cos

domain

^

2

+ y2

that

= exp

.2 jc

+y..2

COS

,

J

,

does not contain the origin, Theorem

is

analytic in

= e u(x,y) CQS

some domain D,

1

in Sec.

25

^

y)

then

+ ie u(*,y) sinv (xy)

a composition of functions that are analytic in D, component functions

it

follows from Theorem

its

U(x,y)

2

\x +y..2

harmonic in such a domain.

= u(x,y) + i'v(jc,y)

Since e fU)

.

analytic in every 1/z

x

Izl

it

2

x-iy

z 2

zz

= e* cosy,

1/Z

_

z

\x +y

13.

(n

l

y = ^- + nn(n = Q,±\,±2,...). Thatis, Inu =

12.

= 0,±1,±2,...).

roots of the original equation are, therefore,

z=nn

(b)

(n

= e u(x y) cosvU.y), -

V(x,y) = e

u(x ' y)

sinv(jc,y)

1

in

39

are harmonic in

D. Moreover, by Theorem 2

in Sec. 25, V(x,y) is a

harmonic conjugate of

U(x,y).

14.

The problem here

is to

establish the identity

(expz)"

(a)

To show

that

= exp(nz)

(n

when n = 0,1,2,..., we = when n 0. Suppose that it is

it is

true

obviously true nonnegative integer. Then (expz)

(b)

m+l

= (expz)"

1

Suppose now that n view of part (a),

(expz)"

is

(expz)

when n - m, where

true

(expz)

m

= -1, -2,

. .

.

),

and write

1

1

exp(znz)

exp(-nz)

1

^expz;

use mathematical induction.

m = -n =

= exp(nz).

SECTION 30 1.

2.

3.

= \n\-ei\+iArg(-ei) = lne-^i = l-^i.

(a)

Log(-ei)

(b)

Log(l-

fa;

loge

= lne + i(0 + 2n7f) = l + 2/i7Fi

(b)

logi

=

(c)

log(-l + V3i)

(a)

Observe that

= lnll - i1+i'Arg(l -

lnl +



= In V2

(n

= |ln2 - -ji.

= 0,±l,±2,...).

i^ + 2nnj = ^2n + ^m = ln2 +

(n

= 0,±l,±2,...).

i^ + 2n^ = ln2 + 2^n + |jm

Log(l + 1)

2

m

It is is

any

= exp(mz)expz = exp(mz + z) = exp[(m + l)z],

a negative integer (n

=

= 0,±1,±2,...).

= Log(2i) = In 2 + ~/

and

2Log(l + 1) = 2^1n V2

+ ijj = In 2 +

Thus 2

Log(l + i) =2Log(l + i).

(n

= 0,±1,±2,...).

1,

2, .

. .

.

Li

(b)

On the other hand, 2

Log(-l + 1) = Log(-2/) =

In 2

- -/

and

2Log(-l + i) =

2(

lnV2

3n

+i^J = ln2 + y/.

Hence 2

Log(-l + /) *2Log(-l + /).

(a)

Consider the branch

logz

= lnr + /0

r>0,

— <6< 4

— 4

Since

2

log(i

)

= log(-l) = lnl + i;r=7B

and

21og/

= 2flnl +

= m, /-|J

we

(b)

find that log(/

2 )

= 2 log/ when this

branch of logz

is

taken.

Now consider the branch logz

= lnr + /0

r

> 0,

— 4

< 9< 4

Here

log(i

2 )

= log(-l) = In 1 + in =

m

and

2 log/

=2

In

1

+ /,5^ = 5m. 2

Hence, for

(a)

2

this particular branch, log(/

The two values of

i

m are

log(€'*

W4

e'

and e

i5tt/

)

,

* 2 log/.

\ Observe that

K

/4

)

l\ ( = lnl + /|-^ + 2«7r| = ( 2n + - \jti

= 0,±1,±2,...)



and log(*

,9W4 )

5n = lnl + i|^ + 2/«rh

(2n

+ l) + -

7n

(n

= 0,±l,±2,...).

(n

= 0,±l,±2,...).

4 Combining these two

sets of values,

log(/

1/2

)

we

find that

= |n +

nm

^

41

On the other hand, 1

1, -logi=.

2

Thus the

lnl

+ i|

of values of log(i

set

= \n + —\m

^ + 2nn

2

) is

the

= 0,±l,±2,...).

(n

1

same

as the set of values of

— logi,

and

we

2

may write log(/

(b)

Note

1/2

)

= |log/.

that

log(i

2 )

= log(-l) = In 1 + {it + 2mt)i = {2n + \)iti

(n

= 0,±l,±2,...)

but that

21ogi'

=2

^ + 2mt

lnl + i'l

= {An + Y)Td

Evidently, then, the set of values of log(/

2 log i. That

2

To

solve the equation logz

2

is

)

= 0,±l,±2,...).

not the same as the set of values of

is,

log(i

7.

2

(/i

= iitll,

)*21ogi.

write exp(logz)

= exp(/7r / 2),

2

or z

= e'*n =

i.

) is the real component of any (analytic) branch of 2 logz, it is harmonic in every domain that does not contain the origin. This can be verified directly by writing

10. Since ln(jc

u{x,y)

+y

= ln(* 2 + y 2 ) and showing

that

^(jc.y) + u (x,y)

= 0.

SECTION 31 1.

Suppose

that

Rez > t

and Rez2 >

zt

0.

Then

= r exp i'0j t

and

z1

-r1 exp i'0 2

where

— it

2

~ it < 0, < — 1

2

,

it

and 2

it ^ — <©,< 2

2

,

,

42

-% <

The fact that

Log(ztz2 )

t

+

2

= Log[(r

x


enables us to write

r2 )exp /(0 t

+

2 )]

= lnfe) + i{Q + x

2)

= (In r + i'0i ) + (In r2 + 1'0 2 ) = Logfo exp i©, ) + Log(r2 exp i0 2 ) {

= LogZ! + Logz2

3.

.

We are asked to show in two different ways that

log

(a)

One way is

= ln

Another way

is

+ iarg

to first

show

(z,

t

to refer to the relation arg

log

(bj

=logz -logZ2



= argZ[ - argz2

in Sec.

* 0, z2

- = -logz (z * 0). To do this, we write ]

z

and then

log^ = log^e"'8 j = ln^ + i(-0 + 2imt) = -[lnr + i(0- 2imt)] = -logz, where n

= 0,±1,±2,.

. . .

This enables us to use the relation

log(z z2 ) 1

= logz +logz2 1

and write

f

log

log \ Z2J

= logz, + log \2 = logz,- log z2

0).

7 and write

= (lnl^l+zargz!) - (lnlz2 l+i'argz2 ) = log^ - logz2

that log[

?t

.

= re

j

43 5.

The problem here

is to

verify that

1/n

=exp -UogzJ

z

given that

1/ e

valid

when n = 1,2,...

m

Then, since

integer. z

it is

= e~

z

is

.

To do

(n

= -l,-2,...),

we put m = -n, where n is a negative we may use the relations z' = 11 z and

this,

a positive integer,

1

to write

p(^ =

1,

iog

4

exp(llogz)] = exp(-llogz) = expQlogz).

SECTION 32 1.

In each part below,

(a)

(1

+

)' 1

n=

0,±l,±2,....

= exp[i log(l + 1)] = exp< i lnV2+i(^ + 2n^

=

n — ln2-[ — + 2nn = exp^--^ - 2n^jexp^ln2j. {i

exp|

1

Since n takes on

all integral values, the

term

-2nn here can be

replaced by +2nit.

Thus

(1

(-1) '* 1

(b)

2.

+

/)''

=exp^-| + 2n^exp^ln2j.

= exp^log(-l)J = exp|^[ln 1 + i(n + 2nn]^ = exp[(2n + 1)/].

= exp(/Logi) = exp^ln 1 +

(a)

P. V.

/'

(b)

P.V.

[j(-l - V3/)]

"

/

-|

exp

-n

= exp |3^Log[|(-l - V3/)| = exp = exp(2/r2 )exp(/3^) = -exp(2;r 2 )

3ni\

)ne-i

In

44

P.V.

(c)

(1

- i) 4 =

exp[4/Log(l - 1)]

'

= exp 4iflnV2-/j

= e*[cos(41n V2) + isin(41n V2)] = «*[cos(21n2) + isin(21n2)]. 3.

Since

- 1 + V3i = 2e 2w
(-1 + V3/)

3 '2

=exp |log(-l + V30] = exp|| ln2 +

/|

^Y

+ lnn

= exp[ln(2 3/2 ) + (3n + l)*ri] = 2-^2 exp[(3/i + where n = 0,±1,±2,....

Observe

that

if

n

(-1 + V3i)

5.

odd, 3n+l

l);n']

3/2

©

<

According

it.

according to Sec. 32, that value

exp(

— Logz = exp |

(-1 + V3/)

is

odd; and so

is

even; and this means that

3/2

arise.

Specifically,

=±2V2.

We consider here any nonzero complex number where -it <

3n+l

even, then

is

= -l. On the other hand, if n is exp[(3/i + Y)iti] = 1. So only two distinct values of exp[(3n +

z

in the exponential

to Sec. 8, the principal value of z

Un is

form

Zq

= rQ exp i'0 o

^/r^exp^i— j;

,

and,

is

= ^ (in r + i'0 o ) exp(ln ^/r~) exp^z'

= nfc exp^j

^

These two expressions are evidently the same.

7.

Observe that when c = a + bi

power

c i

is

any fixed complex number, where c *

0,±1,±2,..., the

can be written as

T = exp(clogi) = exp< (a + bi)

]sil

+ i^— + 2nit

— + 2mt \+ia\ — + 2mt Vl .12 )

= exp -b

it

it

(n

= 0,±l,±2,...).

Thus

C I/

and c

it is

clear that

* 0,±1,±2,...

c lz

I

is

l= exp -b\

— + 2nit

multiple-valued unless b

ensures that

c i

is



=

0, or c is real.

multiple- valued even

when b = 0.

Note

= 0,±1,±2,...),

that the restriction

.

SECTION 33 1.

The desired

derivatives can be found

by writing

-'~ k ^

k

e--e- = j_(d —sin = ±f' dz

dz

2i

ie

= Yi\

iz

e-

)

2i\dz

)=

——

+ie

_d—-e -

iz

dz

j

= cosz

2

and

d d(e* + e-*} lfd =- — — cosz = — dz dz\ 2 ) 2\dz

= -[ie -ie ")- = V 2

2.

From

e

k

—d

+

e

_ fe

dz

= -sinz. 2i

i

the expressions

s'mz

= e*-e'*

,

and

cosz

=

k

e

+ e~

iz

2

2i

we see that cosz + ismz

=

e

k

K

+e-'

+ ,

2

3.

Equation

(4), Sec.

33

Zi

k

-e~ k

= ek

.

2

is

2 sin zt cos z2 Interchanging

e

= sinfo + z2 ) + sin^ - Zj ).

and z 2 here and using the 2 cos z sin z2 x

fact that sin z is

we have

= sin(z + z1 ) - sin(zi - z2 ). t

Addition of corresponding sides of these two equations

2(sin zx cos z^

an odd function,

+ cos z

v

sin Zj )

now yields

= 2 sin(z + Z2 ), t

or

sin(z!

+ z2 ) = sin z, cos z2 + cos z

{

4.

sin 22

Differentiating each side of equation (5), Sec. 33, with respect to z v

cos(zt

+z2 ) = cosZ[ cosz2 -sinz

t

sinz2

.

we have

46

7.

(a)

From

2

the identity sin z

+ cos 2 z = 1, we have

—— + —— = —— 2

cos z

ft)

1

cos z

or

'

1

+ tan

z

= sec

z.

cos z

Also,

sin z .

sm

9.

,22

2

cos z

sin z

From

+

,

cos z ,

.

=

sin z

z

1 ,

.

or

,

1

+ cot

z

= esc

z.

sin z

the expression sin z

we find

= sin x cosh y +

1

cos x sinh y,

that

= sin 2 xcosh 2 y + cos 2 x sinh 2 y

2

Isinzl

= sin 2 x(l + sinh 2 y) + (1 - sin 2 jc)sinh 2 y = sin 2 x + sinh 2 y. The expression cos z

on the other hand,

= cos x cosh y + i sin * sinh y,

us that

tells

Icoszl

2

= cos 2 jccosh 2 y + sin 2 jtsinh 2 y = cos 2 *(1 + sinh 2 y) + (1 - cos 2 x) sinh 2 y

= cos 2 x + sinh 2 y.

10.

2

Since sinh y

is

never negative, 2

(a)

Isinzl

it

follows from Exercise 9 that

> sin 2 x,

or

Isinzl

> lsin*l

and that (b)

11.

In this problem

Icoszl

2

> cos 2 x,

or

Icoszl

>

I

cos xl.

we shall use the identities I

sin zl

2

= sin 2 x + sinh 2 y,

I

cos zl

2

= cos 2 x + sinh 2 y.

y

.

47

(a)

Observe that 2

sinh y =lsinzl

2

- sin 2 x < Isinzl 2

and 2

Isinzl

= sin 2 x + (cosh 2 y - 1) = cosh 2 y - (1 - sin 2 x)

= cosh 2 y - cos 2 x < cosh 2 y. Thus 2

sinh y
(b)

2

< cosh 2 y,

lsinhyl
or

On the other hand, 2

sinh y

=

cos zl

I

2

- cos 2 x <

I

cos zl

2

and Icoszl

2

= cos 2 x + (cosh 2 y - 1) = cosh 2 y - (1 - cos 2 x) = cosh 2 y - sin 2 x <, cosh 2 y.

Hence sinh

13.

By writing

/(z)

2

2

y < Icoszl ^ cosh

= sinz

2

y,

lsinhyl< Icoszl^ coshy.

or

= sin * cosh y -i cos* sinh y, we have

=sin(;c- iy)

f(z)

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

where u( x,

If the

)

= sin x cosh y

and

Cauchy-Riemann equations wx = vy

=

cos;ccoshy Since coshy

x = — + nn

is

(n

never zero,

it

= 0± 1,±2,...)-

Wj, ,

v(x,y)

= -v,

are to hold,

sin*sinhy =

and

follows from the

= -cosx sinh y

first

it is

easy to see that

0.

of these equations that cosx

Furthermore, since sin x

is

= 0;

that

is,

nonzero for each of these values

2 of x, the second equation tells us that sinhy equations hold only at the points

z

= 0,

or y

= 0. Thus

= - + n7t

the

Cauchy-Riemann

(n

= 0±l,±2,...)-

2

no neighborhood of any point throughout which /

Evidendy, then, there

is

we may conclude that

sinz

is

is

not analytic anywhere.

The function /(z) = cosz = cos(jc -/y) = cos x cosh y + isinx sinh y can be /(z)

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

where u(x, y)

and

analytic,

= cos x cosh y

and

v(x, y)

= sin xsinh y.

written as

48

If the

Cauchy-Riemann equations ux = vy uy ~ -vx ,

sin x cosh y

The

first

of these equations

cos nn *0, it follows that equations hold only when

tells

=

16. (a)

is

cos x sinh y

= 0.

= 0, or x = n7t(n = 0,±l,±2,...). Since = y 0. Consequently, the Cauchy-Riemann

us that sin;c

sinhy = 0,

z

So there nowhere

and

hold, then

or

= nn

(n

no neighborhood throughout which /

is

analytic,

and

= 0±l,±2,...).

this

means

that cosz is

analytic.

Use expression

(12), Sec. 33, to write

cos(iz)

= cos(-y + ix) = cosy cosh x — i sin y sinh x

and cos(i'z)

This shows that

(b)

Use expression

= cos(y + ur) = cosycoshx-isinysinh*.

cos(i'z)

= cos(iz)

for all z.

(1 1), Sec. 33, to write

sin(i'z)

= sin(-y + ix) = -sinycosh x - i cosysinhx

and sin(i'z)

= sin(y + ix) - sin y cosh x + cosy sinh jc. i

Evidently, then, the equation sin(zz)

= sin(i'z)

sinycoshx = 0,

Since cosh*

is

sin(/z)

tells

= sin(/z)

= nn

equivalent to the pair of equations

cosysinhx

first

= 0.

of these equations

tells

us that siny

= 0.

= 0,±1,±2,...). Since coshtt = (-1)" ^ 0, the second = So we may conclude that us that sinhx 0, or that x = 0. if and only if z = + inn = nni (n = 0,±1,±2,. .).

Consequently, y equation

never zero, the

is

(n

.

17. Rewriting the equation sinz

= cosh4

as sin;ccoshy

+ icos;tsinhy = cosh 4, we

need to solve the pair of equations

sin x cosh y

= cosh 4,

cos x sinh y

=

see that

we

49

for x

and

any*

y. If

since

cos* =

0.

y

= 0,

the

equation becomes sin*

first

sin*^l and cosh4>l.

= cosh 4, which cannot be

So y*0, and

satisfied

by

the second equation requires that

Thus

x=

~- + n7t

(n

= 0±l,±2,...).

Since

sin^Y + »wj =

the first equation then is

even,

it

becomes

follows that y

= ±4.

The problem here

is to

coshy = cosh 4, which cannot hold when n

(-1)"

Finally, then, the roots of sinz

Z

18.

(-l)",

=

(f

+ 2n7C ±4i

odd. If n

are

= 0±l,±2,...).

(n

]

find all roots of the equation cosz

equation as cosxcoshy-/sin;tsinhy

= cosh 4

is

We

= 2.

= 2. Thus we need to solve

cos*coshy =

2,

sinjrsinhy

since cos x = 2 x and y. We note that y * second in the pair of equations to be solved for

if

y=

start

by writing

that

the pair of equations

= and

0,

that is impossible.

So

the

us that sin;c = 0, or that x = nn (n = 0±l,±2,...). The first equation then tells us that (-l)"coshy = 2; and, since coshy is always positive, n must be even. That is, x = 2nn {n = 0±1,±2,...). But this means that -1 coshy = 2, or y = cosh 2 Consequently, the roots of the given equation are tells

.

z

To express cosh" y

= cosh"

1

2, or

1

coshy =

2, 2.

= 2n^ + /cosh _1 2

which has two values, This

tells

(e

y

us that e

y

y

+ e~ - 4;

apply the quadratic formula to obtain e y the observation that

at this alternative

(2-V3)(2 + V3)

form of the

z

way, we begin with

and, rewriting this as

= 2 ± V3,

= ln

2 + V3

we arrive

in a different

= 0±1,±2,...).

f-4(e y ) + l = 0,

we may

ln(2-V3) = ln

(n

or y

!

2 + V3

= ln(2 ± V3).

Finally, with

= -ln(2 + V3),

roots:

= 2n;r±tln(2 + V3)

(n

= 0±l,±2,...).

e

50

SECTION 34 1.

To find

we

the derivatives of sinhz and coshz,

d(e -e' z

d

.

,

z

)

d

1

write

,

.

z

+ e' z

_ ZN

e

_,. x

__

and

d + d _d coshz= _^__j = __ (e!+

2

3.

z

(

,

Identity (7), Sec. 33, is sin z

that

2 i'

2

sinh z

\

1

+ cos 2 z = 1.

sin(z'z)

we find

z

e

,

z2

by

iz 2

identities that

we have

2

z

z

-e

.

,

=sinlu

.

& here and using the identities

= coshz,

= 1.

cos^ + Zj) = cosz^oszj -sinzjsinzj. Replacing Zj by iZj and cosf/^ +z2 )] = cos(/z )cos(/z2 )-sin(iz )sin(iz2 ). The same 1

were used just above then lead

1

to

+ z2 ) = coshzj cosh^ + sinl^ sinhz2.

cosh(z1

6.

e

or 2

here,

)=

cos(iz)

cosh z - sinh z Identity (6), Sec. 33, is

,

Replacing z by

= isinhz and

+ cosh 2 z = 1,

,

We wish to show that lsinh.*l
in

(a)

two

cosh*

different ways.

Identity (12), Sec. 34, is lcoshzl this tells

us that sinh

2

= sinh 2 x + cos 2 y. Thus

x < Icoshzl

2

2

2

2 ,

or sinh x\< Icoshzl. I

2

Icoshzl -sinh

On

2

x>0; and

the other hand, since

= (cosh x - 1) + cos y = cosh x - (1 - cos y) = cosh x - sin 2 y, we know 2 2 2 2 Icoshzl -cosh x<0. Consequently, lcoshzl
(b)

Exercise U(b), Sec. 33, recalling that cosi'z

7.

(a)

2

Observe

tells

2

2

2

us that lsinhyl
= coshz and

iz

= ~y + ix, we

obtain the desired inequalities.

that

sinh(z +

m) =

that

=

=

=

= -sinhz.

— 51

(b)

Also,

cosh(z +

(c)

From

7ti)

parts (a)

=

and

u/tanh(z

=

L

f

we find

(b),

= + /n) _

— = ——— = -

1

'-z—

-

= -coshz.

that

—— + sinh(z

+ ?n) = -sinhz = -

cosh(z

9.



7n)



sinhz

-coshz

coshz

=

tanhz.

The zeros of the hyperbolic tangent function t

.

=

tanhz

sinhz

coshz are the

same

tanhz are the zeros of coshz, or z

15.

(a)

Observe

which are z = nm (n =

as the zeros of sinhz,

that,

since sinhz

to solve the pair

= ^-j + rucji

=/

x = 0,

= 0,±1,±2,...).

can be written as sinh x cos y + /cosh xsiny

= i, we need

of equations

sinh x cos y

If

(n

The singularities of

0,±1,±2,...).

= 0, cosh x sin y = 1.

the second of these equations

becomes siny = l; and so y =

— + 2nn it

2 (n

= 0,±l,±2,...). Hence

(n

If

(/i

x*0,

the

first

equation

= 0,±l,±2,...). The second

Rewriting cosh z

=^

as

cosh x cos y 4- i sinh x sin y

,

,

sinhxsiny

= 0.

see that

y=

or

= l. But there

= — we 2

2

= 0,

we have no additional roots

2 the pair of equations

coshxcosy = —

cosy

that

then becomes (-l)"coshjr

value of x satisfying this equation, and

(b)

requires

= 0,±l,±2,...).

is

of sinhz

—vmt 7T

no nonzero

= i.

x and y must satisfy

52

x = 0,

If

cosy =

1



.

the second equation

Thus y = cos

1

%

1

— = ±— + 2nn

z

If

x

becomes all x.

0, the

=

Thus no additional

(n

first

= 0,±1,±2,...), and

equation becomes

this

means

= {ln±^7ti

second equation

(-1)" cosh x

and the

satisfied

is

But

tells

this

roots of

(n

us that y

= nn

(n

that

= 0,±l,±2,...).

= 0,±1,±2,...). The first then

equation in x has no solution since cosh*

coshz =



>1

for

are obtained.

2

16. Let us rewrite

coshz = -2 as cosh* cos ;y

+ isinhjcsin;y = -2. The problem

evidently to

is

solve the pair of equations

cosh x cos y If is

If

* = 0,

the second equation

Since cosn^

is satisfied

sinh x sin y

and the

- 0.

first

reduces to cos y

no roots of coshz = -2 arise. find from the second equation that siny = 0, or

no y satisfying

x*Q, we

= -2,

= -2.

Since there

this equation,

= (-l)",

equation can hold only

it

follows from the

when n

z

is

first

1

2

x = cosh

-1

2.

+ (2n + l)m 1

= ±ln(2 + V3) + (2n + l)ni

= 0,±l,±2,...).

= -2. But

this

Consequently,

(n

Recalling from the solution of Exercise 18, Sec 33, that cosh" 2 these roots can also be written as

z

(n

equation that (-i)"coshx

odd, in which case

= cosh"

y-nn

= 0,±1,±2,...).

= ±ln(2 + V3), we note that

(n

=

0,±1,±2,...).

.

Chapter 4 SECTION 37

2.

jQ^J*=j(£-l)<*^

(a)

Te

dt

=

V3 —T = —ircos—^.•f + -1 = — + 1

|v 2

Since le~**l= e~

fcj

3

2/L

L 2/

o

bx

3

6 fe

IV" T"*

udt =

lira

L

The problem here

is to

\e

*•

mB

e-

Mde =

r

27T

1

*

*

when

m*n,

when

m = n.

we write 2*

/

and observe that when

=

J

e

Me~ M dO = 2wje'^'de

m#n, 2ic

i(m-n)B

i(m — n)

When

1

= - lim(l - e - bt ) = - when Re z > 0.

Jj=0

o

this,

4'

verify that ?s?

To do

4

J

/

-;

we find that

,

e""
3.

,i

*sin—

-

m = n,

1

1

i(m-n)

i(jn-n)

= 0.

/becomes 2*

/= \dd = 2n; and the verification

4.

First

is

complete.

of all, 7*

J But

7t

e

(I+,)x

etc

=

x

J

e cos * dtc

+

J e * sin x dx. 1"

also,

,(!+<>

l

+

I

e*e'"-l l

+i

-e*-l l-i l + i l-i

l

+ e*

+

e" A—+ —

i

-

x

x

Equating the real parts and then the imaginary parts of these two expressions,

je'cosx dx = - ^

+e

jVsin;c_* =

and

+ e*

2

o

Consider the function w(t)

= e" and observe that

lie

2tc

jw(t)dt=je

it

dt

=i-i=o.

=

i

i

K Since \w(c)(2n - 0)| =\e \2n c in the interval

l

we find that

= 2%

for every real

number c,

it is

clear that there

is

no number

< t < 2n such that lie

jw(t)dt=w(c)(2n-0).

(a)

Suppose Thus

that w(t) is even. It is straightforward to a

a

a

show

a

and

that u(t)

v(f)

must be even.

a

jw(t)dt=ju(t)dt+ijv(t)dt=2ju(t)dt + 2ijv(t)dt

a

a

= 2 j u(t)dt +

(b)

z

J

a

v(t)dt

Suppose, on the other hand, that w(t) a

J

odd.

is

a

w(t)dt

=2

J

It

w(t)dt.

follows that u(i) and

v(f) are

odd, and so

(n

= 0,1,2,...),

a

= ju(t)dt + J v(t)dt =

+ / = 0.

1

Consider the functions

1

P„(x)

where -1 < x <

1.

K

Since

2 x + i^l-x cosd = V* 2 + (1 -

it

"

_____

= ~j(x + iVl - x 2 cos ej d6

2

)cos

2

d<^x

2

+(l-

follows that 1

T.

P„(*)|<-f Lc

|

+ iVl-x 2 cos0

X

^<-fj0 =

l.

2 )

= 1,

55

SECTION 38 1.

(a)

Start

by writing

—a

I

=

J

-a

w(-t)dt

=

-b

The

substitution

r = -t

a

-b

two

integrals

b

on the right then yields

b

b

= -)u(T)dT-ijv(T)dz = ju(t)dr+ij v(t)dr = jw(r)dt. b

That

j v(-t)dt.

i

-b

in each of these

a

I

J

-a

ui-t)dt +

b

a

a

a

is,

-a

b

jw(-t)dt = jw(r)dt. -b

(b)

a

Start with b

1

b

b

= jw(t)dt = ju(t)dt+ijv(t)dt a

a

and then make the substitution

t

=

a


each of the integrals on the

is

P

I

P

a

That

P

= ju[(T)]'(T)di:+ ijv[(T)]' (r)dr = a

J

{x)dx.

a

is,

b

P f

jw(t)dt=jw[(/)(x)]<}> (x)dx.

3.

The slope of the

line

through the points (a, a) and (fi,b) in the ft plane

b-a

m=— p-a So

the equation of that line

.

is

b-a

t-a = -

p-a

,

(t-a).

is

right.

The

result

+

56 Solving this equation for

t,

one can rewrite

b-a

t

Since

t

=

If

B-a

as

aB-ba T+-^

B-a

.

0(f), then,

...

4.

=—

it

Z(t) = z[^(T)] where zit) ,

b-a

aB-ba

B-a

B-a

= x(t) + iy(t) and

Z(T)

t

= 0(t), then

= 4^(i:)] + /y[0(T)].

Hence

Z'(t)

=

=

5.

If

w(t)

at

x'MtWW + iy'miWW

wcd = z'[
vim]

= f[z(t)] and

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

f(z)

w(f)

The chain

=

+ i-f

ax

z(t)

= x(t) + iy(t), we have

= u[x(t),y(t)] + Hx(t),y(t)].

rule tells us that

du — =

,

ux

x+u y y

.

,

and

dv — =

,

Vj

(

x'

,

+ vy y',

and so

w'(0 = In

(ux x'

+ uy y') + i(yx x' + vy y').

view of the Cauchy-Riemann equations ux = vy and uy w'(t)

= -v,,

then,

= (ux x' - vx y') + i(yxx' + uj) = (ux + ivx )(*' + iy').

That is,

w'(0

when

f

=f

-

= {ux [x(t),y(t)] + ivMt)Mt)])[x'(t) + //(f)] = f'[z(t)]z'(t)

SECTION 40 1.

(a)

C be the semicircle z = 2e'°

Let

(0

<

-2

<

n),

2

x

shown below.

Then

+e

= 2/

(b)

Now let C be the semicircle

This

is

the

same

= 2e i0

z

f£±2
2.

(a)

The

arc

is

C:z = l + e

i0

2jt

Jc

= 2i(-i + 2n-i-7t) = 4 + 2m.

+

f JC

the value here being the

(z

- \)dz =

z

= 2e i0

(0 <

< 2k).

In this case,

^-^-dz = Am,

sum of the

7

values of the integrals in parts (a) and

(n< 6<2n). Then Ijc

.''29

w w ne J(l + « - l)fe <*0 = ije dO = i 2/

=

Thus

2k

C denote the entire circle

Finally, let

(n
as part (a), except for the limits of integration.

.'9

(c)

= 2j'(j + # + 1) = -4 + 2m.

I(^_^) = I(1_1

)

=

.

(b).

58 (b)

Here C: z

= x (0 < x < 2). Then

Jc

3.

(z-l)dz = J(jt-l)<£c =

In this problem, the path

C is the sum of the paths Q C C 2,

,

The function

to

= 0.

-jc

3

,

and

C4

that are

be integrated around the closed path C is /(z) = ne Kt We observe that and find the values of the integrals along the individual legs of the + .

c = c + C2 + C3 i

Q

square C. (i)

Since C,

is

z

=x

(0

< x < 1),

^^ = ^^ = ^-1. i

Jc

1

o

(ii)

C

Since

2 is

z

= 1 + iy (0 < y < 1), i

(Hi)

C

Since

3 is

= (1 - x) +

z

1

(0

i

< x < 1),

i

i

o

(ivj

Since

C4 is

z

= i(l - y) (0 < y < 1), i

= J *»*
i

4^^'HMy = ^JV^y = -2. o

Finally, then, since

we find

shown below.

that l

\c

ne*~ dz

= 4{e* -1).

The path

C is the sum of the paths 3 Q:z = x + ix (-1<*<0)

3

C2 :z = x + ix (0
and

Using

/(z)

= lonC,

and

=[

f(z)dz

/(z)
f(z)

= 4y = 4x 3 on C2

,

we have 1

jc

f(z)dz

+JVJ C2

= Jl(l + i3x 2 )<£c + -1

The contour z(fc)

C

has

4j^ +

t + [*t +

some parametric

+ i3* 2 )
1

-1-10

= Wlli + i[* 3

(l

o

1

= Jit + 3/jA; 2 dbc +

3

J4*

5

12ijA: dtc

6

2i [* ]J :

representation z

= 1 + i + 1 + 2/= 2 + 3/.

= z(t) (a
z(a)

= z and x

= Z2. Then

Jc


= \z\t)dt = [ Z (0]* = z(b) - z(a) = z2 -zv a

To

integrate the branch

around the circle C: z

Let z

=e

C

= e ,e

(0

be the positively

,e

(0

<

<

2;r), write

oriented

circle

1*1=1,

with parametric representation

< B < lit), and let m and n be integers. Then

lc

m z z"dz

But we know from Exercise

=

m

i6

j

{e

)

n

ie

(e-

)

e

i£ dd = i je'<"

+1)

3, Sec. 37, that

lK

je

Me- Md0-- [0 [lit

when

when

m*n, m-n.

VM


60 Consequently,

m+1 * n

when

frt\fc-j° lc

m + l = n.

\2m when

8.

Note that

C is

suggests the parametric representation C:z

With

2 2 x +y =

the right-hand half of the circle

that representation,

-

So, on C,

4.

* = ^4-y 2

2

= ^4-y +iy (-2 < y < 2),

to

.

This

be used here.

we have

dy

2

(

2

I= + ^y 4%

= j(-y + y)dy + ij\t

iW

-2 2

2

^

4

..2

= 4i J -2

= 4/ [sin"

10. Let

(a)

C

be the

f

circle

=

z

1

(1)

- sin"

1

(-1)]

= 4i

= z + Re ie (-Jt<6<

= Am.

n)

-^Rie dd = i]dd = 2m. ie

f

-K

(b)

When n = ±1,±2,...,

=



m (e *

- e-'" *) = / 1

— 2R°

.

sinn/r

=

n

11. In this case,

where a

is

any

real

number other than

zero, the

with a instead of n, yield the result

a~ l

f Jc a

(z-Zo)

dz

=

i—-sin(a^) a

same

steps as in Exercise 10(b),

61 12.

The function f(z)

(a)

representation

is

continuous on a smooth arc C, which has a parametric

z = z(t) (a
Exercise 1(b), Sec. 38, enables us to write p

b

\f[z(t)]z'(t)dt

=f

/[Z(T)k'[0(T)]0'(T)
a

a

where Z(T) = Z[0(T)]

But expression

(14),

Sec 38,

tells


us that

zWt)W'(t) = Z'(t); and so P

*

jf[z(t)]z\t)dt =jf[Z(r)]Z'(t)dr.

f(z) is piecewise continuous on C. Since C can be broken up into a finite chain of smooth arcs on which f(z) is continuous, the identity obtained in part (a) remains valid.

Suppose that

(b)

C is any contour and that

SECTION 41 1.

Let

C be the arc of the circle

\z\=

2 shown below.

o

Without evaluating the

2

integral, let us find

x

an upper bound for j

note that if z

is

a point on C, 2

|z

Thus

-l|>|lz

2

l-l|

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

jc ~T~J

To do

this,

we

<

z

62

Also, the length of

2.

The path

C is

^(47r)

=

So, taking

n.

C is as shown in the figure below.

M = ^ and L = n

The midpoint of

,

we find that

C is clearly the closest point on

V2 C to the origin. The distance of that midpoint from the origin is clearly -y-,

the length of

being V2.

Hence

if

z

is

any point on C, Id >

V2 —

This means

.

that, for

such a point

4 Izl

Consequendy, by taking

M - 4 and L = V2 ||c

3.

The contour

To

^|

,

we have

S ML = 4V2.

C is the closed triangular path shown below.

find an upper

bound

for

\jc

z

\e

(e

-

- z)dz

z

l I

\e

\

,

we let

z be a point on

+ \z\ = e x + jx 2 +y 2

.

C and observe that

C

63 x

But e <

1

less than or

Hence, by

x < 0, and

since

equal to 4. Thus

writing

le*

M = 5 and L =

- zl < 5 when 12, we have (e

Note

that if lzl=

R (R > 2),

^x 2 + y 2

the distance

z

of the point z from the origin

on C. The length of

is

C is

then

2

l2z -ll<2lzl

2

2 +l = 2/? +l

and

4

lz

2 2 2 + 5z + 41 = lz + II lz + 41 >

2 Izl |

2

-1

1

Izl

-4 = (tf 2 - 1)(/? 2 - 4). |

1

Thus 2

2z 4

z

when

lzl=

R

(R >

2

-l

2 +5z +4

4

z

+ 5z 2 +4l

Since the length of

2).

2

2z -l

W+5z

I,

2

+4

2* 2 + l

I2z -ll

teR(2/?

^

CR

2

2

2

(R -l)(R2 -4)

is

^R,

then,

+ 1)

R2

/?l

2

~(/? -l)(/? -4)

R 2 J} and

it is

Here

CR

is

the positively oriented circle

Logz

Izl

lln/? + /0l

=R(R>

<

ln/?+l0l 2 ?

since

-n
Q

is,

1).

If

< "

R2 )

R tends to infinity.

clear that the value of the integral tends to zero as

z

^r

is

a point on

CR

,

then

+ ln/g

R2

of course, 2nR. Consequently, by taking

M = ———

and

L = 2kR,

always

evidently 12.

-z)dz < ML = 60.

z

is

we see

that

Logz

n + k\R R

< ML = 2n

dz z

*

Since it

iim it

+ \xiR _ =

lira

1//? — — = u, -

follows that J c*

Let

Cp

z

be the positively oriented circle \z\=p (0

suppose that /(z)

is

analytic in the disk Id

<


shown

in the figure below,

and

1.

y

We let

= exp^-

bounded disk in that disk.

observe that

Izl

< 1,

We if

z

is

there

is

V

1

\

Jppx

we

note

that, since

are asked to find an upper

a point on

Cp

jc

inasmuch as

M

is

Cp is z-

bound

/(z)

m

M such that for

= \z- m \\az)\<

I

z

M

2np, we may conclude

m f(z)dz

limf

that

<^j=2np = 27tMjp.

independent of p, U2 z-

it

follows that

f(z)dz

|

is

,

U2

Since the length of the path

exp^-i-j

a nonnegative constant

\z-

that,

1

^ log z j = exp|^- ^ (In r + id) =

of the power function here; and

Note

I' /a

represent any particular branch

z

-1/2

I

= 0.

(r

> 0,

a < < a + 2ft)

continuous on the closed l/(z)l^

m f{z)dz

M for each point z .

To do

this,

we

— 65

SECTION 43 1.

The function

z" (n

=

0,1,2,...)

has the antiderivative z"

Jc

I

n+1

^

_»+l

n

.

to

/(n

+ 1) everywhere

in the finite

a point z2 , 1

+1

/i

+l

n + 1^

2

1

'

i/2

2

"r

fa;

x

Zl

.

2.

C from a point z

any contour

plane. Consequently, for

+1

z

e*

—e

e

=
j

it

+1

+

l

it

i

it

>r+2i

cosl

(b)



l
= 2S

J

=

~
U-2)

«

= —lie

i„(f + ,)=:

e

—e

e

I

2/

= « + -. e

3

4

1

1

1

4

4

0.

4 i

1

3.

= ±1,±2,...) always has an antiderivative = z Zq. So, by the theorem in Sec. 42,

Note the function (z -Zq)"' (n does not contain the point

L(z-zo

5.

C

for

any closed contour

Let

C denote any contour from

real axis.

any domain that

r dz = l

does not pass through

that

in

Zq.

z = -1 to z = 1 that, except for This exercise asks us to evaluate the integral

its

end

points, lies

above the

t

I

=

jz'dz, -i

where

z'

denotes the principal branch

z'

= exp(/Logz)

(lzl>0,-^< Argz<

it).

66

An z

cannot be used since the branch But the integrand can be replaced by the branch

antiderivative of this branch

= -1.

z'

since

can

it

= exp(z'logz)

not even defined at

is


lzl>0,-

agrees with the integrand along C. Using an antiderivative of this

new

branch,

we

now write _('+!

*

-

i+l

[(l)

+1

1

_ _£_("^(i+l)(Inl+iO) _ i +

j

(

(i+lXlnl+i*)"! J

-

^ j~

£

('>D'ogl

£

(M)\og(-D

+e = _J_ h _ g-*^*) - ^ ' l+ i + V

j

1 i

.

1

L

1-/

d-o.

SECTION 46 2.

The contours

C

t

and

C2

are as

shown

in the figure below.

In each of the cases below, the singularities of the integrand lie outside

so the integrand

is

analytic

C

x

or inside

on the contours and between them. Consequendy,

imdz=if(z)dz.

C2

;

and

67

(a)

When

f(z)

=

(b)

When

f(z)

=

/(z)

=

(c)

(a)

When

-

the singularities are the points z

z .



+2 .

the singularities are at z

,

=

= 2nn (n = 0,±1,±2,...).

sm(z 1 2)

-, the singularities are at

z

= 2nm

(n

= 0,±1,±2,.

In order to derive the integration formula in question, around the closed rectangular path shown below.

Since the lower horizontal leg e~

l*

along that leg

is

represented by z

we

.

.).

integrate the function e

= x (-a < x < a),

the integral of

is

xl

je-

xl

dx = 2je- dx.

-a

Since the opposite direction of the upper horizontal leg has parametric representation z

= x + bi (-a £x
-]e'

{x+bi)1

dx = -e

the integral of e'

bl

xl

\e

iUx

e

dx

=

1*

along the upper leg

V

is

2

cos2bxdx + ie" \e

j
xl

sm2bxdx,

or simply u 2

-2/ jy* Since the right-hand vertical leg e~'*

along

is

2

cos2fct<£t.

represented by z

= a + iy

it is

O

je-

B

(a+iyf

idy

=

!«-* je

yl

e-

i2ay

dy.

(0

< y < b),

the integral of

68 Finally, since the opposite direction of the left-hand vertical leg has the representation

z

= -a + iy

(0

< y < b),

the integral of e~ l * along that vertical leg is b

b

-fe

H - a+iy)1 idy = -ie- al je y *e i2ay dy.

According to the Cauchy-Goursat theorem, then, a

a

b

xl

b 2\e~ dx - 2e \je-

V_

xl

b

cos2bxdx + ie** \e

y \-'lla

a nay > dy - ie~ je y *e dy = 0;

f :

and

this

X

reduces to

4 xl

]e~

(b)

We now

let

xl

coslbxdx = e~ b \e~ dx + e

a -»

°°

- (a

^ bl) ]e yZ sinlaydy.

in the final equation in part (a), keeping in

mind

the

known

integration formula

xl

\e' dx and the fact that

(al+bl)

e-

yl

\e

sinlaydy < e

- (a2+bl)

yl

je

dy ->

as

a ->

oo.

The result is

xl

e

cos2bxdx = ^-e- b2

(b>0).

2

6.

We let C denote the entire boundary of the semicircular region appearing below. It is made up of the leg C, from the origin to the point z = 1, the semicircular arc C2 that is shown, and the leg C from z = -1 to the origin. Thus C = C + C + C 2 3

t

y

3

.

69

We also let

f(z) be a continuous function that

=

by writing /(0)

is

defined on this closed semicircular region

and using the branch

/(z)

of the multiple-valued function z

m .

= V7e ie/2

r

The problem here

is to

C

>o,-f<*
evaluate the integral of f(z)

by evaluating the integrals along the individual paths C„ C ,and C3 and then 2 adding the results. Li each case, we write a parametric representation for the path (or a related one) and then use it to evaluate the integral along the particular path. around

(i)

C

z

:

v

= re i0 (0
JCi /(z)^=jVF.lJr

(ii)

C2

:

z

=1

I

B •

e'

(0

f(z)dz

<6<

ie

ie

2

dd = ije^ n d0 = if^e na

o

(Hi)

-Cy

z

= re ix

(0

<

r

1).

*

'2

]

L3i

o

<

[V^=|.

Then

it).

= ]e wn

=

Jo

- 1) = - 1 (1 + /).

= 3

3

Then

mdz = -\_ c f{z)dz = -)^e *'\-\)dr = i\^dr = i\-r vll i

\c5

o

The desired

L3

o

Jo

=j 1

f(z)dz + j

Cj

f(z)dz+i f(z)dz = l-h+i) + "-J

3

The Cauchy-Goursat theorem does not apply since f(z)

is

3

= h 3

0.

not analytic at the origin, or even

defined on the negative imaginary axis.

SECTION 48 In this problem,

3

result is

jc f(z)dz

1.

=-/.

\

*" J

we let C denote the square contour shown in the figure below. y 2i

c i

-2,

2

r

-2/

Jt

6 (a)

(b)

f

Jc

Jc

.

,.



,

.

=2m \e~

f

^2

z (z

+ 8)

2z +

1

r

coshz

,

tan(z/2)

f

Jc

Jc z

Jc— * r

=

dz

z \

^

= 2m(-i) = 2n.

,

.

+ 8>

7


z-0

coshz

d3

2tb* f

,

tan(z/2)

r

2

U;

8.

L2J z= _ 1/2

Jc b^sr* "IT

,

iJ^L + Lz

-(_i/2) r

=

I

,

4j

2

=

60,1,1

4

= 0. f(0)

Jz=0

2«f

.

Z=*o

= 2«i"|

4-sec

2

U

Let

(aj

1

= »rsec 2 f

C denote the positively oriented circle

The Cauchy

dz

c

+4

\z



I

when

-2
- il = 2, shown below.

J
z _2/)(z +

r

20

l/(z + 2i) dz

Jc

= 2m\

= 2m

z-2/

z

+ 2ij z=2i

Applying the extended form of the Cauchy integral formula,

f

Jc z 2 (

<2.

integral formula enables us to write

dz

JCrz 2

(b)

—2

& + 4)

2

_ f ~ Jc

2

dz ffc (z

_ 2/)

2 (

z

+ 2i)

2

1/ U + 2Q _ rf l/(z 1+1 ~ J c (z-2i) (* -

Z

=

we have

2£ 1!

1
(z

+ 2i)

71 3.

Let

C be the positively oriented circle

I

z

I

= 3 and consider the function ,

g(w)=\

We wish to find g(w) when w = 2

dz

and when

\w\

>3

(lwl*3).

(see the figure below).

y

Cm )(

i

w = 2/3

V

x

We observe that #(2)

= Jc

_2~

On the other hand, when

5.


z Iwl

>

3, the

= 2 ^'[ 2z2 " z ~ 2 ]

is

/ is analytic inside inside C, then

f£&*»2«rk)

= 2^1(4) = 8m.

Cauchy-Goursat theorem

Suppose that a function not on C. If z

z=2

and

tells

us that g(w)

and on a simple closed contour

f^>*

/ft)* =

f

= 0.

C and that z

is

^

Thus f

J^

The Cauchy-Goursat theorem to C,

7.

f\z)dz = f /(g)
tells

us that this last equation

each side of the equation being

ie

C

f

it),

and

— dz=[ -^-dz = 2m\e —

Jc z

is

also valid

when

zq is

exterior

0.

be the unit circle z = e (-n£d< Cauchy integral formula reveals that Let

'

Jc z

let

at ]

J * =0

a denote any

= 2m.

real constant.

The

On the other hand,

the stated parametric representation for

—dz = ]^~Qie w de =

\c Z

-It

K

It

=i

J

jexp[a(cos0 + isin 6)]d6

/

e

-K

C gives us

e

acos

V

flSin

e

acose d0 = ije [cos(asm 6) + isin(asin 6)]dd -It

-ft

K

It

= - je acos0 sin(asin 8)dd + ije acos9 cos(asin 6)dd. -It

-1t

Equating these two different expressions for the integral It

— e

r I

az

c z

dz,

we have

It

- je

acose

sin(asin d)dd

-It

+ i je acosB cos(asin d)dd = 2m. -It

Then, by equating the imaginary parts on each side of this

last

equation,

we see that

it

je

tt<:ose

cos(asmd)dd =

27f,

-« and, since the integrand here

is

even,

je

(a)

The binomial formula enables us

We

cos(asin 6)dd

=

Jt.

to write

note that the highest power of z appearing under the derivative

differentiating

(b)

acose

it

n times brings

it

down

to z".

So

Pn (z)

is

2

is

z

",

a polynomial of degree

and n.

We let C denote any positively oriented simple closed contour surrounding a fized point z.

The Cauchy

integral formula for derivatives tells us that

rtf-tf^fSfzlLj, dz"

Hence

the polynomials

K '

Pn (z)

Im^is-z)^

1

in part (a) can be written

(„

= 0,1,2,...).

73 (c)

Note

that 2

C?

-l)"

_

(j-1)"(j +

n+l

(s-iy

+ S-l

1)"

(s-iy

Referring to the final result in part (b), then,

we have

Also, since 2

(s (j

- 1)"

+ ir

+1

=

-

(s

+ 1)" _

(*

+ l)"

(s

+1

5

- 1)"

+1

'

we have

2" 2to* j c

9.

+i

s

We are asked to show that 1

rf(s)ds

— m Jc (t ni (s-z)

(a)

In view of the expression for /' (z) in the

/'(z + Az)-/'(z)

_

1

lemma,

1

r

2ml

Az

*

3

f(s)ds

1

(.y-z-Az)

2

(s-z)

2

Az

2(s-z)-Az 2ni[(s-z- Az) (s-z) 1

f

Then

/'(z + Az)-/'(z)

Az

1

f

f

/(j)
7ri|(5-z)

3

_

1 i

r

2^4 1

2(j-z)-Az 2 2 (s-z-Az) (s-z) 3(j-z)Az-2(Az)

(s-z)

2

^' 2 3 /( J f.c — 2iTi 2^/J(,-z-Az) (5-z) f

"

3

74 (b)

We must show that r

3(.s-z)Az-2(Az)

2

2 (3DIAzl+2IAzl )M < jf(s)ds

c(s-z-Azf(s-z?

Now D,d,M, and L

(rf-IAzl)V

The

are as in the statement of the exercise in the text.

triangle

inequality tells us that

I3(j

- z)Az - 2(Az) 2 l< 31* - zl

I

Azl

+ 21 Azl 2 < 3D! Azl + 21 Azl 2

we know from the verification of the Is - z - Azl > d -I Azl> 0; and this means that

Also,

I

(s

expression for /' (z) in the

- z - Az) 2 (s - z) 3 > (d l

.

1

Azl )

2

lemma

3 d > 0.

This gives the desired inequality.

(c)

If

we let Az

tend to

in the inequality obtained in part (b)

we find that

This, together with the result in part (a), yields the desided expression for /"(z).

that

75

Chapter 5 SECTION 52

We are asked to show in two ways that the sequence

1.

(-1Y

z„=-2 + A-f

(n

= l,2,...)

(n

= l,2,...)

n

One way is

converges to -2.

=-2

xn

Another way

is to

2.

is

0, respectively,

=

observe that \z„- (-2)1

~ (- 2)| < e

z„

where n

two sequences

y„=t-9-

and

-2 and

of real numbers converge to 51.

to note that the

if z„

= - 2 + i^-

(n

= 1, 2,.

.

.

n

),

to

apply the theorem in Sec.

Thus for each e>0,

n>n

whenever

,

1

any positive integer such that n >

Observe that

and then

then

/•„=lzJ=i/4 +

— -»2.

But, since

2fl

the sequence

3.

Suppose \zn

that

0„

=Argz2„-»^

(n

limzB

= 1,2,...)

.

is,

for each

= Argz^,

(n

e >0, there

is

a positive integer

In view of the inequality (see Sec. 4)

lzn

it

2 „_,

= 1,2,...),

does not converge.

= z. That

- z\< £ whenever n > n

and

follows that llzj-lzlke whenever

-zl>llzj-idl,

n>n

Q

.

That

is,

limlzj=lzl.

such that

=

76 4.

The summation formula found

in the

example

when

If

we put

z

=

<

where

jr(re'*)"

r

< 1,

52 can be written

in Sec.

lzl
the left-hand side

becomes

= jr rV" e = j^V cosnfl +

u=l

n=l

n=l

r

n

sinn0;

n=l

and the right-hand side takes the form

re

w

- re ie

1

w 2 - re~ w _ re -r _ ie 2 w ie ~ 1 - r(e + e~ ) + r ~ 1 - re~

rcos

1

'

1

- r 2 + irsin

- 2rcos + r 2

Thus

^ > r Zl

.

-

.

_



rcos0-r 2

rsin0

.



cosnd + iy r smnd = T+ T 2 l-2rcos0 + r l-2rcos0 + r 2 t& .

i

Equating the real parts on each side here and then the imaginary parts, summation formulas

E„r

Q= cosn0

B=1

where

6.

Suppose

<

r



rcosd-r 2 -r l-2rcos0 + r 2

< 1. These formulas

that

^z

B

=5.

clearly hold

To show

that

V> r

j

and

«

^

when



a=smnd •

rsin0

appeal to the theorem in Sec. 52. First of

r

=

^z„ =S, we

5>„=X

all,

write zn

=xn +iy„,

5>

B

=7.

OO

Then, since

^(— y„) = —Y,

it

follows that

n=l

jU = 2ta - % = 2>„ + )



«*(->.)]

T 2

arrive at the

,

too.

we note that

and

we

l-2rcos0 + r

n=l

n=l

.

x - iY = j.

S=

X + iY

and

77 8.

Suppose

£z„ = S

that

and

^w„ = T.

In order to use the theorem in Sec. 52,

we write

b=1



=*,+%> S = X + iY

and

w„=un

2>b = * f,yH =Y

and

5>„ = t/,

+ivn

,

T = U + iV.

Now B=l

B=l

|>

B

=

V.

B=l

Since

2(^+"„) = X + t/

it

X^+vJ^+y,

and

follows that

J,[(xn +u„) + i(yn +v„)] =

X + U + i(Y+ V).

That is,

+%) + ("„ + J'vJ] = X + iY+(t/ + /V), B=l

or

B=l

SECTION 54 2

1.

Replace z by z in the

known series ln

~

coshz =

X

z (Izl<

~)

to get «>

cosh(z

Then, multiplying through

this last

2 )

=

2

_4b

Z

(lzl< ~).

equation by z, **

zcosh(z

2 )

= ]£

z

we have the desired result:

4n+1

tS(2n)!

(Izl<~).

78 2.

Replacing z by z -

(b)

1

in the

known expansion : z

n

= 5>7

e

"

n=0


;

we have

~Z

(lzk°o).

„, n!

n=0

So e

^rt =e £(£zi)l

(ld
r.=0

3.

We want to find the Maclaurin series for the function

4

z

To do

this,

we first replace

z by -(z

z

z

+9

9

1

l

4

/

9) in the

+ (z 4 /9)'

known expansion

r-^ly as well as

Then,

if

its

Ozl
condition of validity, to get

we multiply through this

last

equation by

|-,

we have the desired expansion:

n

2

6.

Replacing z by z in the representation

2n+l

sinz

z

=

(lzk~),

Zo

(2n + D!

we have 4b+2

sin(z

2

) = £(-1)"

*j

"„

Qz\<

oo).

Since the coefficient of z" in the Maclaurin series for a function f(z)

shows

The function

4n)

(0)

=

/

and

—^— has a singularity at z =

1.

<2B+1)

(0)

So

l-z \z

- < V2, i\

/

(0)/n!, this

that

f 7.

(n)

is

=

(n

the Taylor series about z

=i

is

= 0,1,2,...).

valid

when

as indicated in the figure below.

y

i

1

To

\V2

J

we start by writing

find the series,

This suggests that

i

1

1

l-z

(l-i)-(z-O

we replace z by

(z

l-i

l-(z-/)/(l-0"

- 1) / (1 - i)in the known expansion

(ld
and then multiply through by

:•

*

n=0

The desired Taylor series

is

then obtained:

(z-iT.

i-z

9

The

identity sinh(z

+ ni) = -sinhz sinhz

So, if we replace z

sa-o*

and the periodicity of sinhz, with period 2 m,

= -sinh(z +

by z - ni

in the

(lz-/I
ni)

known

= -sinh(z - ni).

representation

tell

us that

80

we find

and then multiply through by -1,

sinhz

13.

Suppose

<

that

<

Izl

4.



= -V i£_JEL! (2» + D!

S

<

Then

that

Iz / 41

(

< 1, and we can use

the

|

Z

known expansion

7^- = 2y

4z-z

2

(IzKl).

< Izl < 4,

To be specific, when

1

_ m\< ~).

111-lym ^fz =y£_ 4z~W ^4 fl+l

!_£

4z

= J_ + 4z

= _L + y. y^_ ~4" ^4" +1

4Z

+2

'

B

4

SECTION 56 1.

We may use the expansion

™ 2n+l

sin

to see that

when 0<

Izl

%.

3.

Suppose

(UI< ~>

< °°,

jn

that 1
-S

( ir

and

y

(-!>•

1 ,

recall the

,

f

(-ir

Maclaurin series representation

= Z^ T^'2 1-z

( |Z|<1 )-

±! „=o

This enables us to write

1 1

+Z

1

^

'r-iiRT-i^ — Z n=0 v

1h

„=o z

Z

Replacing n by n -

1

1

in this last series

and then noting

that

81

we arrive at

the desired expansion:

v

1

4.

The

singularities

of the function /(z)

there are Laurent series in

=

are at the points z

powers of z for the domains


=

Y z"

(lzl< 1)

Z

„ =2

„_

1

=

and

and z = 1




1.

Hence

(see the

figure below).

To

find the series

when

<

\z\<

Z

Z

„=0

A

2

As

for the

domain

< lzl<

1

/(z)=

5.

(a)

The Maclaurin

<»,

1,

recall that

Z

n=0

note that

II / zl

=

"7|(z")

series for the function

Z

Z

< 1 and write

=

"7'I^17I)

and write

z+

~%^~%7-

l

is

valid

when

z-1

\z\<

1.

To

find

it,

we

recall

the Maclaurin series representation

= TIT 5>"

(ld<1)

and write

for

1-z z

+l

fziA

~

-<« +

•°

"=1

1

»izr 12 «=0

<-<

- »X »=0

11=1

-

-X^ n=0

1

- X*' «=0

(b)

To

find the Laurent series for the

Maclaurin series for

same function when

—— that was used in part (a). 1

-z

<

Since

1+ZJ n= i)\ZJ z

l,

n =oZ

1 here,

n =0

we

recall the

we may write

Z

z

(l
1

The function /(z) = z(l

the figure below.

Hence

=

and z = ±z, as indicated

there is a Laurent series representation for the

and also one for the domain

To

has isolated singularities at z

+ z2 )

1


°°,

find each of these Laurent series,

which

is

exterior to the circle lzl=

domain


For the domain

/(*>

we recall the Maclaurin series representation

Z

1

+Z

2

n=0

< Id < 1, we have

= ~'7~~t = 1

t HT = tc-DV-

Z B=0

On the other hand, when

B=0

1

1

= -+ £(-1)

V"

1

= £(-l)

2n+l

n+1

z

Z

Z


11

1

^ r__l_V _ ^ (-If _ v (-1)" n=0

£

n=l

+1

.2n+l £~

'

2

z In this second expansion,

1

1.

(lzl< 1

in

we have used

1 the fact that (-1)""

= (-l) fl-1 (-l) 2 = (-1)" +1

.

1).

83 8.

(a)

Let a denote a real number, where -1

< a < 1.

Recalling that

j— = |>"

(IzKl)

»=0

enables us to write

_a

a

_v

1 i

fl

z-a"7"l-(a/z)"^7Tr

'

or

a

(b)

Putting z

= e'° on each

^ a"

= 2j—

(lal
side of the final result in part (a),

a

n

e

we have

-inB

But

a e

ie

-a

(cos0-a)-isin0

(cos0-a) + i'sin0 (cos0-a)-i'sin0

acos0-a 2 -iasin0 l-2acos0 + a 2

and

XflV" 9 n=l

cosn6-i^a n smne. b=1

n=l

Consequendy,

2,a"cosnd =

when -1 < a <

10. (a)

-—

and

r-

l-2acosd + a 2

Ya" smn9 = r ~ l-2acosd + a 2

1.

Let z be any fixed complex number and plane. The function

C the unit circle w = e'*

{-it

<



^ n)

in the

w

/(w) = exp 2v has the one singularity w = in the C, as shown in the figure below.

w

w>

plane. That singularity

is,

of course, interior to

.

w

Now

the function

plane

f(w) has a Laurent

According to expression

series representation in the

domain

I<

<

(5), Sec. 55, then,

exp

(0
[II

where the coefficients J„(z) are

exp

Using the parametric representation expression for

Jn (z)

J" i z) -

1

r

That

|

T7; 2m 'J -

dw

w = e'*

= 0,±l,±2,...).

(n

(-n <$
for C, let us rewrite

as follows:

exp

=

]

Wn+l

2m Jc

this

w——

K

2 v~ v JC» e

*

- iV'd0 =

1

2m

|

exp[izsin ]e~

m *d

_

is,

K J„ (z)

The

last

=

1

^

J

exp[-i

- z sin 0)]
(n

= 0,±l,±2,...).

(h

= 0,±1,±2,...).

expression for /„ (z)in part (a) can be written as *

= ^i(^)

1

|[cos(n0-zsin0)-ism(n0-zsin0)]d0 T~ 2;r J

\cos(n-zsin 2tc j„

=

— 2% 1

*

2 cos(«0 1

i

- zsin )d

f sin (n-z sin )d

2?r j

i

2k

85 That

is,

*

= — J cos(«0 - zsin 1

J„(z)

(n

= 0,±l,±2,...).

iti

11.

fa)

The function f(z) unit circle

is

C.z- e'*

analytic in

(-it

<



<

some annular domain centered

it) is

at the origin;

and the

contained in that domain, as shown below.

For each point z in the annular domain, there

n=0

is

n=l

a Laurent series representation

<

where

(n — ?T

-*?T

and

— 7F

Substituting these values of a„ and b„ into the series,

or

/(a.J-J/e-w.J-lJ/,,.,



we then have

= 0,1,2,...)

86 (b)

Put z

= e' e

in the final result in part (a) to get

or

/(«")

If

k(0)

= ^- J/(e* )d* +

= Ref(e' e ),

w(0)

1£ J/(^)cos[n(0 -

then, equating the real parts

=



J«(0)
on each side of this

last

equation yields

+— 2 J"(0)cos[n(0-0)]
SECTION 60 1.

Differentiating each side of the representation

1-z we find

n=0

that

U

z;

az „ =0

n=0

az

„ =1

n=0

Another differentiation gives

(i

2.

— z;

"Z n=0

Replace z by

as well as in

1 / (1

its

„=o

- z) on

«z

B=1

n=0

each side of the Maclaurin series representation (Exercise

1)

condition of validity. This yields the Laurent series representation

4 = £<=!>>ZI>

«
87 3.

Since the function /(z) valid in the

To

= 1/z

has a singular point at z

Taylor series about z

=2

is

1111 2 + (z-2)

z

it

its

open disk \z-2\<2, as indicated in the figure below.

find that series, write

to see that

= 0,

2

l

+ (z-2)/2

can be obtained by replacing z by -(z - 2) / 2 in the

known expansion

1

(lzl<

1).

Specifically,

l_ly|-

(Z-2) T (Iz-2I<2),

z

ifA.

Z

R-0

2

J

or

(lz-2l<2).

Differentiating this series term

z

by term, we have

- 2r = E^tH" + J J = l^n(z y l

7

1

-~l

l)(z

- 2)"

(lz

- 2I< 2).

n=0

Thus

-| 4.

(

_ 1)>+1)

^y

Consider the function defined by the equations

V-l f(z) = 1

when

z

* 0,

when z = 0.

(lz-2l<2).

When

z

* 0,

/(z) has the power series representation '

/(*) =

2

\

3

+ £ + l. + £. + .^ -1 = 1+ ,

l

z

2!

1!

Since this representation clearly holds

z —z +— 2!

3!

when

z

=

too,

it is

+••

3!

actually valid for all z.

Hence/

is entire.

Let

C

point

be a contour lying

w=1

in the

w = z,

to a point

as

open disk \w -

shown

w According to Theorem

1

II

<1

in the

w

plane that extends from the

in the figure below.

plane

in Sec. 59,

we can integrate the Taylor series

1

representation

(lw-u
w

„=0

term by term along the contour C. Thus

dw "

n=0

f

— = [— =

n=0

But

[Logw] z = Logz-Logl = Logz 1

and n+l

4

J

(w-l)"=J(w-l)"rfw =

n+

l

(z-1) Jl

n+l

n+l

Hence

(-D

^n + and, since (-1)"

-1

^

l

= (-ly-^-l) 2 = (-1)" +1 Logz =

,

this result

X Lii— («-!)" n=l

(Iz-lkl);

n becomes

(Iz-lkl).

— 89

SECTION 61 1.

The

singularities

of the function /(*)

= -^L-

are at z

=

find the Laurent series for/that is valid in the punctured disk

,

± L The problem
here

is to

below.

y i

°

\

x

/

4

-i

We begin by recalling the Maclaurin series representations

2!

1!

Qz\<

3!

oo)

and 1

l-z which enable us

= l + z + z 2 + z 3 +---

Oz\
to write

e

z

= l + z + ^-z + -z 3 +2

2

(\z\< oo)

6

and 1

(\z\
Multiplying these last two series term by term,

2 l

z

-

+\ ,

we have

the Maclaurin series representation

= l + z + ±z 2 +*z 3 +2

-z

6 2

-z

3

-

4

z +-

5

1.2

which

is

valid

side of the

when

\z\<

The desired Laurent

1.

series is then obtained

by multiplying each

above representation by -:

1

z(z'

+ l)

1 = -+l--z

z

,

2

5

2

z

6

+••

(0
90 4.

We know the Laurent series representation

1117

1 2

3

z sinhz

from Example tells

6

z

Expression

2, Sec. 61.

-

of

us that the coefficient

(0
Z

360

z

(3), Sec. 55, for

the coefficients b„ in a Laurent series

in this series can be written

z

dz

2m where

C is the circle

lzl=

1,

taken counterclockwise. Since bx

'

6.

The problem here

is to

is

= - \,

then,

6

c z sinhz

use mathematical induction to verify the differentiation formula

[/(^(z)]

The formula

lc z sinhz

(n

"=E Af \z) g

(n)

when n = 1

clearly true

[/(z)s(z)]'

k

k)

(n

(z)

since in that case

it

= l,2,...).

becomes

= /(z)s'(z) + f(z)g(z).

We now assume that the formula is true when We start by writing true when n = m +

n=

m

and show how, as a consequence,

1.

UWz)]

(m)

(m+1)

={[f(z)g(z)]'}

=mz)g'(z)f

=1 *=0

m)

(m)

,

=[f(z)g

(z)

+ nz)g(z)]

l(m)

+[f(z)g(z)]

(t)

r / (^

(m - t+i)

/

(^)+s

(m - t+1)

(t)

(z)^

(z)

V* /

= /(z)*

(m+1)

(z)

+

£ *=1

'm^

(

m

N

/W( z )^('» +1 -^(z) +

n+1)

/

('

(z)^(z).

it is

But

m

frrC

ml

+ 1)! kl(m + l-k)l

ml

(ro

|

k

,

k-l)

(*-l)!(m-* + l)!

kl(m-k)l

'm + V <

k

>

and so

inz)g(z)r

+i)

n+i

=nz)g°

y

\z)+

z

m

or

[/(^(z)]

The

/(z)

is

Write g(z)

It is

m +r f\z)t«-

,

k

\z).

an entire function represented by a series of the form

f(z)

(a)

=iT

now complete.

desired verification is

We are given that

(m+i)

= z + a2 z 2 +ajz 3 +-'-

(lzl<°°).

= f[f(z)] and observe that

straightforward to

show

that

/(z) = /'[/(z)]/'(z),

= /" [/(z)][/'(z)] 2 + f'[f(z)]f" (z),

g"(z)

and

3

g"\z)

= /'"[/(z)][/'(z)] + 2f(z)f"(z)f"[f(z)] +

f

t/(z)]/' (z)/" (z) + /'[/(*)]/"'(*)

Thus

S(0)

= 0,

^'(0)

= 1,

g"(0)=4a2

,

and

g'"(0)

= 12(a22 + a3 ),

and so /[/(z)]

= z + 20.Z 2 + 2(4 + a,)z 3 +-



(lzl< °°).





)

.

92 Proceeding formally,

(b)

we have 2

/[/(*)]

= f(z) + fljC/Cz)] + «,[/(z)]

3

+- •



= (z + fljZ 2 + OjZ 3 +•••) + (<*2Z 2 + 2^z 3 +-

2^ +

=z+

2(a*

+ a3>z 3 +-

• •

)

+ (fl3z 3 +•

• •

• •

Since

(c)

sinz

=z-



+• • •= z

3!

+ Oz 2 + f - - V+-

=

the result in part (a), with Oj

and a3 =



(IzKoo),



6)

V

-—

,

tells

us that

6

sin (sinz)

8.

= z-— z 3 +---

(lzl< ~).

We need to find the first four nonzero coefficients in the Maclaurin series representation

— coshz

This representation

is

(\z\
= E^z" t^n\

2

\

numbers

valid in the stated disk since the zeros of coshz are the

The series (/i = 0,±l,±2,...)» the ones nearest to the origin being z = ±^-i2 J contains only even powers of z since coshz is an even function; that is, E2tt+l =0 z

= \— + n7t\i

(n

U

= 0, 1, 2,

. .

.

).

To

find the series,

coshz =

1

+— + — + —+•••= 2!

into

1.

The

4!

l

+ -z

series

2

+—z 24

2

6!

4

+"+—z 720 6

(lzl< °°)

result is

,

=1 coshz or

we divide the

1

2

z

2

+

— 5

24

4

61

z

6

z

720

+••

I

V

n>

i^ lzl<—

(i

2

\ I,

93 1

1

,

=1

2

z

coshz

2!

+—z 4

61

4!

6!

5

6

f.

z +•••

I

.

lzl<

V

—n 2

Since

_!_ = £ coshz this tells

+

^ 2!

+ Jl z 4 + JSl z «+... 4!

f|

6!

V

us that

£0=1,

E2 = -1, £4 = 5,

and

E

6

=-61.

z |
2

.

94

Chapter 6 SECTION 64 1.

(a)

Let us write

_i- = I.-i- = I(i- z + z z

+z

The residue

(b)

z

at

z

1

+z

2

3 2 -z + ...) = i-l + z -z +...

z

(0
z

= 0, which is

the coefficient of -, z

is

clearly

1.

We may use the expansion = l-

cosz

2!

4!

1

1

6!

to write

ZC

m

1

osH = z^l--. ? + -. 7 --. 7 + 1

1

1

111111

"\

= z--.- + -.-3---

...J

7

-f.

(0
The residue

(c)

at z

= 0,

or coefficient of -, z

is

now seen to be -— 2

Observe that

z-sinz

1.

1

= -(z-smz) = -

z

.

.

Since the coefficient of

-

T4

3!

5!

(0).

*'j

r

z

z

T2

5!

3!

in this Laurent series

is

0, the residue at z

=

is 0.

z

(d)

Write

cotz 4

z

and

cosz

1 4

'

.

sinz

z

recall that

cosz

=1 1

zT

2

2!

and

_ ~

T4

+z ,

4!

,

=1

T z

2

2

+

-T z

4

24

(lzl<

~)

95 Dividing the series for sinz into the one for cosz,

cosz

z

z

1

we find that

3

Thus cotz 4

z

Note z

_

I

"z

4

fl

z

U

3

+

T5

11 11 + '-)-7~37~V~z -

that the condition of validity for this series is

= nit(n = 0,±l,±2,...).

now evident that

It is

4

due

(e)

to the fact that sinz

has residue

z

(0

—45-.

at z

<&<*).

=

when

= 0.

Recall that

=z+

sinhz

|y

^+

+

... (|

z
and 1

= l + z + z 2 + -.i

(lzl<~).

1-Z There

is

a Laurent series for the function

sinhz

that is valid for

and

1-z


1.

To

1

find

(

.

it,

we

first

1

^

multiply the Maclaurin series for sinhz

2

1

z + -z +

1

6

120

3

,

5

z +•

:

3

z

1

+-z

5

+•

5

z +-

7

z

+ tz o

3

+•••

(0


.

96

We then see that sinhz 4

z

(l-z

)

z

In each part,

(a)

To

1

3

•••

47—^ —z (1

at

z

=

Jc

—^

dz,

is

— 6

J

C denotes the positively oriented circle

evaluate

(0
6 z

z

This shows that the residue of

2.

—17 + -•- +

=

2

lzl= 3.

we need the residue of the integrand at

z

= 0. From

the Laurent series

exp(-z) 2

_ —

1

^

1

_2

z

2

+

1!

^

we see that the



3

z

z

2!

1 ) +...=_ 2

™P^ dz = 2n -l i



jc

11

,

z

2

valid for


1

)

= -2jti

11

1

—+ 1

1

1

1!

2!

3!

z

°°, tells

A

.

.

,

(0
3!

.

exp^jdz, we must

Z

= Z2 + — +

is

2!

,

+...

find the residue of the

= 0. The Laurent series

,

which

z

1!

z

1

+

-1. Thus

is

Likewise, to evaluate the integral integrand at z

z

J

1

1

,

3!

required residue

lc

(c)

,

+

1

4!

11

1 2

z

us that the needed residue

is

— 6

.

Hence





z

97 z -=

r

(d)

As

for the integral

JC 7*l

z

+1 —dz, we need the two residues — 9r -2z z+1 z -2z

z

2

one

at z

=

and one z

at z

z(z-2)'

z-2)



z+

To

.

{

2A

z

z

2

2

(z-2) + 3

1

%

1 2

-

that the coefficient of

z-2

2

+ (z-2)

when

z-2j

2V

(z-2)

l

+ (z- 2)/2

2

|

z-2

2\


- 21 < 2, and

note that the coefficient of

in this product

z-2 is

—3

.

Finally, then,

by the residue theorem,

Jc

In each part of this problem,

(a)

If/(z)

-,

l-z

1 _2 / when

-

2

z

-2z

V


2

2)

C is the positively oriented circle

lzl= 2.

then

:

—4=-—-;

=-

\zj

z

in this last

1

z-2

valid

z) l-(z/2)

= 2, we write

obtain the residue at z

1

is

can be found by writing

z

z(z-2)

which

=

when 0
valid

is

2 z)

2

\

product

z

+1

f

is

+1

= 2. The residue at

z(z-2)

which

of

1.

-z

This

z

1-z

tells

us that

f(z)dz

r

= -- v l+z 3 +z 6 +••• y

= 2mRes <=°

z

=--— z

= 2ni(-l) = -2 m. \ f(-) \z) z

z

2

-

(b)

When

f(z)

1

= 1

+z

>

2

we have

Thus

= 2mRes 4"/f- ] = 2m(0) = 0.

f

(c)

If

/(z)

= -,

it

follows that

f ic

Let

~ /[-) = -• 2

Z

C denote the circle

f{z)dz

lzl=

VZ/

= 2m Res -^/f - ] = 2/ri(l) = 2m. z z =o

1,

z

z

=

T—

The Maclaurin

(b)

Referring to the Maclaurin series for e

Now

the

-

\ZJ

taken counterclockwise.

(a)

series e

Evidently, then,

Z

(lzl< °°) enables us to write

in this series occurs

z

once again,

when

let

n-k = -l,

us write

or k

= n + l.

So, by the residue

z

theorem,

l> expf-ljz = 2m-i—

Jc

The

vz/

final result in part (a) thus reduces to



+ l)!

(/j

= 0,1,2,...).

99 5.

We

are given

two polynomials P(z)

= a +a z + a2 z 2 +--- + a„z''

<2(z)

= b + blZ + b2 z 2 + --- + bm z m

(a„*0)

l

and

(bm

* 0),

where m>.n + 2. It is

show

straightforward to

z

'"" 2

pQ 1 z> -

J_ 2

"

that

Qz

fl

+ a z "" 3 + a2* m

V" +

2(1 / z)

~4

i

1

fri*"*"

+

+

V"

2

w """ 2

+ fl^ + +K •





1

• • •

Observe that the numerator here is, in fact, a polynomial since m -n - 2 > 0. Also, since bm * 0, the quotient of these polynomials is represented by a series of the form 2

do+diZ + dzZ + ••-. That is, P(l/z)

1

1

and

L —r we see that

z

P(\lz)

that

C is

^

.

rt

z

2

(0
= 0.

2(1 /z) all

of the zeros of Q(z)

positively oriented.

plane except at the zeros of <2(z), obtained that

P(z) dz ,c

j

,

has residue

Suppose now that assume

j

Q(z)

lie inside

Since P(z)/Q(z)

is

a simple closed contour C, and analytic everywhere in the finite

follows from the theorem in Sec. 64 and the residue just

it

P(l/z) = 2mRes J_2 = 2^*0 = 0. z=0 z 'ea/z)

C is negatively oriented, this result is still true since then

If

Piz) J__

icQ {z)

P(z)

f

J-c

dz

= 0.

Q(z)

SECTION 65 1.

(a)

From

the expansion

(Izl<~), 1!

we

2!

3!

see that

ii + ,n =z+1+ ^,^11 _._ + _._

zexp|

_j

..

(0
100

-

The principal part of zexpl

at the isolated singular point z

I

=

is,

then,

z

+ 2!'z and z

(b)

The

=

is

+ "'

3!Y

;

an essential singular point of that function.

isolated singular point of

is at 1

involves powers of z 2

z

z

+z

+ 1, we begin by

= -1.

Since the principal part at z

= -1

observing that

= ( 2 + 1) 2 - 2z - 1 = (z + 1) 2 - 2(z + 1) + 1.

This enables us to write

e

_ (z+lf

Since the principal part

is

,

The

point z

t

The

1

z+1 the point z

,

+1

is

= -1

a (simple) pole.

is

the isolated singular point of £ -£ ) and z

we can write

^- — + — -•••1 = 1- — + — -•••

1

3!

z\,

5!

(0
5!

3!

J

principal part here is evidently 0,

and so z =

is

a removable singular point of the

smz

.

function

(d)

+!

m

=

z

-

—^— z

sinz

The

!)

z+1

1+z

(cj

-% +

.

isolated singular point of

cos z is

z

= 0.

Since

z 4

2

cosz

1

—I

the principal part is

(

z

z

I

2!

4!

.

= I_± + £l_... ;

This means that z

z

=

(0
4!

2!

COSZ is

a (simple) pole of

(e)

Upon

writing



^—-j

(2

=

— z)

isolated singular point z

(of order 3).

.

z

z

we

find that the principal part of

(2-z)

(z-2)

=2

is

simply the function

itself.

That point

is

3

at its

evidently a pole

..

101 (a)

The singular point is z = 0. Since

1-coshz

(

1


°°,

6

we have

4!

2!

V

when

,4

_2

1- !+£.+£.+£_+.. 6!

Here the singular point

l-exp(2z)_

is

also z

= 0.

(

2z

I

1!

1

= _2_ J__2^ ll'z

when


°°,

we have

The

3

Since

2

V

_1

and

4!

3!

2j_

2

exp(2z)

= e 2(z -V=e

1

+

js

z

2V 2V

3

2 73

2!

2!*z

m=3

ex P^ z ^ (z-1)

singular point of

6!

2

2

B=

"\

5!

_j__2^_2^

3!'z

IT*"

4!

3

=

— 4 3

3!

(c)

4!

z

m = l and B = —— = -— 2!

(ft)

2\

J

= \ jhe Taylor series 2

2(z-l)

l

2 (z-l)

2

3

.

2 (z-l)

2!

1!

3

(Izl< oo)

3!

enables us to write the Laurent series

exp(2z)

(z-1)

Thus

2

m = 2 and

Since / is analytic at Zq,

f(z)

Let

2

_^

+— L(z-lr

1!

1 •

z-1

2

2

2!

3!

2 2 + — + -J-(z-l)+-

(0
2

= e 2 — = 2e 2

it

= f(z

.

has a Taylor series representation

)+^-(z-z

g be defined by means of the equation

)

+ ^^-(z-z

2

)

+-

(Iz-ZoK^,).

102 (a)

Suppose that /(z

g(z)

*0. Then

)

1

=

/<0 + ^(*-*) + ^U-*o)

Z-Zn

This shows that

(b)

£ has a simple pole

at Zq

Suppose, on the other hand, that /(z

)

+-

(0
°;

2!

1!

2

,

with residue /(z

l
).

= 0. Then

2!

1!

(0<\z-z \
is

just 0, the point z

=

is

a removable singular

point of g.

4.

Write the function 3_2

=

/(Z)

(a>0)

(?W

as

/

_ =

(z)

4Kz) /

-n3

3 3_2

where

0(z)

=v

8a z

.

(z

Since the only singularity of 0(z)

z

*M(Z -

Hz) = ««0 + about z

is at

= -ai,

ai)

+

n3

-

+ «T

0(z) has a Taylor series representation

^(z- ai? +

(lz-ai'l<2a)

= ai. Thus

/(z)

1

=

(z-m)

3

0(ai) + ^rr^Cz

- fl +

Now straightforward differentiation reveals

^ (z) =





4

3

16a /z-8fl z ;

(z

„4

+ ai) ,

2

- ai) +

(0
that

16a (z) = ,

.

and

(z

2!

1!



2

3

(z

(z

-4fl/z-a

+ ai)

3

2 )

Consequently,

Q(ai)

= -a 2 i,

0'(a/)

= ™, and

"(ai)--i.

This enables us to write

/(Z)

The

=

(z-aif [~

a2i

principal part of /at the point z

ill

- ai

z

~

fll) (Z m)2 + " (z f ~ ~i " ']

= ai

is,

(0

< lz - ml < 2a).

then,

all 2 (z - ai)

a (z

2 i

- ai) 3

'

SECTION 67 1.

(a)

The function f(z) where that

(b)

If

z

(/>(z)

=1

is

+2 z~l

Z

= z 2 +2, and

has an isolated singular point at z

observing that 0(z)

a pole of order

analytic

Writing /(z)

and nonzero

at z

=

z-1 = 1, we see

m = 1 and that the residue there is 5 = 0(1) = 3.

we write f(z)

we

Hz)

=

zrf,

lz + 1

where

0(z)

z- -

=

8'

1

see that z

point,

= --

is

a singular point of /. Since 0(z)

/has a pole of order

m=3

there.

The

B _ r(-V2) _

residue

is

analytic and nonzero at that

is

3 16'

2!

fc)

is

= 1.

The function expz z has poles of order

2

expz

+ ;r

2

m = 1 at the two points B

exp TO

i

2 to

and the one

at z

(z-/n)(z +

= -ni

is

z

7ri)

= ±m. The residue at

_ -1 _ ~ In 2 to i

'

z

=

to is

104 1/4

2.

(a)

Write the function /(z)

=

(I

z

/(z)

Hz) =z ^7. z+1

The function 0(z)

where

is

zl

+1


> 0,

< are z < 2 it)

= z m =e 4

analytic throughout

its

as

(lzl>0,0
domain of

definition, indicated in the

figure below.



Branch cut

o

-1

Also,

jl

/

J'osH) Ja«i+<*) it it 1+i , n = (-1) n i/4 _ =e 4 = £4 = e iwM = cos— + zsin— = —f=-*0. .

.

<9(-l)

4

4

This shows that the function /has a pole of order being

5 = 0(-l) =

1

V2

m = 1 at z = -1, the residue there

+/

V2'

(b)

Write the function /(z)

From

= Logz 2 2 (z + D

as

/(z) has a pole of order

this, it is clear that

differentiation then reveals that

Res

Logz & 2

(z

+ l)

2

=

=

tf

+ 2i 8

m=2

at z

=

i.

Straightforward

105 (c)

Write the function .1/2

/(*) =

2

(ld>0,0
2

+D

(z

as .1/2

f(z)

= (z-i)

=

where

2

(z

2

+

Since

f

W

2(z

+ /) 3

and

r 1/2 - *-''*/4 -

_i_

_L

7

Res

f 2

*=<~(z

(a)

,-i/2

_

1/2 1

„„2

+ l)

j*/4



= ft'(t>

_

1

' .

'

1

8V2"

We wish to evaluate the integral 3

+2

3z ,c

where

C is the circle

the singularities z

- 21 = 2 taken in the counterclockwise direction. That circle and ± 3i of the integrand are shown in the figure just below.

\z

= 1,

dz,

2

(z-l)(z +9)

,

Observe that the point z and that

= 1, which

is

the only singularity inside C,

the integrand

3

3z + 2 Res2 (z-l)(z +9)

3z z

3

2

+2 +9

1

2

According to the residue theorem, then, 3z

3

+2

l z-\Z^9) dz = 1M {l) =ni i

-

is

a simple pole of

106 (b)

C is changed to be the positively oriented circle

Let us redo part (a) when in the figure below.

= 1, ± 3/

In this case, all three singularities z

already

know from

Izl

= 4, shown

of the integrand are interior to C.

We

part (a) that 3

3z + 2 Resz * =1 (z-l)(z +9)

It is,

moreover, straightforward 3

to

+2 (z-l)(z' + 9) 3z

Res <= 3i

1

2

show that 3

+2 (z-l)(z + 3/). 3z

15 + 49/

12

and 3

3z

Resz =-3,-

3

+2

_ (z

The residue theorem now

+9 ) tells

+2 _ i) (z _ 3/ ) (z

+2 _ /l 5 dz = 2m\- + 2 } c( z -l)(z +9) V2

4.

(a)

Let

C

,

15 + 49/

12

denote the positively oriented circle r

integral

dz

-rJc z (z

12

j=-3i

us that

3

3z

f

15-49/

3z

— has singularities at

+ 4) y

z

=

Izl

+

15-49/"\

"

= 2, and

=6, m.

note that the integrand of the

and z = - 4. (See the figure below.)

.

107

To

find the residue of the integrand at z

= 0, we recall the expansion

(lzl
and write

_

1

1

1 3

z\z + 4)

4z

_l

_n-3

+ (z/4)

Now the coefficient of -

here occurs

when n =

2,

and

(0
we see that

z 1

1

z\z + 4)

64

Res *=°

,

Consequendy,

Jc

(b)

Let us replace the path at

-2 and with radius

z\z + 4)

C in part 3.

It is

(a)

*=°

-4,

by the positively oriented

^

us that z

0(-4) = -1 /

64.

= -4

centered

— 64

z\z + 4)

_

(z)

Tw-T+Ty tells

+ 21 = 3,

we write

1

This

circle \z

that

Re S

the residue at

32

shown below.

We already know from part (a)

To find

\64)

is

...

.

where

a simple pole of the integrand and that the residue there

Consequendy,

J<^ 5

1

Hz)= 7-

(z+4)

{64

64J

is

108

cosh 7tz dz

f

:

Let us evaluate the integral

J^ C

TIT Z^Z

—+ *

7T"» ij1

= 0,±i of

All three isolated singularities z

C is

wnere

the positively oriented circle lzl= 2.

1

the integrand are interior to C.

residues are

=— — +1 z(r+l)

_

s

z=o

_

cosh nz

cosh nz

Res

Res

=1,

5

z

—+

cosh 7cz =

2

z(z

l)

.

= cosh tfz z(z + i)_

1

.~2'

and



_ cosh nz = cosh tez Res 2 +l)

*=-<"z(z

_ 1 .~2'

z(z-i)

z=-i

Consequendy, coshflzdz

r

=

,c z(z

6.

In each part of this problem,

(a)

It is

straightforward to

f„\ = lf/(z) This function

2

/,

„ = 2m

+l)

1

+—1 +—1 2

V

\

.

.

= 4m.

2,

C denotes the positively oriented circle

show

Oz + 2)

lzl= 3.

that

/n_ + 2z) ZllJ-zd-^ + Sz)2

2

(3

1

111611

z(z-l)(2z + 5)'

-y /^-j has a simple pole at

z

= 0, and



+ 2)2 dz = 2/ri Res *=° Icz(z-l)(2z + 5) (3z

(b)

,

= 9ni.

= 2/ri <2,

z/J

Likewise,

w

if/(z)= '

The function

z\l-3z) (l+z)(l + 2z

\

)

Jc(i + z )(l

)

*-3 2 -*-M-l z

U;

has a simple pole at z

Z(1 " 3z) f

±en

4

+ 2z


4 )

= 2;riRes *=°

z(z + l)(z

= 0, and we

V

1

4

+2)

find here that

= 27ri| -- = -3;ri. |

The desired

109 (c)

Finally,

zV"

The

point z

=

=

+ z3 )

.

The

residue

is

0'(O),

where

\z

z


^ =7f

1 Jl -jf -

a pole of order 2 of

is

±f

then

-

+ z3

1

'

Since (l

+z

3

z

)e

-e l 3z 2

\z)

the value of

0'(O)

f

'c 1

+ z3

So

is 1.

dz

3\Z

= 2m Res

7 fil)

=2jti

W =2m

-

SECTION 69 1.

(a)

Write

esc z

=

= -^4, sinz

where

/?(z)

= 1 and

= sin = 0,

and

tf'(0)

q(z)

= sin z.

q(z)

Since

p(0) z

-

= l*0,

^(0)

must be a simple pole of cscz, with residue P(0)

=- = 1

q'(0)

(b)

= cos0 = l*0,

From Exercise

2, Sec. 61,

CSCZ

we know that

^

= - + — Z+ z

1

1

.

3!

1

1

T 2

L(30

3 z +-

1

Since the coefficient of residue being

1.

-

here

is 1, it

(0
it).

5!.

follows that z

=

is

a simple pole of cscz, the

(a)

Write

z- sinhz = 2

p(z) 6

^)'

z sinhz

^) = *- sinh *

and

tf(z)

=z

2

sinhz.

Since

p(Ki)

it

= m*0,

and

(m) = j? * 0,

follows that

Rc „ *=*«'

(b)

q(m) = 0,

z-sinhz 2

=

z sinhz

p(ni) q'(,ni)

= m__i_ n1 k

Write exp(zr)

p(z)

=

6

sinhz It is

~q(zj'

^= )

exP(*')

?(z)

= sinhz.

easy to see that

«-*»

sinhz

r

q\7ti)

*=-*/

s jnhz

P

g'(-»0

Evidently, then,

Res

exp(

£

+ Res

expU0 = _ «Pff*) + CTpW*) 2

sinhz

(a>

sinhz

2

Write

/(z)

= ^7T'

where

P(z)

= z and

o(z)

= cosz.

Observe that

+ n«:j = «(f .2

(n

= 0,±l,±2,...).

Also, for the stated values of n,

p\- + n7c\ = - + n7t*

and

^| + «^ = -sin^| + n^ =

(-l)

fl+1

^0.

Ill

So

the function f(z)

=

has poles of order

cosz

m = 1 at each of the points r

z„=-+nn ~"

(n

2

The corresponding

(b)

= 0,±l,±2,...).

residues are

Write

tanh ^

Bothp and q

=

where

= sinhz and

p(z)

q(z)

= coshz.

are entire, and the zeros of q are (Sec. 34)

=

z

I

^- + nn nw

2

In addition to the fact that

J ~+

= 0,±l,±2,...)

= 0, we see that

jf

(

(n

\i

f

J

P

i[j

+ nn

)*)

= sinh

= icosnn = /(-!)" *

(^ +

and

= Sinh

+

(f '

*'((f

So

the points z

=

+



= 0,±1,±2,...)

residue in each case being

p\

[

+ n?n =

— + nn

}i

I

)

* °-

are poles of order

m=1

of tanhz, the

Let

C be the positively oriented circle

shown just below.

lzl= 2,

TO

(a)

To

evaluate the integral

tanz

and

Jc

= £^£l

tan z dz ,

72

we write the integrand as

where

)

recall that the zeros of cos z

p(z)

are z

namely z = ±tt/2, are interior tanz interior to C. Observe that

zeros,

Res tanz = z=t/2

n(nll)

=-1

' ,

x

= sin z and

= -^ + n;r to C,

and

tf(tf/2)

q(z)

(n

= cosz,

= 0,±1,±2,...). Only two

and they are the isolated

p(-x/2) _ Res tanz _ „ qX-npi)

_



of those

singularities of

L

Hence tanz*fe

f

(fe)

The problem here

is to

= 27ri(-l-l) = -47ri.

evaluate the integral

.

Jc sinh2z

To do

this,

we

write the

integrand as

— — = £Q \

sinh2z

Now

sinh2z

=

t

where

p(z)

= l and

g(z)

= sinh2z.

q(z)

when 2z = nm (n = 0,±1,±2,...),

or

when



(/z

= 0,±l,±2,...).

2

Three of these zeros of sinh2z, namely singularities

show

and±-^-> are inside

of the integrand that need to be considered here.

that

Res *=o

_J_ = £W = _J_ = I2' sinh2z

q'(0)

2cosh0

C

and are the isolated

It is

straightforward to

113

p(m/2) _

1

Res

1

q'(m/2)

*=»/2sinh2z

1

1

2cosrc

2cosh(;ri)

2'

and

= PtE!2± =

1

Res z =-«v2

S inh2z

1

2cosh(-;rz)

q'(-jti/2)

1

=

2cos(-;r)

=

*

2'

Thus

Jc

5.

CN

The simple closed contour

Within

Q,

is

the function -=

z

z

To

find the residue at z 2, Sec. 61, and write

1

z sinz

U

s inh2z

as

shown

-to.

2

2

in the figure below.

has isolated singularities at

smz

=

and

z

= ±nn

(n

= l,2,...,AT).

= 0, we recall the Laurent series for

- + —z + = -t-cscz = -ri 2 2 1

1

z

z

1

-= + 3

z

1

1

6 z

+

1

1

3!

\z

1

3

z +-

z

(3!)

5!

1

z+-

2

(3!)

1

1

cscz that was found in Exercise

5!_

(0
114

This

tells



-=4 z smz

us that

1

Res «=°

As for the points

z

=

has a pole of order 3 at z

= ±nn

(n

= 1,2,



= ^tt, ~Y~ z smz q(z)

. .

.

,

z

i

and

that

1

smz

6

N), write

where

p(z)

= 1 and

q(z)

= z 2 sinz.

Since

p(±rut)

it

= 1*0,

q(±nn) -

q\±nn) = n 2 n 2 cosnit = (-l)"n z 7F z

and

0,

56

0,

follows that

1

Res

*=±»* z sinz

(-1)"

(-1)"

1

2

(-1)"«V

~~

(-1)"

nV

So, by the residue theorem,

cfe J,c"

Rewriting

this

=

2m 6

z sinz

equation in the form

A (-1)" il and recalling from Exercise tends to infinity,

n

+1

2

=

2

points

4i'-kz sinz

7, Sec. 41, that the value

of the integral here tends

to zero as

N

we arrive at the desired summation formula:

it

C

dz

f 2

12

f (-ir The path

g

tt

here

is

±2 and ±2 +

n

1

2

2 = n

12

the positively oriented boundary of the rectangle with vertices at the i.

The problem

is to

evaluate the integral

dz c (z 2 -l) 2

+3'

115

The

polynomial

isolated singularities of the integrand are the zeros of the

= (z 2 -l) 2 +3.

q(z)

Setting this polynomial equal to zero

the property z also the

2

= l±V3i.

It is

2

and solving for z

,

we find that any

zero z of q(z) has

straightforward to find the two square roots of

two square roots of 1-V3i. These are the four zeros of

q(z).

1+ V3i and

Only two of those

zeros,

z,.^-.^*L

d*+L,

and

he inside C. They are shown in the figure below.

To

and -Zq

find the residues at z

_

1 2

(z -l)

2

P(z) ,

+3

This polynomial q(z)

,

we write the

integrand of the integral to be evaluated as

where

= 1 and

p(z)

q(z)

= (z - 1) +3.

q(z)

is,

of course, the

same q(z)

p and q are analytic at Zq and that p(z 2 q\z) = 4z(z - 1) and hence that ?'(*<,)

We may conclude, then, that

z

)

* 0.

as above; hence q(z

Finally,

it is

)

= 0.

Note, too, that

straightforward to

show

that

= 4z (4 - 1) = -2V6 + 6V2/ * 0. is

a simple pole of the integrand, with residue

P(z

)

q'(z

)

_

1

-2V6+6V2i'

Similar results are to be found at the singular point -Iq.

To be

specific,

q X-Zo ) = -q '(z„ ) = -7(^) = 2V6 + 6V2i * 0, the residue of the integrand at -Zq being

q'i-Zo)

2V6 + 6V2/*

it is

easy to see that

(

Finally,

by the residue theorem,

^

We

Z

2

2

- 1) + 3

are given that /(z)

1

= 2ml

f

\-2sf6 + 6V2i

= l/[?(z)] 2 where ? ,

is

1

+

=

2V6 + 6V2i J

analytic at z*,

__ 2^2

= 0, and m = l at

?(z

These conditions on 4 tell us that q has a zero of order #(z) = (z - Zq)^(z), where £ is a function that is analytic and nonzero

)

?'(z

v

at z

;

+ 2g'(z

).

and

So /has a pole of order 2

=

-^-T

at z

,

= (z -z

q'(z)

Then, by setting z

=z

)g'(z)

+ g(z) and

in these last

q'(z

-M^.

we know that

)#(z),

= (z-z

—^.

and

Res/(z) = 0'(zo ) =

But, since q(z)

=

where

,

)

?"(z)

= (z-z

two equations, we find

= g(z

and

)

?"(z

)

,,

)g (z)

that

= 2s'(z

).

Consequently, our expression for the residue of /at z can be put in the desired form: Q

Res/(z) = - q "^\

(a)

To

find the residue of the function esc 2 z at z

csc z

~ r

<

sn2

»

.

= 0, we write

wnere

^(z)

= sinz.

W(z)J Since q

is entire,

#(0)

= 0, and

q'(0)

= 1 * 0,

the result in Exercise 7 tells us that

2 Rescsc z = - q

[
3

= 0.

Hence

this enables

us to write

f(z)

)*0.



(b)

The residue of the function

^-—j

(z

+z

=

at z

can be obtained by writing

)

1

-,

{z

Inasmuch asqis

+z

2

)

-

q(z)

= z + z2

[q(z)f

entire, q(0)=0,

Res

where

and q'(0)

=-j£<°>_ = - 2

1

(z+z

= 1*0, we know from Exercise 7

1 )

[«W

.

that

.

118

Chapter 7 SECTION 72

1.

To



evaluate the integral

we

-, J— x +1

integrate the function f(z)

closed contour

= Z

n

shown below, where

R>L

y c.

We see that

J

_s x

2

+l

J

^

2

z

+l

where

5 = Res —r—— = Res «'

+1

z

(z-f)(z +

Z + l'_U

Thus

f


f

Now if z is a point on Q, 2

lz

+ ll>llzl 2 -ll=rt 2

-l;

and so

E.

R 2 -l

J,

-»0

as

—dx

n =—

i_J_ R2

Finally, then

1 J _

dx 1 —7 = * + 2

1

1 tt>

or

:

Jjc

2

+1

2

2i

+1

around the simple

-

119 °°

2.

The

integral

dx

— — (x +1) 5

J J

can be evaluated using the function

rrj"

_((jc

—+ 1

2

+ l)

5-

and the same

1)

Here

1.

2+

= 27riB,

i\2 fc( z 2 + i) .2

,

1

B = Res

*"<~V + 1)

Since

.

2

Hz)

1

.

= 12—^ ^' + l) 2 TT (z-i) 2

.

where

2

,

x

0( z )

=

B = '(i) = -7-

that

,

— 1

U+

(z

we readily find

=

(z

simple closed contour as in Exercise

where

f(z)

and so

4i

fife

J „0t 2 ?

If

z

is

a point on

CR we know

+ l) 1)

(*

2

^(z J<

2

from Exercise

,

2

lz

2

+l)

2

1 that

+ ll>i? 2 -l;

thus

b The

desired result

<—inR

dz az

t

2

(z

is,

2 + l)

We begin the evaluation of the fourth roots of -1,

_%

dx

i(x

[

2

+i)

2

f*

~>0

R-*°°.

as

r

or

"2'

dx

%_

JcTTif'T

by finding the zeros of the polynomial z*

and noting that two of them are below the real

shown below, where R>1,

two roots

k

~ €jit/* _

1

1

+

"V2

V2

and J3X/4

+ 1, which are

+1

consider the simple closed contour the

,2

,

then,

°r

3.

J? 3

t=

iJT/4„iJt/2

1

.

/

V

1

i

axis.

In fact, if

we

that contour encloses only

r

120

Now

where

R - Res— — ^

and

= Res—

2?,

*=*2

The method of Theorem 2

in Sec.

69

tells

us that

z,

Z

+1

and z1 are simple poles of

— :

z that

= -7^=rr-4zj 4

311(1

= -T' =TT-4 4z z

fi

2

3

Zi

since

4 z,

= -1 and

zt

= -1.

2

Furthermore,

B +B2 =-j(z +z2 ) = l

l

2

~ 2V2"

Hence t

dx

}R x*

_ n

+ l~j2

dz

t

Jc,

z*

+ i'

Since

J,

c«z

4

+l

we have °°t

dx

n

iy + l"V2'

°r

°r

dx

L

4

n

+ l"2V2-

and

+1

121

— —x—dx— —— 2

r

We

wish

to evaluate the integral

=

=

2

.

We

use the simple closed contour

l(x +l)(x +4)

shown below, where R>2.

must find the residues of the function /(z) =

We



,

:

z

= i and

z

= 2i. They

at its

+ l)(z + 4)

(z

are

B = Res/(z) = l

(z

+ /)(z 2 + 4).

6/

and fi2

=Res/(z) = *=2<

(z (7

2 Z

3/'

+l)(z + 2z). ?=2i

Thus 2

*A

f

2

2

+ l)(x' K(x +l)(x ijtf

+ 4)

+fJJc. c

z dz

+

+ 4)

or

(x R

z

is

a point on 2

lz

CR

,

2 n r z dz = --f 2 J +l)(x +4) 3 c(z 3 + lXz 2 + 4)

x dx

J

If



2

* 2

then

+ ll^llzl 2 -ll = /? 2 -l

and

2

lz

2

+4l2>lld -4l

= /? 2 -4.

Consequendy, 2

z 2

(z

and


+ l)(z 2 +4)

2

-»0 as

2

(R -l)(R -4)

we may conclude that x dx i(jf"

+l)(^ + 4)

n 3

dx J

2

(jt

2

+ l)(* + 4)

it

6

simple poles

'

122

x dx— ———5 — —5- can be evaluated with the aid of the function {(x + 9)(x + 4) f

5.

The

5

integral f

2

2

2

f(z)

2

2

+9)(z +4)

(z

2

R > 3.

and the simple closed contour shown below, where

y

We start by writing 2

x dx

z dz

+ 9){x 2 + 4) 2

J {x 2 R

Jct Z 2 (

+ 9)(^ + 4) 2

where z

2



A = Res Ti—r~n «-* (z + 9)(z +4) z

z

and

z

5,

z

= Res—z

2

5

=

+9)(z

(z

z

=-.

+4r

Now 3

t 2

1

(z

To

find

+ 3i)(z + 4)

2

50/' J?=3<

B2 we write ,

2

__0(z)

z 2

(z

2

+9)(z + 4)

x

._,

-,

2

(z-2/)

where

,

,

0(z)

:

Z"

=

2

(z

+ 9)(z + 2/)

2

Then 2?2

This

tells

13

= 0'(2/) =

200/

us that A J

R

2

x dx 2

(x +9)(x

2

+4)

z

=JL_fJ

2

100


a 2

2


a\2' + 9)(z.2a +4) .

:

Finally, since 2

3

z dz 1,c

7r/?

2

2

*(z +9)(z +4)

we find

2

2

j -»

2

as

(R -9)(R -4)

R -> 00,

that

*

2

d!x

2 2 z f 0c +9)(;c (* +

*

7T

+ 4)

2

or

100

1

U

2

etc

+9)(a:

2

+4)

2

200'

.

.

7.

In order to

show that

P we introduce

V

xdx f

= -—

2

'^(x + l)(x 2 + 2x + 2)

5'

the function

2

2

+ l)(z +2z + 2)

(z

and the simple closed contour shown below.

y

/ x

*

>

X

R Observe that the singularities of /(z) are z

= -1 - i

lower half plane. Also,

in the

if

at

i,

z„=-l + i and

their conjugates

R > V2, we see that

K

jf(x)dx + j f(z)dz = 2m(B

+B

l

),

where 2?

*

=Res/(z)=

z

«-«o

(z

= _JL + _L-I

+l)(z-z

)

io

io

Z=Zo

and

= _L_I-

B =Resf(z) =

1'

l

(z

+ /)(z +2z + 2)

10

5

Evidently, then,

£<&

f

4U

2

2

+ 1)(jc + 2x + 2)

zrfz zdz

= _£_f 5

2

2

^(z + l)(z + 2z + 2)

Since

Z*fe

r

h as

2 z

(z (z

R -»

+ l)(z l)( z + 2z + 2)

°°, this

means

zdz

r

2 i

~

(z (z

z 2

+ l)(z-z

)(z-z^)

that ,.

hm

r



xdx ;

2

2

*-*~L(x +l)(x + 2x + 2) This

is

the desired result.

=

— n 5

2

(tf

-!)(/? -V2)

2

->0

124 8.

The problem here closed contour

There

is

establish the integration formula

is to

shown below, where R >

1

= 3

z

when R > dz

JclZ 3

where the

.

1

dz

r

3

Jc«

+l

According

.

z

=r

(0

<

<

r

using the simple

that is interior

,

dz

JC2Z 3

+1

+1

iicn namely z - e

to the residue theorem,

r

.

,

+1

z=Zo

3

Z

+l

legs of the closed contour are as indicated in the figure.

representation z

=

-

1.

only one singularity of the function f(z)

to the closed contour



J

Since

Q

has parametric

R),

^z = f dr kz + l~{ r3 + l' f

3

and, since

-C2

can be represented by z J

dz

f

JcJZ 3

dz-

r

J-c2Z 3

+i

= re ann

_

+1

(0


,K*n J2 e dr

*f

J(re

,2,c/3

3

R),

_

n ,/3

Jr 3 + 1'

+l

)

f

Furthermore,

5S.V + 1

3z

2-

~

3*

i2 *' 3

Consequently,

V

U

+l

3^

2ff/3

Jca2 >

+r

But 1

2^i?

R -l

3

<. 3

•h z + i

>

3

n as R->.

This gives us the desired result, with the variable of integration r instead of x

2

dr f

JV + 1 »

t

A

3(e

a *n

-e

m

2ni

l

-e

)

3(e

—e

)

:

n

2%

3sm(2^/3)

3\3

— 125

m

and n be integers, where Let formula

(a)

x

2n

m < n.

The problem here

-dx = —esc In \

[—5i

<

+1

+1

The zeros of the polynomial z

(-l)

1/(2B)

.(2k

2n

- -1.

Since

+ i)n (k

In clear that the zeros of z

it is

2n

+1

in the

derive the integration

In

occur when z

=exp

is to

= 0,1,2,.. .,2n-l),

upper half plane are

(*

and that there are none on the

(b)

With the aid of Theorem 2

real axis.

in Sec. 69,

2m

Res-^— = <-»V + l

^

.

Putting

a=

2m + l

we can

n,

= 0,l,2,...,n-l)

c

we find

2m

that

,

= J_ c *(— )+> \_ z ''2n-l 2wc, 2n .

(*

,

1

= 0,1,2... .,n-l).

write

In

2(m-n)+l

" eX

|. (2fc

+ l)7r(2m- 2w + i)l

T

In

J

[\(2* + l)(2m + l);rl '—

= expl

^-

i-i

,

exp[-/(2fc

.

N = + l);r] ,

Thus -2m

Res*=<*z

In

(2t+l)o 2

"

+l

view of the identity (see Exercise n-l

k=0

(*

In

10, Sec. 7)

n

i

1

~

= 0,1,2

/i-l).

126 then,

—m

,2m

n-l

f ^z

2n

+

n-l

l

n

m

e

n

(c)

l-e i2an l-e :2a

e

iitm+l)

*-l -e

e

Consider the path shown below, where

n n

e'

-e

-ia

,a

nsma

R>1.

y c.

The residue theorem

x

r

tells

us that

2m

z

r

2m

54

+
r 2m

^T' +1

or

}R x Observe

2n

+l

that if z is a point

2m lz

l

k

nsina

CR

on

= /?2m

z

2n

+l

then

,

2n

and

\z

+ \\>R 2n

-I.

Consequently,

1

W'+l dz

R2m

-2»

R 2a -l

2(n-m)-l

->0;

-2n

"

/?

1-

1 2«

and the desired integration formula follows.

10.

The problem here

is to

evaluate the integral

dx 2

2 Kx -a) + l] 2

Ji

where a

is

any

real

number.

- ia

n

2i

a

e

'

We do this by following the steps below.

m n

e

i2an

ia

'

e

-l

-e~

ia

127 (a)

Let us

first

find the four zeros of the polynomial

= (z 2 -a) 2 + l.

q(z)

Solving the equation q(z)

=

for z

2 ,

we

obtain z

2

=a±i. Thus two

a + i, and the other two are the square roots of two square roots of a + i are the numbers

the square roots of 5, Sec. 9,

the

Zq

where

A = 4c? + 1.

=-j=>(^A + a+rtA-a)

Since (±z

2

=z\

)

-z

and

=a + i = a-

the

i,

a-

of the zeros are

By

i.

Exercise

,

two square roots of a - i, are

evidently

The four zeros of q(z) tells

us that z

-z

and

z

.

just obtained are located in the plane in the figure below,

and -Zo

lie

above the real axis and

that the other

two zeros

lie

which

below

it.

y •

• Zo

X

(b)





-Zo

Zo

Let q(z) denote the polynomial in part (a) and define the function ;

f(z)

1

=

[q(z)f

which becomes the integrand developed specific,

q\z) 2'(z

we

be evaluated when z = x. The method To be is a pole of order 2 of /.

note that q

is

and

entire

recall

from part

(a) that q(Zo)

= 0.

Furthermore,

= 4z(z 2 - a) and z\=a + i, as pointed out above in part (a). Consequently, ) = 4z„ (Zq -a) = 4iz * 0. The exercise just mentioned, together with the relations

zl=a + i

and

1

+ a1 = A2

g"(Zo)

WW As

in the integral to

in Exercise 7, Sec. 69, reveals that z

for the point

-z

,

=

,

also enables us to write the residue

12z

2

- 4a _

2

3zp

2

3

(4iz

q'(-z)

16iz z

)

we observe

-a _

+ i)-a a — i 16i(a + i)zo' a-i 3(a

that

= -q\z)

and

of /at

q'\-z)

= q'Xz).

Zq'.

2

__

a — /(2a + 3)

16A

2

Zo

Since q(-z

)

=

and #'(-z

2 off. Moreover,

if

B2

_

)

= -q'do) = 4i'z

0, the point

-Zq

also a pole of order

is

denotes the residue there,

_


_


3

g»frb)

f

[^'(-^,)]

_ -5 =

1

3

3

[
.

x

l^'(^o)]

Thus

B.+B2 = B.-B.= 2r'Im5. = —Vim 1

l

1

-g + f(2fl 2 +3)

1

8A2/

We now R >\z

\

Zn

integrate /(z) around the simple closed path in the figure below,

and

CR

where

denotes the semicircular portion of the path. The residue theorem

us that R

jf(x)dx +

= 27ri(A + B2 ),

f(z)dz

or

dx

Im

I;

In order to

show

-a + i(2a 2 +3)

dz 2



[
that

dx lim

f

*^~ Jc«

2

= 0,

[?(z)]

we start with the observation that the polynomial

q(z) can be factored into the form

q (z) = (z - Zq )(z + Zo )(z ~ Zo )(z + z

lz±Zol>llzl-IZoll

= /?-lzol

and

\z±Zq\>\

)•

Izl-lz^l

I

= R-\z

\.

tells

a

129 This enables us to see that

\q(z)\

4

^

(/?-lz„l)

CR

when z is on

Thus

.

1

(R-\z for such points,

and

\)'

we arrive at the inequality nR 8

8

'

(tf-IZoO

which

tells

us that the value of this integral does, indeed, tend to

as

R

tends to

<».

Consequently,

-q + /(2a 2 + 3) n Im 4A 2 ^0

dx P.V.

2

j J

_ .[(*

But the integrand here

is

-a) 2 +l]

2

even, and

z

Im

-a + i(2a +3)

= Im V2

Zo

-a + i(2a 2 + 3)

-jA + a

- i-yjA-a

VA + a +/VA-a VA + a -i^A-a

So, the desired result is

k where

2

2

-a) + l]

2

=^[(2a

2

+ 3)VA^ + .VA^],

A = Vfl 2 + 1.

SECTION 74 r

1.

The problem here

is to

evaluate the integral



=

_*.(*+ this,

we

introduce the function f(z)

=

— (z

inside the simple closed contour

course, in the lower half plane.

+a

)(z

— X*+&)

cosx dx 2

—n~' whose +b

shown below, where

where a > b > 0. To do

singularities ai

and bi

lie

)

R > a. The

other singularities are, of

130 According to the residue theorem, a

e

jj? + a

2

dx

)(x

2

J/(z)e

ft

= 2»XA + ^).

£fe

+b 2 -+ / r.

where

= Res[/(zK<] =

B,

*=<«

v 2 +b 2 ) + ai)(z

-

.

(z

2a(*

.

:=oi

2

-a 2 )/

and z

52

=Res[/(z)e'

]

e-»

=

2b(a

=bi

2

-b 2 )i-

That is,

n 2 2 a -b

e"dx

Is

b

(e

_£1 - \Az)e*dz,

b

\

a

or

(x

Now,

if

z

is

2

+ a 2 )(x 2 + b 2 )

a point on

CR

iz

a

2

-b

Rejf(z)e k dz.

2

,

l/(z)l<

md\e

MR fi

where

M» = *

— (R

it

i

=

5-

=

-a 2 )(R 2 -b 2 )

nR (R

2

-a

2

)(R

2

-b 2 )



as

R -» 00.

cosjtd!*

This problem

2 (jc

is to

+ a 2 )(jc 2 + fc 2 )

a

2

-& 2

evaluate the integral



=—

- has the singularities ±1; and so z +1 contour shown below, where R > 1.

/(z)

2

follows that

r

2.

=

\=e~ y <1. Hence

ReJc/(z)^&|<|jCj[ /(z)e'Vz <>MB nR = So

-a\

( -b

cos*dfr

y

(a>b>0). l

a

ft

f

COS — =

we may

fl.y

dx,

where a > 0.

The function

integrate around the simple closed

<

131

We start

with

)-1-—dx+ X +1

-R

jf(z)e

iai

dz

= 2mB,

C,

where UK = ^-: B = Res[f(z)e ] 1 J

z+i

*="

2i

Hence lOX

\—^dx = ne-

a

-

\f{z)e

iai

dz,

or

cosax

_

,, _ = ^ a -Re /(zy
,

r

.

,„ T

,

Since

L-

\f{z)\
we know

that

icR

Re

\f{z)e^dz

R 2 -V

and so cos ax

J x2

dx = ne~

+l

That is, r

cosax —5

,

J* 2 +l ox

4.

To

^

XS evaluate the integral

\

I

2

X dx,

x +3

2

z

where

z,

= V3i. The

=—

z

Zl

(a>0).

2

we first introduce the function

+3

point zt lies above the

fi&P^M. -

it

(z-zOiz-zJ x

where

axis,

and

zi lies

Hz) =

below

,

Z-Zt

it.

If

we write

.

132

we see that

zx

is

a simple pole of the function f(z)e i2z and that the corresponding residue

V3/exp(-2V3) _ exp(-2V3) B = (j>(z ) = l

1

2V3i

2

Now consider the simple closed contour shown in

the figure below,

where

R > VJ

y

Integrating f(z)e'

2i

around the closed contour, R

we have

ilx

jf—dx = 2mB

l

-if(z)e

i2i

dz.

Thus

j

Now, when

z

is

-^T^dx = Im(2^)-Im jcJ(z)e i2i dz.

l/(z)l<

and

so,

by

CR

a point on

M

R,

,

where

M =—

>

R

as

R -» °°;

limit (1), Sec. 74,

limf f(z)e

iu

R-*~JC,

dz

= 0.

Consequendy, since

Jmj J(z)e i2t dz\<\j nz)e iU dz c c we

arrive at the result

|^
+3

or

j™H.,fr = fexp(-2V3).

„*+3

J

2

is

133

The

integral to

be evaluated

is

* f^"* dx, where a > x +4

f

L /(z)

= ~Tj7^' and b y computing the fourth »

^=V2e te/4 =l + both

and

i

z2

below the

tell

we find that the singularities

shown below, where

in Sec.

-

f

x

3

e

tax

7+4 *

+

k f(z)eiKdz = 27n'^ + ^

where giaZ

5 - Res^g* 4 1

t-tt

z

z

+4

4z

a(1+,)

' '

i

-

g

=

3

4

4

= ggH ~ 4

and

o7 — Kes — z=z

—=

=

lZ * + 4

=

=

4

4z*

4

4

Since

27n'(A+£2 ) =

we are now able

.

e

+e

= i7te

flu; \

cos a,

to write

r

J

R > V2

.

The

69 for finding residues

us that R

define the function

other two

real axis.

The residue theorem and the method of Theorem 2 poles

-4,

We

= V2V 3 *' 4 = V2V*' V*' 2 = (1 + i)i = -1 + i

the simple closed contour

lie inside

singularities lie

roots of

0.

x smax

r

-^r^pfr = « "cosa-Im J .

/Uy*.

at

simple

— 134 Furthermore,

if z is

l/(z)l

and

this

means

CR

a point on

<

then

,

according to limit

ehZdz

(1), Sec. 74.

7 J

\

* \fCR

->

R -> <»;

as

2

^ox

me

dZ -»

as

/?

->

°°,

3

x sinax — —ax = ne x +4 ,

3

consider the simple closed contour

.

cos a

(a>0).

* Jf

(x

2

+ l)(* 2 +9)

Its singularities in

-

iaz

Finally, then,

In order to evaluate the integral

2

-

4

that

R

f(z)=

MR = —R—

where

^ L fW

8.

— R 3

MR

,

we

introduce here the function

the upper half plane are

R > 3.

shown below, where

Since

Res[/(z)e*]=

z

1

'f2 (z+0(r+9)

16e

and 3

Res[/(z)e

iz

«= 3 ' 1

the residue

theorem

tells

x

t

,(

3

] J

-z e*

= 71z (z

+ l)(z + 3i)

Z=3i

3 »

us that

e

a

dx

(^ + l)(^ + 9)

+

/'i

r

Jc/^

=

H-I?

16
or

r

I6e

j:

sin x dx

n

(9

\

t

16e

3

'

i

and

3i,

and

we

135

Now if z is a point on CR \f{z)\
So, in

view of limit

(1),

Im and

this

means "r

,

then

MR*

-

R

l(x

The Cauchy

R-> oo.

Sec. 74,

J

\c

(z)e*dz\

< |

J

\c

->0

(z)e*dz

as /?-> oo;

that

2

+\){x +9)

8e{e

2

i

x smxdx

)'

°

principal value of the integral

T

it

°f

J

(x

2

+ l)(x +9)~ 2

(9

I6e \e

nx
f

J

function /(z)

as

n

2

(R -l)(R 2 -9)

_jc^nxdx_ _ n_(9__\ 2



M* = —

where

f x +4* + 5

can

2

|,

e foun(j

2

^ih

^

^

1

=

r +4z + 5

and the simple closed contour shown below, where

Using the quadratic formula to solve the equation z 2 singularities at the points zl

=-2 +

i

and zx

+4z + 5 = 0, we

=-2-i. Thus

f(z)

=

interior to the closed

The residue theorem

contour and z

tells

x

is

below

the real axis.

us that

where

S = Res and so

smxdx 2me -Imf r = Im l J „ x + Ax + 5 L(zi-^i). r I

,Zl

—5

k

f(z)e dz,

R > V5.

find that

——

3

(z-zjiz-zj is

f

,

/ has

where

z^

136 or

%

sinxdx

r

Now, if z is a point on Q, l/(z)l
lz

then \e \=e~ y

where

M



.

,

f

< 1 and

=

R

Hence

Im

f

/(z)^|<|f

M nR = —

<

R

(/?-V5) and

To

as

R -» oo,

we may conclude that .

10.



2

find the

Cauchy

7

V

J J

~—

sin;t<£t

2

7

=

x + 4x + 5 'a

#



.

sm2

-

e

principal value of the improper integral

^

the function /(z)

=

z+l r + 4z + 5

same simple closed contour as



z+1

,

x + 1 ^ cosx

f

J

.

x

2

+4x + 5

dx,

we

where z = -2 + /, and z = -2 - 1, and x

(z-zJiz-zO

x

in Exercise 9. In this case,

where

(z

5 = Res

+ l)e*

(Zt+l)g

_

(z-^Xz-zi)

fel

=

(-l

(z-zi)

+ Qg-2

''

2ez

Thus

or

r

Finally,

we

(x

observe that

\f(z)\
+ Dcosjc

if

where

z

is

.

tt,

a point on

Ms =

v

.

Cs

,

shall use

r

then

R+l

R+l

(R-izMR-m

(/?-V5)

-» 2

as

R -» oo.

the

137 Limit

(1),

Sec. 74, then

us that

tells

RcJCa /(z)«*
as /?

->

oo,

and so

_,, pv

12.

(aj

r -

Since the function /(z) integral

(* + ^ —

l)cos;t

n,

_

r^ = — (sin 2 -cos

:

= exp(i'z 2 )

,

is entire,

.

2).

the Cauchy-Goursat theorem tells us that

around the positively oriented boundary of the sector

has value zero.

A parametric

The closed path

is

its

0£r
shown below.

from the origin to the point segment from the origin to the point

representation of the horizontal line segment

R is z = x (0
for the

A

je

ixl

dx

+

lll

jc

e

dz-e

iK

rl

dr

'*\e-

=
or

)e

ixl

dx

rl

= e "'*]e- dr-\ i

e

c

By

e dz.

equating real parts and then imaginary parts on each side of this last equation,

see that

1*1

R 2

r jcos(x )dx = -j=je~ *dr-Rej

C*

e*dz

o

and R

|

2

fsin(jc )dbt

1

R

= -H
we

138 (b)

A parametric representation for the arc CR

is

it/* izl

e

jc

Since

.iR

z

cosZ0

1

dz= \e*

= 1 and

iB

'

\

ie

'Rie

it

d9 = iR\e-

(0

<

<

7r / 4).

Hence

RlA * 26 iRla" ie i6 e e de.

follows that

ic/4

,

Jc

e*dz
Then, by making the substitution (3), Sec. 74, of Jordan's inequality,

je-*

lsia26

= 2d in this we find that


2

(c)

= Re ie

a/4

a

= 1,

\e

z

dd.

last integral

2¥~4R



as

and referring

to the

form

R -» oo.

In view of the result in part (b) and the integration formula

z

o

it

follows from the last two equations in part (a) that

2

]cos(x )dx =

^^

and

2

Jsin(*

)d!*

=

SECTION 77 1.

The main problem here

is to

-

derive the integration formula

cos(ax)

- cos(foc) 2

using the indented contour

shown below. y

,

dx = —(p - a)

(a>0,b>0),

139

Applying the Cauchy-Goursat theorem to the function

/(*)

=

we have f J

M

f(z)dz+[ f(z)dz+l f(z)dz+i f(z)dz = 0, JC JL *

«C.

1

or

Since

^

have parametric representations

and

i0 L :z = re = r(p
we can see

- L2 :z = re" = -r(p
and

l

that

^

^ /<«)& + j f(z)dz = jj(z)dz-l_J{z)dz = \

S

wr

&

ibr

~e

dr +

Li

e

— iar

~e

\

—ifer

dr

- (e* + e"*') (g^+g^ ) -5 J cos(ar) - cos(fcr) -5 dr. dr = „ r

f

J

,

,

2J

Thus

p

In order to find the limit of the first integral

fiz)

=

\ z

^

this

2!

+ •••

we see that

lim

z

f

p—»0 JCp

(/az) (

=

|

3!

right here as

N

3

(iaz) |

1!

/(a-fc)

From

2

iaz |

on the

(1H I

p -» 0, we write

ibz(ibz)\(ibz)\ ••• H

1

1!

2!

h

3!

(0
is

f(z)dz

a simple pole of f{z), with restdue~Ik = i(a - b). Thus

= -B

7ti

= -i(a - b)m = %{a - b).

140

As

CR

for the limit of the value of the second integral as

R -» «>, we note that if z is a point on

then

,

rt

^

\e"*\+\e

M

e-v

\

+ e -* ^1 + 1

2

Consequently,

<—t kR = R2

now

It is

p -»

clear that letting

—R -*0

as

and /?-><» yields

^7 cos(ar)-cos(frr )L J dr= n(b-a) j

2J

o

This

is

the desired integration formula, with the variable of integration r instead of x.

Observe

that

when a -

and b =

2, that result

7 1 - cos(2;t)

3

J

But

cos(2jc)

A

=

n-

= 1 - 2 sin 2 x, and we arrive at

2

*

I

2.

,

dx

becomes

Let us derive the integration formula

r

x fa

|(7TIF' where x"

= exp(aln x) when * > 0. _

whose branch cut is the path shown below.

=

(l-a)TT

,

(

4cos(a,/2 )

- 1
'

We shall integrate the function

z"

exp(alogz)

.

_

3;^

origin and the negative imaginary axis, around the simple closed

p Branch cut

By Cauchy's residue

f JL

f(z)dz+i f{z)dz+[ f(z)dz+i f(z)dz = 2mResf(z). JL JC

f

f(z)dz+

i

That

theorem,

z

i=i

p

is,

f

f(z)dz

= 2mResf(z)~

f(z)dz-

f

f(z)dz.

f

Since

L

l

:z

= re i0 =r(p
-I2 :z = re = -r(p
the left-hand side of this last equation can *

be written R

a(lnr+i"0)

«

o(lnr+/)r)

*

a

= J1 71— 2 (r

-4-

2

dr + «"*

1)

a

\ ,i 2 J(r +1)


= (1 +

*

a

f

/+

'J(r 2

. 2

dr.

1)

Also,

Res/(z) =

where

(z)

= (z

the point z

=

i

+

2

'

being a pole of order 2 of the function f(z). Straightforward differentiation

reveals that

'(-\

0'(z)

a(z

_ = e-(a-l)logz L

+ i)-2z

(z

+

3

142

and from

this it

follows that

-iaKll Res f(z) = -ie

fl-a

We now have

Once we show

that

limf f(z)dz =

and

lira f JC.

p-*0 iC„

f(z)dz

= 0,

we arrive at the desired result:

„a _r J

o(^+l)

The

first

2

^_

_/i

JaKll

_\ — a)

_e

e

+ e m* iaic

2

l

of the above limits

is

e

-iaKll

^(1 - a) \

—iaKll

shown by

(l-a)ff

4

e

ia*'2

+e

-ia*l2

AcQS ( a7t / 2 )

writing a+l

f(z)dz

I and noting

that the last

(l-p 2 )

term tends to

as

np = 2r

p -»

Tip

(1-p

2 )

a+1>

since

_

1

\jc

3.

last

mdz

2

(tf

-l)

2

""

2

(R -l)

as /?-»<» since 3

term here tends to

The problem here

is to

1

2

1

|1-

- a > 0.

2

n jr

i,

2

by

for the second limit,

derive the integration formulas

-yxln* "(Mxlnx f l

As

Ra R4

and the

0.

x +l

_ .

. ,

2

6

7f

VI Vjc

J* 2 + l

,

7t

V3

integrating the function

H

,

f(r )

z

m logz e^logz _ 2

z

+l ~

z

2

+l

lzl>0,






143 around the contour shown in Exercise

f(z)dz +

f

f(z)dz

f

2.

As was

= 2mResf(z) -

the case in that exercise,

f(z)dz -

f

f(z)dz.

f

Since

/W-M z-i =

the point z

is

i

=

where

z+i

a simple pole of f(z), with residue

Res/(z) = 0(O = ye'*/6

.

The parametric representations

L :z = re i0 =r(p
-L :z = re = -r(p
1

can be used to write

and

J.

=

f

^.J^l^fr.

Thus

Vrlnr

^ „ J^TT^ + f

r2 I p

By equating

Ifrlnr+in Vr ^

gl /3

(

1

real parts

m^dr +

1 ; p J

r

>

+i

dr=

2

;r

T

e

/^z-jc mdz.

-i c

on each side of this equation, we have

cos(;r / 3)

l^fdr - nsin{7t

1

3)\-^-dr =

-—

sin(;r / 6)

-Ref /(zVfe-Ref /(*)&; and equating imaginary parts yields * 3/~~ i

sin(;r / 3)

f-^-f J

r+1

* flfr

3/~"

= + *rcos(w / 3) \——dr 2 J

r

+l

2



cos(;r / 6)

2

•Imf /(z)
Now

sin(^/3)

= ^, cos(^/3) = ^-, z

show

sin(7r/6)

Z

= ^-, cos(^/6) = 2

p->0 Jc, '

and

lim f /(z)
— 2

that

limf f(z)dz =

/(z)
= 0.

and

it is

routine to

144 Thus

3f V7lnr

2

r

*

2

W3

r

J

r

+l

2

W 2

2 ;r

.

+l

4

That is,

_

3

W3 2

l

2

2

1

2

2

Solving these simultaneous equations for

n2

_ ~

T*

2

4

/j

and 72

we

,

arrive at the desired integration

formulas.

4.

Let us use the function v

=

/(z)

and the contour

in Exercise

f

2

to 2

3

j

{x +l

f

n I* n lzl>0,--
f,



-5

J

8

Q

2

,

dtc

'

2

=n 0.

x +l

around the closed path shown in Exercise

f(z)dz + j

i

V

} Inx

and

2

Integrating /(z)

(logz)

show that

n ——-<& = — (lnx)

2

-2— Z +1

2,

we have

nz)dz = 2mResAz)-[ f(z)dz-\

f(z)dz.

Since

/W-M Z-i the point z

=/

is

where

z+i

a simple pole of /(z) and the residue

Res/(z) = 0(0 = *=•'

^

=

(lnl

2/

is

+ t-;r/2 ) 2 2/

= _*1

'

8i

Also, the parametric representations

L :z = re i0 =r{p
and

-L :z = re 2

iic

= -r(p
2

enable us to write

J,/W*"f$£*

y^,]^f

and

P

P

Since

p

p

p

then,

p

p

p

Equating real parts on each side of this equation,

p

we have

p

and equating imaginary parts yields r

,

27c\^-dr = lmi f(z)dz-lm[

f(z)dz.

p It is

straightforward to

show

lim

f

that

f(z)dz

=

and

lim

Hence

and

Finally,

inasmuch as (see Exercise

1,

Sec. 72),

r

dr 2

Jr + 1

we arrive at the desired integration

formulas.

_ = % 2'

f

f(z)dz =

0.

dr

.

146 5.

Here we evaluate the integral (x

We

a>b>0.

dx, where J

+ a)(x + b)

consider the

function

expl

.1/3

(z

+ a)(z + b)

(z

^logz

+ a)(z + fc)

and the simple closed contour shown below, which is similar to the one used in Sec. 77. The numbers p and R are small and large enough, respectively, so that the points z = -a and

= -b

z

are between the circles.

A parametric (p

<

r

<

R),

representation for the upper edge of the branch cut from

and so the value of the integral of /along that edge

r f

o

exp

(ln/-

_LL :

(r + a)(r

+ iO)

^

r

+ b)

=

A representation for the lower edge from value of the integral of / along that edge from

£

f

{( r

p

R

+ a)(r + b)

to is

to

p

R

is

z

L3

\ dr

= -e i2 *n

to the residue theorem, then,

R

is

z

= re i0

^

= re n * (p
is

\

i(r + a)(r + b)

According

to

is

*exp -(lnr + /2;r) f

p

dr.

R).

Hence

the

147

where

exp|j(lna +

exp|jlog(-a)j

3=Res/(z)

-a+b

gM3

/;r)J

a-b

^

a-b

and

1

exp |log(-t)

B2 =Resf(z) =

exp

(\nb + in)

n

Mb — a b

i" -

-b + a

Z=-b

-b + a

Consequently,

2«-^) . J/(^, J/( ^=

^



V7

(\-e™)\

--

dr

^

Now

L

f(z)dz

(a-p)(b-p)

(a-p)(b-p)

and

/(z)& <

V* (R-a)(R-b)

D= 2nR



2*/?

2

— •-

1

n >0

(R-a)(R-b) MR 2

as



Hence

2me * /3 e/a~-Mb) - 2" -

,

r

o

(r + fl)(r

+ ft)

3

(1

e''

^

n(Ma-Mb) sin(;r/3)(a-fc)

*)

(

V*

'


'*/3

/3

(e'*

_ n{Ma~-Mb) = 7~~ ^/3~

Replacing the variable of integration r here by x,

{(x + a)(x + b)

2^

e™

i

f

fc^n V3

-y(«-*)

-

-Mb) - *)

2ar

V^-Vfr

V3~"

a-fc

'

we have the desired result: Ma'-Mb

a-b

(a>b>0).

148 6.

(a)

Let us

first

use the branch

exp^--logz

-i/2

z z'

2

+l

z

and the indented path shown below

lzl>0,--^
+l

to evaluate the

improper integral

P Branch cut

Cauchy's residue theorem

f

M

tells

f(z)dz+i f(z)dz + -' (

-*

us that

f '^2

f(z)dz+l f(z)dz = 2mResf(z), •'Cp Z=i

or

f

/(z)
+J

f(z)dz

= 2mResf(z)-

f

/(z)rfz-

f

/(z)
Since

I^-.z^re'

=r(p
and

-l^.z-re^ = -r(p
we may write dr 1)

Thus

149

Now the point

Res f{z) = J=<

z

=i

is

evidently a simple pole of f(z), with residue

m

z

exp -I^inl + f.5

exp[-|logij

—. + z~

2i

i

2i

2iVV2

2i

Furthermore,

no

Jp

= 72^ n -TvTT 1-p Vp(i-p .

L

f(z)dz

it

2

2

as

p->0

)

and

\\c

Finally, then,

k4r

m
we

is

the

->

as /?->oo.

have (1

which

n

same

'

VKr 2 + l)

V2

as

r

_

dbc

it

J-vW+1) ~V2' To

evaluate the improper integral

J[ Q

-1/2

-j=



^x(x 2 + l)

,

we now

use the branch

expf-^-logz z

z

(kl>0,0
+l

and the simple closed contour shown in the figure below, which Sec. 77.

C„ and

We stipulate that p < 1 CR .

and

R > 1,

is

similar to Fig.

so that the singularities z

= ±i

99

in

are between



1

Since a parametric representation for the upper edge of the branch cut from

z

= re' (p£r<

R), the value of the integral of /along that edge

— 1

«exp „_.. r

(lnr + i'O)

A representation for the lower edge from

p

to is

value of the integral of/ along that edge from

-f

R

2

2

R

to

z

is

p

-U = -e-*

2

R

= re' 1 " (p
is

dr

=

f

_i

dr.

1

where

exp -i-flnl-H-*

exp --log/

-1/2

2l

B =Res/(z) = [

z

+i

2/

-;«/4

2

2/

2i

and

^ .-1/2

52 = Res/(z) = *=-'

That

exp

z-

4

exp

iogH)



.3* If, , lnl + i

2V

1

-2i

2i

-2i

J

„Vr(r +1)

J

J

Since

VP(i-p

2 )

2k Jp l-p 2

and 2;cR

2;r

-»0

-i3*r/4

2

is,

In p

is

is

Hence, by the residue theorem, R

R

+ l)

~(\nr + iliz) L

to



V^(r

«exp

p

as

oo,

so the

151

we now find that i

,

J J V^(r + l) 2

dr



—e

e

=n

it

=n

When

x, instead of r, is

+e

e

e

= Kco&

urvr

used as the variable of integration here,

we have

the desired

result:

_

dx

7

it

SECTION 78 1.

Write 2

dd _ r ]5 + 4sin0~-k r

C is the positively

where

J z -z~ \

5+4

= -il 2

? J o

2.

5

is

1.

The

quadratic formula tells us that the

far right here are z

a simple pole interior to C; and the point z

= -i/2 and z - -2i. The point

= -2i

is

w-u

M artier'— + 4sin0 *=--42z +5iz-2j

To evaluate

2

L4z + 5i_L_ 1/2

exterior to C.

Thus

-wfi)-^. 3 U»7

the definite integral in question, write

f

dd

Jl + sin

_ 2

r

dz

1

0~Jc 1

+

(7

-7 2i

where

~^2z + 5iz-2'

iz

oriented unit circle \z\=

on the

dz

r

2

l

singular points of the integrand

z

dz_

1

C is the positively oriented unit circle

-i\ 2

'

z

i

_

t

4izdz

~Jc z *_6 z 2 + r

) \z\=

1.

This circle

is

shown below.

r

.

152 Solving the equation (z

2

2

)

- 6(z 2 ) + 1 =

find that the zeros of the pol ynomial z

Those zeros

are, then,

4

for z

2

with the aid of the quadratic formula,

2

- 6z + 1

numbers z such

are the

that z

2

= 3 ± 2V2

= ±^3 + 2-^2 and z = ±V3-2>/2. The first two of these zeros are and the second two are inside of it. So the singularities of the

z

exterior to the circle,

integrand in our contour integral are

z

;

=V3-2V2

means

indicated in the figure. This

and

z1

=-zv

that

where

By

= Res—4 *-* z

_

4iz 2

-6z + l

4izt

/

4^-12^

zf-3

(3

-2V2) -3

2-V2

and

_

4/z

B2 = Res

4

-4/z,

2

3

-4z + 12z

-6z + l

*=-*.z

1

1

zf-3

2V2"

Since

2k V2

\

i

2^(A-H52 ) = 2^|--^j = ±|.-l| = V2^ the desired result is

J l

7.

Let

+ sin 2

C be the positively oriented unit circle

lzl=

In view of the binomial formula (Sec. 3)

1.

In

Jsin

2"

BdO =

IJ

2

2

we

sin

2"

ddd = ± 1

(

z-z

J 2M

i^/c|(T)

2 " +i

2 " +1

(-l)"i

ft

^ *

2i

z2

-l\2«

dz iz

" -z (

"

1)iz

2

2 " +1

"^

(-1)"/ J c (-l)

dz

153

Now each of these last integrals has value zero except when 1

jc

z dz

k = n:

= 2m.

Consequently,

(2w) !(-!)"

1

jsm 2n ddd = 2

"

2»+i

2m =

(2n)!

2

(n! )2

2

2fl

7T.

(«!)

SECTION 80 5.

We are given a function / that is contour C, and we assume inside C,

where each zk

The object here

To do

this,

is to

is

show

we consider the

that

analytic inside

of multiplicity

zf'iz)

/(z)

is

mk z 1

z-zk zg'(z)

Since the term

— —

.

= l,2,...,n)

(See the figure below.)

kth zero and start with the fact that

= (z-zk ) m >g(z),

analytic and nonzero at zk

_ —

mt

(&

that

f(z)

where g(z)

and on a positively oriented simple closed

/has no zeros on C. Also, / has » zeros zt

zg\z) — _ g(z)

m

k

.

From

this, it is

(z-zk ) + mk zk

zg\z) _

_ -WlH

1

z-zt

straightforward to

,

zg\z)

g(z)

g(z)

here has a Taylor series representation at zk ,

g(z) has a simple pole at zk and that

/(«)

An application of the residue theorem now yields

the desired result.

, 1

show

that

m zk k

.

z-zk zf'iz)

it

follows that

Hz)

.

154 6.

(a)

To detennine lzl=

1,

number of zeros of the polynomial z 6 - 5z 4 + z 3 - 2z

the

inside the circle

we write /(z)

= -5z 4

and

= z 6 +z 3 -2z.

g(z)

We then observe that when z is on the circle, l/(z)l

=5

6

and

ls(z)l
+ lzl 3 +2lzl = 4.

Since \f(z)\>\g(z)\ on the circle and since /(z) has 4 zeros, counting multiplicities, inside

it,

the theorem in Sec.

80

tells is that

the

sum

6

Az) + g(z) = z -5z*+z 3 -2z also has four zeros, counting multiplicities, inside the circle.

(b)

Let us write the polynomial 2z

=9

/(z)

Observe

that

and

when z is on

l/(z)l

=9

4

- 2z 3 + 2z 2 - 2z + 9

g( z )

sum

/(z) + g(z), where

= 2z 4 -2z 3 + 2z 2 -2z.

the circle lzl=

1,

4

and

as the

l#(z)l<2lzl +2lzl

3

+ 2ld 2 +2ld = 8.

Since l/(z)l>l#(z)l on the circle and since /(z) has no zeros inside /(z) + g(z)

7.

Let

4

2

3

= 2z - 2z + 2z - 2z + 9

C denote the circle

I

zl

sum

=2

The polynomial z + 3z 3 + 6 can be /(z)

On

the

has no zeros there either.

4

(a)

it,

= 3z 3

written as the

and

$(z)

sum of the polynomials

= z 4 +6.

C,

l/(z)l

= 3lzl 3 = 24

and

l^(z)l

= lz 4 +6l^izl 4 + 6 = 22.

Since l/(z)l>l#(z)l on C and /(z) has 3 zeros, counting multiplicities, inside C, follows that the original polynomial has 3 zeros, counting multiplicities, inside C. 4

(b)

The polynomial z - 2z 3 + 9z 2 + z - 1 can be /(z)

On

= 9z 2

and

g(z)

written as the

sum of the polynomials

= z 4 -2z 3 +z-l.

C,

|/( z )|

= 9lzl 2 = 36 and

U(z)l=lz

4

3 -2z + z-ll
it

155

C

and /(z) has 2 zeros, counting multiplicities, inside C, the original polynomial has 2 zeros, counting multiplicities, inside C.

Since l/(z)l >\g(z)\ on follows that

2

3

5 The polynomial z + 3z + z + 1 can be

(c)

/(z)

On

= z5

and

$(z)

written as the

sum of the polynomials

= 3z 3 +z 2 + l.

C, 5

l/(z)l=lzl

= 32

and

= l3z 3 +z 2 +ll<3lzl 3 +lzl 2 +l = 29.

I#(z)l

Since \f(z)\>\g(z)\ on C and /(z) has 5 zeros, counting multiplicities, inside C, follows that the original polynomial has 5 zeros, counting multiplicities, inside C.

10.

it

The problem here

is to

give an alternative proof of the fact that any polynomial

P(z)

where n >

1,

= a + a,z +

+ a^z"" + aH z" 1

• • •

(a.

* 0),

has precisely n zeros, counting multiplicities. Without loss of generality,

may take an =l

it

we

since

r

P(z)

= a„

Ea.

+ BL z + ... + EB=L^ +z ^

Let

/(z)

Then

let

= zB

and

g(z)

= a +o z + - + an _ z"' 1

l

z

is

.

R be so large that /?>l + la + lo + ---+laB _

If

1

l

a point on the circle C: lzl= R,

1

l.

l

1

we find that -1

lg(z)l fS la

l

+ l^llzl + ••• + la^llzl" = la + iaJR + ••• + la^ltf-

1

l

1

< Iflol/?"- + Ifl!!/?" 1

-1


+ ••• + t^.J/?"

-1

= (la + laj + - + l

= l/(z)l.

Since /(z) has precisely n zeros, counting multiplicities, inside arbitrarily large, the desired result follows.

C and since R

can be made

-

156

SECTION 82 1.

The

singularities of the function

3

2s

5-4 are the fourth roots of 4.

They

are readily found to be

or

V2, See the figure below, where


-V2,

and

y > -4l and R > V2 +

-V2*.

y.

Y+iR

The

function

—-4

2?V'

=—

e

s

has simple poles at the points

j

= V2,

jt

= V2/,

j,

= - V2,

and

*3

and

V

X Res^FW] = 2 Res ^il 2

3

=

f

2

2



+

= cosh V2f + cos V2f.

o„3*.r

2j£

2

=

157

Suppose

now that s

\s\=\y+Re'°\

It

is

CR

a point on

,

and observe


that

ie \s\=\y+Re \>\y-R\ = R-y>«j2.

and

follows that

\2s

3 \

= 2\s\ 3 <2(R+yf

and s

I

4

- 41 > lid 4 -4I> (/? - y) 4 - 4 > 0.

Consequendy,

lF(*)l<

(J?+ y)

^ (/?-y)

.

4

'

->0as/?->

-4

This ensures that

/(/)= cosh V2r + cos V2f.

2.

The polynomials

in the denominator of

2s

F(s) (s

have zeros at s

= -1 and

s

= -1 ± 2i.

e"F(s)

t

= -1 + 2/'. The

-2 + 2s + 5)

Let us, then, write

e*(2j-2)

= (j

where j

+ l)(s

2

+ IXj-^X^-J!)

points -1, su and s{ are evidently simple poles of e"F(s) with the

following residues:

i_ ^=Resfe"F(j)] L J

g

Jt

z=-l

52 = Res[*"F(,)l =

B =Res[e«F(s)]= WJ L 3

~,

^'^^ (J,

+1X51-5.)

(2^-2) -Si)

" g <(2

^- 2)

Sl e '(2 Sl

= fi - 1

-2)

(5,+lX^-J!)

W<,

158 It is

easy to see that

i2t

— —e -t + e -t ,

f e i2 '-e- i2

Now let s

+

n

'

+ e- i2

''

=
2

2/

v

e

'

be any point on the semicircle shown below, where y >

and R >

-\/5

+ y.

Since

\s\=\y +

we find

Re

ie

\
and

i6

\s\=\y+ Re \>\y- R\=

R- y> V5,

that

I2j-2I<2IjI+2<2(/?+7) + 2,

\s

+ l\>\\s\-\\>(R-y)-l>0,

and 2

\s

+2s + 51=1 s - Sl \\s - s,\ > (\s\-\sj) 2 > \{R - 7) 2 - V5

f > 0.

Thus

I

and

m

^=1

=

<

2

Is

+ II \s + 2j + 51

we may conclude

+2 2 [(* - y) - !][(/? - 7) - V5]

.

Y)

that

/(/)

= e"'(sin It + cos It - 1).

as

R -> oo,

159 4.

The function „2

s

—a 2 (a>0)

(r+a 2 ) 2 has singularities at s

= ±ai. So we consider the

simple closed contour shown below, where

Y>0 and R>a+y.

Upon

writing

F(s)

yv

=

.

(s

we

see that

F(j).

is

- a/) 2

where

0(s)

analytic and nonzero at s

Furthermore, F(s)

=

a pole of order 2 of F(j); and

at points

=-

= ai. Hence

where F(s)

we know from

where

is analytic.

expression

Res [e"F(s)] + Res [e^Fis)] = 2Re[e

sQ is a pole of order

(2), Sec.

ia,

(fc

1

+ b2 t)]

m=2

Consequently, s

is

of

also

82, that

,

and b2 are the coefficients in the principal part

- ai

s

of F(s) at

ai.

(s

- ai) 2

These coefficients are readily found with the aid of the

Taylor series for

(s)

about s

=

ai:

1 (s

- ai)

(s)

1

= (s

- ai)

first

two terms

in the

160

=

It is

(ai)

^)

straightforward to

0'(ai)

r + 7^- + -

show

,

that

(ai)

(0

= 1 / 2 and

\ai)

= 0, and we


find that

2a).

^=

and

b2 =l/2. Hence

Res [e*F(sj\ + Res [e"F(s)] = 2Re

^(j'j = tcosat.

We can, then, conclude that f{t)

= tcosat

( fl

provided that F(s) satisfies the desired boundedness condition. z

is

a point on

CR

this

2

\z

means

for that condition,

when

,

\z\=\y+Re and

As

> 0),

ie

\£y+R = R+y

and

ie

\z\=\y+Re \>\y-R\=

R- y > a;

that

-a 2 \<\z\ 2 +a 2 <(R+y) 2 + a 2

and

2

\z

+a 2 \>\\ z 2 -a 2 \>(R-y) 2 -a 2 \

>0.

Hence

6.

We are given „.

which has isolated

singularities at the points

-n = °>

_ (2n-l)^

„ *.=

2

This function has the property F(s)

f(t)

To

sinh(xs)

.

_

.

311(1

,f

=

J"

= ~ (2n-l)n.

,

(n

*

„ = l,2,...). ,

2

F(s), and so

= Res [e*F(sj\ + J^Res [«*F(j)] + Res [e"F(j)]j.

find the residue at j

= 0, we write

3 _ xy + (xy) / 3 + 2 2 2 j cosh* 5 (l + * /2! + ...)"

sinh(xy)

!

• • •

+ jcV/6 + — s + s /2 + -

oc

0
3

{

2

.

161 Division of series then reveals that s

is

a simple pole of F(s), with residue x; and,

according to expression (3), Sec. 82,

= Res F(s) = x. Res le^Fis)] L 1 s=s

s=s„

As

= 1,2, ...), we write

for the residues of F(s) at the singular points a; (n

=

F(s )

= s 2 cosh j

= sinh(jcs) and

where

-

.

We note that .

.

.

—^—

(2n-l);a

.

= isin^

_

5*0

,

and

„ = 0;

v

q(sn )

furthermore, since q'(s)

we find



that

?W .

=

(2n-l)V.

.

ism

(2w-1)tf

.(2n-l)

= -,i

-r

4

2

^V.smnttcos— jt

'='.

1

7r

Sec. 82,

(4),

J

art

,

n\

( sm\nn-.

now gives



2 2 n = v(2n-l) ^ (-l)"/*0.

2) is

'

,

4

a simple pole of F(s), and

(2n-l)

?'(*.)

Res

2

2

-costt7rsin—

2

V

view of Theorem 2 in Sec. 69, then, sn

Expression

-J.

J

.(2n-l)

In

= 2 j cosh j + j 2 sinh a,

J

2

us

He. L[e"F(s)]J - 2 Re(-± 2 '='"»

.

[;r

8



-^ (2n-l)

(-1)"

2

i^^expTifi^l] 2



(2n-l)*ct

= —?•— ^—rsinn 1 (2«-l) 2 .

sin

-

2

2

L



(2n-l)nt cos-

2

JJ

,

162

Summing all of the above

-x

f(t)

7.

residues,

+— —

we

arrive at the final result:

-j sm

2_,

cos-



The function F(s)

=

1 -

JCOSh(5

where

it is

agreed that the branch cut of s

=

isolated singularities at s

in

= 1,2,

The point

...).

1/2

.ycosh(.y

and dividing expression

)

does not

and when cosh(j

l/2

)

a simple pole of F(s), as

is

this last

denominator into

1.

is

s

along the negative real axis, has

or at the points s„

seen by writing

+ s 2 /2 + s 3 I2A+-

Sec. 82, tells us that

for the other singularities,

F(s)

=

^

l.

we write

where

p(s)

= 1 and

q(s)

= scosh(s m ).

Now p(sn )

= 1*0 and

q(sn )

= 0;

also, since

q\s) = ±s

it is

straightforward to

x

4

(*„)

show

m sinh( 5

1/2

+ cosh(* 1/2 ),

)

that

(2n-l);r n\ (2n-V)7t = -^-^-sinl(nit - = ±J±-^L(-Vf±^ .

I

= _( 2n Jj n

In fact, the residue is found to be

Resfe" F(s)] = Res F(s) =

As

lie

= 0,

4l + (j 1/2 ) 2 /2! + (/ /2 ) 4 /4!+-]

)

(3),

s

m

'

1/2

1;

and

1

,

163

So each point sn

is

a simple pole of F(s), and

pw = i.i=«l

ResF(J)=

#

Consequently, according to expression

= Res \e"F(s)] J L

e'"'

'=',

K 2n-l

CO

(3), Sec. 82,

Res F(s) = - £-^-exp ;r 2n -

(2n-l)Vf (n



= l,2,...).

Finally, then,

/(/)

= Res [e

st

F(s)]

£ Res

+

[**F(j)1 ,

or

(2n-l)Vf nZi 2n-l

Here we are given the function

.

.

_

coth(^y / 2) s

2

cosh(^y/2)

_

~(/ + l)sinh(^s/2)'

+l

which has the property F(s) = F(J).

We

consider

first

the singularities at s

= ±i. Upon

writing

F(j)\

u where

=

\

jl /•

s-i

we

find that, since 0(i)

3(b), Sec. 65];

the (/i

residue

and the same of e"F(s)

= 0,±l,±2,...),

F(s ) =

(s

= 0,

£yl q(s)

the point is true

is

or at the points s

where

p(s)

+ i)sinh(ns f 2)

a removable singularity of F(s) [see Exercise

i is

of the point -i.

The other

0.



coshers/ 2)-

=

= 2ni

= coshf



singularities

= 0,±1,±2,...). To

(n

\2

At each of these

and q(s) = J

J

2

(s

points,

it

occur when

follows that

nsl2 = nni

find the residues,

+ l)sinh[



\2

we write

and note

that

p(2ni)

= cosh(nm) = cos(nn) = (-1)" *

and

q{2ni)

= 0.

Furthermore, since

q\s) =

2

(s

+

1)|

cosh^H + 2,sinh(-0 j

we have q\2ni)

= {-An 2 + l)-^cosh(n^i) = (-4n 2 + l)-cos(n^) = ^

/r(4w

2

- 1)

(-1)"*0.

2

Thus *= 2 »<-

Expressions (3) and (4) in Sec. 82

now tell us

Res[e"F(5)l



n An 2 -I

q\2ni)

= 0,±1,±2,...).

that

= ResF( y) = 1

and

A_

Res[e"F(j)]+ Res \e"F(sj\ = 2 Re e i2 "'( 2

is,



n An 2 -\

n An -I The desired function of t

cos 2nt

= 1,2,...).

then,

4^cos2/if

2

*r*\

The function F(V) K '

where

it is

= 1,2,

.

.

.)•

sinh(x? s

2

sinhjxs

The point 1/2

)

m smh(s

)

_

s

J sinh(^

s

=

=

m

is

)

(0<*<1),

1/2

)

does not

and when

sinh(j

l/2

)

lie

along the negative real axis, has

= 0,

a pole of order 2 of F(s), as

m + (xs U2 1 3! + (xs m [s + {s m f / 3! + (s m 3

xs 2

«

1 '2

2

agreed that the branch cut of s

isolated singularities at in

-

)

1 '2

)

5

)

/ 5!

+•

or at the points is

seen by

first

s

= -n 2 it2

writing

x + x sl 6 + xV 1 120 +• 3

• •

5

1 5! +•••]

~

s

2

+ s 3 1 6 TP 1 120 +•••





-

165 and dividing the series in the denominator into the series in the numerator. The result sinh(xy 2



.

1/2

)

/

rsinhCs In

view of expression

1/2 '2 1

\

_ —

1

1

3 =X x— -(x X ~X) -x] ~T 2 + ~T\

6

s

)

— 1

h'

is

(0<\s\<

2 7T ).

s

Sec. 82, then,

(1),

= Ux'-x) + xt = \x{x % - 1) + xt. Res [e"F(s)] 1 1 "°°

As

6

for the singularities s

F(s)

=

4t q(s)

Observe that p(-n

2

- -n 2 n2

n2 )*0

= sinh(^ 1/2 ) and

2

2

q{-n n

and

= 2s

easy to see that q'(-n

= 1,2, ...), we write

(n

where

q'is)

it is

6

sinh(5

1/2

)

n 2 ) * 0. So

2

= Q.

)

q(s)

= s 2 sinh(j 1/2 ).

Also, since

+ ijj 1/2 cosh(j 1/2 ),

the points s

= -n 2 ^2

(n

= 1,2, ...),

are simple poles

and

of

2sinh(xs >=-« 2 jt j

£



(j)

1/2

1/2

cosh(5

= —5*

1/2

.1-1

)

+1

— «

(-1)"

2

)

5

sinn;cc

(/i

= 1,2,...).

(n

= l,2,...).

Thus, in view of expression (3), Sec. 82,

=

B+1

4 ^f-*"" V ^nnx 2

(-1)

Res

le"F(s)]

f(t)

= Rcj[e^F(s)] + ±Re

'



Finally, since

we arrive

at the

AyF(

expression

fit)

—n

2v(-ir',,v,, 1 g " sin

= ~x(x 2 -l) + xt 6

10.

S )],

The function

F00=—

**

*

-3

jr

— jsinhj

5

has isolated singularities at the points s

=

and

j„

= nm,

s„

= -nm

(n

= 1,2,.

.

.).

166

Now jsinhj

3

= 5i 5|

s + —s +---)

)

6

and division of this

= s 2 + —s 4 +-

(0


oo),

6

series into 1 reveals that

in

1

(0<\s\<7t).

This shows that F(s) has a removable singularity at s

have a removable singularity

there;

and so

s Res \e 'F(s)] = L

J

s=s

To

find the residue of F(s) at s n

^0) = and observe

-—

= nm

where

p(.y)

(n

= 1,2,.

.

.),

we write

= sinhj-j and

= -nm * 0,

q(nm) = 0,

and

Consequently, F(.s) has a simple pole at s B

Res F(.) =

4^

=

tf'(n;n)

Res

0.

=

2 ,y

sinh.s

that

p{nni)

Since F(j)

s

Evidently, then, e 'F(s) must also

.

= F(s ), st

\e F(s)]

2 2 n+l q'(nm) = n 7t (-l) *0.

and

,

= -rrar nV(-l) n+1

^



= U,...).

n;r

the points sn are also simple poles of F(s); and

+ Res \e"F(sj\ = 2 Re

(

l

T ie inn = 2Re

Ml

=2

we may write

(-1)"

(icos/i7#-sin/Mtt)

sinnnt. nit

Hence the desired

result is

/(0 = Res [e"F(sj\ +

f (Res [e*F(j)l + Res

or

-smnm.

[^'F(.y)l},

7

167 11.

We

consider here the function „.

.

F(s)

a>

where

and cd*G)„

= 0,

5

Because the

first

=

sinh(xj)

s(s

+co)coshs

2



=

h

(n

= ±tui,

5

— and

(0
.

= 1,2,...). The singularities ^

= ±6)J

(n

of F(j) are at

= l,2,...).

term in the Maclaurin series for sinh(xs)

xs,

is

it is

easy to see that s

=

a removable singularity of e"F(s) and that

= 0.

Res[«*F(j)l

To find

= coi, we write

the residue of F(s) at s

cv

x

F(5)

from which

= ^^: s — coi

follows that s

it

rv \ Res F(s) = t>

Since fjs)

= F(J),

=

^/

1

1

s=s

u where

coi is

-\

=

(0)i)

;

,

+ cm)coshs

simple pole and

——

sinh(jcflw)

/sin ox

coilcoi cosh(o)i)

-2co cos o

j

for the residues at s

= 2^-

sinh(xy)'

then,

j

F(s)

=s{s

= 2Re|" j Res [e*F(5)l Res \e"F(s)]+ l l As

v

jl

(s)

= ©„/

where

Si

L-2tf)

(n

= 1,2,.

p(j)

.

.),

"^

iVl = 2



J

2fiJ

-

cos©

we put

= sinh(*.y) and

sinfitt =

2

sin G)f COSfl)

F(s) in the form

q(s)

= (s 3 + G) 2s)coshs.

q(s)

Now

p(fl)„i)

= swh(x(Q„i) =

q'(s)

we find

i

sin

fi)„.x

*

and

q(0)„i)

= (s 3 + 0) 2 s)sinhs + (3s 2 +

= 0. 2

fiJ

Also, since

)cosh s,

that

q'(o)J)

= (-coli + Q) 2 G)J)smh(o)n i) = -Q)„ (m 2 -
Hence we have a simple pole

at s

= an i,

q'(o)n i)

with residue

-con (a)

2

- G) 2 )sm G)n

=

sin oar sin g» 2

G)

cosg>

is

168 Consequently,

Resfe* F(s)]+ Res \e"F(s)] = 2Re

2

= 2-

2

-G)n ((Q -G)n )sin(Dn

But

sin o)„

n;r - -j

= sin(

j

= (-l)" +l and this means ,

1

>

-

2

0)n (a}

2 -Q)„)smo) n

that

2^^>^*

Res[e"F(*)l + Res f g 'T(j)1 =

j-*.!

smw.xsinco.t

2

6>_

a)

sin6)"'

-co

2

Finally,

/(f)

That

= Res [e"F(s)] + (Res [ e "F(*)] + Res

[e"F(*)]J +

J)f Res

is,

_

sinffit*sin6tt [

CO

COSO)

2

y

^

sina)„;csinfly 0) n

© - a:

[e"F(*)]

+ Res

[e"F(s)]\