Solucionario del Chrurchill-Variable Compleja

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

Vic
Higher Education

Boston: Gur Ridge, k_ Dubuque.lA Madison. WI NewYork San Francisco St Louis
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Table of Contents

Chapter 1.
Chapter 2.

Chapter 3...

Chapter

Chapter 5

Chapter 6

Chapter 7...

"COMPLEX VARIABLES AND APPLICATIONS" (Je) by Brown and Churchill

Chapter 1
SECTION 2

1. (@) (2-D-i0-V20=N2-i-i- V2 =-=24;

© 2-3-2

© wen)

2. (a) Relic) =Reli(x+iy)] =Re(-y+ix)=

(0) Imdiz) = Imfi(x + iy)] = Im(-y+ ix) = x=Rez.

EIERN a4) +2)= (+2) 14(l+a)z +2) +242)

Sierra etre

4 Met, then 2 -2242= (140? 24 D+ 2242-2724 220.

5. To prove that multiplication is commutative, write

a “ae Je +)
PMA N= Bee

ae = Co 0?
Ca YM Pa HAY)

6. (a) To verify the associative law for addition, write

rt Nee)
(AAA AS ET EEE IE]
AIDA Ay IE HG) + CI]
ite te).

(0) To verify the distributive law, write

24,42) = (GOI) + I= GIN Ha HID)
(2m, IN D D HY + +)
6 A D HHH +)
= Ge FAN) = +9)
SGM +n) =a 2%.

10. The problem here isto solve the equation 2+2+1=0 for z=(x,y) by writing

Grey) + Cy) + (40) = 0,0)

Since
GP txt] 2ay+y)= (0,0),

it follows that

ytxt150 and 2xy+y=0,

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

al

Poi HIGH , GDS)
3-4 BR IE

El Si EN
0-02 -D 13060 =107

0)

(o) U-0'=(1-DA-D E = CA) =-4.

2. (a) (ie=-z since 2+(-Nz=al+(-D)=z

CE

=: (#0)
ie 2 (#0).

3 Galata malaria) = alee) = (eee).

AS ero
AO

(E (2) ea (440,240),

a

a

O =D, 2 =(5,0)

() 42-30, 2=0,4)

(dat ad

2. Inequalitics (3), Sec. 4, are
RezSIRedsld and ImzSlImalsia.

‘These are obvious if we write them as

seiner and ys is VF.

3. In order to verify the inequality 4/21 2IRez1+lImzi, we rewrite it in the following ways:
RER halt,
PIE EI SET NEIN
Lal = 2ially| +P 20,
(iO.

This form ofthe inequality to be verified is obviously tru since the left hand sde is a
fect square,

4. (0) Rewrite e—1+ii=1 as fe(—0]=1. This is the circle cemered at 1-3 wih adios 1
Nis shown below

Va

5. (a) Write I2-dü+lz+4i=10 as I2- dil41z—(~4i)= 10 10 see that this isthe locus of all
points z such that the sum of the distances from z to 4i and -4i is a constant. Such a
curve isan ellipse with foci +4.

(0) Write lz= Ll=tz-+il as lz—l1=1z~(-i)l to see that this is the locus of all points z such
that the distance from z to 1 is always the same as the distance to ~i. The curve is,
then, the perpendicular bisector ofthe line segment from 1 to ~i.

SECTION 5

(0) E

() TENER = 2-9" Ria mi

dd) EAST —Al=I2z + St ie Tr SIZE T = V3 122 +51.

2. (a) Rewrite Re(2— ‘This is the vertical line

‘rough the point z

as Relx+i(-y—I)]=2, or x
2, shown below.

(0) Rewe Dei 4 2 2. Tis the circle entered at À wih

radius 2, shown below.

3. Write g =, +i and z, =, +, Then
ADS A HO
EEE = (0d) Ce dy) =

DD = Ge HO FAM

(ks = 92) “HOU + 2,92) = (ni “HOH 01) = BE

==

lol tal
Tze) le

8. In this problem, we shall use the inequalities (see Sec. 4)
IReasid and le +2,+2]<lel#lel#lel
Specifically, when [45 1,

Reaezecd|sinezecis 2s se 24e 2414

10, First write ¿42 +3= (2? =? -3). Then observe that when I2l= 2,

1 aereos ji? 1]=14-10=3

12321213 =|1af-a|=14-31=1.

‘Thus, when lal=2,

424 ei M ~3123-1=3.

ir
‘Consequently, when z lies on the circle l= 2,

leo LL
Aral ar 3

1. (a) Prove that z is re > 222.

(=) Suppose that Z=z, so that x=
Thus 22 x +02 x, or z is real.

x+iy. This means that i2y=0, or y=0.

Then

—i0 = x +10.

(2) Suppose at zis el so ht 2=

(b) Prove that z is either real or pure imaginary ¢> 2° = 2.

(=) Suppose that Then (rh) = (x+y), or ¡4xy=0. But this can be
only if either x=0 or y=0, or possibly x=y=0. Thus z is either real or pure
imaginary.

(>) Suppose hat z is ihr real or pure imaginary. If zi el so that z= hen
If z is pure imaginary, so that z= iy, then 2° = (-iy)? = (iy)

12. (a) We shall use mathematical induction to show that

A ead @=23

“This is known when n=2 (Sec. 5). Assuming now that i is true when n= m, we may
write

LE
GFR FFE) Zoe
Hart
BAe, tun

(6) In the same way, we can show that

ave (n=2,3,..).

‘This is true when = 2 (Sec. 5). Assuming that tis true when n = m, we write

Boa Sabet = Ges En Ven = arte Ea) Zot

aan

14. The identities (Sec. 5)

(e fe-2)=2,

A

BARS) Hla = Re.
15. Sino 1 can be written inthe following
ways:
(E) ES
2)
¿art ¿tar
4 3
SECTION 7
1. (0) Since

= argi—arg(-2~2i),

2), ac E. cost, pine vis

€ value of ar
one 4

(0) Since
argt/3 nf = Garg(v3 à,

or m. So the principal value is -#+ 27, or m.

‘one value of arg(v3 Dis d- 2)

of the equation le = 2 in the interval OS 0 <2# is geometrically
1 and that le” — 11 is the distance between

‘The solution 8=
evident if we recall that €" lies on the circle I
the points €” and 1. See the figure below.

y

e
° a
‘We know from de Moivre's formula that
(€0s0+isin6)’ = c0s30 +isin30,
«
cos” @+3cos" Olisin 6) +3cos O(isin 8)? + (isin 8)’ = cos30 + isin38.
‘Thatis,

(cos? 6 3c0sBsin? 6)+ i(3cos* Bsin O sin’ 8) = c0s30 + isin36.

By equating real parts and then imaginary parts here, we arrive at the desired trigonometric
identities:

(a) cos38 = cos? 6—3cosOsin* 8; (b) sin30 = 3cos* OsinO —sin* 0.

Here z=rel is any nonzero complex number and 7 a negative integer (n=-I.
Also, m==n=1,2,.... By writing

ey

ey ER A

Tee

Thus the definition 2"

we see that (2) can also be written as

fey!

10

9. First of all given two nonzero complex numbers 2, and z,, suppose that there are complex
‚numbers c and c such that 2, = je, and 2, = ¢%,. Since

Ille; and te,

Gisele

it follows that Izllz;l.
Suppose, on the other hand, that we know only that I 112,1. We may write

a =rexp(ié) and 2, =r, explid).

I we introduce the numbers

asno 4) and are

2
we find that
EINEN k
.n i lex] expli
an ano) = expire
and
se 9,+8, 8,-8,
aa po au,
Tratis,

400 and 2268,
10, E S=1+2+24-=+2, then

Si

1-2
Hence $= =,

provided 231, Thatis,

(ab.

Now the real part of the left-hand side here is evidently
1460584 c0s204---+c0sn8;

and, to find the real part of the right-hand side, we write that side in the form

1-explit + ne], CP

T=expG@) al

which becomes

o
in y sin 22408 2 nt De
¿sin +i] cos 5 — cos nr
2si
‘The real part of tis is clearly
@n+18

&

2sin

and we arrive at Lagrange's trigonometric identity:
sin +08

a

2sin.

140058 +c0s20+---+cosn8 =

u

(0<8<2m).

2
SECTION 9

1. (0) Since 2i= 203 +242)] (£=0,1,42,...) th desired 10015 are

CRE)

a = vie coseno) rra

ON

Thatis,

and

«y being the principal root. These are sketched below.

0 Oven a 19200 (-G+260)] (E=0,41,42,... Hence

E E]

‘The principal root is

are"

and the other root is
a = (ze er

‘These roots are shown below.

(k=0.D.

13

2. (a) Since —16=16expli(x +2km)] (k= 0,41,22,...), the needed roots are

cos em ( $442)

‘The principal root is

cone {cos isin) oof

‘The other three roots are

a = Qe = oi = VFL

CAC ner),
and
= ee n= VIF) = VÍA =D.

‘The four roots are shown below.

© Fit wie 8-051 = 161g (-22 242) «=0,4142...). Then

CALE colo +) &=0123.

“The principal root is

14

‘The others are

‘These roots are all shown below.
»

3. (a) By writing

D = exp

‘The principal root is

‘The other two roots are

and

zn

All three roots are shown below.

E

Ge et un,

QE) © may

BEE = qi VB,

0-0,

=1expli(=+2Km)] (k= 0,t1,42,...), we see that

eos! isin
3

(0) Since 8=8exp[i(0+2km)] (k=0,21,42,...), the desired roots of 8 are

the principal one being

‘The others are

a niet" lacio)

15

015) (=0.12,34,5),

3
nite ye VA cos isin
cy = Vie" =-42,
are er == A
ETA a e, ae

All six roots are shown below.

4, The thee cube roots ofthe number 2,

ee ze] &=012)

In panicular,

= 20n(2) ATA +.

16

(3-0-3 +i
ar

4 =40} =(6,0,)0, +

203 ++ (3 - Dil 1443
CS

5. (a) Leta denote any fixed real number, In order to find the two square roots of ati in
exponential form, we write

A=lail= Vars and a= Arg(at).

Aexpli(ac+ 2m] = 041,22.)

we see that

(asi? Ven (5-12)] &=0n.
“Thats the desired guar root are

VAe*" and JA

Vie”,

(0) Since a +i lies above the eal axis, we know that O-<ar< x. Thus 0 < À, and this

stc coo and slo. Se cu

Consequently,

Vet” =A (cos ris)

eit À
AVATAR a)

7

6. The four roots of the equation z*+4=0 are the four fourth roots of the number 4. To
find those roots, we write -4=4expli(x+2km)] (k= 0,t1,42,...). Then

cat a &=0123.

CCE ONE

‘This enables us to write

=G-aXe-aXe-aXe-6)

(20 Xe= cs) (ec ec)
NT
=[(+0*+1)[6-0*+1]

= (2 42242) -22+2).

7. Let c be any nth root of unity other than unity itself. With the aid of the identity (see
Exercise 10, Sec. 7),

Mare Gen,

we find that

9. Observe first that

¡(0+2km)]" pit! 0-2kx)_ 1 pi Den pin
A a O ap

18

where £=0,1,2,....m-=1. Since the set

is the same as the set

&=0.

but in reverse order, we find that ("y

SECTION 10

1. (a) Write lz-2+i151 as 12-(2-DS1 to see tha this isthe set of points inside and on the
circle centered atthe point 2—i with radius 1. Ibis not a domain.

©) Wate era e-(

Joa ost de enn tt

exterior to the circle with center at ~3/2 and radius 2. Iris a domain,

19

(e) 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. Itis a domain,

(à) The set Imz=1 is simply the horizontal line y=1. Itis not a domain.

,

(O The set 12-421 can be written in the form (x-4)° + y 2 +7, which reduces to
+82. This set, which is indicated below, is nor a domain. The set is also
geometrically evident since it consists of all points z such thatthe distance between z
{nd 4 à greater han or equal tothe distance between z and the origin.

20

4. (a) The closure of the set = <argz <x (z #0) isthe entre plane,

(8) We first write the set IReclcid as l</ ey”, or xa, But this last

inequality isthe same as y*>0, or L>0. Hence the closure ofthe set [Re dl<lz is the
entire plane.

(6) Since Le Eee bte et Re( E) can be ten as se
(4 -2x)+y"20. Finally, by completing the square, we arrive at the inequality
(=? + y? 21°, which describes the circle, together with its exterior, that is centered

With radius 1. The losue of this ets sell

5

a

(à Since = (+) = y +i2xy, the set Re(2")>0 can be writen as ’<x?, or
Iylelat. The closure of this set consists of the lines y-= 4x together with the shaded
region shown below.

Since every polygonal line joining z, and z, must contain atleast one point that is not in 5, it
is clear that Sis not connected.

‘We are given that a set contains each of its accumulation points. The problem here is to
show that S must be closed. We do this by contradiction. We let z, be a boundary point of
'$ and suppose that it is not a point in S. The fact that z, is a boundary point means that
every neighborhood of 2, contains at least one point in S; and, since z, is notin S, we see
that every deleted neighborhood of $ must contain at least one point in S. Thus 2, is an
accumulation point of S, and it follows that z is a point in S. But this contradicts the fact
that z, is notin S. We may conclude, then, that each boundary point z, must be in S. That
is, Sis closed.

Chapter 2
SECTION 11

1. (a) The function f(2) is defined everywhere in the finite plane except at the

points 2=#i, where 2? +1=0.

() The function f(e)= alt) is defined throughout the entire finite plane except forthe

= is defined everywhere in the finite plane except for the
imaginary axis. This is because the equation z+ 7 = is the same as x= 0.

1

(d) The function f(2)=- is defined everywhere in the finite plane except on the

3 Using x= 225 and y

OR =y ~2y+iQx-2ay)
ey (axe

2

HDD

2 2
aie =P 42ie
5 tie +

SECTION 17
5. Consider the function.
fos

6 &) #0,

where z= x+iy. Observe that if z= (x,0), then

10-20)

x,

1052) a1.
(207

23
Butif 2=(x,2),

“This shows that f(z) has value 1 at all nonzero points on the real and imaginary axes but
value -1 at all nonzero points on the line y=x. Thus the limit of f(z) as z tends to 0
‘cannot exist.

10, a) To show tat kin =

‘we use statement (2), Sec. 16, and write

ya

we refer to statement (1), Sec. 16, and write

is it lim
(©) Toesabish he mit tim —

lim =lim @~
ERE

(0 Tovey im E we apy men es 16 ve

ma

ai ;
<2

im Ar = EE

Sn una

11. In this problem, we consider the function

+b
TSG (ad-be #0)

(a) Suppose that c=0. Statement (3), Sec. 16 tells us that lim 7(z) == since

im aim et €
TW) arb a

(©) Suppose that € # 0. Statement (2), Sec. 16, reveals that lim 7()

la » im Othe a
ee

Also, we know from statement (1), Sec. 16, hat im T(z) == since

Im td
AUTO hard

SECTION 19
1. (a) Mt) =3 ~2244, then

d dd

PQ) =z -2r+4)= age AEH AIDA = 62-2.

e

(0) If f()=(-42), then

F@)=30-42F GA) IA 8 = 24 42).

¿1
(o (es

@x+Ze-n-G-vSac+n MOT 3

ae Gary Ge) Gee
(o FZ 40), then
fu DARE pag rr gare 2e

Er 5 er

a2 V4 ee EN FON
A a.

3. = Ve (2 #0), then
w= fG+ AG

Hence

LO rape 7

4, We are given that /(z,)= g(z,)=0 and that f"(%) and g’(zq) exist, where g/(e,) #0.
According tothe definition of derivative,

Similarly,

8/Go)= li

fim LO fg LOM uy) _ 2/0/72. pe)
ERIC)" lim SE-B) $)

€
=
(D) f(z)=2-2=(x+iy)-(x-iy) =0+i2y. So u=0, v=2y.
yok tc Condy ammeter ae presse,

lc) f)=2x+ ip”. Here u=2x, y:
45,8 2-2 9-1

we have O=1. Thus the Cauchy-Riemann equations do

(Os ee? = e"(cosy~isiny) =e" cosy~ie"siny. So u=e‘cosy, v=—e" siny.
4, =¥, = "cosy =—e" cosy = 26° cosy =0 = cosy =0. Thus

y= hana (n=0,41,42,...)

y, => ~e'siny =e" siny=> 2e'sin y =0= siny=0. Hence

y=nx (n=0,41,42,...)
Since these are two different sets of values of y, the Cauchy-Riemann equations cannot.
be satisfied anywhere.

Since
(Pty #0,

J’) exists when 240. Moreover, when 2#0,

29
CES

ayy AY

Pau, +, =

So fz) exists only when y= x, and we find that

SRH) un) = 2x Fil

(0) fle)=zlme=(xtiy)y=ay+i92, Here w= ay and =. We observe that

2 y=2y=3 y =0 and u,=-v, = x=0.

Hence f’(e) exists only when z=0. In fact,

F/O) =U,(0,0)+ iv, (0,0) = 0+i0= 0.

4 (a) fs

4en040)+(-Zsinde) (#0), Sine

pant ma dummen,

{is analytic in its domain of definition. Furthermore,

ne D)

4 i (60640—isin40) = Leve
a 7

© ONE? =reosS+inrsin$ (1>0,0<0<a+25) Since

8
sind =-n,

ru, zen

FR 3
2

Lo, and a
e

‘fis analytic in its domain of definition. Moreover,

ae tu ie cs Lesin®
SO" +iv) Erazo 9)

() F@)=E*costinr)+ie*sinier) (7>0.0<8<27) Since

ru, =—e*sin(lnr) =v, and uy =—e"*cos(lnr)=—rv,,

is analytic in its domain of definition. Also,

*costinr)+ie*sinttnr] =i LO.

+i), e have w=? and v=(1—y). Observe that

y 23x?

= SH 20 and à, =r, = 020.

2

2

Evidently, then, the Cauchy-Riemann equations are satisfied only when x=
‘Thatis, they hold only when z = i, Hence the expression

Fe)=u, Hi, =3x +10 23x
in which case we see that /'()=

is valid only when z

6, Here u and y denote the real and imaginary components of the function f defined by means

of the equations
ia
when
so 240,
[O wien z=0
Now
ey
Et ety

when z #0, and the following calculations show that

2,(0,0)=v,(0,0) and 1,(0,0)=-»,(0,0):

i ORAL 0)-U0,0) y, AE
100) m OO pin LE,
14 (0,0) = lim ¿0.04 Ay) -u0,0) _

a ay

1.0.0) = im 20+ B80) — W040)
(0,0) = fim ee.

»,0,0)= lim "00+ 49)= »60.0)
(0,0) = lim

ay

17, Equations (2), Sec. 2, are

u,cos8-+u, sind:

-ursin®+u,rc0s0:

9.

Solving these simultaneous linear equations for u, and u,, we find that

0050 20 and u, =u,sind-+u, 222,

Likewise,

IAE

‘Assume now that the Cauchy-Riemann equations in polar form,

(a) Write f(z) =u(r,8)+iv(r,6). Then recall the polar form

ru,

wei)

for the derivative of fata point 2, =(7,,8,) in the following way:

29

iemann equations, which enables us to rewrite the expression (Sec. 22)

30
(8) Consider now the function

fe

with

ur 9= and A

the final expression for f’(z,) in part (a) tells us that

(as) ares

4)

when 20.

10. (a) We consider a function F(x,y), where

and y=

Formal application of the chain rule for multivariable functions yields

125)

à

MENTA ze 2 1)

AXE ay a a

(6) Now define the operator

suggested by part (a), and formally apply it 0 a function f(z) = (x, y) +iv(x,y)

Hi

= Mu) fus) eu]

is tells us that

If the Cauchy-Riemann equations u

af/a=0.

==, are satisfied,

3

SECTION 24

L (a)

7)

@

2 0

(o)

1 (a)

S(e)=3x+y+iGy—x) is entro since

and el:

{f(e)=sinxcoshy + icosxsinhy is entire since

oo ya, ad a,

in xsinh y ==»,

fle) =" sin x ie cos x = ¢7 sin s+ i(~€7? cosx) is entre since

-e?sinx

Sercosxey, and 4,

FO =(@ - Dee” is entre since itis the product of the entire functions

2 and ha)=ere'

$0 (cosy ~isiny) =¢Tcosy +i(-e sin)

‘The function g is entire since itis a polynomial, and his entire since

(= xy ty is nowhere analytic since

y= y=1 and u,=-v, ==:

which means that the Cauchy-Riemann equations hold only atthe point z = (0,

*(cosx + isinx) = €” cosy + ie’ sinx is nowhere analytic since

=0

spe’ sin = 6 sin x = 2e sin x = 0=9 sin:

y, € CO8x = 6” cos x = 2e” cos =0= cosr=0.

More precisely, the roots of the equation sinx=0 are nx (1=0,+1,42...), and
1" #0. Consequently, the Cauchy-Riemann equations are not satisfied

D. Since f(2) is tal-valued,ithas the form f(2)=u(x,))+i0. The Cauchy-Riemann
equations 4, =v,.4, ==, thus become u, =0,u, =O; and this means that u(x,y
where ais (ral) constant. (See the prof ofthe theorem in Sec. 23.) Evidently, then,
J6)=0, Thatis, f is constant in D.

2
(©) Suppose that a function f is analytic in a domain D and that its modulus fis
constant there. Write |f(z)]=c, where ¢ is a (real) constant. If c=0, we see that
‘Fl@)=0 throughout D. If, onthe other hand, c +0, write STE)
TOF.
Since f() is analytic and never zero in D, the conjugate 772) must be analytic in D.
Example 3 in Sec. 24 then tells us that f(z) must be constant in D.

SECTION 25
1. (a) Teis straightforward to show that 4, +4, =0 when u(x,y)=2x(1-y). To find a
harmonic conjugate v(x,)), we stan with u,(x,3)=2=2). Now

4, =, =, =2-2y=9 Way) =2y~y +900).

22229005 Ha) =I Ox) = He.

Consequently,

v(a,y) =2y~y +(x? ye,

(0) is sraighforward 0 show that 4, +4, =0 when u(x,y)=2x—2? +39". To find a
‘harmonic conjugate v(x,y), we start with 1,(x,9)=2- 3x? +3y*. Now

1322-30 43) => v(x,y) = 2y 3: y+ y + 600).

¥, = 69 = 6xy~ $"(3) = 9) =0= AR) =e.

Consequently,
va) =2y ye e

when u(x,y)=sinhxsiny. To find a
=coshxsiny. Now

(6) Skis sraightforward to show that 4 +4
harmonic conjugate ¥(x,)), we star with (|

cosh.xsiny = v(x,y)=—eosh xeosy + PH).

4, =-v, = sinhxcosy =sinh xcosy—9"(2)=9 6x) =0= O(a) =e.

Consequently,

V3) = —coshacosy+c,

3

3

(@) It is straightforward to show that u, +4, =0 when u(x,y). To find a

—
rr

vay

* harmonie conjugate v(x,y), we star with u,(x,y) = — Now

FF = vs +60).
y >

wry

z V(x) = 09 61x) =.
===

HT
Consequently,
way)

Suppose that y and V are harmonic conjugates of u in a domain D. This means that
and u,=V,,

V,==4 tu, =

Hence w(x,))=c, where c is a (real) constant (compare the proof of the theorem in Sec.
23). Thats, v(x,3) —V(%,y

‘Suppose that u and v are harmonic conjugates of each other in a domain D. Then
and =.

Ik follows readily from these equations that

and v,=0, y,

Consequently, u(x,y) and v(x,y) must be constant throughout D (compare the proof of the
theorem in See. 23).

‘The Cauchy-Riemann equations in polar coordinates are
ru =v, and

Now

Uy ==, tog

Puy Hr Hu ST = Pa

Pu, +, + yy

whic is he polar form of Laplace's equation. To show ha y sais the same quan,
we observe that

‘This tells us that the function u=Inr is harmonic in the domain r>0,0<@<2m. Now it

follows from the Cauchy-Riemann equation ru, =v, and the derivative u, =2 that vy

thus v(r,9)= 0+ 9(7), where Pfr) is at present an arbitrary differentiable function of r.
‘The other Cauchy-Riemann equation u,=—rv, then becomes O=-r6(r). That is,
9()=0; and we see that $(r)=c, where c is an arbitrary (real) constant. Hence
-+c is a harmonic conjugate of u(r,

35
Chapter 3
SECTION 28
1. (0) expl2#3ni)=eexp(#3xi) =—é, since expl3i)= 1
lazo)
fr) Euro

(c) exple + mi) = (expz)(exp mi)

expz, since exp i

3. First write

exp@) = exp(x—iy) =e'e? =e" cosy—ie" siny,

where 2=x+iy. This tells us that exp()=u(x,y)+iv(x,y), where

u(x y) =eFcosy and v(x,y)=—eFsiny.

Suppose that the Cauchy-Riemann equations u,=v, and x, = -v, aro satisfied at some
point z= x +. Its easy 0 see tha, forthe functions wand v hee, these equations become
cos y =O and sin y= 0. But there is no value of y saying this pac of equations. We may
conclude that, since the Cauchy-Riemann equations fail to be satisfied anywhere, the
funcion exp) isnot analytic anywhere

4. Te function exp‘) is entire since it isa composition of the entire functions 2’ and expz;
and he chin rl for derivatives tls us that

Zonal)

4 à = 220xp(2')

entire by writing

alo

en

and sing the Cauchy-Riemann equations. To be specifi,
2xenpfe? y")oos(29) ~2yenp(x?—y") sin

Ayexpx? y Joos(20)=2xexp(e -3")sin(2x)

36

We first write
lexp(2z +0) =|expl2x+i(2y+=e*
and
Expte = lexpl-2ay + ie? - ze”.

‘Then, since

Iapt2z +) + expCie*)|slexp(2z+i}+exptic*),
it follows that

fexp(2z+i)+explic*)|se* +e".

First write

exp6e*)]=lexplCx +? f = apa? y) +20] = expe? -y*)

expile’) = exp + 3°).

Since x y" 5x2 + y°, it is clear that exp(x—y#) Sexp(x? + y"). Hence it follows from

the above that

fetes exp,

‘To prove that lexp(-22}< Les Rez>0, wire

exp(~22)|=fexp(-2x ~i2y)]= exp(-2x),

Lis then clar chat the statement to be proved i the same as exp(-2x)< 1 ¢2.x>0, which is

obvious from the graph of the exponential function in calculus.

37

8. (a) Write e*=~2 a5 ete? =2e™. This ells us that
&=2 and y=r+2nr n=0,4182..)
Thats,
x=in2 and y=Qn+)e (n=0,41,42,..)
Hence
2=I02+Qn+Dmi : &=04122,

ei, from which we see that

(b) Write e=1+ Bias ete”

ad y=Z 4208 m=02142,
That is,
x=In2 and y (rd) (n=0,41,22,...
Consequently,
¿=m02+(2m +2) (n= 041,42...

(c) Write exp(2z—1)=1 as ete = Lei and note how it follows that

etal and 2y= 0420 (n= 0.41.82,
Evidenty, then,
xo} and yon (= 04142...
and this means that
¿Jena (n=0,41,42,..)

9. This problem is actually to find all roots ofthe equation

38

Todo ths, setz

+ iy and rewrite the equation as

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

© and -x=x+2ne,

where n may have any one of the values n=0,41,42,... Thus

and x=nr (n=041,42,..).

‘The roots of the original equation are, therefore,
ent (n= 041,42...)

10. (a) Suppose that ef is real. Since e* = e*cosy+ie'siny, this means that e*siny=
Moreover, since e* is never zero, siny=0. Consequently, y= nm (n=0,41,42,
thats, Imz = nn (n= 0,41,42,...).

(6) On the other hand, suppose that e* is pure imaginary. It follows that cosy =
+n (n=0,41,42,...). Thatis, Imz= 1,2...)

or that

Because Re(e')= e* cosy, it follows that

moral)

Since e"* is analytic in every domain that does not contain the origin, Theorem 1 in Sec. 25
ensures that Re(e*) is harmonic in such a domain.

13. If f(2)=u(x,y) + iv(x,y) is analytic in some domain D, then
ef =e eos) Hier sin v(x).

Since eis a composition of functions that are analytic in D, it follows from Theorem 1 in
Sec. 25 that its component functions

VAN Ps VGny)= et sinv(x,y)

39

are harmonic in D. Moreover, by Theorem 2 in Sec. 25, V(x,y) is a harmonic conjugate of
Ut»)
14, The problem here i to establish the identity
(exp2)" =exp(nz) (n=0,41,42,..).
(a) To show that it is true when n=0,12,.... we use mathematical induction. It is

obviously true when n=0. Suppose that it is true when m= m, where m is any
rnonnegative integer. Then

pm + Dez).

(expz)" (expz) = exp(mz)expz =exp(mz +2)

(pay

(0) Suppose now that i a negative imteger (n=—1,~2,..), and write m=-n=12,... In
view of part (a),

« (= 1-1 1 expire).
a CHENE =

SECTION 30

1. (a) Logl-ei)=Inl-eiltiArg(e)=Ine--

(6) Log(t—i) = Init ite ing) = in V2 nai

2. (a) loge=Ine+i(0+2nn)=142nni (n=0,#1,42,..).

A most22.

ES n+ Da (n=
(9 tol) Lean) 102320 +3)o 202142,
3, (a) Observe that
Log(t +i)? =Log(2i)=In2+ Fi
and
aLog(l +0) =2( nv +12) 102+

Thus
Logi +i)? = 2Log(l +i,

() On the other hand,

Log(-1+ i? # 2Log(=1+5).

4. (a) Consider the branch
loge=tnr +10 (r>0.5<0<22),
rar

Since

years ant dogo

we find that log(F”

= 2logi when this branch of loge is taken.

(6) Now consider the branch
3m 9 ME
loge =Inr+i0 101912
B= In, (poesie)
Here

log(?)=log(-l)= In +in= ni and 2togi =

Hence, for this particular branch, log(i?)# 2logi.

5. da) The two values of i! are € and e*/, Observe that

(n=0,41,42,...)

logce®

‘Combining these two sets of values, we find that

tog”) =(n+

7

a

‘On the other hand,
Liogi = 1in1+i{ £+2nr)]=(n+1 n=
beeietfmefzemje[nl)e mensisn
mos to ts of 1 de ane a are of Ya we
may write
(0) Note that
Log) = Jog(-1) = Int + (r+ 2m) = (On + Di @=04142.)
bat that
atogé=2fnt+i($-+20n)] = (Ans (n= 041,42.

Evidently, then, the set of values of log(?) is nor the same as the set of values of
2logi. Thats,

log?) # 2logi.

To solve the equation logz=in/2, wie exp(loge) = explin /2), or 2= 6"

10. Since In(x” + y*) is the real component of any (analytic) branch of 21082, itis harmonic in

every domain that does not contain the origin, This can be verified directly by writing
(ay) Ina? +3”) and showing that u, (2,9) +4, (3,9) =0.

SECTION 31

L

Suppose that Rez, > O and Rez, >0. Then

a=nexpie, and 2 =mexpie,,
where

Zo <Z E dad

$<0,<% and -£<0, <8

“2
‘The fact that —r <0, +0, < m enables us to write
Logan) = Logl(1,1,)exp i(@, +0,)]=In(7,)+1(0, +0,)
= (nr +10,) + (nr, +10,)=Log(r, exp 19,) + Log(r,cxpi0,)

= Logs, +Log2,

3. Weare asked to show in two different ways that

fs)

3} toes “lose, G #02 #0)

(9 seva toto main a] area, nes Tan wie
2) E af a
a) la

(0), Anoter way ito first show tat log?) = log (#0). To do this, we wate 22

is = Giniz+éargz,)- (Inlz,I+iargz,) =1087, - 1082.

and then

af) Le) «tte 06 2) <tr sie},

where n=0,+1,42,.... This enables us to use the relation

opt) =!

a +ogz,

dk)

and write

43

5. The problem here is to verify that

exer) ee ws

To do this, we put m==n, where n is a negative
2 and

given that it is valid when »
integer. Then, since m is a positive integer, we may use the relations 2

acer]

fenton

SECTION 32
1. Incach part below, n= 0,41,42,..

(a) (+0 =erpliloglt+D] =f fins Ean

=omfzin2-(5+20m)]=cs9(-E-2n5)e(¿1n2)

Since m takes on all integral values, the term -2nx here can be replaced by 42n%.
Thus
z i
exp -L +20 Jexp( Lin:
+i mes a Fm ) ze 2)

tf a
© ("= ex plost-0]=exp{ tnt site 2nn}}

2. @ PN. # serps sen {int 1%] (2)

01] cto

= exp(2a?Jexp(i3n)

7

(9 PV. olas = en e tn -iZ)) acre
=e*[cos(4InV/2)+ isin(41n-¥2)] =e*[cos(2In2) + isin(21n2)).

Since 14-431

Cua” eno Donc] = e005 (eo)

Sexpfin(2"?)+ An + 1)ri] = 2/Zexplän + mil,

26”, we may write

where n=0,41,42,.... Observe that if n is even, then 3n+1 is odd; and so
explGn+1)xi]=-1.. On the other hand, ifn is odd, 3n+1 is even; and this means hat
expín+Dai]=1. So only two distinct values of (1% (30?” arise. Specifically,

(A+ N39"? = £22.

We consider here any nonzero complex number z in th exponential form 2, =7,6xp Ou.
where -#<@, S m. According to Sec. 8, the principal value of 2"* is Viens 2) and,
according to Se 32, at vale is

cxf See) =enf ori, ete en( 2-2)
neo ro exis ely one

Observe that when c= a+bi is any fixed complex number, where c%0,41,42..., the
power i* can be written as

* =esnctoso seat int Zenr)]}

ee]
[er] 202142...

and itis clear that ll is multiple-valued unless 6 = 0, or cis real. Note that the restriction
c#0,41¢2,... ensures that i* is multiple-valued even when b= 0.

SECTION 33

1. The desired derivatives can be found by writing

2. From the expressions

sinz

we see that

cosz+isinz

3. Equation (4, Sec. 33 is
2einz 0062, =sin(z+2,)+sin(a 2).
Iterchanging 2 and z here and using the fact that sin zis an odd function, we have
2eose,sinz, =sin(z, + %,)—sin(z,~2,)
‘Addition of corresponding sides ofthese two equations now yields

Asinz, cosz, +cosz,sinz,)= 2sin(z, +24),

Sina, +2,)=sinz,cosz, +cosz sing,.
4. Differentiating each side of equation ($), Sec. 33, with respect to 2,, we have

costa, +2,) = 0062 0062, sing sing,

45

46

7. (a) From the identity sin? z-+cos?z= 1, we have

, or Iran?

(0) Arso,

Tyr ot too z= sc” z

9. From the expression

sinz=sinxcoshy+icoszsinhy,

we find that
Isinaf = sin? xcosh' y+cos’ xsinh y
= sin? x(L+ sinh? y)+(1-sin? x)sinh? y
= sin? x+sinh’ y.
The expression
cosz= cosxeoshy+isinxsinhy,
‘on the other hand, tells us that

y

cos! x-+sink y.

10, Since sinh? y is never negative, it follows from Exercise 9 that

@ Isinafzsin’x, or Isina2lsinxl
and that
© Icosafzcos"x, or Icosd>lcasıl.

11. In this problem we shall use the identities

x+sinh? y, leoszf = cos? x + sinht y.

a

(a) Observe that
sinh? y =lsina? sin? Sisinz?
and
in? x + (cosh* y 1) = cosh? y~(1~sin? x)
=cosh y cos? x $ cost y.
Tus

sinh” ysisinzf’Scosh’y, or IsinhyiSIsinzis cosh y.

(6) On the other hand,

sinh? y =lcos cost x <icos af

and

leosaf = cos? x + (cosh? y~1)= cosh y (1 cos? x)
= cosh’ ysin' x 5 cosh’ y.

sinh? y Slcos a < cosh’ y, or Isinhyl<lcoszls coshy.

in xcosh y —icos xsinh y, we have

UY) HIM),

ux,y)=sinzcoshy and v(x)

cos sinh y.

Ifthe Cauchy-Riemann equations u, =¥,, u, ==v, are to hold, itis easy to see that

cosxcoshy =

and sinxsinhy =0.

Since coshy is never zer, it follows from the fist ofthese equations that cos = 0; thats,
Z +nx (n=04142,.). Furthermore, since sin x is nonzero for each of these values

2
of x, the second equation tells us that sinhy=0, or y=0. Thus the Cauchy-Riemann

‘equations hold only atthe points

tne (n= 041,42...)

Evidently, then, there is no neighborhood of any point throughout which f is analytic, and
we may conclude that sinz is not analytic anywhere.
‘The function f(2) = cosZ = cos(x —iy) = cos xcosh y + isin xsinh y can be written as

FO = ul),
where

u(x,y)=cosxcoshy and v(a,9) =sinzsinhy.

Ifthe Cauchy-Riemann equations u,

sinxcoshy=0 and cosxsinhy=0.

‘The first of these equations tells us that sinx=0, or x=na(n=0,41,42,...). Since
cosna #0, it follows that sinhy=0, or y=0. Consequently, the Cauchy-Riemann.
equations hold only when

zen (n= 04142...

So there is no neighborhood throughout which f is analytic, and this means that cos? is
nowhere analytic.

16. (a) Use expression (12), Sec. 33, 10 write

ESC) = 205 =y FR) = cos ycosh x ~isin ysinhx

cos(iZ) = cos(y + ix) = cos yeosh x~isin ysinhx.

(6) Use expression (11), Sec. 33, 0 write

Sind) = nt FH) = sin ycosh x — cos ysinhx

Sin()= sin(y + ix) = sin yeosh x + icosysinhx.

Evidentiy, then, the equation Sin) = sin(iZ) is equivalent to the pair of equations

sinycoshx=0, cosysinhx =0.

Since coshx is never zero, the first of these equations tells us that siny=0.
Consequently, y=nr (n=0,4142,..... Since cosnr=(-l) #0, the second
equation tells us that sinhx=0, or that x=0. So we may conclude that
Sin) = sin(Z) if and only if z= 0+ ine = ni (n=0,41,42.....

17. Rewriting the equation sinz = cosh4 as sinxcoshy+icosxsink y =coshd, we see that we

need to solve the pair of equations

sinxeoshy=cosh4, cosxsinh y=0

18.

49

. If y =0, the first equation becomes sinx = cosh4, which cannot be satisfied by
So y#0, and the second equation requires that

= Etre G=04142.)

sin(Z-+n0)=

Y

the first equation then becomes (-1)"cosh y = cosh4, which cannot hold when n is odd, If
itfollows that y = #4. Finally, then, the roots of sinz = cosh4 are

(Eran) (n= 04142...)

‘The problem here is 0 find all roots of the equation cosz =2. We start by writing that
equation as cosxcosh y —isinxsinh y= 2. Thus we need to solve the pair of equations

cosxcoshy=2, sinxsinhy=0

for x and y. We note that y +0 since cosx=2 if y=0, and that is impossible, So the
second in the pair of equations to be solved tells us that sinx=0, or that x=nm
(n=0+1,42,...). The first equation then tells us that (—1)"cosh y= 2; and, since coshy is
always positive, n must be even. That is, x=2nx (1=0+1,22,..... But this means that
coshy =2, or 2. Consequently, the roots ofthe given equation are

2=2nx +icosh"2 (M=041,52,

To express cosh'2, which has two values, in a different way, we begin with
y=cosh!2,orcoshy=2. This ells us that e +6” = 4; and, rewriting this as

(YA) 1=

we may apply the quadratic formula to obtain e? = 2+~3, or y=In(24+/3). Finally, with
the observation that

n2-13=

[eG fr) AN

‘we arrive at this alternative form of the roots:

2=2nrtiln(2+ V5) = 04142...)

50

SECTION 34
1. To find the derivatives of sinhz and coshz, we write

3. Identity (7), Sec. 33, is sin? z+cos*z = 1. Replacing z by iz here and using the identities
sindiz)=isinhz and cos(lz)=coshz,

we find that Psinh? z+ cosh? z =

sh? z- sin’ z= 1.
Identity (6), Sec. 33, is cos(z, + z,) = c0sz,cosz, -sinz,sinz,. Replacing z, by iz, and
2 by iz, here, we have cosa +2)]= cosiz)eos(z,)~ sin(i)sin(z). The same
ens that were used just above ten lead co

cosh(g, +2,) = coshz,coshz, +sinh, sinh z,.

6. We wish to show that
Isinh atsleoshals coshx

in two different ways.

(a) Identity (12), Sec. 34, is Icoshaf'= sinh? x + cos’ y. Thus lcoshal’-sinh?x20; and
this tells us that sinh? xSicosha, or sinhxi<icoshzl. On the other hand, since
Icoshzf = (cosh? x—1)-+c0s" y = cosh# x (1 cos? y) = cosh? x sin” y, we know that
Icoshzf ~cosh* x < 0. Consequently, Icosh fs cosh? x, or leash IS cosh.

(8) Exercise 11(b), Sec, 33, tells us that Isinh yisicoszis cosh y. Replacing z by iz here and
recalling that cosiz = coshz and iz =—y+ ix, we obtain the desired inequalities.

7. (a) Observe that

sinh(e + mi)

si

@) Also,

cosh(e+ mi) =

(c) From pars (a) and (b), we find that

inh + mi)

arm)

9. The zeros of the hyperbolic tangent function

are the same as the zeros of sinhz, which are z = ni (n=0,+1,+2,
(Eram)i m=02182...

tanhz are the zeros of coshz, or

15, (a) Observe that, since sinhz:
to solve the pair of equations.

can be writen as sinh xcosy+icoshxsin

sinh xcosy=0, coshxsin,

i
(1=0,£1,42,...). Hence

=0, the second of these equations becomes sin.

2 -(2m+3)o (n=0,41,42,..)

If x#0, the first equation requires that cosy=

, or yaztnz

(n=0,41,22,..). The second then becomes (-1)"coshx =1. But there is no nonzero
value of x satisfying this equation, and we have no additional roots of sinhz = i.

0) Being he Las tec aia, ess xn y mui

the pair of equations

, the second equation is satisfied and the first equation becomes
= + nm (n=0,41:42,..) and tis means that

‘Thus y= cos"

=(2nt 3) (n= 041,42,

10, the second uan els us at y= nz (n= 041 82,.) Teint ten
becomes (—1)" coshx

But this equation in x has no solution since coshx2 1 for

7
all x. Thus no additional roots of coshz =

are obtained.

16. Let us rewrite coshz=-2 as cosh xcosy-+isinhxsiny=~2. The problem is evidently to
solve the pair of equations

coshxcosy=-2, sinhxsiny=0.

), the second equation is satisfied and the first reduces to cos y
ing this equation, no roots of coshz = -2 arise

If x#0, we find from the second equation that siny=0, or y=nm (n:

Since cosnx =(-1), it follows from the first equation that (~1)"coshx

‘equation can hold only when n is odd, in which case x = cash 2. Consequently,

22 cosh 24 (2n+ Iai (n=0,th42,..).

Recalling from the solution of Exercise 18, Sec 33, that cosh" 2 =:tIn(2 + V3), we note that
these roots can also be written as

zetin+¥3)+Qn+Dai (n=0,81,42,

SECTION 37

“le

or uE]
(c) Since le"*l=e"*, we find that

when Re z >0.

jeansfrau]

3. The problem here is to verify that

5 £ when man,

a Shen

To do this, we write

= fees
and observe that when m:n,
12 [ e

When m=n, T becomes

and the verification is complete.

4 Fistofal,
ftom ace fetcosadeife'sinxde,

But also,

jewel]

171

3

54

Equating the real parts and then the imaginary paris of these two expressions, we find that

Leet leet

2

[rias E us fre

2

‘Consider the function w(1)=e* and observe that

Since w(c)(2n-0} =f*|27 = 2x for every real number e, itis clear that there is no number
ein the imerval 0 << 27 such that

Ja =w(cy2n-0),

(@) Suppose at) is even, His straightforward to show that us) and v() must be even

=a[facoieifooa af

() Suppose, on the other hand, that w(1) is odd. I follows that u(t) and vs) are odd, and so

Juve

Consider the functions

PAG) 12.)

je IF cos! DI a

where -1Sx 51. Since

[nenn canola ET
itfollows that
1} 5 Es
lecolsfperoli=a cose tos [40-1

ss
SECTION 38
1. (a) Start by writing

1= Jupe furor iu

‘The substitution = + in each of these two integrals on the right then yields

Junar-ijunar

woarsifsnar= war

sy
Tmena= [werde
sus
¿jua juana

and then make the substitution £= p(*) in each of he integrals on the right. The result

A ‘ 7
1= Joe’ ari Mane mdr= menear.
‘That is,
2
Jude =[wtocene’(oae.

3. The slope ofthe line through the points (a,a) and (8,b) in the Tr plane is

So the equation ofthat line is

56

Solving this equation for , one can rewrite it as

b-a „„aß-ba

Since 1= 94), then,

IE Z(1)=219(0)), where 26) = x(0) + (0) and £ = Ar), then

ZO ADA.
Hence

zu) Late OR O)
7 ae
DANI = LAD.
IE WO = J(2(0] and CR) = ut, 3) + 1,9), 200 =x(0.+1000, we have

W(t) = lato), OL HEX, 0)
‘The chain rule tells us that

Bear and str,

de

andso
WO EN.

In view of the Cauchy-Riemann equations

J, and 4, =—v,, then,

WE GX = yy) FiO HY = N)

Thatis,
WOOD OOO + O1= FORO

when 128

SECTION 40

1. (0) Let Ce the semicircle z= 2e" (0< 9 m), shown below.

‘This is the same as part (a), except forthe limits of integration. Thus

(© Finally, let C denote the entire circle 2=2e" (05 0525). In this case,

42 apna,
dr

‘the value here being the sum ofthe values of the integrals in part (a) and ().

2 (a) Thearcis C:z=1+e* (m<OS2m). Then

[tend faser-neraonifeman {ET
que
-¢

1
¿Up =0

se

(0) Here Ciz=x (05152). Then

Le-n&-fa-né le | o

‘The function to be integrated around the closed path C is f(z) = me”. We observe that
C=C+G+G,+C, and find the values of the integrals along the individual legs of the
square C.

(i) Since G is 2=x(0Sx51),
[red wfetac= et 1

Gi) Since Cis z=1+iy OS ys),

[ataca je jerga.
(ii) Since Gis z= (1-x) +i 0S x51),

f nerd afer

(ir) Since Cis <= 11-9) OS ys,

Lrésnjes

y= fera
Final, then, since
[pretidem | redes |, modes [mets | net,

we find that
fred ae 0.

4

6.

39

‘he path Cis the sum of the pas

Gizexti’ -1s x50) and C:2=x+i0 OS x81).
Using

F@)=10nG, and fe) bx on Cy,

we hve

[ 104, teodes[, 104=[ ona jerarca

¡AAA
pel ea] receto dim,

The contour Chas some parametric representation 2 = 0) (a #5), where za) = and

2(b) = z,. Then
Jae [rod RON =20)~ ka) = 2-2,
To integrate the branch

z

(a> 0,0<arge<2m)

around the circle C:z=e (0< <2), write

Let C be the positively oriented circle |;
2=e* (05 8< 2%), and let m and n be integers. Then

Tere”

|, with parametric representation

Leze= “40.

40 = fete

But we know from Exercise 3, Sec, 37, that

Teen wien men,

} [22 wien men.

60
Consequently,

275 when m+1=n.

Lera-? wien m+len,

8. Note that Cis the right-hand half of the circle x*+y*=4, So, on C, x=4=y". This

suggests the parametric representation C:z=J4—y" +iy(-2<y $2), to be used here.
‘With that representation, we have

(0) When n=21,42,...,

Jonny den (net) nierao=ir jee

=

11, In this case, where a is any real number other than zero, the same steps as in Exercise 10(b),
‘with a instead of n, yield the result

Lese

LE sinçar).
a

sl

12. (a) The function f(z) is continuous on a smooth are C, which has a parametric

representation 2=2(1) (a $1 < b). Exercise 1(b), Sec. 38, enables us to write
, :
Jf’ Wat =[ IZDA,

where . “

2) =) (astsp).
But expression (14), Sec 38, tell us that

CADIZ
and so

: A
JAetonewae={ AZOZ md.

(6) Suppose that Cis any contour and that f(z) is piecewise continuous on C. Since Ccan
be broken up into a finite chain of smooth arcs on which (2) is continuous, the
identity obtained in part (a) remains valid,

SECTION 41

1. Let Che the are of the circle lal=

shown below.

Without evaluating the integral, let us find an upper bound for Ik E | To do this, we

note that iz is a point on C,

-1j2 ie

4

1
3

año eat of Cis Lor Song Md and La, efi at

(Ejea

2. Tepu Cis as shownin fur below. Te midpoint of Cis cleat elsa onto
being V3.

Hence if is any point on C, lel222, This means that, for such a point

Consequently, by taking M = 4 and L= V3, we have

|S lee»

3. The contour Cis the closed triangular path shown below.

To find an upper bound for lí. =D], we let bea point on C and observe that

tet -ZIstel+lz

6

But e* $1 since x0, and the distance 37 FF of the point z from the origin is always
less than or equal to 4. Thus le‘ ~215'5 when zis on C. The length of C is evidently 12.
Hence, by writing M = 5 and L = 12, we have

liceo.

Note that if lal=R (R> 2), then

CRE CEE ES)

and
lef 4527-441 =1z? + 11127 +412[le? 1] [la 4] = (RE CRE ~ 4),
Thus
21
Fase

1. ae warn
e452 44 | ORR 4)

and itis clear that the value of the integral tends to zero as R tends to infinity. |

Here Cy is the positively oriented circle Id= R(R> D. If zis point on Cy, then

Logz
?

since -r<@< m. The length of Cy is, of course, 2xR. Consequently, by taking

a+ ink

M and L=2xR,

‘we see that

Since

it follows that

Let €, be the positively oriented circle Id=p (0<p< 1), shown in the figure below, and
suppose that (2) is analytic in the disk Ia <1

We let 2°!” represent any particular branch

con(-fioee) =exf-fanr+ia)]=-hen(-i2) @>a.acocas2m
of the power function here; and we note tha, sine /() is continuous on the closed
bounded dik A, there i anonnegative constant M such ht 1(2}'sM for ach point
in tat die We are asked to find an upper bound fr | ce] To do his, we
obser ati isa point on Cy,

Since the length ofthe path C, is 2p, we may conclude that

lí, oa] Ham =2m0

Not that inasmuch as M is independent of p, it follows that

ando.

65
SECTION 43

1. The function 2* (n = 0,1,2,..) has the antiderivative 2"* /(1-+1) everywhere in the finite
plane, Consequently, for any contour C from a point 2, 10 a point 27.

an |e
LT TT Er

3. Note the function (z-2,)""(n=#1,#2,.) always has an antiderivative in any domain that
does not contain the point z=2,. So, by the theorem in Sec. 42,

[e-2)"&=0

for any closed contour C, that does not pass through 2,

5. LetCdenote any contour from 2=—1 to z=1 that, except for its end points, lies above the
ral axis. This exercise asks us to evaluate the integral

‘where z' denotes the principal branch

‘pCiLoge) (d>0,-<Argz<m).

66

An antiderivative of this branch cannot be used since the branch is not even defined at
=1. But the integrand can be replaced by the branch

eet) (er0-&canec®)

since it agrees with the integrand along C. Using an antiderivative of this new branch, we
can now write

Ce peon en

ae RE

SECTION 46

2. The contours G and C, are as shown in the figure below.

In each of the cases below, the singularities ofthe integrand lie outside G, or inside C,; and
so the integrand is analytic on the contours and between them. Consequently,

Lfo«=[ fa

a

(a) When Fe) = sr the singularities ae the points

242

2, the singul at 2=2nn (n=!
Facer" Singulries are at z= 2 (n= 81

(0) When f(2)=:

(c) When JG)

the singularities are at z = 2nni (n= 0,41,2,...).

ra

4. (a) In order to derive the integration formula in question, we integrate the function
around the closed rectangular path shown below,

Since the lower horizontal leg is represented by z= x (-a£x< a), the integral of

€ along that leg is

Since the opposite direction of the upper horizontal leg has parametric representation
2=x+bi (-aSxS a), the integral of e* along the upper leg is

Jer cosabede +e! fe sin2beds,

ae fe nds

Since the ighthand vertical legis represented by z= a-+iy (05 yb) the ira of
e” along itis

fertigte jaca,

68
Finally since the opposite direction of the left-hand vertical leg has the representation
a+ (OS ySb), the integral of €" along that vertical lg is

ei rs

According to the Cauchy-Goursat theorem, then,

(0) We now let a>= in the final equation in part (a), keeping in mind the known
integration formula

fra
and the fact that ‘

ehe nase gone

The resultis

Les 6>0)

6 Welet C denote the entire boundary of the semicircular region appearing below. Itis made
up of the leg C; from the origin Lo the point ¿=1, the semicircular arc C, that is shown, and
the leg C, from z=~I tothe origin. Thus C=C, +C, +G,

9

We also et f(z) be a continuous function tha is defined on this closed semicircular region
by writing (0)=0 and using the branch

IO (>0-5<0<35)

of the multiple-valued function 2/*. The problem here is to evaluate the integral of f(z)
around C by evaluating the integrals along the individual paths G, C,,and C, and then
adding the results. In each case, we write a parametric representation for the path (or a
‘lad one) and then us it to evaluate the integral long the paria a e P

@ Gra=re"(0SrSD, Then

[104 [rid E

fi) G:z=1.e* 0SO<m). Then

Isar fet? erao=ifenao

(ii) -G; z=re* S181). Then

[,fodn-[., ford J rear

“The desired result is

[Foden | serdar |, fords |, 1094

defined on the negative imaginary axis

SECTION 48
1. Inthis problem, we let C denote the square contour shown in the figure below.

7

A 22
cose CT +9 or
oa a)

1 ee].

Le %
= 2ni( sect 22) = imsec!( 2) when 2<x, <
ze à) (2) men 2<x, <2.

2. Let C denote the positively oriented circle 1z- il

„shown below.

Mere. 2 (a

(0) Applying the extended form of the Cauchy integral formula, we have

heat heating af]
era GA le a de

E

LL. TS

n

3, Let Che the positively oriented circle Izi=3, and consider the function

wle3),

We observe that

e [2%

=2mi[24'—2-2),,, = 2ni(4) = Bi.

2-2

(On the other hand, when Iw > 3, the Cauchy-Goursat theorem tells us that g(w)=0.

5. Suppose that a function fis analytic inside and on a simple closed contour C and that zo is
noton C. If za is inside C, then

Led _ | _fledde__ 2m a
LES rl)

Lid
LEE

LÉO. | 10%
ee

u Hea)

‘The Cauchy-Goursat theorem tells us that this last equation is also valid when 2 is exterior
10 C,each side ofthe equation being 0.

7. Let C be the unit circle z=e* (#8), and let a denote any real constant. The
Cauchy integral formula reveals that

n
On the other hand, the stated parametric representation for C gives us

[Ea fore iene espta(eos + sin 140

à fe***#{cos(asin 9) + isin(asin Od
= fet intasin8)40 +i fe costasin 8)48.
Equating these two different expressions forthe integral [La we have
— fe? sin(asin 6)d8 + i e**"* cos(asin 6)d0 = 2
“Then, by equating the imaginary pats on each side ofthis last equation, we see that
fet" cos(asin 8)d8 =2n;

and, since the integrand here is even,

fet" cos(asin 6)d0 = x.

8. (a) The binomial formula enables us to write

Lota ap RN as
road Sey =e LS} cy.

We note that the highest power of z appearing under the derivative is 2%, and
differentiating it times brings it down to 7°. So P,(2) is a polynomial of degree n.

() We let C denote any positively oriented simple closed contour surrounding a fized point
2. The Cauchy integral formula for derivatives tells us that

012.)

E er

Im)
Hence the polynomials P,(e) in part (a) can be written

1 Gyr

Er Pre

B
(e) Note that

CENT

FCM ACM (= 012.)

feray-f@ 1

Az
Then
Auer £@)_ 1p fords _ 1 As-2)-Az 5) 6
E <2) 2ni le OS E = pon

2136-24-24)
la oO

7
(©) We must show that

GDiaai + 21d YM
CAE

Now D, d, M, and L are as inthe statement of he exercise in the text, The triangle
‘inequality tells us that

13(5=2)d2—2(Az)*is 31s ~ lA + 21027 S IDA + DA.

‘Also, we know from the verification of the expression for f"(z) in the lemma that
ls 2-42 d -1Ad>0; and this means that

I(s-2- A2) (5-2) 12 (d -1dd))*d > 0.
‘This gives the desired inequality.
(6) Ifwelet Az tend to Oin the inequality obtained in part (6) we find that

Az - (A2)
a ro =

‘This, together with the result in part (a), yields the desided expression for £"(2).

15
Chapter 5

SECTION 52

1. Weare asked to show in two ways that the sequence

ai

a

converges to 2. One way is to note thatthe two sequences

„en

x=-2 and y, (=L2..)

of real numbers converge to ~2 and 0, respectively, and then to apply th theorem in See.
St. Another way is to observe that e, =(-2) > ‘Thus for each £>0,

-C2}<e whenever non,

‘ery is any postive ger such that ny 2

2 ornato qi

reste set 2

O, = Arey, and Oui = Age, (2212.0,

then

But, since
the sequence 6, (n=1,2,...) does not converge.

3. Suppose that limz, =z. That is, for each £>0, there is a positive integer n, such that
lz, =21< e whenever > ny. In view of the inequality (see Sec. 4)

lala,

itfollows that Hi e whenever n>n,. Thatis, limit.

26
4. The summation formula found in the example in See. 52 can be written

when Izicl.

If we put z=re, where 0 <r < 1, the left-hand side becomes
Zoe

and the right-hand side takes the form

y cosn8+ ¡Sr sind;

ret ire 1080-7? +irsin®

Tore Imre Tore" Hemer 1270004

FETES rsinó.

Pa een

Equating the real parts on each side here and then the imaginary parts, we arrive at the
‘summation formulas

5 = reg" a sind
Per ET ETS

where O<r<1. These formulas clearly hold when 7 = 0 100.

6. Suppose thar Y,

appeal tothe theorem in Sec. 52. First of al, we note that

3, we write 2,

+ S= XY and

5 Tostow dat Fe

Lauer a ner

am ace A

i.) = Dlx, +i(-y, = Xi =.

n
8. Suppose that Sr, =S and Zw, = 7. In order to use the theorem in Sec. 52, we write

went, SHX+iY and veut, THU+iV,

ner and du

Fo. +u)=X+U and Iontwertv,

it follows that

Die, +4) +10, Hy) x HUHN).

Thats,
Die, ++ HKM,
Fa +m)=S+T.
SECTION 54

1. Replace zby 2* in the known series

cosa (ic)
wgx

cose (aca).
‘Then, mulpling trough his at equation by z, we have the deste result

zeosh(e?)= Ÿ ET (ae).

Beni

78
2. (0) Replacing zby z-1 in the known expansion

(d<@),
veuve
(d< >).
»
Ses, ser (<a.
3. Wewantio find the Maclurin series forthe function
E
14479)"
“To do this, we first replace z by -(z* /9) in the known expansion
1
mi (a<D,
as wel ass condition of valid, o get
i &
nem» u

‘Then, if we multiply through this last equation by > ‘we have the desired expansion:
f@= ie AY ann ta< 3).

art

6. Replacing z by 2° in the representation

oe fee
wehave
sin) co À Qué)

nin

»
Since the coefficient of z* in the Maclaurin series for a function f(2) is (0)/ nl, this
Shows that

"@)=0 and f*(0)=0 CU

7. Te fanion 7,
lei < V2, as indicated in the figure below.

has a singularity at z=1. So the Taylor series about z=i is valid when

ni

‘This suggests that we replace z by (z~i)/ (1- in the known expansion

Le aac
E
and en multiply through by —L-. The desired Taylor eres sen obtained:
(e-a< V2),

9 The identity sinh(z-+ mi) =—sinhz and the periodicity of sinh z, with period 2, tll us that

sinhz = -sinh(e+ xi

inh( = Mi)

So, if we replace ¿by z— in the known representation

Ei
Dern u

80
and then multiply through by —1, we find that

sinh = E Games).

(zen,
1 a Sas
tee era hea he
SECTION 56
1. We may use the expansion
a
2% Yen (d<=)
tose that when Da <
2, 1 ech 1 pay.
E AC AA
E
3. Suppose that 1 land recall the Maciucn series representation
(acy

Allee).

aL
wwe arrive at the desired expansion:

cy
Fer Dra (<ld<),

1. The singularities of the function f(z)= Fag we a he points z= 0 and 2=1 Hence
sings Za

ter a Lara sis in pones of o ie domains Ode and 1 id (ue de
figure below).

Te ti re non ld when cl. efi we ea
the Maclaurin series representation

(act)

(ac).

2
(0) To find the Laurent series for the same function when 1<lalé, we recall the

Maclaurin series for —— that was used in part (a). Since

einer ve may wie
z

bbs

SL

me ae

(ide).

Pa
+e)

the figure below. Hence there is a Laurent series representation for the domain O <Izi<1

7. The function f(z) =: has isolated singularities at

and also one for the domain 1 <l2l< =, which is exterior o the circle Izl=1.

(den.

cyte

f=

In this second expansion, we have used the fact that (1)! = (-1)"(-)

83
8. (a) Let a denote a real number, where -1< a <1. Recalling that

Se esco

‘enables us to write

ríe (laictal< 0),

(b) Puting z=e* on each side of the final result in part (a), we have

But

gi a (60800) ~isind _ acos0—a'—iasinO
a (cos0—a) isin (c0s0—a)—isind 1-2acosd+a*

Es cosnd i$ sinnd,

10, (a) Let z be any fixed complex number and C the unit circle w =,

109= 1-3]

has the one singularity w = 0 in the w plane. That singularity is, of cours, interior to
Gas shown in the figure below. Bun

84

AN
NS
da

Now the function f{w) has a Laurent series representation in the domain O <Imi< =.
According to expression (5), Sec. SS, then,

«lie Sue Damen,

u

Using the parametric representation w = €* (-FS $$ 1) for C, letus rewrite
this expression for J,(2) as follows:

iea0 =. Jerntzsinole "ao

Fe oxtt-ind ~csin pes (M=0,+1,42...).

(b) “The last expression for J,(2)in part (a) can be written as

Jaa

Jieosing - zsin 9) —isin(nó —zsing)1dó

ar

#35 fentes canoas juncos

28

= 7 2fcostné -zein ap - 5 (n= 01,82...)

85

Thats,
1 eosind sin yd (n= 0,422,

11. (@) The fonction f(e) is analytic in some annular domain centered at he origin; and the
unit circle C:z =e (-1 < $< x) is contained in that domain, as shown below.

y

Foreach point z in the annular domain, there is a Laurent series representation

Dare Eh

where

= ef LOH ah [HOD raga Len) 4

a lreneas
Substituting these values of a, and b, into the series, we then have

10 limar ra

fot I fe elle] {2}

in the final result in part (a) to get

set I per) dor LS if perfor? à mena,

f(e*)cosIn(8 - 6))dé.
If w(@)=Re f(e”), then, equating the real parts on each side of this last equation yields

= una, LS Jugrcotnco- on.

SECTION 60

1. Differentiating each side of the representation

(<D,
we find that
Lorne zie,
Another diferentiation gives =
arado Erroneo duch

Te as

2. Replace zby 1/(1-2) on each side ofthe Mactaurin serie representation (Exercise 1)

(cb,

Q<lz- liées),

87
3. Since the function f(2)=1/z has a singular point at z= 0, its Taylor series about z, =2 is
valid in the open disk Iz —21<2, as indicated in the figure below.

To find that series, write

DIE

(ae,
Specifically,
(e~21<2),
(2-22).
Differentiating this series term by term, we have
Ent -27 e-21<2),
Tus
Fora (2-2<2).

4... Consider he function defined by the equations

88

When 20, f(z) has the power series representation

EN PREPA TA
(o) 1 Mets

Since this representation clearly holds when z = 0 too, it is actually valid for all z. Hence f
is entire.

Let € be a contour lying in the open disk Iw-l<1 in the w plane that extends from the
point w=1 to. point w =z, as shown in the figure below.

According to Theorem 1 in Sec. 59, we can integrate the Taylor series representation

=Fenw-ır Qu II)
term by term long the contour C. Thus
E [0 dv.
But
Loge—Log!=Logz
and
toy DT ey
E foward [ a ] Fr
Hence

Ce N a-ıy CE

aged
ae mone El

AD? = (NÉ, this result becomes

Loge FS an y del.

89
SECTION 61

1. The singularities of the function f(

ae +)
find the Laurent series for fthat is valid in the punctured disk 0 <lzl<1, shown below.

Foray Meat 2=0,£i, The problem here is to

(déc)
and
(a<b,
which enable us to write
ei (dc)
and
att (en.

a+
Multiplying these last two series term by term, we have the Maclaurin series representation

e 1
Hate los.
Fer $

which is valid when lel The desired Laurent series is hen obtained by mulipying each
side of the above representation vy 1

el

Wan: Oxide.

as

We know the Laurent series representation

Ze Oasen

from Example 2, Sec. 61. Expression (3), Sec. 55, for the coefficients b, in a Laurent series

tells us that he coefficient dy of + in this series can be written
z

PR SE u
NT 2m tee sinh”

‘where Cis the circle Ial=1, taken counterclockwise, Since b, =

E)

‘The problem here is to use mathematical induction to verify the differentiation formula.

UVeser= Errore 12...

‘The formula is clearly true when n = 1 since in that case it becomes
LOG) = O8 + O8.

We now assume that the formula is true when n= m and show how, as a consequence, it is
true when n = m +1. We start by writing

E OA RS

SOFT" WS”

= DC TE morro

(rar

- vo raros

9
But

CH) mi mt tI (mer
A JUL) Km) (k= Dim b+ DE Klm+i-k)! ke}

and so

(OEI EY rose FE),

Lest“

MEN, un. geh,
o.

‘The desired verification is now complete.

. We ae given that f(e) is an entire function represented by a series of the form
JOR baste (ac).
(a) Write g(2)= FU] and observe that

(aseo),

ae 080.020

ID 3!
Itis straightforward to show that
SOLO“.

£O=PUOI OF + SU" @.
and

COS UMD +26 OF OF UO LOTTO").
‘Thus
8(0)=0, g(0=1 g"(O)=4a,, and g(0)=12(a3 ta).

and 0
SLAC =2+2a,2" +20} +0)0+ Qd<e).

(6) Proceeding formally, we have

ASN FO + GUE + al SN +

ta ta+-)ta (tag tat) ta(etag! tag h) be

Sera tarte) + tad + (ate)
meta +2 +a),

(©) Since

(d<=)

(dc),

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

2 (4)

a” Zn
is representation is valid in de stated disk inc the zeros of coshz aro he numbers
2H) (a 2044122. ie ons nearest 10 he origin being =F

i. The series

2
contains only even powers of z since coshz is an even function; that is, Eu
(n=0,1,2,...). To find the series, we divide the series

tg, À
coshe= 142+ deste Ne)
zen tt dc)

into 1. The result is
£
ee
(a 2

(eek

Since

Be

‘his tells us that

94

Chapter 6
SECTION 64
1. (0) Lotus wie
Lia a
el lo 4
lee (<<.
The side at 2=0, which sth coefficient of 1, is leary 1
(0) We may use the expansion
fies
aaa (leo)

to write

&
Odd<~).

‘The residue at z

(c) Observe that

(O<tzi< =)

z

Since the coefficient of + in his Laurent series i, the residue at «= 150.

(d) Write
se
sin
and recall that
(ace)
and

(alc).

9
Dividing the series for sinz into the one for cosz, we find that

Odkl< m)

Odden),

2=nx (n=0,41,42,..). ILis now evident that SE has residue
?

(e) Recall that

(dc)

(de),

‘There is a Laurent series forthe function.

sinhz

Faye oy)

thatis valid for O<tal<1. To find it, we first mulüply the Maclaurin series for sinhz

ds Lo Mis as à

Jeter ee Je +24
ds

poros

2 yds
CES

cedo ated

Work oat
(0 <lzl< 1).
is tows tte esi of BE, u 220 2.
des 3

2, Ineach part, C denotes the positively oriented circle la=3.

1 Too [SID a, none de ras fe ineganda 250. Fm

the Laurent series

uud décor,

22

a

we see that the required residue is —1. Thus

LED = 2ni(-1)

(e) Likewise, to evaluate the integral, c'e jr, we must find ie reidue ofthe
aac
inegandat +

Em (tt

2/2/11
sé+£slil
MT Mz

‘The Laurent series

+)

(@ sorte inet [ELL vecs mo reins of
gel est
2-2 4-2)

fone at ¿=0 and one at 2 = 2. The residue at z =

wa ES

can be found by writing

whichis valid when O-del<2, and observing that te coefficient of À inthis last

pénis À rocade date wwe

241 (@=243, 1 Las 3 ) 1

de) 22 2D A 2) GD

dal

which is valid when 0 <=21<2, and note thatthe coefficient of —L in his product
=

i 3 ay en y ai een,

ze { 1 3)
2-24
per 2°2

3. In each part of this problem, € is the positively oriented circle 1dl=2.
(a) If fe)
24)

z

when 0 daté 1. This tells us that

[rod =2mRes 4 7(+)=20i-9=-2.

98
1
©) When f(a) =r. we have

0:

Lio =inres 5 /(2)=2010)=0

(9 à oh atom Lal, a te

J.10d:=20805 22) 202.

4. Let C denote the circle I=

, taken counterclockwise.

Se

(a) The Maclaurin series e* = Ÿ 2 (zi< =) enables us to write

ime)

E

(6) Referring to the Maciaurin seres for e* once again, let us write
PONS Let

Now the 2 in this series occurs when n-k
theorem,

Si fees

1

Veen

‘The final result in part (a) thus reduces to

honfe+!)a=2ui

(De

Odach.

(n=012...

, or k=n+L. So, by the residue

(n=041,

We are given two polynomials
PO taztaz tra: (a, #0)
and
MS be + het be ++ be 6,40),
where mz 42.

ILis straightforward to show that
a ae tat
Bethe tae tb,

‘Observe that the numerator here is, in fact, a polynomial since m—n-220. Also, since
b, #0, the quotient of these polynomials is represented by a series of the form

drdard alt. Thatis,

1 Pale,
TES

(#0).

Pale)
72)

1
7

dy tdet dye +. OddcR)s

=]

1 Pasa) z
and we ce hat 27 O7, Nas ride 2=0.

Suppose now that all of the zeros of Q(e) lie inside a simple closed contour C, and
assume that Cis positively oriented. Since P(z)/ Q(z) is analytic everywhere in the finite
plane except atthe zeros of Q(), it follows from the theorem in Sec. 64 and the residue just
obtained that

PO p< opiRec| 1. Pall)
ASES quel

If Cis negatively oriented, this results sul true since then

Pe)
0)

PQ),
d=- | Demo,
Leo

SECTION 65
1. (a) From the expansion
(ds),

‘we see that

Old).

100
“The principal part of ces) at the isolated singular point z =0 i, then,

1

Br
Are

and ¿=0 is an essential singular point of that function.

(0) The isolate singular point of ¡E ia z

involves powers of z+1, we begin by observing that

Since the principal part at z

(e+ = 22-1

= +) = 22+) +L
‘This enables us to write

2 Gena
Te En

Cry
=

Sine the principal pat is le point 2=-1 ia simple) poe
(c) The point z=0 is the isolated singular point of 24, and we can write

id=).

‘The principal pat here is evidently 0, and so z=0 isa removable singular point ofthe

funcion 4

@

Ode),

the principal partis = 2. mis means that 2=0 is a simple) poe of SE

1
(0 pon wine GE ay

isolated singular point z = 2 is simply the function itself. That point is evidently a pole
(of order 3).

5 we find that the principal part of "> at its
es

101

ESO]
isz=L The Ta
agp FEL The Taylor series

of 2-1)

(c) The singular point of

exp = ee

PEPA ls
Fe] tee

$. e
Gene Odr-I<e).
3. Since fis analytic at 2 it has a Taylor series representation
$= Nha) Le a+ LD gg} + (ea l<R).

Let g be defined by means ofthe equation

Lo
%

s@=

102
(a) Suppose that f(z,) #0. Then

[ner De een! +]

La Le La aye. Ode-zicR).

This shows that g has a simple pole at 2, with residuo f(2,).

(6) Suppose, on the other hand, that f(2,)=0. Then

1 fre) en
ee Dear]

LE Le aye (Ode sleR)

Since the principal part of g at 2, is just 0, the point 2=0 is a removable singular
point of 4

4. Write the function.
Be

10 (a>0)
=
EL where pt e HE
za er

Since the only singularity of 9(z) is at z=—ai, $(z) has a Taylor series representation
$= a+ PE a a +. (= ail<20)
about ¿=al. Thus

ay LD a+ 8
(ec + Te an + LE

102 Gare] Ocle-aite2e,
e

Now straightforward differentiation reveals that

103
Consequently,

(ai) =

Fi, ga)=—5, and #'(ai

‘This enables us to write

f= cal

EN (=a) Lea + ] (O<iz-ail<20).

2
“The principal part of fat the point z= ai is, then,

i ald

SECTION 67
1. (a) The function f(2) =

where $(z)=2? +2, and observing that 6(2) is analytic and nonzero at z= 1, we see
that z=1 isa pole of order m=1 and that the residue there is B= 9(1) =3.

© Eve write

ST where 0

fo

wo see that 2=- isa singular point of f Since $(2) is analytic and nonzero at that
point, fhas a pole of order m=3 there, The residue is
pa) 3
2 16
(c) The function
expe expz

em Gamera)

has poles of order m=1 at the two points z =i, The residue at z= mi is

104
2. (0) Weitthefnction JG =~ 0d>0,0<age<2m a

10-22, where 9G) =2"

* 0d>0,0<argz< 2m.

‘The function 9(z) is analytic throughout its domain of definition, indicated in the
figure below.

#0.

= ent mens isin T

”

being

it the funcion f(z) = HOB
0) Wee unción Se) = ¿E as

102 where

From this, it is clear that f(z) has a pole of order m=2 at z=i. Straightforward
differentiation then reveals that

ata

=9=

er E + ear

105
(c) Weite the function

(d>0,0<argz<2m)

Since

3. (a). We wish to evaluate the integral

LE
Keane)

where Cis the circle 12—21=2, taken in the counterclockwise direction. That circle and
the singularities z= 1, +31 of the integrand are shown in the figure just below.

Observe that the point z =1, which is the only singularity inside C, isa simple pole of
the integrand and that

P42 304
AA

According 10 he residue theorem, then,

32°42 (1
Lise)

106
(©) Lotus redo part (a)when Cis changed to be the positively oriente circle li 4, shown
in the figure below.

In this case, all three singularities z= 1, +3i of the integrand are interior to C. We
already know from part (a) that

32 +2 #
Sere "2

Lis, moreover, straightforward to show that

3242

3242] _ 154491

2

‘The residue theorem now tells us that

32° +2
tal

4 (0) Let € dena poly ond ce 122, and not at he integrand ofthe

integral J_—- has singularities at z=Oand ¿=-4, (Sec the figure below.)
cz FE

107
To find the residue of the integrand at z = 0, we recall the expansion

(ae
and wire
—_- 11. Fz) Fo
Fern aa] 223 Er O<ti<a).
Nor ect ot oc wen 12, ane ta

Rest =
= Fern
Consequently,
aaa
eed)

(b) Letus replace the path C in part (a) by the positively oriented circle lz + 21
at ~2 and with radius 3. It is shown below.

ÓN
NA

We already know from part (a) that

Res 4.
DETTE ET:

To find the residue at —4, we write

1 oz)
A MR Marz

‘This tells us that z= —4 is a simple pole ofthe integrand and that the residue there is
$(-4) =-1/ 64, Consequently,

de
al,

108
$. asus rt ingl [SE u ithe pestle i 2
A tos fold sigularidos ¢= 0-4 of the integrand ar intro to C. ‘The desire

residues are

cosh az ee]
Res
Se 241

cosh me _ cosh mz
en.

i
a
com _entæ] „1
Reste se] „1,
el.
Consequently,

6. Ineach par of this problem, C denotes the positively oriented circle Izl= 3.

(a) is straightforward to show that

Get then +0 Br

O" Dar

mistico (8) basins

Garay
Les **

2-32)
Trade)"

1
ao “ies +2

‘he facon 7 (2) asa imp pot at 220, and we find or hat

La

109
(c) Finally,

1
1) Zatz)
The point 2 0 isa poe of oder20f (2). ‘The residue is (0), where
Az

re

eo

the value of 6/(0) is 1. So

SECTION 69
L (a) Write
1-29 E ai
de © where p(z)=1 and g(z) =sinz.
Sus

PO=1#0, 40)

2=0 must be a simple pole of cscz, with residue

201,
gO 1

(0) From Exercise 2, Sec. 61, we know that

erlegen

1 dee (O<tel< a).

Gy St

Since th coeticient of À here is 1, it follows that z=0 is a simple pole of cscz, the
residue being 1.

10

2. (a) Write
FO, where pt =2-sinhe and glo me
ne: P@)=z-sinhz and 9) => sinbz.
Since
PO=mRO, a(ni)=0, and g'(ni)=n" #0,
itfottows that
zosinbe _ pi) _ mi
RS Paine 9 Ga) 07
© Wie
EL PO where pla) = =
She Tag Where PORRO and g() = sinhz,
Is easy to see that
Res MEN. DE) ia and Reg PEO PE _
E TN To E
Evidently, hen,
PDC y pe OC) - y expla) + exp(-imt)
Res PED 5 Res BEN), TC O
SE inne 888 sinh? 2 Dos
3. (@) Write
= LO) ah = =
E Where pl = and g(c)= cose
Observe that
dö+nr)-0
Also, forthe stated values of,

{gone)eZenre0 ma of

am
So the function /(2)=_ E haspoles of order m=1 at each ofthe points
2

(=0,+1,42.....

‘The corresponding residues are

LORS

ae)

(0) Write

ane» 22, where p(e)=sinh and q(z) = cosh,

Both p and q are entire, and the zeros of q are (See. 34)

(1=0,41,42,..

In addition o the fact that

(goa)

So the points 2=(Z-4nn) (n= O12.) ae poles of order m=1 ofthe, the
res in each es Being

2
4. Let C be the positively oriented circle t=

(a) Toevauae the integral [anse we wi he integrand as

10) is
z= 22, where ple)! (2) =0082,
7) p(e)=sinz and q(z) = cos:
and recall thatthe zeros of eosz are 2= hna (m

+1,82...) Only two of those

zeros, namely z=%x/2, are interior to C, and they are the isolated singularities of
tanz inierior to C. Observe that

crie ED ui Regen ee
Regine cgay M BRE Ca)

Hence
f¿tanzdo =2ril-1-)=-4mi.
de

E. To do this, we write the

(6) The problem here is to evaluate the integral |. sant

integrand as

LA

Fare Fy’ “em =H and alo) sinh

Now sinh2z=0 when 22= ri (n

#1,42,..), or when

‚mi
= (n=0,41,82,..).
Three of these zeros of sinh2z, namely Oand:t =, are inside C and are the isolated
singularities of the integrand that need to be considered here. It is straightforward to
show that

A
RS Sande TO Tod 2

18

ad
ec. ee NS
msinh2z g’-mil2) 2cosh(-m) 2cos(-7) 2
Thus

de
Kenne

5

71 has isolated singularities at
Zsinz

Within Cy, the function

2=0 and z=ine (n=12,...,N).

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

), we recall the Laurent series for cscz that was found in Exercise

LL eae Ee
Fle ey 5

(O<la< a).

14

This tells us al 7

Has a pole oforder3 at z=0 and that

1

Res
DETTE

AM waite

As forthe points 2 =-tnx (n= 1,2,

120,

where pt = sine.
© pl) =1 and q(z)=2'sinz,

Since

pitnn)=140, atm)

(inn) = n°? cosnm = (nr #0,

it follows that

Res

res

Er oy

So, by the residue theorem,

ae
Men 2? sinz
Rewriting this equation in the form
Cyt
A 12 die ¿sinz

‘and recalling from Exercise 7, Sec. 41, that the value of the integral here tends to zero as N
tends to infinity, we arrive atthe desired summation formula:

6. The path C here is the positively oriented boundary of the rectangle with vertices at the
points +2. and +2+i. The problem is to evaluate the integral

de
La

us
‘The isolated singularities ofthe integrand are the zeros of he polynomial

a =(@ 143

‘Setting this polynomial equal to zero and solving for 2”, we find that any zero of q(z) has
the property z* =1:-V/3i. It is straightforward to find the two square roots of 1+-V3i and
also the two square roots of 1~¥/3i. These are the four zeros of (2). Only two of those
e108,

Bi
“Er

ie

PER A ae = =

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

‘To find the residues at z, and —2,, we write the integrand of the integral to be evaluated as

1 2

= wi (2) =: e?
Era whee ml =

“This polynomial g(z) is, of course, the same q(2) as above; hence g(2,)=0. Note, too, that
and q are analytic at 2, and that p(z,)#0. Finally, it is straightforward to show that

4{(@)= 4e(2! 1) and hence that
1 (e)= 40-1) =-206 +603 #0.
‘We may conclude, then, that z, is a simple pol ofthe integrand, with residue

PG) 1
YG) WOH

Similar results arc to be found at the singular point ~Z,. To be specific, ts easy to see that

ge)

the residue of the integrand at -Z, being

2). 1
CEE ET EE

16
Finally, by the residuo theorem,

fé à 1 Pi
lar EN en)

he a

7. We are given that f(z)=1/ {q(2)F, where q is analytic at 2, g(2,)=0, and q/(2,) #0.
‘These conditions on q tell us that g has a zero of order m=1 at %. Hence
(2) = (—z,)g(@), where g is a function that is analytic and nonzero at 2,; and this enables
us to write

fae + where (=

a KG Fo

So fhas a pole of order 2 at 24, and

Res f(@)= 00) == 280),
Res f= 9) = E

But, since g(2) = (2 2,)8(€), we know that
dO=G-2)8 +80) and OCULAR.

‘Then, by setting z= z, in these last wo equations, we find that

HAIE) and 9"(e) =28'(e)-

Consequently, our expression for the residue of fat 2, can be put inthe desired form:

Res f()= ive”

8. (a) To find the residue of the function csc*z at z= 0, we write

i A

An, wien ging.
or are

Since q is entire, 4(0) =0, and g/(0)= 1 # 0, the result in Exercise 7 tells us that

Re go _
RS op =o

um

(©) The resdve ofthe function F7 at 2 = 0 can be obtained by writing
ae

N
er) “a

Inasmuch asg is entire, g(0)=0, and 4/(0) =1 #0, we know from Exercise 7 that

mare Mh osa.

1 ro
Res —L
ON

18
Chapter 7

SECTION 72

1. Tocvatate de integral 5%. we negra the funcion FO = 7 around ie simple
closed contour shown below, where R> 1.

We see that

where

Now if zis a point on Cy,

and so

Finally, then

no

y q 1
minga JE era el n fos a
2 Meinen [Ly cant vat wing te con JL adhe a

simple closed contour as in Exercise 1. Here

fa de
ly

j de de
AY et

Ifzisa pointon Cy, we know from Exercise 1 that

i242 R=1;
sn
à
Le 5 Rao
The desir resis, te,
j de a e. | dez
IT er

3. We begin the evaluation of JE by finding ie zeros ofthe polynomia +1, which are

the fourth roots of —1, and noting that two of them are below the real axis. In fact, if we
consider the simple closed contour shown below, where R>1, that contour encloses only
the two roots

and

wom semen aft

120

Now

where

Since

wwe have

Eau

2
deal

“fy

E7A

Fr

A We mit ta te eg! |

shown below, where R>2.

We must find the idues of the function f()=—
vanes ess (Far
z=i and 2=2i. They are

|
ARO TT
and
. à
HE di rl
Ts
Í ELA Éd _
Lee MCS CEST)
«
jazz Bde
Janes RENTE ETS

Wzisa point on Cy, then
RHIN RL and 1 +412 ml RP.

Consequently,
2 z
de aR
Le SES ms
E R

and we may conclude that

de y A
LA as

121

We use the simple closed contour

at its simple poles

12
¡Fa ï
Jesse can be evaluated it the aid ofthe funcion

5. The integral

E
NOTE

and the simple closed contour shown below, where R>3.

We start by writing

ce dd
foot hope Rat
where
sá Res
x Sipe A

ETES
To find B,, we write

Fie 2
EET A IO a
Bea E,

‘This tells us that
fae a

le FO +47 100,

ode
I. COEUR

Tod
le +97 +4)

ade

OS

Finally, since

we find that

7. Inorder to show that

ade

"res 5

we introduce the function

z

en

#6 +e" +2242)

and the simple closed contour shown below.

13

In

Observe that the singularities of f(z) are at i,

2=-1#i and their conjugates -i,

i in the lower half plane. Also, if R> 2, oma

preeemese +B),

where
BR

and

Evidenty, ten,
je ede
E RR

Since

N ade +f. de ER

RN CE DR VIS

as Re, this means that

tim [HE —
Boe +22)
“This is he desired result

124

t far _2r
The problema here it establish e integracion formal sing he simp
8. The problem ron formula FA = ZF win sine
closed contour shown below, where R>1.
Te inl oe ingly te aston J) = gg manly = tac nr
?

to the closed contour when R> 1. According to the residue theorem,

de de de +
A A
Ad
where the legs of the closed contour ar as indicated in ie figure. Since €, has parametre
representation =r (OSS R),
LE
PIPA

and, since ~C, can be represented by z=re"*” (OSr SR),

de
¿AP ET
Furthermore,
Le UN
330373
Consequently,
part a 1
a-e De PL
But

“This gives us the desired result, with the variable of integration y instead of x:

¡A na ee nie
¡a qe aa 3

125

Letm and n be integers, where 0<m<n. The problem here is to derive the integration
formula

Te LA 2m+1
EA
een)

(a) The zeros of the polynomial 2” +1 occur when 22

eo ep Be] 12.20),
“clear that the zeros of z°" +1 in the upper half plane are
song DE] (=0,12,..n=1)

and that there are none on the real axis.

(0) With the aid of Theorem 2 in Sec. 69, we find that

La
a m“
Putting an 2th we can write

(2k+1)x(2m~2n4:

(HD,

‘The residue theorem tells us that

2m u
[el em

Observe that fz is a pointon Cy, then
Wet RÉ and 124112 R=,

‘Consequently,

zu
rare

and the desired integration formula follows.

10. The problem here is to evaluate the integral

Fa de
I ea"

where ais any real number. We do this by following the steps below.

127
(a) Letus first find the four zeros of the polynomial

ea.
Solving the equation q(2)=0 for 2”, we obtain 2’ =a:ti. Thus two of the zeros are

the square roots of a-+i, and the other two are the square roots of a-i. By Exercise
5, Sec. 9, the two square roots of a-+ are the numbers

1
ae FENTE) and =

where A= Va’ +1, Since(t2,)* = =@+i=a~i, the two square roots of a-i, are
evidently

% and -%,
‘The four zeros of (2) just obtained are located in the plane in the figure below, which
tells us that z and —Z, lie above the real axis and thatthe other to zeros lie below it.

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

1
OS

which becomes the integrand in the integral to be evaluated when <=x. The method
developed in Exercise 7, Sec. 69, reveals that 2, is a pole of order 2 off. To be
‘specific, we note that q is entire and recall from part (a) that q(z,)=0. Furthermore,
q'(z)=42(z — a) and ¿=a+i, as pointed out above in part (a). Consequently,
de) =42(z3 — a) = dix, #0. The exercise just mentioned, together with the relations
2 = a +i and 1+ a? =At, also enables us to write the residue B, of fat zo:

d'&) __2G-4a_3eh-a_Hatij—a ani _ ania +3)
We "ir léi(a+ De aci 164

As forthe point 2, we observe that

a-)=-¥@ and 9-2

128

=) = 4iz, #0, the point
1, denotes the residue there,

(c) We now integrate f(z) around the simple closed path in the figure below, where
Ro>lzy| and C, denotes the semicircular portion of the path. The residuo theorem tells
us that

Jforde+ |, ede =2ni(B, +8),

«
j de tiges) | de
Nea oF % OS
In order to show that

de
TOS

‘we start withthe observation tha the polynomial g(z) can be factored ino the form

-a\e+a)e-2Xe+2)

Recall now that R> Iz. If is a point on Cy, so that lel= R, then

Ietglalld-lall= Rela! and 12 Zi2lld-IZll= Rat

129
‘This enables us to see that Ig(e)1> (R-Iz,)* when z is on C,. Thus

a

for such points, and we arrive atthe inequality

1 aR rl
lor dr a x
Consequently,
je)
But the integrand here is even, and

E)
er
ic tes u thatthe value ofthis integral des, indeed, tend to O as R tends 10 =.
E de x
fr
inf catia? 43) pal ati" +3) NFS WAR
pa VatatWa-a Vira=Na=

So, the desired result is

Sr à
[ele NAEH ana),

where Asa’ +1.

SECTION 74

Te proben bee oral he img | SEE where «> 20. Todo

this, we introduce the function f(2): ; whose singularities ai and bi lie

Crane roy
inside the simple closed contour shown below, where R> a. The other singularities are, of
course, in he lower half plane.

a

130

‘According to the reside theorem,
3 e de je,
Le Jo de = 2ni(B, + B,),
where “
A = Res al” 2a ai
and
Reste — | et
ROA ern ET Te TE
Thatis,
ed 8
le Fa als ) Jara,

¡A
AORTA
Now, if zis pointon Cy,

IS My where My=

1
RARE)

and le'i=0"? $1. Hence

href, seoetal sf, seta a= 0 Re

So it follows that

¡E
LAA FF

(a>b>0).

SS dr, where a20. The function

2. This problem is to evaluate the integral {SE
+

oy; ht sign nd 10 we my ie cd ein ced
nou tou mi RL

131
We start with

J He + [red =2ni8,

#

where
Hence

Sa

f cosax = Fa

ass ae [100 “de,
Since

IS()ISM, where My

we know that

froee are
andso

ER
Ira

Thatis,

(20)

J xsin2x

Tocata e nea [28224 4, we cinta eon

an
LOTS ae
here a = V5. The point lis above he x axis, and Z is below it. we vite

se),

where 6(e)=

120

132
‘we see that z, is a simple pole of he function f(2)e"* and that the corresponding residue is

=) SCA en

Now consider the simple closed contour shown in the figure below, where R > V3.

Integrating f(z)e"* around the closed contour, we have

Í ¿Gu 208, foe de,

f xsinx

agde = lm) Im), fade de.

Now, when z isa point on Cy,

IF(NS My, where My =:
and so, by limit (1), Sec. 74,
Him J, f@e™de=0.

Consequently, since
inf. fetes], o"

we arrive atthe result

TA e perl Gesine y Bexo
roca, or [Ea eg

133

6 The img o be evaluated is [LÍA wee 230. We define the fonction

+ and, by computing the fourth roots of ~4, we find that the singularities

f@)

+4

VTA and 2, 2 ern Vee id

both lie inside the simple closed contour shown below, where R>VZ. The other two
singularities lie below the real axis.

‘The residue theorem and the method of Theorem 2 in Sec. 69 for finding residues at simple
poles tellus that

Se att [,,F@ede=2miB +8),
where
and
Since

ine“*cosa,

a ft)
sven loi

ea

‘dr = ne"*cosa~ImJ_ f(zede.

134
Furthermore ifz is a point on Cy, then

LENS My where My =
and this means hat

028 Re

fim, sere sf, os] 0 as B+,

according to limit (1), Sec. 74, Finally, then,

5 aa (a>0).
8. In order to evaluate the integral [EME ve introduce here the function
2 + D +9)

10 er: US singularities in the upper half plane are ¿and 3i, and we
consider the simple closed contour shown below, where R>3.

Since

EC eee
tiros sera,” d

the residue theorem tells us that

Î wet de

a da

air), sore

135
Now if zis a point on G then

IGN M, where My as Re,

eh u
RARA

So, in view of limit (1), See. 74,

fmf, foetal, foetal v0.0 a

and this means that

is eG)

‘The Cauchy principal value of the integral [SEE can be found with the aid ofthe
Frs

and the simple closed contour shown below, where R>~/5.

fonction £0)= rs

Using the quadratic formula to solve the equation z*+42+5=0, we find that f has

| i
241 and 4=-2-1. Thus 0 —— wie
* Pa

singolarities at the points,

is interior to the closed contour and Z, is below the real axis.

‘The esd tere el at
ER o
LE, sortean,
vice

tela]
mu (22 (22),

nl mie“
(a3).

Gna
and so

f _sinzde
[tres

- Im, feed,

136

1 1
YOM, where M= e
sr M E GS

fmf, teeta sf, ea] me OR

and we may conclude that

DATE
Lars

+ Dcosx

10. To find the Cauchy principal value of the improper integral is Mens. we shall use

ar ze
NE)

same simple closed contour as in Exercise 9. In this case,

+ Where 2, =-2+i,and Z =-2-1, and the

(+Detde ee
je Saree |, SO d=2mB,
where
ref pe] ene crane
“ea en) 2e
Thus
fee Dcosx y, i
J pares ~ReCmiB) [, fede,
«

(+Doosx a
jeden: Fae eg di = Fein? -c062)-, Hed.

Finally, we observe that if z is a point on Cy, then

Rel RA

a REO SRO

GSM where M, =

137
Limit (1), See. 74, then tells us that

[ref eretael sf, secrete 90 as Rr,

and so

y] Etes,

a,
Lang ein? 0062)

12. (a) Since the function f(z) = exp(iz*) is entre, the Cauchy-Goursat theorem tells us that its
integral around the positively oriented boundary of the sector 0Sr<R, 0S OS #/4
has value zero. The closed path is shown below.

A parametric representation of the horizontal line segment from the origin to the point
Ris =x (OSxS.R), and a representation for Ihe segment from the origin to the point

Reis zero" (OSs R). Thus

fra anomeric,

fra wether Led

By equating real parts and then imaginary parts on each side of this last equation, we
see that

Jroscerar= Jo je“dr=ref, ee

ed.

Genen jean

138
(©) A parametric representation for the arc Cy is z= Re” (OSOS 1/4). Hence

y Ce Tore neta nin fora met

‘Then, by making the substitution $ = 20 inthis last integral and referring to the form.
), See. 74, of Jordan inequality, we find that,

sad RT AI
ask orang sh 2 LT a
ll. al 71 METI

(c) In view of the result in part (2) and the integration formula

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

Jesse LE ans fainceyae

SECTION 77
1. The main problem here is to derive the integration formula.

¡EA
ye

z0-0) (a20,b20),

using the indented contour shown below.

139
“Applying the Cauchy-Goursat theorem to the function

euer

S@=

wwe have

[foar|, sods), 10d+f, fode=0,

[fod], fode=-

_ Sede f, Fae.

Since 1, and ~L, have parametric representations

e] Manco graf 104-[, flode.

In order to find the limit of the first integral on the right here as p = 0, we write

af, dez, Gas | liz?) (dbz, bey? | Gba)
PACE CES |

O<tai<w).

From this we see that z=0 is a simple pole of (2), with residue

| [fe = B= tab = ab)

y

140
‘As forthe limit of the value ofthe second integral as R > =, we note that if zs a point on
Ca, then

Consequently,
2 au,
|p, feds Beat 2% 90 as Rs.
is now clear tat eting p 90 and R> = yells

JO ay ae)

‘This is the desired integration formula, with the variable of integration r instead of x.
Observe that when a=0 and 6=2, that result becomes.

[0500 ae

But cos(2x)= x, and we arrive at

2. Letus derive the integration formula

Ir ddr
AO Feos(ax/2)

Cica<3,

where x"=exp(alnx) when x>0. We shall integrate the function

z*_ _exp(alogz)

O

(oa-Fane)

whose branch cut is the origin and the negative imaginary axis, around the simple closed
path shown below.

141

By Cauchy residue theorem,
J po], fords |, order |, rod -2mRes re.
Thatis,
J fooder, serde=2nites s@-[, fode-[, fode

Since

(@SrSR) and ~Liz=re*=-r(psrsR),

the left-hand side of ths last equation can be written
gn) rs

a ee e ar
Ir
En

Pr A N BE GB
free

Res f= 9D where 9(2)=

the point z=i being a pole of order 2 ofthe function (2). Straightforward differentiation

reveals that

ya ef e+)
COS EEE

142
and from this it follows that

Res ft) = ie"

We now have

r
Fry

ae). dr = ED un _ Sf @de- |, Fae
Once we show that
Jim J, fle)de=0 and lim |, fode=0,

we arrive at the desired result:

E er CE PE lome
Par NN TOS

‘The first of the above limits is shown by writing

ee
I. road er oF

and noting thatthe lat term tends to 0 as P +0 since a+1> 0. As for the second limit,

and the last term here tends to O as R-> = since 3-a>0.

3. The problem here is to derive the integration formulas

by integrating the function

eo

no Bp F cage dE

ze

)

143
around the contour shown in Exercise 2. As was the case in that exercise,

[ford], roro -[ fede |, rod

#2 ee 1ogz

FO ;
ati

where (2) =

the point z = isa simple pole of (2), with residue

She promi prenne
Lia=rel=r(pSrSR) and ~L:z=re"=-r(psrsR)
arras
Lo «¡rta and [Node er rin
mus

Je sen ars

ee Eier [, pde, rod.

By equating real parts on each side ofthis equation, we have

Fe
—Ref, fde-Re], fledde:

[tt scons | Gitta sin QUE

and equating imaginary pats yields
z=

sora] ar oro] E

Im [t,t

Y 8

.cotemal, anti. car an itis ine o

Now sin(=/3)
show that

Jim, f@de=0 and Im, = 0.

144

Thatis,

Solving these simultaneous equations for 1, and 7,, we arrive at the desired integration
Formulas.

Let us use the function
ÉS

|
(0-2)

Integrating f(z) around the closed path shown in Exercise 2, we have

[fod], 10d =2mReS5()—Í, forde-f, sede

£0) =22 where 90=

the point = is a simple pole of (2) and the residue is

(logiy* _ nt+in/2yt

Res CO = =" we

Also, the parametric representations

Liz=re”

@SrSR) and —Liz=rét=r(psrsR)

145
enable us to write

[rot [gee and Pi el
Since
J floder [ ro] Lee ren je PT à
then,
dede ré

and equating imaginary parts yields
nr
alae =imf fede IJ, fede
tis straightforward to show that

fim [,,f(@)de=0 and Jim | fle)de=0.

Hence
don? ®
an E
a
raf ar =0

Finally, inasmuch as (see Exercise 1, Sec. 72),

dr
Pal

we arrive at the desired integration formulas,

146
Ve

5. Here we crate the inept fae, where 03050. We comider the

1
zu exp( lose
f@= (a> 0,0<arge<2n)

TNT)

function.

and the simple closed contour shown below, which is similar o the one used in Sec. 77. The
numbers p and R are small and large enough, respectively, so that the points 2=-a and
2=-b are between the circles.

A parametric representation for the upper edge of the branch cut from p to R is
(OST SR), and so the value ofthe integral of falong that edge is

jot (nr +10)
dr

(EN) fes Fran

A representation forthe lower edge from p tois R is ¿=re*" (PSrSR). Hence the
value ofthe integral off along that edge from R to p is

Lonard \
zeliean) ae on uy

V+aXr+b) aXr+b)

‘According to the residue theorem, then,

far zent Nr
art Jos e res [ron iB, + B,),

where
_ rfi] exo zane in
and
exp] =log(-b) Unb+im)
ante Eee], fine]
ra “bra
‘Consequently,
Lt E
Marat an a sted
Now
amp Be
INC ls GED ar CE et
and
288
lí, sena Ra (R=aXR=b) gs Rom
Hence

VE y 2 a AND) e*° la E
AT O E E TE)

AGA ONO) 2 Na
ea" VB yy 5 amb

Replacing the variable of integration r here by x, we have the desired result:

Iara

147

(a>5>0)

148
6. (a) Letus frst use the branch

in_exa(—loge ai
102 (190.-E<ug:<3E)

EH 241
and the indented path shown below to evaluate the improper integral

Ik
Ira

y

Cauchy residue theorem tells us that

J, foods [, Feder |, seddes |, fla)de=2niRes fle),

J, fode+f, sorde= amines se) |, 10d |, fled.

La hdr
[fot 10 a Iren”

Thus

ab Ter ARO, of Forde

149

Finally, then, we have
oof

which is the same as
Ré
+ NZ
: E
(6) To evaluate the improper integral er

esos
(id>0,0<arge<2a)

2+1

and ie simple closed contour shown inh gue below, which similar 1 Fig, 9 in
Sec. 77. We stipulate that p<1 and R>1, so that the singularities z = +i are between
Gand Ci

f@=

150
Since a parametric representation for the upper edge of the branch cut from p to R is
2=ré" (pSrSR), the value ofthe integral of falong that edge is

1
o =

1
[a ee

A representation for the lower edge from p to is Ris 2 = re
value of the integral off along that edge from R 10 p is

1
pola] A
flag ladon

Hence, by the residue theorem,

3 (prs R), and so the

laa + rod [pe + [de = 2708, +B),

where

BRO

En

| otal stains)

ee )- [stod- Jos
Since
2p 2
Lrofs er)
and

2m 2e
[rel Zr es a Ro,

151
‘we now find that

Frost

‘When x, instead of r, is used as the variable of integration here, we have the desired

result
Ds
IN CE
SECTION 78
1. Wie
==
ca + Siz—2"

where Cis the positively oriented unit cicle Iel=1. The quadratic formula tells us thatthe
singular points ofthe integrand on the far right here are z=—i/2 and z=~2i. The point
112 is a simple pole interior to C; and the point z =~2i is exterior to C. Thus

JA nami
al...

2. Toevaluate the definite integral in question, write

v0 1
fée]

¡2
rs {
a 14

‘where Cis the positively oriented unit circle Izl=1, This circle is shown below.

,
ON

152
‘Solving the equation (2) —6(2*)+1=0 for z* with the aid of the quadratic formula, we
find that the zeros of the polynomial z* 62? +1 are the numbers z such that z* =3+ 2,2.

‘Those zeros are, then, z= 43+ 22 and 2=+y3- 2,2. The first two of these zeros are
exterior to the circle, and the second two are inside of it. So the singularities of the
integrand in our contour integral are

a ad =,

indicated in the figure. This means that

where
and
ic “hi i
Re EE D
BREST ada 2-3 WT
Since

ana nee)

the desired result is

ja _
fre r.

7. Let C be the positively oriented unit circle Izi=1. In view of the binomial formula (Sec. 3)

153

Now each ofthese last integrals has value zero except when 4:

Leeds 2m
Consequently,
Tan 040 = 1. En) 27 Cm)
js 008 = EG (ay Pay
SECTION 80

5. We ar given a function dat is analytic inside and on a positively oriented simple closed
contour C, and we assume that has no zeros on C. Also, fhas zeros 2, (E=1,2,...n)
inside C, where each z, is of multiplicity m,. (See the figure below)

“The object here is to show that

LO 29
a

Todo this, we consider the Ath zero and start with the fact chat

mes.

PORC CN

‘where (2) is analytic and nonzero at 2,. From this, it is straightforward to show that

LO me ‚so memaeme CONO
J@ 2-4 8) 2-4 E) FOR

a here has a Taylor series representation at 2, it follows that DE

has a simple pole at z, and that

Since the term

vo.
O

‘An application of the residue theorem now yields the desired result.

154
6. (a) To determine the number of zeros of the polynomial 2*—Sz* +2? -2z inside the circle
I=], we write

NOS and ge)= +222,

‘We then observe that when z is on the circle,
I/GA=S and Ig(NSIA + te? + 2iat=4,

Since |/(z}>Ig(2)| on the circle and since f(z) has 4 zeros, counting multiplicities,
inside it, the theorem in Sec, 80 tells is that he sum

PORC PEER
also has four zeros, counting multiplicities, inside he circle.
(6) Letus write the polynomial 22° -22° +22 -22+9 asthe sum f(z)+g(2), where
F@)=9 and g(z)=2e4~22' 4224-22.
Observe that when z is on the circle La 1,
IF@N=9 and 1g(2ys ASA NL + 2iet + ied =8,

Since 1$(A>Ig(2H on the circle and since f(z) has no zeros inside it, the sum
I) + (2) =22" ~22? +22? ~22-+9 has no zeros there either.

7. Let C denote the circle id= 2.

(a) The polynomial z* + 32° +6 can be written asthe sum of the polynomials

F(2)=32? and g(2)

onc,
IFGA=3I2'=24 and Iglelziz‘+6lsidt+6=22.

Since If(2X>Ig(e) on C and f(z) has 3 zeros, counting multiplicities, inside C, it
follows that the original polynomial has 3 zeros, counting multiplicities, inside C.

(0) The polynomial 2* 22? +92* + 21 can be writen as the sum of the polynomials

FO=9 and g(z)=24~22? 42-1,

onc,
let = 22° +2-Ils elt + ll =35,

If A=9I2'=36 and 1g(2V

155

Since If(eN>Ig(2) on C and f(z) has 2 zeros, counting multiplicities, inside C, it
follows thatthe original polynomial has 2 zeros, counting aultipliciis, inside C.

(c) The polynomial 2° +32? +2" +1 can be written as the sum of the polynomials

SOS and ser
onc,

Id and ISIAH SNS + leP 41 = 29.

Since IS(X>Ig(2) on C and f(z) has 5 zeros, counting multiplicities, inside C, it
follows that the original polynomial has 5 zeros, counting multiplicities, inside C.

10. The problem here is to give an alternative proof of the fact that any polynomial
Poza tarta tas (a,%0),

where n 21, has precisely n zeros, counting multiplicities. Without loss of generality, we
may take a, =1 since

Pe) CES)
a 4,
La
FT and gs a tat tar.
‘Then let R be so large that

R> Ital la += Ha,

If 2 is a point on the circle C:1zl= R, we find that

g(a Lalla la lle slat lat Rta, RT
CR HLL ++, IR la +++, RO
< RR = Rt ald = (EX

Since f(e) has precisely n zeros, counting multiplicities, inside C and since R can be made
arbitrarily large, the desired result follows.

156
SECTION 82

‘The singularities of the function

ag?

a
are the fourth roots of 4. They are readily found to be

s= Ten

V2, Vii, -V3, and - Ai

123),

See he igre below, where y> V7 and R> VE à y.

The function
tas simple poes atthe points

SEE VÍ V2, and 5,
sá

= cosh ¥21+cos-V2t.

157
Suppose now thats is a point on C,, and observe that

I=iy+REIST+R=RAY and sl=l7+Re*Z1y—Rl=R—7> V3.

follows that,
sm 2is? s 20R+ Y
and
let ares —a12 (Rp 430.
Consequently,
AUR+Y?
IRs LED 0 as Rom,
VE ERS
This ensures that

= cash. + cos V2r.

‘The polynomials in the denominator of
polynomi

2-2

FO Tang ease

have zeros at 5==1 and s=-1:21 Lotus, then, write

eos)

FO aa

1+2i. The points —1, ,, and 3, are evidently simple poles of e*F(s) with the
following residues:
a e*(2s-2)
Resle"F] =
cr ul.

Roser] =D a
Ar al

retro. aD MG =D ] 7
o a la ES

158
It is easy to see that

B+B+B,

sins
Istely +Re"iS y+R=R4Y and Idaly+ Refisiy- R= R-7>V3,
12s = 21S 2d +25 2(R+ 7) +2,
Is+ll2llsi-12(R=Y)=1>0,
=
= sfls—512Usl-igh? 2[(R- 7) 43] >0.
e LES
A

10 = "(sin 21+ 0082-1).

159
4. The function

FO= (a>0)

ray

has singularities at s = ai. So we consider the simple closed contour shown below, where,
y>Oand R>a+7.

Upon waiting

of

we see that 9(s) is analytic and nonzero at $, = ai. Hence s, is a pole of order
FG), Furthermore, F(s)=F(5) at points where F(s) is analytic, Consequently, 3, is also
a pole of order 2 of F(4); and we know from expression (2), Sec, 82, that

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

Refe(h +00],

where band b, ar the coefficient in the principal part
+,

a Gay
of F() at ai. These coefficients are readily found with Ihe aid ofthe first two terms in the
Taylor series for 4(5) about 5, = ai:

Hai)

BOS u

=.

[wa + (ai) +: -]

1
Gay

Gai

160
(0 <is aile 20).

Itis taightforward to show that g(ai)=1/2 and $’(ai)=0, and we find that 4 =O and
by =1/2, Hence
Res[etFoo} + Res[e"FC9

We can, then, conclude that
f= teosat (a>0,
provided that F(s) satisfies the desired boundedness condition. As for that condition, when
zisapointon Cy,
la=ly+Re IS y+R=R+Y and ld=ly+ Re"l217—Rl= R—7> a
and this means that

Ie al Sat s(R+7) +a? and Let + al 1l2f ern a >0.

Hence
CÉPECER
FOIS TRE 0 2 Rom
6. Weare given

inh(as)

Fes $ coshs

(O<x<D,

which has isolated singularities at the points

i, and 3, =!

This function has the property FO) FG), and so

HO=Res[e"FCa]+S [Resler Uo} Resle" Po}.

To find the residue at 3, =

TES 0)
52725 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[etFoo}= Res FO) =x.

As forthe residues of F(s) atthe singular points s, (n=1,2,..), we write

F¢s)= 22 where pls) =sinh(xs) and 9(s) =s* coshs.

46)
Wenot that
po) = isin DE 20 and a(s,)=0;
farhemor, since
q'(s)=2scoshs +5" sinhs,
we find that

Se
Fam)

In view of Theorem 2 in Sec. 69, then, s, is a simple pole of F(s), and

Expression (4), Sec. 82, now gives us

Resfe"]+ Resle"ro]=2Re

162
Summing all of the above residues, wo arrive atthe final result:

in =D = Dat
5 a

7, The function

isolated singularities at s, =0 and when cosh(s'”)

=0, or at the points

(m

). The point s, is a simple pole of F(s), as is seen by writing

ia 1 1
CE IES DEEE TETE" res

and dividing this last denominator into 1. In fact, the residue is found to be 1; and
‘expression (3), Sec. 82, tells us that

Res[e"“F(s)] =Res F(s)=1
As forthe other singularities, we write

ro 22 wbere_p(s)=1 and a) = sense

Now
pls) =1%0 and die,

also, since

con scone!

itis straightforward to show that

163
So each points, isa simple pole of F(s), and

ResF(s)=

Consequently, according to expression (3), Sec. 82,

a Gaeta.
Final, then,
10 =Res[eFCo]+ Dres Fo],
= On- Wir
vote

3. Here we are given the function

(gy = 206812). coshtra/2)
+1 G+ Dsinh(as/2)"

which has the property FUG}= FCG). We consider first the singularities at s=:4i, Upon

writing
cosh (ns /2)

id Gr Dane"

22 where 96)=

we find that, since 9(1)=0, the point i is a removable singularity of F(s) [see Exercise

3(b), Sec. 65]; and the same is true of the point ~i. At each of these points, it follows that

occur when ms/2=nai

the residue of e"F(s) is 0. The other singular
(n=0,£1,42...., oratihe points s=2ni (n=0,t1:t2,...). To find the residues, we write

F=f where 76) on) and = + Di)

164
and note that

PO2ni)= cosh(nni) =cos(nz) =(-I)* #0 and g(2ni)=0.
Furthermore, since

{oe +n cou

Jr)

ie
On) = (an? +0) eoshlani) = Can +02 Fosa) == =D. = #0.
‘Thus
pyro Hd neous...

Expressions (3) and (4) in Sec. 82 now tell us that

Resfe'Fis]=Resrn=

and
Res[e"FG)]+ Resle"Fs)]= anden

‘The desired function of is, then,

9. The function
‚sinh(xs"?)

PETE O<x<b,

PO sania) SEEM

where it is agreed thatthe branch cut of 5”? does not lie along the negative real axis, has

isolated singularities at s=0 and when sinb(s"”

=0, or at the points s

(n=1,2,...). The point s=0 is a pole of order 2 of F(s), as is seen by frst writing.

x's) 6428 1100 +
TS

PINE
E E EOS VE

10.

165

and dividing the serie in the denominator into the series in che numerator. The result is
sinh(xs'®) _ 1
BETT

hote Gide)

In view of expression (1), Sec. 82, then,

Reslerro]= 4 tas ga Dear.
‘As forthe singularities s=—n? ( ), we write

roo where p(s)=sinhxs!) and 9(6) =s*sinh(s!”),
Observe that p(-n'r?) #0 and g(-n'n*)=0, Also, since
4/25 sins) +2 costs),

itis easy lo sec that ¢(-n'") #0. So the points s=—n?n? (n=1,2,...) ar simple poles

of F(s) and
Res, Fis) 29) n=
OM Las
‘Thus, in view of expression (3), Sec. 82,
E gate E
Reg leFo]= Te" sanas 12...

Finally, since

HO =Resle"FOo]+¥, Res [er],
we ave atthe expresion

Lig
HO= Ga!

The function

has isolated singularities at the points

50 and 5, =nmi, 5 =-nai (n=1,2.....

166
Now

sinh CES

La
soe
6

and division of this series into 1 reveals that

(O<is< m)

‘This shows that F(s) has a removable singularity at 5,. Evidently, then, e*F(s) must also
have a removable singularity there; and so

Res[e"r6]

To find the residue of F(s) at 5, = nai (n=1,2,..), we write

noto where p(s)=sinhs—s and q(s)="sinhs

and observe that
Ori) =-nmi #0, q(nmi)=0, and Oui) = nr (D +
‘Consequently, F(s) has a simple pole at s,, and

p(n
Res F)=
O a) RRE

Since F(G}= F(), the points 5, are also simple poles of Fe); and we may write

Res[e"F@]+Res[erin]= and EY, YE, se] ane icon sana]

2 sine.
nr
Hence the desired results

SO =Resfe Fe] S [Res [erFCo]+Res|e" Fo}.

167
11. We consider here the function

Fo ne)

ac Pers

e

where 0>0and 020, =: 2% (n=1,2,..). The singularities of F(s) are at

2

s=tof and sata, (

Because the first term in the Maclaurin series for sinh(xs) is xs, it is easy to see that s =0 is
a removable singularity of €") and that

Res[eF]=o.

To find the residue of F(S) at = oi, we write

sinh(xs)
S{s-+ ai)coshs*

from which it follows that s = av is simple pole and

- where (3) =

sinh(xei) __ _isinax
aaa) 2050

Res F(5) = Han)
Since FIG} = FOG), then,

Res[e"Fco]+Res "FL =2R¢ Era ee

As forthe residues at 5= @i (n=1,2,...), we put F(s) in the form

Ea where p(s) =sinh(xs) and g(s)= (5 + 0*s)coshs.

Now p(a,i)=sinh(xo,i) = isin @,x #0 and q(@,

=0. Also, since

= + 0*s)sinhs +s? +0*)coshs,
‘we find that

i+ & ,sinh(o,5) =-02,(0* - o})sino, #0.

q)

Hence we have a simple pole at s = 0, with residue

LPO). inox
RO ai) Tara? oda

168

Consequently,
a inox a]. nono

[er Fin)+ Res — ising |) Orne
Resle"Foo}+ Res [e F6] de ] ETES

Ba anne) C7 and eu at

alero] Reg [rc] 2 CT magias

Finally,

s0.=Resle*FC)}+ [Res fe" Feo]

Fol} +S [Res [ero]+ Res [er].
Thatis,

„Snaxsiner SCD snoxsinos
FO rue De tear

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