Student Solutions Manualfor use with

Complex Variables

and ApplicationsSeventh Edition

Selected Solutions to Exercises in Chapters 1-7

by

James Ward BrownProfessor ofMathematics

The University ofMichigan- Dearborn

Ruel V. Churchill

Late Professor ofMathematics

The University ofMichigan

McGrawHill

Higher Education

Boston Burr Ridge, IL Dubuque, IA Madison, Wl New York San Francisco St. Louis

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Milan Montreal New Delhi Santiago Seoul Singapore Sydney Taipei Toronto

Table ofContents

Chapter 1 1

Chapter 2 22

Chapter 3 35

Chapter 4 53

Chapter 5 75

Chapter 6 94

Chapter 7 118

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

Chapter 1

SECTION 2

1. (a) (V2-/)-/(l-V2i) = V2-/-/-V2=-2i;

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

(0 (3>D(3,-l)(^) = (10,0)(i,

1

L) = (2,l).

2. (a) Re(iz) = Re[i(x + iy)] = Re(-y + ix) = -y = -Im z

;

Im(i'z) = Im[i*(;t + iy)] = Im(-y + a) = x = Re z.

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

= l+z+z + z2 = l + 2z + z

2.

4. If z=4±i, then z2 -2z + 2 = (l±i)

2-2(l±t) + 2 = ±2i-2T2i' + 2 = 0.

5. To prove that multiplication is commutative, write

Z1Z2 = (*i.yi)(*2.y2 )= ~ Wa» y^2 + *iy2 )

= (*2*i -yzVp^ H-x^) = (x2,y2 )(xl,yl )= z2zl

.

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

(Zj + z2 ) + z3 = l(xl ,yl

) + (x2 ,y2 )] + (x3 ,y3

) = (x, + x2 ,v

t+ y2 ) + (*3 ,y3 )

=((*i + x

2 ) + jcj, (y, + y2 ) + y3 )= + (x2 + x

3 ), y, + (y2 + y3 ))

= (^1 » yi ) + (*2 + ^3 . y% + v3) = (*i . yi ) + [(x2 >y* ) + (*3 .^ )]

= z1 +(z2 +z3 ).

2

(b) To verify the distributive law, write

z(zl+ z2 ) = (*,30K*i.yi) + (x2 ,y2 )] = (jc,30(*i + x2 , + y2 )

= (ja, + xx2- yy, - yy2 ,

yjc, + yx2 + xyx+ xy2 )

= (xxl~yy

l+ xx

2 -yy2 ,yx, +xy

l+yx

2+xy

2 )

= ~ yyv yxl+ xy,) + (xx2 - yy2 ,

yx2+ xy2 )

= (x, y)(xl ,yl ) + (x, y)(x2 ,y2 ) = zz

x+zz2 .

10. The problem here is to solve the equation z2 + z + 1 = for z = {x,y) by writing

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

Since

(x2 - y

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

it follows that

x2 -y2

+;e + l = and 2xy + y = 0.

By writing the second of these equations as (2x + l)y = 0, we see that either 2x + 1 = or

y = 0. If y = 0, the first equation becomes x2 +x + l = 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 y2 = 3/4, or y = ±V3/2. Thus

SECTION 3

! (a)1 + 2/

|2 - / = (1 + 2Q(3 + 4Q

|

(2 - Q(-5Q _ -5 + lOj -5-10/= 2.

3-4/ 5/ (3-4/)(3 + 4/) (5/)(-5/) 25+

25 5'

> 5/ 5/ _ 5/ _ 1

(1 - 0(2 - 0(3 - ~ (1 ~ 3/)(3 - ~ ^10/ " ~2

5

(c) (l-/)4=[(l-0d-0]

2=(-2/)

2 =-4.

2. (a) (-l)z = -z since z + (-l)z = z[l + (-1)] = z • = 0;

1 1 z z(b) 77- =— - £ = 7 = z fe*0).

1/z z z 1

3

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

6.

Z3Z4

' Z\Z2

1= z.

— £2(z3 *0,z4 * 0).

7.ZyZ _Z2Z \Z2j

VZ\ ^ |=

KZl)

1 = (z2 *0,z*0).c2 /

SECTION 4

w ^=(-73,1), z2 =(V3,o)

4

(c) ^=(-3,1), z2 =(l,4)

Z. + Z,y

(d) z, = + i>i , z2 = *i-%

2. Inequalities (3), Sec. 4, are

Rez<IRezl<lzl and Imz< llmzl< Izl.

These are obvious if we write them as

x<\x\<^x2 + y2

and y<lyl<V*2 +y2

.

3. In order to verify the inequality V2 Izl > IRezI + llmzl, we rewrite it in the following ways:

V2~V*2 +y2

>l*l + lyl,

2(*2 + y

2)>l;cl

2 + 2|jtllyl + lyl2

,

lxl2-2l;tllyl + lyl

2 >0,

(l*l-lyl)2 >0.

This last form of the inequality to be verified is obviously true since the left-hand side is aperfect square.

5

4. (a) Rewrite \z- 1 + il= 1 as |z - (1 - 1)|

= 1. This is the circle centered at 1 - i with radius 1.

It is shown below.

( IX\I \-i J

5. (a) Write I z - 4il+

1

z+ 4/1= 10 as I z - 4il

+

\z - (- 4i)l= 10 to see that this is the locus of all

points z such that the sum of the distances from z to 4i and -4i is a constant. Such a

curve is an ellipse with foci ±4i.

(b) Write \z - \\=\z + i\ as \z - ll=lz - (-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 of the line segment from 1 to -i.

SECTION 5

1. (a) z + 3i' = z + 3i=z-3i;

(b) iz = iz=-iz;

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

1

2 = 4 - 4/ - 1 = 3 - 4/

;

M I(2z + 5)(V2-0I=I2z + 5IIV2-jI=I2TT5IV2+T = V3I2z + 5I.

2. (a) Rewrite Re(z-/) = 2 as Re[* + i(-y - 1)] = 2, or x = 2. This is the vertical line

through the point z = 2, shown below.

3>

2 x

6

(b) Rewrite I2z - i\= 4 as 2 z --2

radius 2, shown below.

= 4, or = 2. This is the circle centered at — with2

ji/2,

3. Write z, = xl+ iy

land z2 = x2 + ry2 . Then

and

Zr ~z2 = (*, + (y1)-(x2 +%) = (*i-^

2 ) + I'(>'i

-y2 )

= (*i - *2 ) - - ft ) = (*i- Vi)~(x2 - iy

2 ) = zl-z2

ZyZ2 = (xl+ iy

l)(x2 + iy

2 ) = {xxx2- yj2 ) + i(y^

2+ xj

2 )

= (X& -

y

xy2 ) - i{yxx2 + x

ty2 ) = (x1- fy,)^ - iy2 ) = z,z2 .

4. (a) zfaZs = (ziZ2 )z3 = zfo z3 = (*i £2)^3 = *i 22 z3 ;

W z4= z

2z2 = z

2z2 = zzzz = (z z)(z z) = zzzz = z

4.

6. fa)

(b)

ZyZ-2^3 J

z2z3

Z2Z3 Z2 Z3

__lzil_ = Jzi

l_

lz2Z3 l lz2 llz3 l

8. In this problem, we shall use the inequalities (see Sec. 4)

IRezl<lzl and \zx+z2 + z3 |^|£i| + |z2 |

+ M-

Specifically, when lzl< 1,

|Re(2 + z + z3

)|< 12 + z + z

3l < 2+lzl +lz

3l = 2+lzl+lzl

3 < 2 + 1 + 1 = 4.

7

10. First write z4 - 4z

2 + 3 = (z2 - l)(z

2 - 3). Then observe that when Izl=2,

Iz2-ll>|lz

2l-lll| = |lzl

2-l| = l4-ll=3

and

lz2-3l>|lz

2M3l| = |lzl

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

Thus, when lzl=2,

lz4-4z

2+3l=lz

2 -lHz2 -3l>3-l = 3.

Consequently, when z lies on the circle lzl= 2,

1

z4-4z

2 +3 lz4-4z

2+3l 3

1 -<I

11. (a) Prove that z is real <=> z = z.

(<=) Suppose that z = z, so that x - iy = x + iy. This means that i'2y = 0, or y = 0.

Thus z = * + iO = x, or z is real.

(=>) Suppose that z is real, so that z = x + iO. Then z = x - iO = jc + i'O = z.

Prove that z is either real or pure imaginary <=» z2 = z

2.

(^) Suppose that z2 =z2

. Then (jr-i'y)2 =(x + iy)

2, or i4xy = 0. But this can be

only if either x = or y = 0, or possibly jc = y = 0. Thus z is either real or pure

imaginary.

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

z2 = x

2 = z2

. If z is pure imaginary, so that z = iy, then z2 = (-iy)

2 = (iy)2 = z

2.

12. fa) We shall use mathematical induction to show that

Zi+z2 +-"+z„ =z1+z2 +---+zn (n = 2,3,...).

This is known when n = 2 (Sec. 5). Assuming now that it is true when n = m, we maywrite

Zt+Z2 +*"+Zm +Zm+1 - (Z

t+Z2 + -,-+£m ) + Zm+l

= (z1+z2 +-+zm ) + zm+1

= Zt+ Z2 H l"Zm ) + Zm+1

= Zj + Z2 H l"Zm + Zm+1 .

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

^z1 ---zn =zlz1 --zn (n = 2,3,...).

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

^lZ2' '

'£mZm+l ~ (Z1Z2 "'Zm )Zm+l ~ (^2 "

'Zm )

= (^1^2"

'2m )^m+l

= ^2 '*

' ^nAr

14. The identities (Sec. 5) zz =\z\2and Rez = -^-^ enable us to write lz-z l= R as

2

(z-Z )(z-Z ) = i?2

,

zz-(z5, + z!Q+Zo5> =/?2'

lzl2-2Re(zz ) + lz l

2 = /?2

.

15. Since x - -^-^ and y = the hyperbola x2 -y2 = 1 can be written in the following

ways:

. _n2 / _\2z+z\ fz-z

= 1,

2 ; 1 2i

z2 + 2zz + z

2. z

2 -2zz+z 2

= 1,

2z2 + 2z

2_

z2+z

2=2.

SECTION 7

1. (a) Since

argl _2 -2i '

= 3X8

1

~ afg(~2 ~ 2/) '

one value of argj*' —

jis

y ~f~~£~J'or Consequently, the principal value is

9

(b) Since

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

one value of arg(V3 -i*)6

is 6f™\ or -it. So the principal value is -it + 2n, or it.

4. The solution d = n of the equation \e'e -11= 2 in the interval < < In is geometrically

evident if we recall that e'e

lies on the circle \z\= 1 and that le'* - II is the distance between

the points eieand 1. See the figure below.

5. We know from de Moivre's formula that

(cos + isin 0)3 = cos30 + isin30,

or

That is,

cos3 + 3cos

20(/'sin 0) + 3cos 0(isin 0)

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

(cos3 - 3cos 0sin

z0) + i'(3cos

z0sin - sin

30) = cos30 + isin30.

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

identities:

(a) cos30 = cos3 0-3cos0sin2

0; (b) sin30 = 3cos20sin0-sin

30.

8. Here z = re'9

is any nonzero complex number and n a negative integer (n = -l,-2,...).

Also, m = -n = l,2 By writing

and

(z-l

)

m

=[l^~6)

J =[-JeK-mB) =\e i(-~me

\

we see that (zm

) =(z )

m. Thus the definition z" = {z )

mcan also be written as

z"=(zmT

l

-

10

9. First of all, given two nonzero complex numbers zi and z 2 , suppose that there are complexnumbers c

xand c 2 such that Zj = qc

2and z^ = c

tc2 . Since

IzJHqllcJ and IzjIHcJIqIMcJIql,

it follows that IzJ^ZjI.

Suppose, on the other hand, that we know only that 1^1=^1. We may write

Zi = r{exp(i6

l ) and z2 = r, exp(/02 ).

If we introduce the numbers

q = rxexpl i

2

Jand c2 = expl i

*

we find that

2 J'

c,c2 = rt exp^i-^-Jexpl

J= r

texp(i^) = z,

and

i+ilvi ,.0i-0qc2 = rtexp| i

-^y*Jexp^-i

|= n exp 2

= z2 .

That is,

Zi = qc2 and z2 = c,c2 .

10. If 5, = l + z+z2

+"-+z", then

5 - zS = (1 + z + z2+• • .+z" ) -

(Z + z2 + z

3 +. . . ) = i _ £"+1

l-z"+1

Hence S =— .provided z*l. That is,

1-z

1 — 7"+1

H-z + z2+-+z"=— (z*l).

1-z

Putting z = e'e(0 < < 2tt) in this identity, we have

1 J(n+l)6

l + em +ei20+-+e ine =

l ~ e

\-e ie

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

l + cos0 + cos20+-+cosn0;

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

11

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

l-exp(/0)

exp

exp(-!)

exp^-if)

-exp\(2n + l)0"i- —

2

expH) -exp®which becomes

9 . . (2n + l)0 . . (2n + l)0cos— i sin— cos ism •

2 2 2 2 1

-2i'sin—2

or

r . , (2n + l)0l J (2n + l)0lI sm— + sin -

—

^ j+ 1 1 cos— - cos -

—

^ j

2sin—

The real part of this is clearly

. (2/i + l)0

i +sin

-ir-2 2sin^

2

and we arrive at Lagrange's trigonometric identity:

(2n + l)0

1sm

1 + cos + cos20+- • -+cosn0 = — + ^

—

2 2sin—

(0< 0<2;r).

12

SECTION 9

1. (a) Since 2/ = 2exp i\^ + 2kn (k = 0,±1,±2,...), the desired roots are

(2/)1/2=V2exp

That is,

and

= i+i

c being the principal root These are sketched below.

(* = <U).

(b) Observe that 1-V3i = 2exp i\ —j + 2Jbr (£ = 0,±1,±2,...). Hence

(l-V3/)1/2=V2exp

The principal root is

i'I -^ + kn

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

,2 2 J

V3-i

V2'

and the other root is

V2

These roots are shown below.

(A: = 0,1).

13

2. (a) Since -16 = 16exp[i(;r + 2*7r)] (k = 0,±1,±2,...), the needed roots are

(-16)1/4 = 2exp ,

it kiti — +—4 2

(* = 0,1,2,3).

The principal root is

c =2,- = 2(cosf + / Si„l] = 2(-^ + -^j = V2(l +i).

The other three roots are

and

cx= {ij^y*11 = c

Qi = V2 (i + /)/ = -V2 (i - o,

c2 =(2O*'* = -c =-V2(l + i),

c3= (2e'*

/4)e»*

12 = c (-i) = V2 (1 + OH) = V2(l - /).

The four roots are shown below.

(b) First write -8 -8V3/ = 16exp ^-~ + 2kn^ (* = 0,+l,±2,...). Then

(-8-8V3i)l/4 =2exp ,

it kit(* = 0,1,2,3).

The principal root is

c =2e-,Vr/6 =2 cos^-/sin- =2

I 2 2= V3-i.

The others are

c2 =(2,-/6)e

i'r

= -c =-(V3-0,

c3 = (2e-'wy 3*'2 = c (-/) = -(1 + V3/).

These roots are all shown below.

(a) By writing -1 = lexp[i(?r + 2kn)] (k = 0,±1,±2,. . .), we see that

(-1)1/3 = exp

The principal root is

., % 2kiti — +

3 3

Jt . . jc 1 + V3ic =e = cos— + jsin— =

The other two roots are

and

c, = e'* = -l

„ _ jsxii _ nx-ixn „ K . . 7t 1-V3ic2= e =e e = cos isin— = ,

All three roots are shown below.

15

(b) Since 8 = 8exp[i(0 + 2for)] (k = 0,±1,±2,...), the desired roots of 8 are

81/6 =V2exp(/^ (* = 0,1,2,3,4,5),

the principal one being

The others are

c =V2.

= V2V- = V2(cos| + /sin|] = Vlfi +^/V^.v 2 2 ,

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

1 V3

2"T

V2

N1-V3/

V2*

c3 =j2e'*=-j2,

c4 =(j2eM3

)eb, = -c

i

=-

and

1 + V3/

V2'

c5 =(V2V

2*V* = -c2=i^.

V2

All six roots are shown below.

The three cube roots of the number z = -A^2 + A-Jli = 8exp^/-^j are evidently

(zor=2exp[I + 2|E)] (* = 0,U).

In particular,

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

16

1 H" "^3/With the aid of the number o3

=, we obtain the other two roots:

v 2 J V2

c2 =c ©3 =(c Q)3)fi)

3= •(V3 + 1) + (V3-1)/

V2

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

2 J V2

5. (a) Let a denote any fixed real number. In order to find the two square roots of a + i in

exponential form, we write

Since

we see that

A = \a + i\ =Var+T and a = Arg(a + i).

a + i = Aexp[i(a + 2kn)]

(a + i)V2 =VAexp

That is, the desired square roots are

ify + Jbr

VaV "2and VaV^V* = -4Aeia,

\

(* = 0,±1,±2,...),

(* = 0,1).

cc it

(b) Since a + i lies above the real axis, we know that < a < n. Thus <— <— , and this

2 2

tells us that cosf ^ ] > and sinf )> 0. Since cosa =— , it follows that

\2 \2 A

a /1 + cosor 1 [. ~a ^fA + a

and

. asin

2 V 2 -7211 A 42-fA'

1 - cos a 1 |^__a _ VA-a

Consequently,

±VAVa/2 =±VA^cos| + isin|j = ±VA^

= ±-^(VA + a+iVA-a).

VA + a . VA-fl ^|

^Va +iV2VA J

17

6. The four roots of the equation z4 + 4 = are the four fourth roots of the number -4. To

find those roots, we write -4 = 4expD'(;r + 2ibr)] (& = 0,±1,±2,...). Then

(-4)1/4 =V2exp

To be specific,

(f + Y)l = V2^/4e^2(* = 0.1,2,3).

c = V2V™ = V2fcos^ + /sin^l = V2 f-L + i

*

V 4 4j VV2 V2j

q =c e"c/2

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

c2 =c e"r

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

= l + i,

c3 = Coei3*/2 = (1 + /)(-/) = 1-/.

This enables us to write

z4 +4 = (z-

c

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

= [(z - q)(z - c2 )] • [(z - c )(z - c3 )]

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

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

2+ l]

= (z2 + 2z + 2)(z

2-2z + 2).

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

.2 , ,l-Z"

we find that

l+z + zz+-+z"-

I =-^- (z*l),l-z

l + c + c2+...+c«-i = l_£l =: J__L =

1-c 1-c

9. Observe first that

(z-r^^exp^^) 1 i(-0

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

m "ilr m m

18

and

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

7=rexp-—-exp- -,

where Jfe = 0, 1, 2, . . . ,m - L Since the set

i{-2kn)exp-i L

(* = 0,l,2,...,ro-l)m

is the same as the set

i{2kn)exp^ L

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

but in reverse order, we find that (z1/m

)1 = (z

_1

)1/m

.

SECTION 10

1. (a) Write I z - 2 + il£ 1 as I z - (2 - i)l£ 1 to see that this is the set of points inside and on the

circle centered at the point 2-i with radius 1. It is not a domain.

O

(b) Write I2z + 3I>4 as > 2 to see that the set in question consists of all points

exterior to the circle with center at -3/2 and radius 2. It is a domain.

19

(c) Write Imz > 1 as y > 1 to see that this is the half plane consisting of all points lying

above the horizontal line y = 1. It is a domain.

y

y =i

X

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

y

y = \

X

(e) The set < argz < — (z * 0) is indicated below. It is not a domain.4

(f)The set Iz - 4l£l zl can be written in the form (x - 4)

2 + y2 1 x

2 + y\ which reduces to

*^2. This set, which is indicated below, is not a domain. The set is also

geometrically evident since it consists of all points z such that the distance between z

and 4 is greater than or equal to the distance between z and the origin.

20

4. (a) The closure of the set -it < argz < n (z # 0) is the entire plane.

1

(b) We first write the set IRezklzl as \x\<^Jx2 +y2

, or x2 <x2 + y

2. But this last

inequality is the same as y2> 0, or \y\> 0. Hence the closure of the set IRezklzl is the

entire plane.

(c) Since - =— = 7-7 = -5

—

-rr, the set Re - < - can be written as —. = < -, orz zz \z\

2 x2 +y2VzJ 2 x

2 +y22

(x2 -2x) + y

2>0. Finally, by completing the square, we arrive at the inequality

(x - 1)2 + y

2 > l2

, which describes the circle, together with its exterior, that is centered

at z = 1 with radius 1. The closure of this set is itself.

21

(d) Since z2 = (x + iyf =x2 -y2

+ ilxy, the set Re(z2) > can be written as y

2 < x2

, or

\y\<\x\. The closure of this set consists of the lines y = ±x together with the shaded

region shown below.

Since every polygonal line joining ztand z2 must contain at least one point that is not in S, it

is clear that S is not connected.

8. We are given that a set S contains each of its accumulation points. The problem here is to

show that S must be closed. We do this by contradiction. We let Zq be a boundary point of

S and suppose that it is not a point in S. The fact that z is a boundary point means that

every neighborhood of zQ contains at least one point in S; and, since z is not in S, we see

that every deleted neighborhood of S must contain at least one point in S. Thus Zq is an

accumulation point of S, and it follows that z is a point in S. But this contradicts the fact

that Zq is not in S. We may conclude, then, that each boundary point Zq must be in S. That

is, S is closed.

5. The set S consists of all points z such that lzl< 1 or lz - 2I< 1, as shown below.

22

Chapter 2

SECTION 11

1. (a) The function f(z) = 2

*is defined everywhere in the finite plane except at the

z +1

points z = ±i, where z2 + 1 = 0.

(fc) The function /(z) = Arg( -]is defined throughout the entire finite plane except for the

point z = 0.

z(c) The function /(z) = is defined everywhere in the finite plane except for the

z + z

imaginary axis. This is because the equation z + z = is the same as x = 0.

(d) The function /(z) = -— is defined everywhere in the finite plane except on the1 — Izl

circle Izl = 1, where 1 -Izl2 = 0.

3. Using x = and y = ^—-, write

f(z) = x*-y2 -2y+ i(2x - 2xy)

. fa±tf + ftdg, + _ + + _ fet Sfe -o

_2 -2 2 -2

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

2+2/z.

2 2 2 2

SECTION 17

5. Consider the function

/w-if l

-'x + iyY (z*0),

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

x + iON2

and if z = (0,y),

23

But if z = (*,*),

/fe) = l,I^ =137.1

This shows that /(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 /(z) as z tends to

cannot exist.

4z2

10. (a) To show that lim -r = 4, we use statement (2), Sec. 16, and write(z - 1)

4 -P2

lim ,kZ

\-, = lim——-? = 4.

V-»i (z-l)J(b) To establish the limit lim ——

—

j = <», we refer to statement (1), Sec. 16, and write

lira -7—^—r = lim (z - 1)3 = 0.

*-»il/(z-l)3

«-»

z2 + l

(c) To verify that lim = °°, we apply statement (3), Sec. 16, and writez-1

i-ilim —^-j = lim =- = 0.

«^fly j^» l+z

2

11. In this problem, we consider the function

7-(z ) = f£±A (ad- be *0).CZ + fi?

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

lim = hm = — = 0.«-»° T(l/z) »-»o a + bz a

24

(b) Suppose that c * 0. Statement (2), Sec. 16, reveals that lim T(z) = — since*-»- c

limT|'il = lima+ fe a

\zJ c + dz c

Also, we know from statement (1), Sec. 16, that lim T(z) = 00 sincel-*-d/c

Um -i-= lim £+4,0.7(2) z-»-rf/c az + &

SECTION 19

1. (a^ If /(z) = 3z2 -2z + 4,then

/'(z) = ^(3z2-2z + 4) = 3^-z

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

fife az az az

If /(z) = (l-4z2)

3,then

/'(z) = 3(1 - 4zT T-d - 4* > = 3(! " 4z2f(-8z) = -24z(l - 4z

2)

dz

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

(2, + 1)(1). (z _ 1)2 _ 3

JK)(2z + l)

2(2z + l)

2(2z + l)

2

(d) If /(z) =(1+

f)4

(z^0),thenz

^An^^_ a + z2yd 2

dz

(z2

)

2(z

2

)

2nz) = ' ^ <1 + Z )~ (l + Z ) g = z

24(lH-z

2)

3(2z)-(lH-z

2)42z

_ 2z(l + z2)

3[4z

2-(l + z

2)] _ 2(l + z

2

)

3(3z

2 - 1)

25

3. If /(z) = l/z (z*0), then

1 1 -AzAw = f(z + Az)-f(z) =z + Az z (z + Az)z

Hence

/'(z) = lim— = lim — =—\r.A*->o *z-*0(z + Az)z z

We are given that f(z ) = g(z ) = and that f\z ) and g'(z ) exist, where g'(z )*0.According to the definition of derivative,

Similarly,

Thus

rfe) = lira /Mzfe) = lim/w.

«-*«o Z - Zq Z-Zq

g-(z ) = lim^)-^ ) = lim-^-.

lim = UrnfWb-**) -_\^Z)I{Z-^

_ f(Zo)

*-*zog(z) s(z)/(z-z ) lim^(z)/(z-z ) ^'(z,,)'

SECTION 22

1. (a) /(z) = z= x — iy. So u = x, v = -y.

Inasmuch as = Vj, => 1 = -1, the Cauchy-Riemann equations are not satisfied

anywhere.

(b) f(z) = z-z=(x + iy)-(x-iy) = + i2y. So u = 0, v = 2y.

Since mx = vy=> = 2, the Cauchy-Riemann equations are not satisfied anywhere.

(c) f(z ) = 2x + ixy2

. Here u = 2x, v = xy2

.

ux = vy=> 2 = 2xy => xy = 1-

w = -vT=> = -y2

=> y = 0.

Substituting y = into Ay = 1, we have = 1. Thus the Cauchy-Riemann equations donot hold anywhere.

(d) /(z) = exe~'

y - e*(cosy - isiny) = excosy - ie

xsiny. So u = e

xcosy, v = -e

xsiny.

«x = vy=> cosy = -e

xcosy => 2e* cosy = => cosy = 0. Thus

y =^ + nn (n = 0,±l,±2,...).

uy= -v* =* -e'siny = e'siny => 2e*siny = => siny = 0. Hence

y-nit (« = 0,±1,±2,...).

Since these are two different sets of values of y, the Cauchy-Riemann equations cannotbe satisfied anywhere.

26

3. (aj /(z) = - = -.- =— =— _ + Soz z z Izl x +y2

x2 +y2

Since

x a -yu = —: t and v = , , .

x2 + y

2x

2 + y2

/'(z) exists when z 0. Moreover, when z * 0,

y2 -*2

. 2xy

(*2 +v2

)

2+V+/)2 "(x

2 +y2)

2

» , . y2 -*2

,. 2xy x

2-i'2jcy-y

2

_ (*-») = (gr _ ay \_

(x2 +y2

)2

(zz)2

(z)2(z)

2z2-

(fc) /(z) = jc2 + /y

2. Hence u = x

2and v = y

2. Now

ux = vy=> 2x = 2y => y = jc and M

y= -v, => = 0.

So /'(z) exists only when y = x, and we find that

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

W f(z) = z Im z = (x + iy)y = xy + iy2

. Here u = xy and v = y2

. We observe that

wx = vy=> y = 2y => y = and «

y= -vx * = 0.

Hence /'(z) exists only when z = 0. In fact,

/'(0) = Mje (0, 0) + iv, (0, 0) = + iO = 0.

4. (flj /(z) = -^ = ^cos4^ +^~sin40j (z*0). Since

4 4ru

r=— cos40 = vg and ue =—-sin40 = -rv

r

/is analytic in its domain of definition. Furthermore,

f'(z) = e-w(u

r+ iv

r )= e-

w^~cos40 + i^-sin4aj

= ~e- ie(cos46-isin4d) = --^"'V'48

r r-4 4 4

r3e,5B

(re,e

f

(b) f(z) = -Jre''2 = Vr cos— + n/r sin— (r > 0, a < < a + In). Since

2* 2*

^ , .

rur=—cos- = ve and ue =-—sm- = -rv

r ,

/is analytic in its domain of definition. Moreover,

f\z) = e-w(u

r + ivr )=MrT--cos| + irVsiiif

-— «"wfcos- + /sin-) = A=e-Wewn2^ I 2 2 J 2V^

1= 1

2^eien ~2f(z)'

(c) f(z) = e~ecos(lnr) + ie-

esin(lnr) (r>0,0< 0<2tt). Since

V „ ' Vy

1

U V

rur=-e~e

sin(ln r) = ve and ug =-e~ecos(ln r) = -rv

r

/is analytic in its domain of definition. Also,

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

= e-w e~ sin(lnr)

|^.

gacos(lnr)

r r

= -^fe~ecos(ln r) + ie~

esin(lnr)l =

re L J z

When /(z) = *3 + i(l - yf, we have k = x 3

and v = (1 - y)\ Observe that

ux = vy=> 3x

2 = -3(1-yf=*x2+ (1-yf=0 and w

y=-

Vje =>0

Evidently, then, the Cauchy-Riemann equations are satisfied only when x = and y = 1.

That is, they hold only when z = i. Hence the expression

f'(z) = ux + ivx = 3x2 + iO = 3x

2

is valid only when z = i, in which case we see that /'(/) = 0.

Here u and v denote the real and imaginary components of the function / defined by meansof the equations

Az) =t— when z*0,z

when z = 0.

Now

xl +y2x

2 + y2

when z * 0, and the following calculations show that

",(0,0) = v, (0,0) and «y (0,0)

= -v,(0,0):

Ax-»0 AjC A*->0 AjC

w(0 + Ajc,0)--«(0,0)

Ax

w(0,0 + Ay)- «(0,0)

Ay

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

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

Ay-»0 Ay Ay-»0 Ay

A"->0 Ax

Ay-»0 Ay Ay->0 £y

Equations (2), Sec. 22, are

ux cos + uysin

-w,rsin0 + iycos0

= "r>

Solving these simultaneous linear equations for ux and uy

, we find that

a sin 6 , . n cos0ux = u

rcosO-ug and u

y=ur

smd + ue

Likewise,

r

a sin0 cos0= v

rcos -

v

e and vy= v

rsm0 + vg .

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

rur=vg ,

ue =-rvr ,

are satisfied at z„. It follows that

a sin 6 cos 6,

. _ . - cos0wx = i/

rcos0-we

= ve + vsin0 = vr smd+ vg = vv ,

r r ry

. a ,cos0 sin0 _ f . sin0\

My= «

rsin d + ue

= ve vrcos0 = -l v

rcos0-ve \

= ~vx-

(a) Write f(z) = u(r, 0) + iv(r, 0) . Then recall the polar form

r«r=ve ,

M„=-rvr

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

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

for the derivative of/at a point z = (r , O ) in the following way:

f'(Zo ) = e~i0(-vg --u(

A =—lg (ue + ivg ) = —(ue + ivg ).

\r r J re zn

30

(b) Consider now the function

With

f{^\ _ 1 _ 1 _ 1 -is _ 1 /„ „ Q . • Q\ cos sin 6j\Z)-- =—ig=-e = -(cos 6 -ism 6) = 1.

z re r r r r

,.(*. Q\ C0S & At as S"VS\Bu{r, d) = and v(r, 0) =

,

r r

the final expression for /'(z ) in part (a) tells us that

£K \~

l ( sin # .cos0/ (z) =— 1

z \ r r

1fcos - /sin 6

z\ r

\ r J zKre'l z2

when z*0.

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

z + zX = and y =

z-z2 '

2i

Formal application of the chain rule for multivariable functions yields

dz dxdz dy dz dx \2 J dy \ 2i)~ 2{dx+l

dy

(b) Now define the operator

dz 2\dx dy j

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

dz 2{dx dy) 2dx 2dy

=fa +*J +fa + ivy)= \[{ux - v

y ) + i(vx + uy )].

If the Cauchy-Riemann equations uz = vv , u = -vr are satisfied, this tells us thaty y *

df/dz = 0.

31

SECTION 24

1. (a) f(z) = 3x + y + i(3y - x) is entire since

U V

ux =3 = vy

and uy=l = -vx .

(b) f(z) = sinx cosh y + icosx sinhy is entire since> „ ' v

v,

U V

ux = cosx cosh y = vy

and uy= sinxsinhy = -vx .

(c) f(z) = e~y sinx - ie~

y cosx = e~y s'mx + i(-e~y cosx) is entire since

' v ' > , ,

V

ux = e~y cosx - vy

and uy= -e~y sinx = -vx .

(d) f(z) = (z2 - 2)e'

xe"y is entire since it is the product of the entire functions

g(z) = z2 -2 and h(z) = e~

xe~

ty = e~x(cosy - isiny) = e~

xcosy + i(-e~

xsiny).

' v ' > v'

U V

The function g is entire since it is a polynomial, and h is entire since

ux = -e~xcosy = v

yand u

y= -e~

xsiny = -vx .

2. (a) /(z) = xy + iy is nowhere analytic since

U V

ux = vy=>y = l and u

y=-vx =>x = 0,

which means that the Cauchy-Riemann equations hold only at the point z = (0,1) = i.

(c) f(z) = eye

lx = ey (cosx + /sin x) = gycosj: + iVsinjc is nowhere analytic since

U V

Mx = vy=* -ey sinx = ey sinx 2ey sinx = => sinx =

and

= -v, => ey cosx = -ey cosx => 2^ cosx = => cosx = 0.

More precisely, the roots of the equation sinx = are nit (n = 0,±l,±2,...), and

cos /iff = (-1)" * 0. Consequently, the Cauchy-Riemann equations are not satisfied

anywhere.

7. (a) Suppose that a function f(z) = u(x,y) + iv(x,y) is analytic and real-valued in a domain

D. Since f(z) is real-valued, it has the form f(z) = u(x,y) + i0. The Cauchy-Riemann

equations ux =vy,u

y= -vx thus become ux =0,uy

= 0; and this means that u{x,y) = a,

where a is a (real) constant. (See the proof of the theorem in Sec. 23.) Evidently, then,

f(z) = a. That is, / is constant in D.

32

(b) Suppose that a function / is analytic in a domain D and that its modulus is

constant there. Write |/(z)| = c, where c is a (real) constant. If c = 0, we see that

f(z) - throughout D. If, on the other hand, c * 0, write f(z)7(z) = c2

, or

/(*) = •

C"

mSince f(z) is analytic and never zero in D, the conjugate f(z) must be analytic in D.

Example 3 in Sec. 24 then tells us that f(z) must be constant in D.

SECTION 25 (Yl

1. (a) It is straightforward to show that uxx +uyy=0 when u(x,y) = 2x(l-y). To find a

harmonic conjugate v(x,y), we start with ux (x,y) = 2-2y. Now

ux =vy=>v

y=2-2y=> v(x,y) = 2y-y2 + </>(x).

Then

uy= -vx => -2x- -<p'(x) => 0'(jc) = 2* = *

2 + c.

Consequently,

v(;c,y) = 2y-y2+ (jc

2 + c) = x2 - y2 + 2y + c.

(b) It is straightforward to show that uxx + uyy=0 when u(x,y) = 2x - x 3 + 3xy

2. To find a

harmonic conjugate v(x,y), we start with w^Oc.y) = 2-3*2 + 3y2

. Now

ux = Vy ^>vy=2-3x2

+ 3y2=» v(*,y) = 2y- 3x

2y + y

3 + <j)(x).

Then

uy= -vx => 6xy = 6xy - <p'(x) => <p'(x) = 0=> 0O) = c.

Consequendy,

v(jc,y) = 2y - 3x2y + y

3 + c.

fcj It is straightforward to show that uxx +uyy=0 when «(jc,y) = sinhjcsiny. To find a

harmonic conjugate v(x,y), we start with ux (x,y) = coshx sin y. Now

= vy

vy= coshxsiny => v(x,y) = -coshxcosy + <p(x).

Then

«j, = -vx => sinhxcosy = sinhxcosy - <j)'(x) => 0'(jc) = => 0(*) = c.

Consequently,

v(*,y) = - coshxcosy + c.

(d) It is straightforward to show that m„+w=0 when u{x,y) = , - , . To find a

harmonic conjugate v(x,y), we start with ux (x,y) = r~^~TT- Now(x

2 + y2)2

Then

= v, => vy= ~

(jc2

2

+ 2)2=> v(*,y) = -^£-5.+ ^(jc).

2 2 2 2

"y= "Vx ^ (x

2+y

2

)

2=

(x2 + y

2)

2 ~ ^ *'(jc) = 0=*

=°-

Consequendy,

v(x,y) =— T + c.

x +y

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

ux = vy ,

uy= -vx and ux = Vy ,

uy= -Vx .

If w = v - V, then,

wx =vx -Vx =-uy+Uy=0 and w

y=v

y-V

y=ux -ux

= 0.

Hence w(jc,y) = c, where c is a (real) constant (compare the proof of the theorem in Sec.

23). That is, v(x,y)-V(x,y) = c.

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

ux = vy ,

Uy = -vx and vx = uy ,

vy= -ux .

It follows readily from these equations that

ux — 0, uy= and vx = 0, v

y= 0.

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

theorem in Sec. 23).

The Cauchy-Riemann equations in polar coordinates are

rur=ve and ue - -rvT

.

Now

rur= ve => run + u

r= v

er

and

Thus

run + rur+ ugg =rv6r

- rvr6 ;

and, since v^. = vr6 , we have

r2un + rur + Ugg = 0,

which is the polar form of Laplace's equation. To show that v satisfies the same equation,

we observe that

1 1 1

"e = -rvr=> v

r=— ug

=> v„ =—ug—

K

frr r r

and

rur=vg =*vgg

= rur0 .

Since uBr= urg , then,

r2vn + rvr + vee = ue - ru^ -ue + rur8 = 0.

If w(r,0) = lnr, then

rV +mr + ugg = r2

[

~J

+ rfi 1 + = 0.

This tells us that the function w = In r is harmonic in the domain r>O,O<0<27T. Now it

follows from the Cauchy-Riemann equation rur= vg and the derivative ur~~ mat ve ~ ^

thus v(r,6) = + <j>(r), where 0(r) is at present an arbitrary differentiable function of r.

The other Cauchy-Riemann equation ug =-rvrthen becomes O = -r0'(r). That is,

<j>\r) = 0; and we see that <j>(r) = c, where c is an arbitrary (real) constant. Hence

v(r, 6) = 8 + c is a harmonic conjugate of u(r, 6) = In r.

35

Chapter 3

SECTION 28

1. (a) exj>(2±3m) = e2ex$(±3m) = -e

2, since exp(±3;ri) = -l.

2 + m ( IV m\ r( n . . n\(b) exp—^— = 1 exp- II exp— 1 = V«l cos- + isin- I

(c,) exp(z + 7ri) = (expz)(exp7ri) = -expz, since exp7ri = -l.

3. First write

exp(z ) = exp(* - iy) = exe~

!y = excos y - ie

xsin y,

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

u{x,y) - excosy and v(x,y) = -ex

smy.

Suppose that the Cauchy-Riemann equations ux = vyand u

y= -vx are satisfied at some

point z = x + iy. It is easy to see that, for the functions « and v here, these equations become

cos y = and sin y = 0. But there is no value of y satisfying this pair of equations. We mayconclude that, since the Cauchy-Riemann equations fail to be satisfied anywhere, the

function exp(z) is not analytic anywhere.

4. The function exp(z2

) is entire since it is a composition of the entire functions z2and expz;

and the chain rule for derivatives tells us that

-|exp(z2

)=exp(z

2)-|z

2 = 2zexp(z2).

Alternatively, one can show that exp(z2) is entire by writing

exp(z2) = exp[(* + iy)

2

j= exp(*

2 - y2)exp(i2*y)

= exp(x2-y

2)cos(2ry) + iexp(jc

2 - y2)sin(2;ty)

>w

< » „/

U V

and using the Cauchy-Riemann equations. To be specific,

ux =2xexp[x2 -y2

)cos(2;ty)-2yexp(;t2-y

2)sin(2xy) = v

y

and

uy= - 2y exp(x

2 - y2) cos(2xy ) - 2x exp(x

2 - y2)sin(2.xy) = -vx .

36

Furthermore,

-y-exp(z2

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

2 -

y

2)cos(2xy) + iexp(x

2 -

y

2)sin(2xy)]

= 2zexp(z2

).

5. We first write

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

2x

and

|exp(iz2

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

2 - y2

)]|= e

2xy.

Then, since

|exp(2z + 1) + exp(iz2

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

2

)|

,

it follows that

|exp(2z + 1) + exp(/z2

)|< e

2x + e~2xy

.

6. First write

|exp(z2

)|= |exp[(x + ry)

2

]|= |exp(j:

2 -/) + i2xy| = expU2 -y2

)

and

exp(lzl2) = exp(jc

2 + y2).

Since x2 - y

2 < x2 + y

2, it is clear that exp(*

2 - y2) < exp(*

2 +y2). Hence it follows from

the above that

|exp(z2)|<exp(lzl

2).

7. To prove that |exp(-2z)| < 1 <=> Re z > 0, write

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

It is then clear that the statement to be proved is the same as exp(-2*) < 1 <=> x > 0, which is

obvious from the graph of the exponential function in calculus.

37

(a) Write el = -2 as eV = 2*'*. This tells us that

e*=2 and y = n + 2nn (n = 0,±l,±2,...).

That is,

* = ln2 and y = (2n + l);r (n = 0,±1,±2,...).

Hence

z = ln2 + (2n + l)«f ' (n = 0,±l,±2,...).

(b) Write *l = 1 + V3 1 as <?V = 2e

,(*/3), from which we see that

e*=2 and y =j + 2nn (n = 0,±1,±2,...).

That is,

* = ln2 and y = ^2» + |j« (n = 0,±1,±2,...).

Consequently,

z = \xi2 + {ln + ^jti (« = 0,±1,±2,...).

(cj Write exp(2z - 1) = 1 as e2x~l

ei2y = le

i0and note how it follows that

<?

2x_l =l and 2y = + 2nx (n = 0,±l,+2,...).

Evidendy, then,

2

and this means that

x = \ and y = n^ (n = 0,±1,±2,...);

z = -^ + nm (n = 0,±l,±2,...).

This problem is actually to find all roots of the equation

exp(iz) = exp(/z).

38

To do this, set z = jc + iy and rewrite the equation as

eye'

a =eye".

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

e~y = e

y and -x = x + Inn,

where n may have any one of the values n = 0,±l,±2,.„. Thus

y = and x = nn (n = 0,±1,±2,...).

The roots of the original equation are, therefore,

z=nn (n = 0,±l,±2,...).

10. (a) Suppose that ez

is real. Since el = e

xcosy + ie

xsin y, this means that e

xsiny = 0.

Moreover, since ex

is never zero, siny = 0. Consequently, y = nn(n = 0,±1,±2,...);

that is, Imz = n^(« = 0,±l,±2,...).

(b) On the other hand, suppose that ez

is pure imaginary. It follows that cosy = 0, or that

y = ^- + nn(n = Q,±\,±2,...). Thatis, Inu =— + /i7r(n = 0,±l,±2,...).21 2

12. We start by writing

1 _ z _ z x-iy

z zz Izl2

x 2 +y2x

2 + y2

x2 +y2

x . -y+ i-

'

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

Re(<?1/Z

) = exp2 . 2\x +y

cos

^2 + y2

= exp.2 , ..2

COS2

, ..2jc +y J \x +y

Since is analytic in every domain that does not contain the origin, Theorem 1 in Sec. 25

ensures that Re(e1/z) is harmonic in such a domain.

13. If f(z) = u(x,y) + i'v(jc,y) is analytic in some domain D, then

efit) = e

u(x,y)CQS^ y) + ie

u(*,y)sinv(xy)

Since efU) is a composition of functions that are analytic in D, it follows from Theorem 1 in

Sec. 25 that its component functions

U(x,y) = eu(x

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

u(x 'y)sinv(jc,y)

39

are harmonic in D. Moreover, by Theorem 2 in Sec. 25, V(x,y) is a harmonic conjugate of

U(x,y).

14. The problem here is to establish the identity

(expz)" = exp(nz) (n = 0,±1,±2,...).

(a) To show that it is true when n = 0,1,2,..., we use mathematical induction. It is

obviously true when n = 0. Suppose that it is true when n - m, where m is any

nonnegative integer. Then

(expz)m+l = (expz)"

1

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

(b) Suppose now that n is a negative integer (n = -1, -2, . .. ), and write m = -n = 1, 2, . . . . Li

view of part (a),

(expz)" =^expz;

1 1 1

(expz)m

exp(znz) exp(-nz)= exp(nz).

SECTION 30

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

(b) Log(l- = lnll - i1+i'Arg(l - = InV2 — = |ln2 - -ji.

2. fa; loge = lne + i(0 + 2n7f) = l + 2/i7Fi (n = 0,±l,±2,...).

(b) logi = lnl+i^ + 2nnj = ^2n +^m (n = 0,±l,±2,...).

(c) log(-l + V3i) = ln2 +i^ + 2n^ = ln2 + 2^n +|jm (n = 0,±1,±2,...).

3. (a) Observe that

and

Log(l + 1)

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

2Log(l + 1) = 2^1nV2 + ijj = In 2 +

Thus

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

(b) On the other hand,

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

and

2Log(-l + i) = 2( lnV2 +i^J = ln2 + y/.3n

Hence

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

(a) Consider the branch

Since

logz = lnr + /0 r>0, — <6<—4 4

log(i2) = log(-l) = lnl + i;r=7B and 21og/ = 2flnl +

/-|J

= m,

we find that log(/2) = 2 log/ when this branch of logz is taken.

(b) Now consider the branch

Here

logz = lnr + /0 r > 0,— < 9 <4 4

log(i2) = log(-l) = In 1 + in = m and 2 log/ = 2 In 1 + /

,5^2 ,

= 5m.

Hence, for this particular branch, log(/2) * 2 log/.

(a) The two values of im

are e'W4

and ei5tt/

\ Observe that

( Klog(€'*

/4) = lnl + /|-^+ 2«7r| =

( l\2n +- \jti (« = 0,±1,±2,...)

and

log(*,9W4

) = lnl + i|^ + 2/«rh5n

(2n + l) +-4

7n (n = 0,±l,±2,...).

Combining these two sets of values, we find that

log(/1/2

) = |n +^nm (n = 0,±l,±2,...).

41

On the other hand,

1, . 1-logi=-2 2

lnl + i| ^ + 2nn = \n +—\m (n = 0,±l,±2,...).

1Thus the set of values of log(i ) is the same as the set of values of — logi, and we

2may write

log(/1/2

) = |log/.

(b) Note that

log(i2) = log(-l) = In 1 + {it + 2mt)i = {2n + \)iti (n = 0,±l,±2,...)

but that

21ogi' = 2 lnl + i'l ^+ 2mt = {An + Y)Td (/i = 0,±l,±2,...).

Evidently, then, the set of values of log(/2) is not the same as the set of values of

2 log i. That is,

log(i2)*21ogi.

7. To solve the equation logz = iitll, write exp(logz) = exp(/7r / 2), or z = e'*n = i.

10. Since ln(jc2 +y

2) is the real component of any (analytic) branch of 2 logz, it is harmonic in

every domain that does not contain the origin. This can be verified directly by writing

u{x,y) = ln(*2 + y

2) and showing that ^(jc.y) + u (x,y) = 0.

SECTION 31

1. Suppose that Rezt> and Rez2 > 0. Then

zt= r

texp i'0j and z1 -r1

exp i'02 ,

where

it ~ it , it ^ it— < 0, <— and <©,<— ,

21

2 22

2

42

The fact that -% <t+ 2 < n enables us to write

Log(ztz2 ) = Log[(r

xr2 )exp /(0

t+ 2 )] = lnfe) + i{Q

x+ 2 )

= (In r{+ i'0i ) + (In r2 +

1'02 ) = Logfo exp i©, ) + Log(r2 exp i02 )

= LogZ! + Logz2 .

3. We are asked to show in two different ways that

log =logzt-logZ2 (z, * 0, z2

?t 0).

(a) One way is to refer to the relation arg — = argZ[ - argz2 in Sec. 7 and write

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

(bj Another way is to first show that log[ -]= -logz (z * 0). To do this, we write z = re

and then

log^ = log^e"'8

j= ln^+ i(-0 + 2imt) = -[lnr + i(0- 2imt)] = -logz,

where n = 0,±1,±2,. . . . This enables us to use the relation

log(z1z2 ) = logz

1+logz2

and write

log

\ Z2J

log = logz, + log

f\2 = logz,- logz2

43

5. The problem here is to verify that

z1/n =exp -UogzJ (n = -l,-2,...),

given that it is valid when n = 1,2,... . To do this, we put m = -n, where n is a negative

integer. Then, since m is a positive integer, we may use the relations z'1 = 11 z and

1/ ez = e~

zto write

= 1,

p(^iog4

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

SECTION 32

1. In each part below, n = 0,±l,±2,....

(a) (1 + 1)' = exp[i log(l + 1)] = exp< i

{i1= exp| —ln2-[ — + 2nn

n

lnV2+i(^+ 2n^

= exp^--^ - 2n^jexp^ln2j.

(b)

Since n takes on all integral values, the term -2nn here can be replaced by +2nit.Thus

(1 + /)'' =exp^-| + 2n^exp^ln2j.

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

2. (a) P.V. /' = exp(/Logi) = exp^ln 1 + /-|

j

exp - n

(b) P.V. [j(-l - V3/)]"= exp|3^Log[|(-l - V3/)| = exp

= exp(2/r2)exp(/3^) = -exp(2;r

2

)

3ni\ )ne-iIn

44

(c) P.V. (1 - i)4

' = exp[4/Log(l - 1)] = exp

= e*[cos(41nV2) + isin(41n V2)] = «*[cos(21n2) + isin(21n2)].

4iflnV2-/j

3. Since - 1 + V3i = 2e2w</3

, we may write

(-1 + V3/)3'2 =exp |log(-l + V30] = exp|| ln2 + /|

^Y+ lnn

= exp[ln(23/2

) + (3n + l)*ri] = 2-^2 exp[(3/i +

where n = 0,±1,±2,.... Observe that if n is even, then 3n+l is odd; and so

exp[(3n + l);n'] = -l. On the other hand, if n is odd, 3n+l is even; and this means that

exp[(3/i + Y)iti] = 1. So only two distinct values of (-1 + V3/)3/2

arise. Specifically,

(-1 + V3i)3/2

=±2V2.

5. We consider here any nonzero complex number z in the exponential form Zq = rQexp i'0

o ,

where -it < © < it. According to Sec. 8, the principal value of zUn

is ^/r^exp^i—j; and,

according to Sec. 32, that value is

exp( —Logz|= exp ^ (in r + i'0

o ) = exp(ln ^/r~) exp^z' = nfc exp^j^

These two expressions are evidently the same.

7. Observe that when c = a + bi is any fixed complex number, where c * 0,±1,±2,..., the

power iccan be written as

T = exp(clogi) = exp< (a + bi) ]sil + i^— + 2nit

= expit

-b — + 2mt \+ia\ — + 2mt.12 ) Vl

it

Thus

(n = 0,±l,±2,...).

I/Cl= exp -b\ — + 2nit (» = 0,±1,±2,...),

and it is clear that lzc

I is multiple-valued unless b = 0, or c is real. Note that the restriction

c * 0,±1,±2,... ensures that ic

is multiple-valued even when b = 0.

SECTION 33

1. The desired derivatives can be found by writing

—sin = ±f'k -'~k ^

dz dz

e--e- = j_(d iz _d- iz

2i )

e-—-e -

j2i\dz dz

and

= Yi\ie +ie )=—

2— = cosz

d d(e* + e-*} lfd k d _fe—cosz =— =- — e +—e

dz dz\ 2 ) 2\dz dz

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

i 2i

2. From the expressions

e*-e'* , ek + e~

iz

s'mz = and cosz =2i 2

we see that

ek+e-'

K,ek -e~k kcosz + ismz = + = e .

2 2

3. Equation (4), Sec. 33 is

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

Interchanging Zi and z2 here and using the fact that sin z is an odd function, we have

2 cos zxsin z2 = sin(z

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

Addition of corresponding sides of these two equations now yields

2(sin zxcos z^ + cosz

vsin Zj ) = 2 sin(z

t+ Z2 ),

or

sin(z! + z2 ) = sin z, cos z2 + cos z{sin 22

.

4. Differentiating each side of equation (5), Sec. 33, with respect to zv we have

cos(zt+z2 )

= cosZ[ cosz2 -sinztsinz2 .

46

7. (a) From the identity sin2z + cos

2z = 1, we have

sin2z cos

2z 1 ,22—— +—— =—— ' or 1 + tan z = sec z.

cos z cos z cos z

ft) Also,

sin z cos z 1

. , +. , =

. , , or 1 + cot z = esc z.sm z sin z sin z

9. From the expression

sin z = sinxcosh y + 1 cosxsinh y,we find that

Isinzl2 = sin

2xcosh

2y + cos

2x sinh

2y

= sin2x(l + sinh

2y) + (1 - sin

2jc)sinh

2

y

= sin2x + sinh

2y.

The expression

cosz = cosxcosh y + i sin*sinh y,

on the other hand, tells us that

Icoszl2 = cos

2jccosh

2y + sin

2jtsinh

2y

= cos2*(1 + sinh

2y) + (1 - cos

2x) sinh

2y

= cos2x + sinh

2y.

10. Since sinh2y is never negative, it follows from Exercise 9 that

(a) Isinzl2 > sin

2x, or Isinzl > lsin*l

and that

(b) Icoszl2 > cos

2x, or Icoszl > I cos xl.

11. In this problem we shall use the identities

I sin zl2 = sin

2 x + sinh2y, I cos zl

2 = cos2 x + sinh

2y.

47

(a) Observe that

sinh2y =lsinzl

2 - sin2x < Isinzl

2

and

Isinzl2 = sin

2 x + (cosh2y - 1) = cosh

2y - (1 - sin

2x)

= cosh2y - cos

2 x < cosh2y.

Thus

sinh2y<lsinzl

2 < cosh2y, or lsinhyl<lsinzl< coshy.

(b) On the other hand,

sinh2y = I cos zl

2 - cos2x < I cos zl

2

and

Icoszl2 = cos

2 x + (cosh2y - 1) = cosh

2y - (1 - cos

2x)

= cosh2y - sin

2x <, cosh

2y.

Hence

sinh2y < Icoszl

2^ cosh

2y, or lsinhyl< Icoszl^ coshy.

13. By writing /(z) = sinz =sin(;c- iy) = sin* cosh y -icos* sinh y, we have

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

where

u(x,

y

) = sinx coshy and v(x,y) = -cosxsinhy

.

If the Cauchy-Riemann equations wx = vy ,

Wj, = -v, are to hold, it is easy to see that

cos;ccoshy = and sin*sinhy = 0.

Since coshy is never zero, it follows from the first of these equations that cosx = 0; that is,

x =— + nn (n = 0± 1,±2,...)- Furthermore, since sinx 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 at the points

z =- + n7t (n = 0±l,±2,...)-2

Evidendy, then, there is no neighborhood of any point throughout which/ is analytic, and

we may conclude that sinz is not analytic anywhere.

The function /(z) = cosz = cos(jc -/y) = cosx coshy + isinx sinhy can be written as

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

where

u(x, y) = cosx coshy and v(x, y) = sin xsinh y.

48

If the Cauchy-Riemann equations ux = vy ,

uy~ -vx hold, then

sinxcoshy = and cosxsinhy = 0.

The first of these equations tells us that sin;c = 0, or x = n7t(n = 0,±l,±2,...). Since

cosnn *0, it follows that sinhy = 0, or y = 0. Consequently, the Cauchy-Riemann

equations hold only when

z = nn (n = 0±l,±2,...).

So there is no neighborhood throughout which / is analytic, and this means that cosz is

nowhere analytic.

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

cos(iz) = cos(-y + ix) = cosy coshx — i siny sinhx

and

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

This shows that cos(i'z) = cos(iz) for all z.

(b) Use expression (1 1), Sec. 33, to write

sin(i'z) = sin(-y + ix) = -sinycoshx - icosysinhx

and

sin(i'z) = sin(y + ix) - siny coshx + i cosysinh jc.

Evidently, then, the equation sin(zz) = sin(i'z) is equivalent to the pair of equations

sinycoshx = 0, cosysinhx = 0.

Since cosh* is never zero, the first of these equations tells us that siny = 0.

Consequently, y = nn (n = 0,±1,±2,...). Since coshtt = (-1)" ^ 0, the second

equation tells us that sinhx = 0, or that x = 0. So we may conclude that

sin(/z) = sin(/z) if and only if z = + inn = nni (n = 0,±1,±2,. . .).

17. Rewriting the equation sinz = cosh4 as sin;ccoshy + icos;tsinhy = cosh 4, we see that we

need to solve the pair of equations

sinx cosh y = cosh 4, cosxsinhy =

49

for x and y. If y = 0, the first equation becomes sin* = cosh 4, which cannot be satisfied by

any* since sin*^l and cosh4>l. So y*0, and the second equation requires that

cos* = 0. Thus

x = ~- + n7t (n = 0±l,±2,...).

Since

sin^Y + »wj = (-l)",

the first equation then becomes (-1)" coshy = cosh 4, which cannot hold when n is odd. If n

is even, it follows that y = ±4. Finally, then, the roots of sinz = cosh 4 are

Z =(f

+ 2n7C]±4i (n = 0±l,±2,...).

18. The problem here is to find all roots of the equation cosz = 2. We start by writing that

equation as cosxcoshy-/sin;tsinhy = 2. Thus we need to solve the pair of equations

cos*coshy = 2, sinjrsinhy =

for x and y. We note that y * since cosx = 2 if y = 0, and that is impossible. So the

second in the pair of equations to be solved tells us that sin;c = 0, or that x = nn

(n = 0±l,±2,...). The first equation then tells us that (-l)"coshy = 2; and, since coshy is

always positive, n must be even. That is, x = 2nn {n = 0±1,±2,...). But this means that

coshy = 2, or y = cosh-12 . Consequently, the roots of the given equation are

z = 2n^ + /cosh_12 (n = 0±1,±2,...).

To express cosh"1

2, which has two values, in a different way, we begin with

y = cosh"1

2, or coshy = 2. This tells us that ey + e~

y - 4; and, rewriting this as

(eyf-4(ey

) + l = 0,

we may apply the quadratic formula to obtain ey = 2 ± V3, or y = ln(2 ± V3). Finally, with

the observation that

ln(2-V3) = ln

we arrive at this alternative form of the roots:

(2-V3)(2 + V3)

2 + V3= ln

2 + V3!

= -ln(2 + V3),

z = 2n;r±tln(2 + V3) (n = 0±l,±2,...).

50

SECTION 34

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

d . , d(e z -e'z

) 1 d , . _ZN ez + e'

z

and

d , d (

e

z + ez

\ 1 d , , _, x ez -e z

. ,_coshz=_^__j =__(e!+ ,.)=__ =sinlu.

3. Identity (7), Sec. 33, is sin2z + cos

2z = 1. Replacing z by & here and using the identities

sin(z'z) = isinhz and cos(iz) = coshz,

we find that i'

2sinh

2z + cosh

2z = 1, or

cosh2z - sinh

2z = 1.

Identity (6), Sec. 33, is cos^ + Zj) = cosz^oszj -sinzjsinzj. Replacing Zj by iZj and

z2 by iz2here, we have cosf/^ +z2 )]

= cos(/z1)cos(/z2 )-sin(iz1

)sin(iz2 ). The sameidentities that were used just above then lead to

cosh(z1+ z2 ) = coshzj cosh^ + sinl^ sinhz2.

6. We wish to show that

lsinh.*l<lcoshzl< cosh*

in two different ways.

(a) Identity (12), Sec. 34, is lcoshzl2 = sinh

2x + cos

2y. Thus Icoshzl

2-sinh

2 x>0; and

this tells us that sinh2x < Icoshzl

2, or Isinhx\< Icoshzl. On the other hand, since

Icoshzl2 = (cosh

2 x - 1) + cos2y = cosh

2x - (1 - cos

2y) = cosh

2 x - sin2y, we know that

Icoshzl2-cosh

2 x<0. Consequently, lcoshzl2<cosh2

;t, or lcoshzl<coshx.

(b) Exercise U(b), Sec. 33, tells us that lsinhyl<lcoszl< coshy. Replacing z by iz here and

recalling that cosi'z = coshz and iz = ~y + ix, we obtain the desired inequalities.

7. (a) Observe that

sinh(z + m) = = = = = -sinhz.

51

(b) Also,

cosh(z + 7ti) = Lf =—

—

'-z—1— =——— = -- = -coshz.

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

u/- _ sinh(z + ?n) -sinhz sinhztanh(z + /n) =—— - = — = = tanhz.

cosh(z + 7n) -coshz coshz

9. The zeros of the hyperbolic tangent function

t . sinhztanhz =

coshz

are the same as the zeros of sinhz, which are z = nm (n = 0,±1,±2,...). The singularities of

tanhz are the zeros of coshz, or z = ^-j + rucji (n = 0,±1,±2,...).

15. (a) Observe that, since sinhz = / can be written as sinhxcosy+ /coshxsiny = i, we need

to solve the pair of equations

sinhx cosy = 0, coshx siny = 1.

itIf x = 0, the second of these equations becomes siny = l; and so y =— + 2nn

2

(n = 0,±l,±2,...). Hence

(n = 0,±l,±2,...).

7TIf x*0, the first equation requires that cosy = 0, or y =—vmt

(/i = 0,±l,±2,...). The second then becomes (-l)"coshjr = l. But there is no nonzero

value of x satisfying this equation, and we have no additional roots of sinhz = i.

(b) Rewriting cosh z = ^ as coshx cosy 4- isinhx siny = — , we see that x and y must satisfy2 2

the pair of equations

coshxcosy = — , sinhxsiny = 0.2

52

If x = 0, the second equation is satisfied and the first equation becomes1 1 %

cosy = — . Thus y = cos1 — = ±— + 2nn (n = 0,±1,±2,...), and this means that

z = {ln±^7ti (n = 0,±l,±2,...).

If x 0, the second equation tells us that y = nn (n = 0,±1,±2,...). The first then

becomes (-1)" cosh x = But this equation in x has no solution since cosh* > 1 for

all x. Thus no additional roots of coshz = — are obtained.2

16. Let us rewrite coshz = -2 as cosh* cos ;y + isinhjcsin;y = -2. The problem is evidently to

solve the pair of equations

coshxcosy = -2, sinh x siny - 0.

If * = 0, the second equation is satisfied and the first reduces to cos y = -2. Since there

is no y satisfying this equation, no roots of coshz = -2 arise.

If x*Q, we find from the second equation that siny = 0, or y-nn (n = 0,±l,±2,...).

Since cosn^ = (-l)", it follows from the first equation that (-i)"coshx = -2. But this

equation can hold only when n is odd, in which case x = cosh-1

2. Consequently,

z = cosh"1

2 + (2n + l)m (n = 0,±1,±2,...).

Recalling from the solution of Exercise 18, Sec 33, that cosh"1

2 = ±ln(2 + V3), we note that

these roots can also be written as

z = ±ln(2 + V3) + (2n + l)ni (n = 0,±1,±2,...).

SECTION 37

Chapter 4

2. (a) jQ^J*=j(£-l)<*^

T |v2T ir ^.•f ,i V3 /e dt = —- =— cos— + *sin— -1 =— + -;

o L 2/ 1 2/L 3 3 J 4 4'

fcj Since le~**l= e~bx

, we find that

6 IV" T"* 1 1

fe""</f = lim f e

udt = lira = - lim(l - e

-bt) = - when Re z > 0.

L *• Jj=0 * *

3. The problem here is to verify that

\emB

e-Mde =

o

To do this, we write

?s? r when m*n,

27T when m = n.

2* 2w

/ =JeM

e~MdO = je'^'de

and observe that when m#n,

i(m-n)B

i(m — n)

2ic

1 1

i(m-n) i(jn-n)= 0.

When m = n, /becomes

and the verification is complete.

2*

/= \dd = 2n;

4. First of all,

7* 7t

Je(I+,)x

etc =Jexcos* dtc +

1"

J e* sinx dx.

But also,

,(!+<>

l + I

e*e'"-l -e*-l l-i l + e* A + e"+ i—-—

.

l + i l + i l-i

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

je'cosx dx = - ^+ e

and jVsin;c_* =o

2

Consider the function w(t) = e" and observe that

lie 2tc

jw(t)dt=je it

dt =

l + e*

=i-i=o.i i

Since \w(c)(2n - 0)| =\eK\2n = 2% for every real number c, it is clear that there is no number

c in the interval < t < 2n such that

lie

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

(a) Suppose that w(t) is even. It is straightforward to show that u(t) and v(f) must be even.Thus

a a a a a

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

= 2

a a a

j u(t)dt + z

Jv(t)dt = 2

Jw(t)dt.

(b) Suppose, on the other hand, that w(t) is odd. It follows that u(i) and v(f) are odd, and so

a a a

Jw(t)dt = ju(t)dt +

1J v(t)dt = + / = 0.

Consider the functions

1K _____ "

P„(x) = ~j(x + iVl - x2cos ej d6 (n = 0,1,2,...),

where -1 < x < 1. Since

x + i^l-x2 cosd = V*2 + (1 -

x

2)cos

2 d<^x2 +(l-

x

2) = 1,

it follows that

1 T.| X

P„(*)|<-f Lc + iVl-x2cos0 ^<-fj0 = l.

55

SECTION 38

1. (a) Start by writing

—a -a -a

I =Jw(-t)dt =

Jui-t)dt + i j v(-t)dt.

-b -b -b

The substitution r = -t in each of these two integrals on the right then yields

a a b b b

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

That is,

-a b

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

-b a

(b) Start with

b b b

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

a a a

and then make the substitution t = <p(x) in each of the integrals on the right. The result

is

P P P

I = ju[<l>(T)]<t>'(T)di:+ ijv[</>(T)]<t>'(r)dr =J

{x)dx.

a a a

That is,

b P

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

3. The slope of the line through the points (a, a) and (fi,b) in the ft plane is

b-am = — .

p-a

So the equation of that line is

b-a ,t-a = - (t-a).p-a

56

Solving this equation for t, one can rewrite it as

b-a aB-bat = — T+-^ .

B-a B-a

Since t = 0(f), then,

... b-a aB-baB-a B-a

4. If Z(t) = z[^(T)] , where zit) = x(t) + iy(t) and t = 0(t), then

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

Hence

Z'(t) = + i-f =x'MtWW +iy'miWWax at

=vim]

+

wcd = z'[<ktW(t).

5. If w(t) = f[z(t)] and f(z) = u(x,y) + iv(x,y), z(t) = x(t) + iy(t), we have

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

The chain rule tells us that

du , . , dv , ,— = uxx+uyy and — = Vj(x' + v

yy',

and so

w'(0 = (uxx' + uyy') + i(yxx' + v

yy').

In view of the Cauchy-Riemann equations ux= v

yand u

y= -v,, then,

w'(t) = (uxx' - vxy') + i(yxx' + uj) = (ux + ivx )(*' + iy').

That is,

w'(0 = {ux [x(t),y(t)] + ivMt)Mt)])[x'(t) + //(f)] = f'[z(t)]z'(t)

when f = f -

SECTION 40

1. (a) Let C be the semicircle z = 2e'° (0 < < n), shown below.

-2 2 x

Then

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

(b) Now let C be the semicircle z = 2ei0 (n<d< 2ri) just below

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

f£±2<fc = 2/

.'9

+

2k

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

(c) Finally, let C denote the entire circle z = 2ei0(0 < < 2k). In this case,

f ^-^-dz = Am,JC 7

the value here being the sum of the values of the integrals in parts (a) and (b).

2. (a) The arc is C:z = l + ei0 (n< 6<2n). Then

2jt Ijc

Jc(z - \)dz = J(l + «

w - l)few<*0 = ije

nedO = i

.''29

2/

=I(^_^) = I(1_1)=

.

58

(b) Here C: z = x (0 < x < 2). Then

= 0.

3. In this problem, the path C is the sum of the paths Q , C2 , C3 , and C4 that are shown below.

The function to be integrated around the closed path C is /(z) = neKt

. We observe that

c = ci+ C

2 + C3 +Q and find the values of the integrals along the individual legs of thesquare C.

(i) Since C, is z = x (0 < x < 1),

i

Jc^^ =^^ = ^-1.

1

o

(ii) Since C2 is z = 1 + iy (0 < y < 1),

i i

(Hi) Since C3 is z = (1 - x) + 1 (0 < x < 1),

i i

o

(ivj Since C4 is z = i(l - y) (0 < y < 1),

i i

J *»*<fe =4^^'HMy = ^JV^y = -2.

o

Finally, then, since

we find that

\cne*~

ldz = 4{e* -1).

Jc(z-l)dz = J(jt-l)<£c = -jc

The path C is the sum of the paths

Q:z = x + ix3 (-1<*<0) and C2 :z = x + ix

3(0<x£l).

Using

/(z) = lonC, and f(z) = 4y = 4x3on C2 ,

we have

1

jcf(z)dz = [ f(z)dz +JC2 /(z)<fe

=Jl(l + i3x2)<£c +

J4*3(l + i3*

2)<it

VJ-1 o

1 1

= Jit + 3/jA;2dbc +4j^ + 12ijA:

5dtc-1-10

= Wlli + i[*3

t + [*t + 2i: [*

6

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

The contour C has some parametric representation z = z(t) (a<t< b), where z(a) = zxand

z(fc) = Z2. Then

Jc<fe = \z\t)dt =

[Z(0]* = z(b) - z(a) = z2 -zva

To integrate the branch

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

Let C be the positively oriented circle 1*1=1, with parametric representation

z = e,e

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

lczmz"dz = j

{ei6

)

m(e-

ie

)

n

i£edd = i je'<"

+1)VM</0.

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

lK[0 when m*n,

[lit when m-n.jeM

e-Md0--

60

Consequently,

frt\fc-j°when m+1 * n -

lc \2m when m + l = n.

8. Note that C is the right-hand half of the circle x2 + y

2 = 4. So, on C, * = ^4-y2. This

suggests the parametric representation C:z = ^4-y2+iy (-2 < y < 2), to be used here.

With that representation, we have

dy

2 2 ( %

= j(-y + y)dy + ij\-tI= +^y2

-2 iW4- ^

2 ..2 4 = 4i

= 4/ [sin"1

(1) - sin"1

(-1)] = 4i = Am.

J -2

10. Let C be the circle z = z + Re ie (-Jt<6< n)

(a) f=

f-^Rie iedd = i]dd = 2m.

-K

(b) When n = ±1,±2,...,

=— (em* - e-'"

1

*) = /— sinn/r =2R° .

n

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

f (z-Zo)a~ l

dz =i—-sin(a^)Jca a

61

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

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

b p

\f[z(t)]z'(t)dt =f/[Z(T)k'[0(T)]0'(T)</T,

a a

where

Z(T) = Z[0(T)] («<T<j3).

But expression (14), Sec 38, tells us that

zWt)W'(t) = Z'(t);

and so

* P

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

(b) Suppose that C is any contour and that f(z) is piecewise continuous on C. Since C can

be broken up into a finite chain of smooth arcs on which f(z) is continuous, the

identity obtained in part (a) remains valid.

SECTION 41

1. Let C be the arc of the circle \z\= 2 shown below.

o 2 x

Without evaluating the integral, let us find an upper bound forjjc~T~J

note that if z is a point on C,

|z2-l|>|lz

2l-l| = |lzl

2-l| = l4-ll = 3.

To do this, we

Thus

62

Also, the length of C is ^(47r) = n. So, taking M = ^ and L = n , we find that

2. The path C is as shown in the figure below. The midpoint of C is clearly the closest point on

V2C to the origin. The distance of that midpoint from the origin is clearly -y-, the length ofC

being V2.

V2Hence if z is any point on C, Id >— . This means that, for such a point

Consequendy, by taking M - 4 and L = V2 , we have

Izl4

||c^|S ML = 4V2.

3. The contour C is the closed triangular path shown below.

To find an upper bound for\jc

(ez - z)dz , we let z be a point on C and observe that

\ez -

z

I

<

\el

\ + \z\ = ex + jx2 +y2

.

63

But ex < 1 since x < 0, and the distance ^x2 + y

2of the point z from the origin is always

less than or equal to 4. Thus le* - zl < 5 when z is on C. The length of C is evidently 12.

Hence, by writingM = 5 and L = 12, we have

(ez-z)dz < ML = 60.

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

l2z2-ll<2lzl

2+l = 2/?

2 +l

and

lz4+ 5z

2+ 41 = lz

2+ II lz

2 + 41 >|Izl

2-1

1

1 Izl2-4 | = (tf

2 - 1)(/?2 - 4).

Thus

2z2 -l

z4+5z

2 +4I2z

2-ll 2*2 + l

z4 + 5z

2+4l (R

2-l)(R2 -4)

when lzl= R (R > 2). Since the length of CR is ^R, then,

I,

2z2 -l

W+5z2 + 4

^ teR(2/?2 + 1)

~(/?2-l)(/?

2 -4)/?l R2

R2J} R2

)

and it is clear that the value of the integral tends to zero as R tends to infinity.

Here CR is the positively oriented circle Izl=R(R> 1). If z is a point on CR , then

Logz lln/? + /0l< ln/?+l0l

<^r + ln/g

?2 " R2

since -n<Q<7t. The length of Q is, of course, 2nR. Consequently, by taking

M =——— and L = 2kR,

we see that

Since

it follows that

Logzdz

* z< ML = 2n

n + k\R

R

it + \xiR _ 1//?iim = lira —-— = u,

J c* z

Let Cpbe the positively oriented circle \z\=p (0 <p < 1), shown in the figure below, and

suppose that /(z) is analytic in the disk Id < 1.

y

I I' /a 1 \

1 V Jppx

We let z represent any particular branch

-1/2 = exp^-^log zj= exp|^-^ (In r+ id) = exp^-i-j

|(r > 0, a < < a + 2ft)

of the power function here; and we note that, since /(z) is continuous on the closed

bounded disk Izl < 1, there is a nonnegative constantM such that l/(z)l^M for each point z

in that disk. We are asked to find an upper bound for

observe that if z is a point on Cp ,

M

I zmf{z)dz . To do this, we

\z-U2m = \z-

m\\az)\<

Since the length of the path Cp is 2np, we may conclude that

jcz-mf(z)dz <^j=2np = 27tMjp.

Note that, inasmuch as M is independent of p, it follows that

limf z-U2f(z)dz = 0.

65

SECTION 43

1. The function z" (n = 0,1,2,...) has the antiderivative z"+1/(n + 1) everywhere in the finite

plane. Consequently, for any contour C from a point zx to a point z2 ,

.Zl _»+l 1

Jc^ n + 1 . n + 1 /i + l n + 1^2 1

'

"r2

2. fa; I e*z<fe

=

—

i/2

e —e j + 1 l + i

it it it

(b)J

cosl — l<fe = 2sin

>r+2i

= 2 Si„(f + ,)=:2/

= —lie e —e e I

=~<H=7 + « = « + -.e

U-2)41

3

1 1

4i

4 40.

3. Note the function (z -Zq)"'1

(n = ±1,±2,...) always has an antiderivative in any domain that

does not contain the point z = Zq. So, by the theorem in Sec. 42,

L(z-zor l

dz =

for any closed contour C that does not pass through Zq.

5. Let C denote any contour from z = -1 to z = 1 that, except for its end points, lies above the

real axis. This exercise asks us to evaluate the integral

t

I = jz'dz,

-i

where z' denotes the principal branch

z' = exp(/Logz) (lzl>0,-^< Argz< it).

66

An antiderivative of this branch cannot be used since the branch is not even defined at

z = -1. But the integrand can be replaced by the branch

z' = exp(z'logz) lzl>0,-— <argz<—

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

_('+!

1 + 1

- *[(l)

i+l

(j- ^

j~£('>D'ogl

£(M)\og(-D

j

_ _£_("^(i+l)(Inl+iO) _ (i+lXlnl+i*)"! = _J_ h _ g-*^*) - ^ + e1 . 1 L

i + Ji + V ' l + i 1-/

d-o.

SECTION 46

2. The contours Ctand C2 are as shown in the figure below.

In each of the cases below, the singularities of the integrand lie outside Cxor inside C2 ; and

so the integrand is analytic on the contours and between them. Consequendy,

imdz=if(z)dz.

67

(a) When f(z) = - the singularities are the points z =

z + 2(b) When f(z) = . .

—, the singularities are at z = 2nn (n = 0,±1,±2,...).

sm(z 1 2)

(c) When /(z) = -, the singularities are at z = 2nm (n = 0,±1,±2,. . .).

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

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

e~l* along that leg is

je-xldx = 2je-

xldx.

-a

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

z = x + bi (-a £x<a), the integral of e'1* along the upper leg is

-]e'{x+bi)1

dx = -ebl\e

xle

iUxdx =V j<f cos2bxdx + ie"

2

\exlsm2bxdx,

or simplyu

-2/2

jy*2

cos2fct<£t.

Since the right-hand vertical leg is represented by z = a+ iy (0 < y < b), the integral of

e~'* along it is

O B

je-(a+iyf

idy = !«-* jeyl

e-i2ay

dy.

68

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

z = -a + iy (0 < y < b), the integral of e~l* along that vertical leg is

b b

-feH-a+iy)1

idy = -ie-al

jey*e

i2aydy.

According to the Cauchy-Goursat theorem, then,

a a b b

2\e~xldx - 2e

b\je-

xlcos2bxdx + ie** \ey\-'lla

>dy - ie~ajey

*enay

dy = 0;

V_:

f

and this reduces to X4

]e~xlcoslbxdx = e~

b\e~

xldx + e

-(a^bl)]e

yZ

sinlaydy.

(b) We now let a -» °° in the final equation in part (a), keeping in mind the knownintegration formula

\e'xl

dx

and the fact that

e-(al+bl)

\eyl

sinlaydy < e-(a2+bl)

jeyl

dy -> as a -> oo.

The result is

exlcos2bxdx = ^-e-b2

2(b>0).

6. We let C denote the entire boundary of the semicircular region appearing below. It is madeup of the leg C, from the origin to the point z = 1, the semicircular arc C2 that is shown, and

the leg C3from z = -1 to the origin. Thus C = C

t+ C

2+ C

3.

y

69

We also let f(z) be a continuous function that is defined on this closed semicircular regionby writing /(0) = and using the branch

/(z) = V7eie/2

r>o,-f<*<f

of the multiple-valued function zm

. The problem here is to evaluate the integral of f(z)around C by evaluating the integrals along the individual paths C„ C2 ,and C3 and thenadding the results. Li each case, we write a parametric representation for the path (or arelated one) and then use it to evaluate the integral along the particular path.

(i) Cv

: z = rei0 (0<r<l). Then

JCi/(z)^=jVF.lJr =[V^=|.

(ii) C2

: z = 1 • e'B(0 < 6 < it). Then

I f(z)dz = ]ewn

ieiedd = ije^nd0 = if^e

na'2]*= - 1) = -1 (1 + /).

2o o L3i Jo 3 3

(Hi) -Cy z = reix

(0 < r < 1). Then

\cmdz = -\_c

f{z)dz = -)^e i

*'\-\)dr = i\^dr = i\-rvll

\

=-/.5 *" J

o o L3 Jo 3

The desired result is

jcf(z)dz = j f(z)dz + j f(z)dz+i f(z)dz =l-h+i) +h = 0.

1 Cj "-J 3 3 3

The Cauchy-Goursat theorem does not apply since f(z) is not analytic at the origin, or even

defined on the negative imaginary axis.

SECTION 48

1. In this problem, we let C denote the square contour shown in the figure below.

y

2i

ci

-2,r 2 Jt

-2/

(a) f

6 =2m \e~z\ .

, = 2m(-i) = 2n.

(b)f^- dz = f^ 7 + 8>

<fe =iJ^LJc z(z

2 + 8) Jc z-0 Lz2 + 8. U; 4

Jc 2z + 1Jc z -(_i/2) L2J z=_1/2 I 4j 2

. ,. r coshz , r coshz , 2tb* f d3

,

<d> Jc—* Jcb^sr*"IT 60,1,1

Jz=0

=f(0)

= 0.

, . r tan(z/2) , r tan(z/2) . 2«f

Z=*o

= 2«i"| 4-sec2— 1 = »rsec

2f— I when -2<xn <2.U 2

Let C denote the positively oriented circle \z - il= 2, shown below.

(aj The Cauchy integral formula enables us to write

dz c dz r l/(z + 2i)

JCr2 +4 J<z2 +4 Jc(z _2/)(z + 20 Jc z-2/

dz = 2mz + 2ij z=2i

= 2m\

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

f& _ f ffc r 1/ U

Jc(z

2 + 4)2 ~ Jc

(z _ 2/)2

(z + 2i)2 ~ Jc (* -

dz _ f l/(z + 2Q2

= 2£(z-2i)

1+1Z

1!

1

<fe (z + 2i)

71

3. Let C be the positively oriented circle Iz I= 3 , and consider the function

g(w)=\ dz (lwl*3).

We wish to find g(w) when w = 2 and when \w\ > 3 (see the figure below).

y

Cm

)( i

V w = 2/3 x

We observe that

#(2) = Jc z_2~ <fe = 2 ^'[2z2 " z ~ 2

] z=2= 2^1(4) = 8m.

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

5. Suppose that a function / is analytic inside and on a simple closed contour C and that z is

not on C. If z is inside C, then

f£&*»2«rk) and f^>* f/ft)* =^

Thus

f f\z)dz = f /(g)<fe

J^ Z-Zo J^(Z-Z )

2'

The Cauchy-Goursat theorem tells us that this last equation is also valid when zq is exteriorto C, each side of the equation being 0.

7. Let C be the unit circle z = eie (-n£d< it), and let a denote any real constant. The

Cauchy integral formula reveals that

f —dz=[ -^-dz = 2m\e at

]= 2m.

Jc z Jc z — J *=0

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

\c—dz = ]^~Qiewde = / jexp[a(cos0 + isin 6)]d6Z -K e

-It

It K

= iJeacosVflSin ed0 = ije

acose[cos(asm 6) + isin(asin 6)]dd

-ft -It

K It

= - jeacos0sin(asin 8)dd + ije

acos9cos(asin 6)dd.

-It -1t

r eaz

Equating these two different expressions for the integral I— dz, we have

c zIt It

- jeacosesin(asin d)dd + i je

acosBcos(asin d)dd = 2m.

-It -It

Then, by equating the imaginary parts on each side of this last equation, we see that

it

jett<:ose

cos(asmd)dd = 27f,

-«

and, since the integrand here is even,

jeacose

cos(asin 6)dd = Jt.

(a) The binomial formula enables us to write

We note that the highest power of z appearing under the derivative is z2", and

differentiating it n times brings it down to z". So Pn (z) is a polynomial of degree n.

(b) We let C denote any positively oriented simple closed contour surrounding a fized point

z. The Cauchy integral formula for derivatives tells us that

rtf-tf^fSfzlLj, („ = 0,1,2,...).

dz"K

' Im^is-z)^1

Hence the polynomials Pn (z) in part (a) can be written

73

(c) Note that

C?2-l)" _ (j-1)"(j + 1)" +

(s-iyn+l

(s-iy S-l

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

Also, since

(s2 - 1)"

= (s - + 1)" _ (s - 1)"

(j + ir+1

(* + l)"+1

5 + 1'

we have

2" 2to* jc s + i

9. We are asked to show that

1 rf(s)ds

m J (t —ni Jc (s-z)

3*

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

/'(z + Az)-/'(z) _ 1 r

Az 2ml1 1

(.y-z-Az)2

(s-z)2

f(s)ds

Az

1f 2(s-z)-Az

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

Then

/'(z + Az)-/'(z) 1f/(j)<fe _ 1

Az

f JV)as i r

7ri|(5-z)3 2^4

2(j-z)-Az

(s-z-Az)2(s-z)

2(s-z)

3

1f 3(j-z)Az-2(Az)

2

2iTi J f.c —2^/J(,-z-Az)2(5-z)

3"/(^'

74

(b) We must show that

r3(.s-z)Az-2(Az)

2

c(s-z-Azf(s-z?jf(s)ds < (3DIAzl+2IAzl

2 )M(rf-IAzl)V

Now D,d,M, and L are as in the statement of the exercise in the text. The triangle

inequality tells us that

I3(j - z)Az - 2(Az)2l< 31* - zl I Azl + 21 Azl

2 < 3D! Azl + 21 Azl2

.

Also, we know from the verification of the expression for /' (z) in the lemma that

Is - z - Azl > d -I Azl> 0; and this means that

I (s - z - Az)2(s - z)

3l > (d - 1 Azl )

2d3 > 0.

This gives the desired inequality.

(c) Ifwe let Az tend to in the inequality obtained in part (b) we find that

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

Chapter 575

SECTION 52

1. We are asked to show in two ways that the sequence

(-1Yz„=-2 +A-f (n = l,2,...)

n

converges to -2. One way is to note that the two sequences

xn =-2 and y„=t-9-(n = l,2,...)

of real numbers converge to -2 and 0, respectively, and then to apply the theorem in Sec.

51. Another way is to observe that \z„- (-2)1 = Thus for each e>0,

z„ ~ (- 2)| < e whenever n>n ,

1where n is any positive integer such that n >

2. Observe that if z„ = - 2 +i^- (n = 1, 2,. .. ), then

n

/•„=lzJ=i/4 +— -»2.

But, since

2fl=Argz

2„-»^ and2„_,

= Argz^, (n = 1,2,...),

the sequence 0„ (n = 1,2,...) does not converge.

3. Suppose that limzB = z. That is, for each e >0, there is a positive integer such that

\zn - z\< £ whenever n > n . In view of the inequality (see Sec. 4)

lzn -zl>llzj-idl,

it follows that llzj-lzlke whenever n>nQ . That is, limlzj=lzl.

76

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

when lzl<l.

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

jr(re'*)" = jr rV"e = j^V cosnfl + rnsinn0;

u=l n=l n=l n=l

and the right-hand side takes the form

rew

1 - re~w_ re

w -r2 _ rcos - r2 + irsin

1 - reie

'

1 - re~w ~

1 - r(eie + e~

ie

) + r2 ~

1 - 2rcos + r2

Thus

^ . - . . _ rcos0-r2. rsin0

> r cosnd + iy r smnd = —T + i—T .

Zl t& l-2rcos0 + r2 l-2rcos0 + r

2

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

summation formulas

E„ Q rcosd-r2

j V « • a rsin0r cosn0 = —

-r and > r smnd = -— —T ,

B=1 l-2rcos0 + r2 ^ l-2rcos0 + r

2

where < r < 1. These formulas clearly hold when r = too.

6. Suppose that ^zB=5. To show that ^z„ =S, we write zn =xn +iy„, S = X + iY and

n=l n=l

appeal to the theorem in Sec. 52. First of all, we note that

5>„=X and 5>B=7.

OO

Then, since ^(—y„) = —Y, it follows that

jU = 2ta -% )= 2>„ + «*(->.)]

=

x - iY = j.

n=l

77

8. Suppose that £z„ = S and ^w„ = T. In order to use the theorem in Sec. 52, we writeb=1

z» =*,+%> S = X + iY and w„=un +ivn , T = U + iV.

Now

Since

2>b = * f,yH=Y and 5>„ = t/, |>B

= V.B=l B=l B=l

2(^+"„) = X + t/ and X^+vJ^+y,

it follows that

That is,

or

J,[(xn +u„) + i(yn +v„)] = X + U + i(Y+ V).

+%) + ("„ + J'vJ] = X + iY+(t/ + /V),

B=l

B=l

SECTION 54

1. Replace z by z2in the known series

coshz =X~ z

ln

to get

cosh(z2) =2

«> _4bZ

Then, multiplying through this last equation by z, we have the desired result:

zcosh(z2) = ]£

**z4n+1

tS(2n)!

(Izl< ~)

(lzl< ~).

(Izl<~).

78

2. (b) Replacing z by z - 1 in the known expansion

: zn

e = 5>7 <W<«).n=0 " ;

we have

n=0

So

~Z „, (lzk°o).n!

^rt =e£(£zi)l (ld<eo)e

r.=0

3. We want to find the Maclaurin series for the function

z z 1

z4 +9 9 l + (z

4 /9)'

To do this, we first replace z by -(z4/ 9) in the known expansion

r-^ly Ozl<l),

as well as its condition of validity, to get

Then, if we multiply through this last equation by |-, we have the desired expansion:

n

6. Replacing z by z2in the representation

Zo (2n + D!

we have

2n+l

sinz =z

(lzk~),

j

4b+2

sin(z2) = £(-1)" * "„ Qz\< oo).

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

(0)/n!, this

shows that

f4n)(0) = and /

<2B+1)(0) = (n = 0,1,2,...).

7. The function —^— has a singularity at z = 1. So the Taylor series about z = i is valid whenl-z

\z - i\ < V2, as indicated in the figure below.

y

1 i i

\V2 J

To find the series, we start by writing

1 1

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

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

1 * n=0

and then multiply through by :• The desired Taylor series is then obtained:

(ld<l)

i-z sa-o*(z-iT.

(lz-/I<V2).

9 The identity sinh(z + ni) = -sinhz and the periodicity of sinhz, with period 2m, tell us that

sinhz = -sinh(z + ni) = -sinh(z - ni).

So, ifwe replace z by z - ni in the known representation

80

and then multiply through by -1, we find that

sinhz = -V i£_JEL!—( | Z _ m\< ~).

S (2» + D!

13. Suppose that < Izl < 4. Then < Iz / 41 < 1, and we can use the known expansion

7^- =2y (IzKl).

To be specific, when < Izl < 4,

1 111 ^fz-lym =y£_ = J_ +y^_ = _L + y.4z-z

24z !_£ 4z~W ^4fl+l

4z ~4"+14Z B

4"+2'

4

SECTION 56

1. We may use the expansion

2n+l

sin-S (-ir™ (UI< ~>

to see that when 0< Izl < °°,

%.j n y (-!>• 1, , f (-ir 1

3. Suppose that 1 <lzl< 00 and recall the Maclaurin series representation

T^- =Z^ (|Z|<1)-1-z ±!'2 „=o

This enables us to write

1

-1 'r-iiRT-i^

1 + Z ^ 1h— Z n=0 v „=o z

Z

Replacing n by n - 1 in this last series and then noting that

81we arrive at the desired expansion:

1 v

4. The singularities of the function /(z) = are at the points z = and z = 1. Hence

there are Laurent series in powers of z for the domains <lzl< 1 and 1 <lzl< <» (see thefigure below).

To find the series when < \z\< 1, recall that =Y z" (lzl< 1) and write

2 A Z Z „=0 n=0 Z Z „=2 „_ Z Z

As for the domain 1 < lzl< <», note that II / zl < 1 and write

/(z)="7'I^17I)

="7|(z")

=~%^~%7-

z + l5. (a) The Maclaurin series for the function is valid when \z\< 1. To find it, we recall

z-1the Maclaurin series representation

TIT = 5>" (ld<1)

for and write1-z

z + l 1~

fzi-<«+ »izr <-< -»X --X^1

- X*'•° A 12 »=0 n=0 «=0

"=1 «=0 11=1

(b) To find the Laurent series for the same function when l<lzl<<*>, we recall the

Maclaurin series for —— that was used in part (a). Since1 - z

< 1 here, we may write

z z

1+-ZJn=i)\ZJ n=oZ n=0 Z

(l<lzl<oo).

The function /(z) =1

has isolated singularities at z = and z = ±z, as indicated inz(l + z

2)

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

and also one for the domain 1 <lzl< °°, which is exterior to the circle lzl= 1.

To find each of these Laurent series, we recall the Maclaurin series representation

1 2 n=0

(lzl< 1).

For the domain <Id< 1, we have

/(*> = ~'7~~t =1tHT =tc-DV- 1

= -+ £(-1)V" 1 = £(-l)n+1z2n+l

Z 1 + Z Z B=0 B=0 Z Z

On the other hand, when 1 <lzl< °°,

11 1 ^ r__l_V _^ (-If _v (-1)"+1

z2

.2n+l'

n=0 £ n=l£~

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

(-l)2 = (-1)"

+1.

83

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

»=0

enables us to write

a a 1

j— = |>" (IzKl)

a _ a i _v fl

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

or

a ^ a"= 2j— (lal<ld<°o).

(b) Putting z = e'° on each side of the final result in part (a), we have

a n -inBe

But

a (cos0-a)-isin0 acos0-a2-iasin0

eie -a (cos0-a) + i'sin0 (cos0-a)-i'sin0 l-2acos0 + a

2

and

XflV"9 cosn6-i^ansmne.

n=l b=1 n=l

Consequendy,

2,a"cosnd =-— r- and Ya" smn9 = rl-2acosd + a2 ~ l-2acosd + a

2

when -1 < a < 1.

10. (a) Let z be any fixed complex number and C the unit circle w = e'* {-it <<t>^ n) in the w

plane. The function

/(w) = exp2 v w>

has the one singularity w = in the w plane. That singularity is, of course, interior to

C, as shown in the figure below.

w plane

Now the function f(w) has a Laurent series representation in the domain <Im>I< <

According to expression (5), Sec. 55, then,

[IIexp

where the coefficients J„(z) are

(0<lwl<<»),

exp w ——]

2m Jc Wn+l dw (n = 0,±l,±2,...).

Using the parametric representation w = e'* (-n <$<n) for C, let us rewrite

this expression for Jn (z) as follows:

1 r

exp| 2 v~

J" i-z) = T7; J JC»2m ' ev

- K

1*

- iV'd0 =|exp[izsin <l>]e~

m*d<t>

.

2m _

That is,

1K

J„ (z) =^ Jexp[-i - z sin 0)] <fy (n = 0,±l,±2,...).

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

1*

^i(^)= T~ |[cos(n0-zsin0)-ism(n0-zsin0)]d0

2;r J

\cos(n<t>-zsin<j))d<l> f sin(n</>-z sin </>)d<j>

2tc j„ 2?r j

1 *i

=— 2 1cos(«0 - zsin <j>)d<j>

2% i 2k(h = 0,±1,±2,...).

85

That is,

1*

J„(z) =—Jcos(«0 - zsin

iti(n = 0,±l,±2,...).

11. fa) The function f(z) is analytic in some annular domain centered at the origin; and the

unit circle C.z- e'* (-it < <j> < it) is contained in that domain, as shown below.

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

n=0 n=l <

where

— ?T -*?T

(n = 0,1,2,...)

and

— 7F—

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

or

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

86

(b) Put z = e'e

in the final result in part (a) to get

or

/(«") = ^- J/(e* )d* +1£ J/(^)cos[n(0 -

If k(0) = Ref(e'e), then, equating the real parts on each side of this last equation yields

w(0) =— J«(0)<ty+—2 J"(0)cos[n(0-0)]<fy.

SECTION 60

1. Differentiating each side of the representation

1-z n=0

we find that

U z; az „=0 n=0 az „=1 n=0

Another differentiation gives

(i — z; "Z n=0 „=o «z B=1 n=0

2. Replace z by 1 / (1 - z) on each side of the Maclaurin series representation (Exercise 1)

as well as in its condition of validity. This yields the Laurent series representation

4 = £<=!>>ZI> «<lz-ll<~).

87

3. Since the function /(z) = 1 / z has a singular point at z = 0, its Taylor series about z = 2 is

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

To find that series, write 1111z 2 + (z-2) 2 l + (z-2)/2

to see that it can be obtained by replacing z by -(z - 2) / 2 in the known expansion

1

Specifically,

or

l_ly|- (Z-2)Tz ifA. 2 J

Z R-0 z

(lzl< 1).

(Iz-2I<2),

(lz-2l<2).

Differentiating this series term by term, we have

J =l^n(z -2r l =E^tH" + l)(z - 2)" (lz - 2I< 2).71 J-~l y

n=0

Thus

-| (_1)>+1)^y (lz-2l<2).

4. Consider the function defined by the equations

V-lf(z) =

1

when z * 0,

when z = 0.

When z * 0, /(z) has the power series representation

/(*) = -z

' 2 3 \

l +£ + l. + £.+.^1! 2! 3!

-1 , z z= 1+—+—+••

2! 3!

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

Let C be a contour lying in the open disk \w - II< 1 in the w plane that extends from the

point w = 1 to a point w = z, as shown in the figure below.

w plane

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

1

w „=0

term by term along the contour C. Thus

dw

(lw-u<i)

" n=0 n=0

But

and

f— = [— = [Logw]

1

z = Logz-Logl = Logz

4

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

Hence

^n + l ^ n

and, since (-1)"-1 = (-ly-^-l)

2 = (-1)"+1

, this result becomes

Logz =XLii— («-!)"

n + l

(-D

n+l

Jl

(z-1)

n + l

n+l

(Iz-lkl);

(Iz-lkl).n=l

SECTION 61

1. The singularities of the function /(*) =-^L- are at z = , ± L The problem here is to

find the Laurent series for/that is valid in the punctured disk <ld< 1, shown below.

89

yi

\°

4

-i

/ x

We begin by recalling the Maclaurin series representations

and

which enable us to write

and

1! 2! 3!

1

l-z= l + z + z

2 + z3+---

ez

= l + z + ^-z2+ -z

3+-2 6

1

Qz\< oo)

Oz\<l),

(\z\< oo)

(\z\<l).

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

2 ,

- = l + z + ±z2+*z 3+-

zl + \ 2 6

-z2

-z3-

1.2 5

z4+-

which is valid when \z\< 1. The desired Laurent series is then obtained by multiplying each

side of the above representation by -:

1 , 1 5 2= -+l--z

—

z +••z(z' + l) z 2 6

(0<kkl).

90

4. We know the Laurent series representation

1 1117z2sinhz z

36 z 360

Z (0<lzkw)

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

tells us that the coefficient of - in this series can be written

z

dz

2m lc z sinhz

where C is the circle lzl= 1, taken counterclockwise. Since bx= - \, then,

6

'c z sinhz

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

(n[/(^(z)]

(n) =E Afk\z)g

"-k)(z) (n = l,2,...).

The formula is clearly true when n = 1 since in that case it becomes

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

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

UWz)](m+1)={[f(z)g(z)]'}

(m)=[f(z)g

,(z) + nz)g(z)]

(m)

=mz)g'(z)fm)+[f(z)g(z)]

l(m)

=1 r /(t)(^(m-t+i)

(^)+s*=0 V* /

/(t)

(z)^(m-t+1)

(z)

= /(z)*(m+1)

(z) +£*=1

'm^ ( m N

/W(z)^('»

+1-^(z) + /

('n+1)(z)^(z).

But

frrC

k ,

m ml|

ml (ro + 1)!

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

'm + V

<k

>

and so

minz)g(z)r

+i)=nz)g°

n+i\z)+

y

z

or

[/(^(z)](m+i)

=iTm

,

+rf\z)t«-k

\z).

The desired verification is now complete.

We are given that /(z) is an entire function represented by a series of the form

f(z) = z + a2z2+ajz

3+-'- (lzl<°°).

(a) Write g(z) = f[f(z)] and observe that

It is straightforward to show that

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

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

and

g"\z) = /'"[/(z)][/'(z)]3

+ 2f(z)f"(z)f"[f(z)] +ft/(z)]/'(z)/"(z)+ /'[/(*)]/"'(*) •

Thus

S(0) = 0, ^'(0) = 1, g"(0)=4a2 , and g'"(0) = 12(a22 + a

3 ),

and so

/[/(z)] = z + 20.Z2+ 2(4 + a,)z

3+- • (lzl< °°).

92

(b) Proceeding formally, we have

/[/(*)] = f(z) + fljC/Cz)]2+ «,[/(z)]

3+- •

•

= (z + fljZ2 + OjZ

3+•••) + (<*2Z

2 + 2^z 3+- • • ) + (fl

3z3+• • •

)

= z +2^ + 2(a* + a3>z3+- • •

.

(c) Since

sinz = z -—+• • •= z + Oz2 + f- -V+- •

•

3! V 6)

the result in part (a), with Oj = and a3= -— , tells us that

6

sin(sinz) = z-— z3 +---

(IzKoo),

(lzl< ~).

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

— =E^z" (\z\<Zcoshz t^n\ \ 2

This representation is valid in the stated disk since the zeros of coshz are the numbers

z = \— + n7t\i (/i = 0,±l,±2,...)» the ones nearest to the origin being z = ±^-i- The series

U J 2

contains only even powers of z since coshz is an even function; that is, E2tt+l

=0

(n = 0, 1, 2, . ..). To find the series, we divide the series

coshz = 1+—+— + —+•••= l +-z2 +—z4+—z6+"- (lzl< °°)

2! 4! 6! 2 24 720

into 1. The result is

,1 2 5 4 61 6 (i i^ n>

\= 1

—

z +— z z +••Ilzl<— I,

coshz 2 24 720 V 2

or

93

1 , 1 2 5 4 61 6 f. . n= 1 z +—z z +•••

I lzl<—coshz 2! 4! 6! V 2

Since

_!_ = £ +^ + Jlz4 + JSlz«+... f|

z|<f|,coshz 2! 4! 6! V 2

this tells us that

£0=1, E2= -1, £4

= 5, and E6=-61.

94

Chapter 6

SECTION 64

1. (a) Let us write

_i- = I.-i- = I(i- z + z2-z

3 + ...) = i-l + z -z2 +... (0<lzkl).

z + z z 1 + z z z

The residue at z = 0, which is the coefficient of -, is clearly 1.

z

(b) We may use the expansion

cosz = l-2! 4! 6!

to write

m 1 1 1 1 1 1 "\ 111111ZCosH = z^l--.? + -.7

--.7 +

...J

= z--.- +-.-3---7 -f.

(0<lzl<°°).

The residue at z = 0, or coefficient of -, is now seen to be -—

.

z 2

(c) Observe that

z-sinz 1. . . 1= -(z-smz) = -

z z z r 3! 5!*'j

T2 T4

3! 5!

(0<lzl<<*>).

Since the coefficient of - in this Laurent series is 0, the residue at z = is 0.

z

(d) Write

and recall that

cotz _ 1 cosz4 ~ 4 '

.

z z sinz

T2 T4 T2 -T4

1 z,z , z z

cosz = 1 + = 1 +2! 4! 2 24

(lzl< ~)

and

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

95

cosz 1 z z3

Thus

cotz _ I fl z 11 11z4 "z4 U 3 T5

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

+- (0 <&<*).

Note that the condition of validity for this series is due to the fact that sinz = when

z = nit(n = 0,±l,±2,...). It is now evident that has residue —-. at z = 0.

(e) Recall that

z4

45

sinhz = z +|y

+^ + ...(| z |

<oo)

and

1-Z

There is a Laurent series for the function

1i 2= l + z + z

2 + -.- (lzl<~).

sinhz 1 . ( 1 ^

that is valid for <lzl< 1. To find it, we first multiply the Maclaurin series for sinhz

and1-z

2

,

1 3 1 5z + -z:

+ z +•6 120

3 1 5z +-z +•

z5 +-

7 3z + tz +••• (0<lzl<l).

o

96

We then see that

sinhz 17 1=— + -•-+ ••• (0<lzl<l).

z4(l-z

2

) z3

6 z

This shows that the residue of 47—^ at z = is —

.

z (1 — zJ

6

2. In each part, C denotes the positively oriented circle lzl= 3.

(a) To evaluateJc—^

—

dz, we need the residue of the integrand at z = 0. From

the Laurent series

exp(-z) _ 1

2—

_2

^ z,z2

z3

, ) 1 1 1 1 z ,A . .,1— + +...=_ + +... (0<lzl<°°),

^ 1! 2! 3! J z2

1! z 2! 3!

we see that the required residue is -1. Thus

lc™P^

dz = 2n i-l)= -2jti .

(c) Likewise, to evaluate the integraljc

z2 exp^jdz, we must find the residue of the

integrand at z = 0. The Laurent series

, • , 11 1 1 11 11

2 Z 1 1 1 1 1= Z +— +— + +

1! 2! 3! z 4! z2

which is valid for <lzl< °°, tells us that the needed residue is — . Hence6

97

r z + 1

(d) As for the integral -=

—

—dz, we need the two residues ofJC 7* — 9rz l

-2z

z+1 z + 1

z2-2z z(z-2)'

one at z = and one at z = 2. The residue at z = can be found by writing

z + 1

z(z-2) z-2) { 2A z) l-(z/2)

\ 2 2 z)

f z z1 %

2 22

which is valid when 0<lzl<2, and observing that the coefficient of - in this last

z

product is — . To obtain the residue at z = 2, we write

z + 1 (z-2) + 3 1

z(z-2) z-2 2 + (z-2) 2V z-2j l + (z- 2)/2

2\ z-21

z-2|

(z-2)2

which is valid when <lz - 21< 2, and note that the coefficient of in this productz-2

3is — . Finally, then, by the residue theorem,

Jc z2-2z V 2 2)

In each part of this problem, C is the positively oriented circle lzl= 2.

(a) If/(z)l-z :

-, then

/ - =-—4=-—-;

—

r = -- l+z3+z

6+••• =--—

z

2-\zj z -z z 1-z z

v yz z

1_2

when <lzl< 1. This tells us that

f(z)dz = 2mRes\f(-) = 2ni(-l) = -2m.<=° z \z)

(b) When f(z) =1

2 > we have1 + z

Thus

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

(c) If /(z) = -, it follows that ~/[-) = -• Evidently, then,Z 2 VZ/ Z

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

z\ Z J

Let C denote the circle lzl= 1, taken counterclockwise.

(a) The Maclaurin series ez =T— (lzl< °°) enables us to write

(b) Referring to the Maclaurin series for ezonce again, let us write

Now the - in this series occurs when n-k = -l, or k = n + l. So, by the residuez

theorem,

l> expf-ljz =2m-i— (/j = 0,1,2,...).Jc vz/ (« + l)!

The final result in part (a) thus reduces to

995. We are given two polynomials

P(z) = a +alz + a2z

2+--- + a„z'' (a„*0)

and

<2(z) = b + blZ + b2z2 + --- + bmz

m(bm * 0),

where m>.n + 2.

It is straightforward to show that

J_ pQ 1 z> - flQz'""2 + a

iz"" 3 + a2*

m~4+ • • • + fl^

w"""2

z2 "

2(1 / z) V" + fri*"*"1

+V"2+ • • • +K

1

Observe that the numerator here is, in fact, a polynomial since m -n - 2 > 0. Also, since

bm * 0, the quotient of these polynomials is represented by a series of the form

do+diZ + dzZ2 + ••-. That is,

1 P(l/z) j , j 2(0<lzl</^);

L 1 P(\lz) . ^ rtand we see that —r has residue z = 0.

z 2(1 /z)

Suppose now that all of the zeros of Q(z) lie inside a simple closed contour C, and

assume that C is positively oriented. Since P(z)/Q(z) is analytic everywhere in the finite

plane except at the zeros of <2(z), it follows from the theorem in Sec. 64 and the residue just

obtained that

P(z)dz = 2mRes

z=0

J_ P(l/z)

z2'ea/z)

= 2^*0 = 0.,c Q(z)

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

Piz) J__ f P(z)

icQ{z) J-c Q(z)

dz = 0.

SECTION 65

1. (a) From the expansion

we see that

1! 2! 3!

,n ^,^11 iizexp| _j =z+1+_._ +_._ + ..

(Izl<~),

(0<lzl<oo).

100

The principal part of zexpl - I at the isolated singular point z = is, then,z

2!'z+3!Y

+ "';

and z = is an essential singular point of that function.

(b) The isolated singular point of is at z = -1. Since the principal part at z = -11 + z

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

z2 = (2 + 1)

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

This enables us to write

e _ (z+lf -% + !) + ! 1

1+z z+1 z+1

Since the principal part is —^— , the point z = -1 is a (simple) pole.,

z + 1

(cj The point z = is the isolated singular point of £m-£

) and we can write

z

sinz 1 ^-— +— -•••1 = 1-— +— -••• (0<lzl<~).z z\, 3! 5! J 3! 5!

The principal part here is evidently 0, and so z = is a removable singular point of the

-t

. smzfunction .

cos z(d) The isolated singular point of is z = 0. Since

z

cosz 1 ( z2

z4

I 2! 4! ;

=I_±+ £l_... (0<lzl<oo),

z 2! 4!

I COSZthe principal part is — . This means that z = is a (simple) pole of .

z z

(e) Upon writing —^—-j = we find that the principal part of3

at its

(2 — z) (z-2) (2-z)

isolated singular point z = 2 is simply the function itself. That point is evidently a pole

(of order 3).

(a) The singular point is z = 0. Since

101

1-coshz 11-

( _2 ,4 6

!+£.+£.+£_+..V 2! 4! 6! J 2\ z 4! 6!

when <lzl< °°, we have m = l and B =—— = -—

.

2! 2

(ft) Here the singular point is also z = 0. Since

l-exp(2z)_ 1 ( 2z 2V 2 373 2V 2V "\

I 1! 2! 3! 4! 5!

= _2_ J__2^ _1 2j_ _j__2^_2^ll'z

3

2!*z2

3!'z 4! IT*"

23

4when <lzl< °°, we have m = 3 and B = =—

.

3! 3

(c) The singular point ofexP^z

^ js z = \ jhe Taylor series(z-1)

exp(2z) = e2(z-V=e 1 +

2(z-l)l22(z-l)

2

.2

3(z-l)

3

1! 2! 3!

(Izl< oo)

enables us to write the Laurent series

exp(2z) _ ^(z-1)

2

2+— •

1 22

22

+— + -J-(z-l)+-L(z-lr 1! z-1 2! 3!

(0<lz-ll<°°).

2Thus m = 2 and = e

2— = 2e2

.

Since / is analytic at Zq, it has a Taylor series representation

f(z) = f(z )+^-(z-z ) +^^-(z-z )

2 +- (Iz-ZoK^,).

Let g be defined by means of the equation

102

(a) Suppose that /(z ) *0. Then

g(z) =1

/<0 +^(*-*) +^U-*o)2 +-

- - 1! 2!° ;

Z-Zn(0<k-z l</?o).

This shows that £ has a simple pole at Zq , with residue /(z ).

(b) Suppose, on the other hand, that /(z ) = 0. Then

1! 2!

(0<\z-z \<Ro).

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

point of g.

4. Write the function3_2

/(Z) =(?W (a>0)

as

_ 4Kz)3_2

8a3z

/(z) =/ -n3

where 0(z) = v .n3-

(z + «T

Since the only singularity of 0(z) is at z = -ai, 0(z) has a Taylor series representation

Hz) = ««0 + *M(Z - ai) +^(z- ai? + (lz-ai'l<2a)

about z = ai. Thus

/(z) =1

(z-m) 30(ai) + ^rr^Cz - fl + (z - ai) +

1! 2!(0<lz-ail<2a).

Now straightforward differentiation reveals that

16a4/z-8fl

3z2

. , 16a3(z

2-4fl/z-a

2)

^ (z) =—; , „ 4

— and <j> (z) = -(z + ai)

4(z + ai)

3

Consequently,

Q(ai) = -a2i, 0'(a/) =™, and </>"(ai)--i.

This enables us to write

/(Z) =(z-aif [~

a2i ~f(Z ~ fll) ~i

(z " m)2 + "']

(0 < lz - ml < 2a).

The principal part of/at the point z = ai is, then,

ill all a2i

z - ai (z - ai)2

(z - ai)3 '

SECTION 67

1. (a) The function f(z) -Z + 2

has an isolated singular point at z = 1. Writing /(z) =z~l z-1

where (/>(z) = z2+2, and observing that 0(z) is analytic and nonzero at z = 1, we see

that z = 1 is a pole of order m = 1 and that the residue there is 5 = 0(1) = 3.

(b) If we write

f(z) =lz + 1

Hz)

z- -zrf, where 0(z) =

1

8'

we see that z = -- is a singular point of /. Since 0(z) is analytic and nonzero at that

point, /has a pole of order m = 3 there. The residue is

B _ r(-V2) _ 3

2! 16'

fc) The function

expz expz

z2 + ;r

2(z-/n)(z + 7ri)

has poles of order m = 1 at the two points z = ±m. The residue at z = to is

Bi

exp TO _ -1 _ i

2 to 2 to~ In

'

and the one at z = -ni is

1041/4

2. (a) Write the function /(z) = (I zl> 0, < arez < 2 it) asz + 1

Hz)/(z) = z^7. where <p{z) = z

m =e 4 (lzl>0,0<argz<2;r).z + 1

The function 0(z) is analytic throughout its domain of definition, indicated in the

figure below.

Also,

-1 o

• Branch cut

jl / n , ni/4 _ J'osH) Ja«i+<*) iwM it . . it 1+i<9(-l) = (-1) =e 4 = £ 4 = e = cos— + zsin— = —f=-*0.

4 4 V2

This shows that the function /has a pole of order m = 1 at z = -1, the residue there

being

5 = 0(-l) =1 + /

V2'

(b) Write the function /(z) =Logz

(z2+ D

2as

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

differentiation then reveals that

Logz tf + 2iRes & = =

(z2 + l)

28

(c) Write the function

.1/2

/(*) =(z

2+ D

2

105

(ld>0,0<argz<2;r)

as

Since

and

f(z) =.1/2

(z-i)2

where =(z + 2

f W2(z + /)

3

r1/2 - *-''*/4 - _L _i_ ,-i/2 _ j*/4 _ 1.

'

71/2

1 — '

Res f „„ = ft'(t>1

*=<~(z2+ l)

2

(a) We wish to evaluate the integral

8V2"

3z3 + 2

,c (z-l)(z2+9)

dz,

where C is the circle \z - 21 = 2 , taken in the counterclockwise direction. That circle and

the singularities z = 1, ± 3i of the integrand are shown in the figure just below.

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

Res-3z

3 + 2 3z3 + 2

(z-l)(z2+9) z

2 + 9

According to the residue theorem, then,

3z3 + 2

1

2

liz-\Z^9)

dz = 1M{l)

=ni -

106

(b) Let us redo part (a)when C is changed to be the positively oriented circle Izl = 4, shownin the figure below.

In this case, all three singularities z = 1, ± 3/ of the integrand are interior to C. Wealready know from part (a) that

3z3 + 2 1

Res-*=1 (z-l)(z

z+9) 2

It is, moreover, straightforward to show that

3z3 + 2 3z

3 + 2Res -

<= 3i (z-l)(z' + 9) (z-l)(z + 3/).

and

15 + 49/

12

Res-3z

3 + 2 3z3 + 2

z=-3,-(z_ +9) (z _ i)(z _ 3/)

The residue theorem now tells us that

j=-3i

15-49/

12

f 3z3+ 2 , _ /l 15 + 49/ 15-49/"\ ,5 dz = 2m\-+ + =6

}c(z -l)(z2+9)

"V2 12

m.

4. (a) Let C denote the positively oriented circle Izl = 2, and note that the integrand of the

r dzintegral -r- — has singularities at z = and z = - 4. (See the figure below.)

Jc z (z + 4)

y

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

107

and write

(lzl<l)

1 _ 1 1

z\z + 4) 4z3

_l + (z/4)

_n-3

Now the coefficient of - here occurs when n = 2, and we see thatz

1 1

Consequendy,

Res ,*=° z\z + 4) 64

Jc z\z + 4) \64) 32

(0<lzl<4).

(b) Let us replace the path C in part (a) by the positively oriented circle \z + 21 = 3, centered

at -2 and with radius 3. It is shown below.

We already know from part (a) that

ReS^ —

.

*=° z\z + 4) 64

To find the residue at -4, we write

1 _ <t>(z) . ... 1

Tw-T+Ty where Hz)= 7-

This tells us that z = -4 is a simple pole of the integrand and that the residue there is

0(-4) = -1 / 64. Consequendy,

J<^ 5(z+4) {64 64J

108

f cosh 7tz dzLet us evaluate the integral :—7T"» wnere C is the positively oriented circle lzl= 2.

JC TIT * 1 1^ Z^Z + ij

All three isolated singularities z = 0,±i of the integrand are interior to C. The desired

residues are

_ cosh nz cosh nzRes—s =—5

z=o z(r+l) z +1 .

=1,

_ cosh 7cz cosh tfzRes—= =

z(z2+ l) z(z + i)_

1

.~2'

and

_ cosh nz cosh tezRes— =*=-<"z(z

2+l) z(z-i)

_ 1

.~2'z=-i

Consequendy,

r coshflzdz „ /, 1 1, . .

= = 2m 1+—+— \= 4m.

,c z(z2+l) V 2 2,

6. In each part of this problem, C denotes the positively oriented circle lzl= 3.

(a) It is straightforward to show that

f„\ Oz + 2)2

1 /n_ (3 + 2z)2

lf/(z) =z(z-l)(2z + 5)'

111611

ZllJ-zd-^ + Sz)-

This function -y/^-j has a simple pole at z = 0, and

I(3z + 2)2— dz = 2/ri Res

cz(z-l)(2z + 5) *=° z/J= 2/ri

<2,= 9ni.

(b) Likewise,

if/(z)=z\l-3z)

±en*-3

'w(l+z)(l + 2z

4)

-*-M-lz2 U; z(z + l)(z

4+2)

The function \ )has a simple pole at z = 0, and we find here that

f

Z(1 " 3z)

4<fe = 2;riRes

Jc(i + z)(l + 2z4)

*=°V 1

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

(c) Finally,

109

zV"then ±f^=7f+

1 Jl

z3

)

The point z = is a pole of order 2 of -jf -. The residue is 0'(O), where

z \z

<p(z) = -

Since

<t>\z)

1 + z3

'

(l + z3)e

z -e l3z

2

3\Z

the value of 0'(O) is 1. So

f'c 1 + z

3dz = 2m Res 7f

il)=2jtiW =2m -

SECTION 69

1. (a) Write

esc z = = -^4, where /?(z) = 1 and q(z) = sin z.sinz q(z)

Since

p(0) = l*0, ^(0) = sin = 0, and tf'(0)= cos0 = l*0,

z - must be a simple pole of cscz, with residue

P(0) =- = 1q'(0) 1

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

^ 1. 1 1

CSCZ = - +—Z+ Tz 3! L(30

25!.

1

z3+- (0 <lzl< it).

Since the coefficient of - here is 1, it follows that z = is a simple pole of cscz, the

residue being 1.

(a) Write

z- sinhz p(z)

z2sinhz

=^)' 6 ^) = *- sinh * and tf(z) = z

2sinhz.

Since

p(Ki) = m*0, q(m) = 0, and (m) = j? * 0,

it follows that

Rc „ z-sinhz = p(ni)= m__i_

*=*«' z2sinhz q'(,ni) n1 k

(b) Write

exp(zr) p(z)

sinhz=

~q(zj'6 ^)

= exP(*') ?(z) = sinhz.

It is easy to see that

«-*» sinhz q\7ti)r

*=-*/ sjnhz g'(-»0P

Evidently, then,

Resexp(£ + Res

expU0 = _ 2 «Pff*) + CTpW*)sinhz sinhz 2

(a> Write

Observe that

/(z) = ^7T' where P(z) = z and o(z) = cosz.

«(f + n«:j = (n = 0,±l,±2,...)..2

Also, for the stated values of n,

p\- + n7c\ =-+ n7t* and ^| +«^ = -sin^| +n^ = (-l)fl+1

^0.

Ill

So the function f(z) = has poles of order m = 1 at each of the pointscosz r

~"2

z„=-+nn (n = 0,±l,±2,...).

The corresponding residues are

(b) Write

tanh ^ = where p(z) = sinhz and q(z) = coshz.

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

z = I ^- + nw \i

2nnf (n = 0,±l,±2,...)

In addition to the fact that J (~ + jfJ

= 0, we see that

Pi[j+ nn

)*)= sinh(^ + = icosnn = /(-!)" *

and

*'((f+ = Sinh

(f '

+ n?n)=

* °-

So the points z = + (« = 0,±1,±2,...) are poles of order m = 1 of tanhz, the

residue in each case being

p\ [— + nn }i I

Let C be the positively oriented circle lzl= 2, shown just below.

TO 72 x

(a) To evaluate the integralJc

tan z dz , we write the integrand as

tanz = £^£l) where p(z) = sin z and q(z) = cosz,

and recall that the zeros of cos z are z = -^ + n;r (n = 0,±1,±2,...). Only two of those

zeros, namely z = ±tt/2, are interior to C, and they are the isolated singularities of

tanz interior to C. Observe that

n(nll)Res tanz = ,

' =-1 and Res tanz • „ _z=t/2 tf(tf/2) qX-npi)

_ p(-x/2) _ L

Hence

f tanz*fe = 27ri(-l-l) = -47ri.

(fe) The problem here is to evaluate the integralJ

integrand as

c sinh2z. To do this, we write the

—\— =£Qt where p(z) = l and g(z) = sinh2z.

sinh2z q(z)

Now sinh2z = when 2z = nm (n = 0,±1,±2,...), or when

— (/z = 0,±l,±2,...).

2

Three of these zeros of sinh2z, namely and±-^-> are inside C and are the isolated

singularities of the integrand that need to be considered here. It is straightforward to

show that

Res_J_ =£W =_J_ = I*=o sinh2z q'(0) 2cosh0 2'

Res

113

1 p(m/2) _ 1 1 1

*=»/2sinh2z q'(m/2) 2cosh(;ri) 2cosrc 2'

and

Res1 = PtE!2± = 1

= 1 = *

z=-«v2 Sinh2z q'(-jti/2) 2cosh(-;rz) 2cos(-;r) 2'

Thus

Jc sinh2z U 2 2-to.

5. The simple closed contour CN is as shown in the figure below.

Within Q, the function -= has isolated singularities at

z smz

z = and z = ±nn (n = l,2,...,AT).

To find the residue at z = 0, we recall the Laurent series for cscz that was found in Exercise

2, Sec. 61, and write

1 1 1 1 1= -t-cscz = -ri - +—z +

z sinz z2

z2\z 3!

1 1

(3!)z

5!z3+-

1 1 1-=+ +z

36 z

1 1

(3!)2

5!_z+- (0<IzI<tt).

114

This tells us that -=4— has a pole of order 3 at z = and thatz smz

1 1Res«=° z

i smz 6

As for the points z = ±nn (n = 1,2, . . . , N), write

~Y~— = ^tt, where p(z) = 1 and q(z) = z2sinz.

z smz q(z)

Since

p(±rut) = 1*0, q(±nn) - 0, and q\±nn) = n2n2

cosnit = (-l)"nz7F

z56 0,

it follows that

1 1 (-1)" (-1)"Res*=±»* z

2sinz (-1)"«V (-1)"

~~ nV

So, by the residue theorem,

J,c" z sinzcfe = 2m

6

Rewriting this equation in the form

A (-1)"+1

= tt2

g f dz

il n2

12 4i'-kz2sinz

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

f (-ir1

= n2

it n2

12

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

dz

c(z

2-l)

2 +3'

The isolated singularities of the integrand are the zeros of the polynomial

115

q(z) = (z2-l)

2+3.

Setting this polynomial equal to zero and solving for z2

, we find that any zero z of q(z) has

the property z2 = l±V3i. It is straightforward to find the two square roots of 1+ V3i and

also the two square roots of 1-V3i. These are the four zeros of q(z). Only two of those

zeros,

z,.^-.^*L and d*+L,

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

To find the residues at z and -Zq , we write the integrand of the integral to be evaluated as

1 _ P(z)

(z2-l)

2 +3 q(z), where p(z) = 1 and q(z) = (z -

1) +3.

This polynomial q(z) is, of course, the same q(z) as above; hence q(z ) = 0. Note, too, that

p and q are analytic at Zq and that p(z ) * 0. Finally, it is straightforward to show that

q\z) = 4z(z2 - 1) and hence that

?'(*<,) = 4z (4 -1) = -2V6 + 6V2/ * 0.

We may conclude, then, that z is a simple pole of the integrand, with residue

P(z ) _ 1

q'(z ) -2V6+6V2i'

Similar results are to be found at the singular point -Iq. To be specific, it is easy to see that

qX-Zo ) = -q'(z„ ) = -7(^) = 2V6 + 6V2i * 0,

the residue of the integrand at -Zq being

q'i-Zo) 2V6 + 6V2/*

Finally, by the residue theorem,

f = 2ml1 1 __

^

(

Z2 -

1)2 + 3 \-2sf6 + 6V2i

+2V6 + 6V2i J

=2^2

We are given that /(z) = l/[?(z)]2

, where ? is analytic at z*, ?(z ) = 0, and ?'(z )*0.These conditions on 4 tell us that q has a zero of order m = l at v Hence#(z) = (z - Zq)^(z), where £ is a function that is analytic and nonzero at z ; and this enablesus to write

f(z) =-^-T , where =—^.

So /has a pole of order 2 at z , and

Res/(z) = 0'(zo )=-M^.

But, since q(z) = (z -z )#(z), we know that

q'(z) = (z-z )g'(z) + g(z) and ?"(z) = (z-z )g,,

(z) + 2g'(z ).

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

q'(z ) = g(z ) and ?"(z ) = 2s'(z ).

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

Res/(z) = - q"^\.

(a) To find the residue of the function esc2z at z = 0, we write

csc z ~r < sn2 »

wnere ^(z) = sinz.

W(z)J

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

Rescsc2z = - q = 0.

[<z'(0)]3

(b) The residue of the function -

—

^-—j at z = can be obtained by writing(z + z

)

1

-, where q(z) = z + z2

{z + z2

) [q(z)f

Inasmuch asqis entire, q(0)=0, and q'(0) = 1*0, we know from Exercise 7 that

Res 1 =-j£<°>_ = -2 .- (z+z1

) [«W

118

Chapter 7

SECTION 72

1. To evaluate the integral J——

-, we integrate the function f(z) = around the simplen x +1 Z +1

closed contour shown below, where R>L

y

c.

We see that

where

_Jsx

2 + l J^ z2 +l

Thus

5 = Res —r—— = Res«' z +1 (z-f)(z + Z + l'_U 2i

f f <fe

Now if z is a point on Q,

lz2 + ll>llzl

2-ll=rt

2-l;

and so

J,

E.

R2 -l i_J_R2

-»0 as

Finally, then

1 dx 1 dx n1—7 = tt> or —: =—

.

_J *

2+ 1 Jjc2 +1 2

119

°° dx 1

2. The integralJ—5—rrj" can be evaluated using the function f(z) =— 5- and the sameJ (x +1) (z + 1)

simple closed contour as in Exercise 1. Here

_((jc2 + l)

2+fc(z

2 + i).2 , i\2

= 27riB,

where B = Res1

*"<~V + 1)2

. Since

1 Hz) . . , x1

12—^ = TT-^' where 0(z) =—

-

U +(z2 + l)

2(z-i)

2

we readily find that B = <j>'(i) = -7- , and so4i

J„0t2 + 1)

22 J<

fife

?(*

2 + l)2

2 ^(z2+l)

2

If z is a point on CR , we know from Exercise 1 that

lz2 + ll>i?

2-l;

thus

dzt az

b (z2 +(z2 + l)

2

The desired result is, then,

nR J? 3<—i t= , ,2 ~>0 as R-*°°.

°r dx _% r dx %_

i(x2+i)

2 "2' orJcTTif'T

3. We begin the evaluation of[ f* by finding the zeros of the polynomial z* + 1, which are

+1

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

jit/* _ 1 1

k ~ €"V2

+V2

and

J3X/4 iJT/4„iJt/2 1 . / V 1 i

120

Now

where

R - Res—^— and 2?, = Res—

r

*=*2 Z + 1

The method of Theorem 2 in Sec. 69 tells us that z, and z1 are simple poles of —:

andz +1

that

^=rr-- = -7- 311(1 fi2 =TT-- = -T'4zj Zi 4 4z2

3z2 4

since z,4 = -1 and zt = -1. Furthermore,

Bl+B

2 =-j(zl+z2 )

=~2V2"

Hence

Since

t dx _ n t dz

}R x* + l~j2 Jc, z* + i'

J,c«z4 +l

we have

°°t dx n °r dx n

iy + l"V2'°r L4 + l"2V2-

121

r x2dx

We wish to evaluate the integral —=———=—— . We use the simple closed contourl(x +l)(x

2+4)

shown below, where R>2.

We must find the residues of the function /(z) =—, : at its simple poles(z + l)(z + 4)

z = i and z = 2i. They are

Bl= Res/(z) =

(z + /)(z2 + 4).

and

Thus

*=2< (z2+l)(z + 2z).

fi2=Res/(z) =

*=2<(7

Z

?=2i

6/

3/'

f*A +f

K(x2+l)(x

2 + 4) Jc

z2dz

or

ijtf + l)(x' + 4) Jc. + + 4)

* x2dx n r z

2dz

=--f —JR(x

2+l)(x

2+4) 3 Jc(z3 + lXz

2 + 4)

If z is a point on CR , then

lz2 + ll^llzl

2-ll = /?

2 -l and lz2+4l2>lld

2-4l = /?

2 -4.

Consequendy,

z2<fc

(z2 + l)(z

2+4)

and we may conclude that

(R2-l)(R2 -4)

-»0 as

x dx n

i(jf" +l)(^ + 4) 3 J(jt

2 + l)(*2 + 4) 6

dx it

122

f x dx5. The integral f—5———5——5- can be evaluated with the aid of the function

{(x2 + 9)(x2 + 4)

2

f(z)(z

2+9)(z

2+4)

2

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

y

We start by writing

J

x dx

R{x

2 + 9){x2 + 4)

2 Jct(Z

2 + 9)(^ + 4)2

z2dz

where

Now

z2

z2

A = ResTi—r~n— and 5, = Res—5 = =-.

«-* (zz + 9)(z

z+4)

z

t1

(z + 3i)(z2 + 4)

2

J?=3<

(zz+9)(z

z

+4r

3

50/'

To find B2 , we write

z2

__0(z) ._, x , , Z"

(z2+9)(z

2 + 4)2

(z-2/):

-, where 0(z) =

Then(z

2+ 9)(z + 2/)

2

2?2 = 0'(2/) =

This tells us that

13

200/

'

Ax

2dx

JR(x

2+9)(x

2+4)

2100 Mza + 9)(z

a+4)

:

=JL_f100 J<Wz2 +

z2</z

.2 . a\2'

Finally, since

we find that

1,

z2dz

c*(z2+9)(z

2+4)

2

7r/?3

(R2 -9)(R2 -4)

j -» as R -> 00,

f 0c2 +

*2d!x 7T

(*z+9)(;c

2 + 4)2

100or

1

* etc

U2+9)(a:

2+4)2 200'

7. In order to show that

P V f

xdx= -—

'^(x2 + l)(x2 + 2x + 2) 5'

we introduce the function

(z2+ l)(z

2+2z + 2)

and the simple closed contour shown below.

y

/ x >*

.

R X

Observe that the singularities of /(z) are at i, z„=-l + i and their conjugates

z = -1 - i in the lower half plane. Also, if R > V2, we see that

K

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

where

and

2? =Res/(z)=*

«-«o (zz+l)(z-z )

Bl=Resf(z) =

Z=Zo

= _JL _L-io

+io

I

Evidently, then,

(z + /)(z +2z + 2)

f £<& = _£_f zrfz

4U2 + 1)(jc2 + 2x + 2) 5 ^(z2 + l)(z

2 + 2z +

= _L_I-10 5

1 '

zdz

Since

r Z*fe r

h (z2 + l)(z

2 + 2z + 2)~

(z2 +(zz + l)(z

i + 2z + 2)

as R -» °°, this means that

zdz

(zz+ l)(z-z )(z-z^)

,. r xdx nhm —; =—

.

*-*~L(x2+l)(x

2 + 2x + 2) 5

2)

(tf2-!)(/? -V2)

2->0

This is the desired result.

124

8. The problem here is to establish the integration formulaJ— - = using the simple

closed contour shown below, where R > 1.

1

There is only one singularity of the function f(z) =

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

dz . r dz . r dz

3 , namely z - eiicn

, that is interior

z +1

JclZ3 +l Jc« z

3 + 1 JC2Z3 + 1 z=Zo Z

3 +l

where the legs of the closed contour are as indicated in the figure. Since Q has parametric

representation z = r (0 < r < R),

fz = f

dr

kz 3 + l~{ r3 + l'

and, since -C2 can be represented by z = reann

(0 < r < R),

dzJ- *

J2*nf dz r dz _ f e

,Kdr _ n,/3 f

JcJZ3 + i J-c2Z 3 + 1 J(re ,2,c/3

)3 + l Jr3 + 1'

Furthermore,

Consequently,

But

5S.V + 1 3z2- ~ 3*

i2 *' 3

U V + l 3^2ff/3 Jca2> +r

•h z3 + i

. 1 2^i? n< > as R-><x>.R3 -l 3

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

fdr 2

JV + 1 3(ea*n -e l

2m 2ni n 2%

» t A 3(e -e -e ) 3(e —e ) 3sm(2^/3) 3\3

125

Let m and n be integers, where < m < n. The problem here is to derive the integrationformula

[—5-

—

-dx = —esci x

2n + 1 In \ In

(a) The zeros of the polynomial z + 1 occur when z2n - -1. Since

.(2k + i)n

In(-l)

1/(2B) =exp

it is clear that the zeros of z2n + 1 in the upper half plane are

and that there are none on the real axis.

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

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

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

2m 2m ,

Res-^— = .

c\_ ,

= J_c*(—)+>

<-»V + l 2wc,z''- 1

2n2n-l (* = 0,1,2... .,n-l).

^ . 2m + lPutting a = n, we can write

In

2(m-n)+l |. (2fc + l)7r(2m- 2w + i)l" eXT In J

[\(2* + l)(2m + l);rl , , N.

= expl i-i ^- '— exp[-/(2fc + l);r] =

Thus

-2m

Res-*=<*z

2" + l In

(2t+l)o(* = 0,1,2 /i-l).

In view of the identity (see Exercise 10, Sec. 7)

n-l i n

k=0 1 ~

126

then,

n-l ,2m n-l

f^z2n + l

m— el-ei2an

e-ia

n l-e:2a -ia

m ei2an

-l

n'

eia-e~

ia

m eiitm+l)*-l n 2i n

n e -e n e'a -e ,a nsma

(c) Consider the path shown below, where R>1.

y

c.

The residue theorem tells us that

r x2mr z

2m54 r2m

J T57TT^ +L 3TTT <fe = 2;R'2 Res^T'Zr x +1 JC,, Z +1 k=o

z=c"Z +1

or

}R x2n +l nsina k z

2n +l

Observe that if z is a point on CR , then

lz2m

l = /?2m

and \z2n + \\>R2n

-I.

Consequently,

W'+l dzR2m

R2a -l "/?

-2»1

-2n

2(n-m)-l

1-1

->0;

2«

and the desired integration formula follows.

10. The problem here is to evaluate the integral

Ji

dx

Kx2-a) 2 + l]

2 '

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

127

(a) Let us first find the four zeros of the polynomial

q(z) = (z2-a)2 + l.

Solving the equation q(z) = for z2

, we obtain z2 =a±i. 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 + i are the numbers

Zq =-j=>(^A + a+rtA-a) and -z ,

where A = 4c? + 1. Since (±z )

2=z\ =a + i = a- i, the two square roots of a - i, are

evidently

z and -z .

The four zeros of q(z) just obtained are located in the plane in the figure below, which

tells us that z and -Zo lie above the real axis and that the other two zeros lie below it.

•

y

•

Zo

X

• •

-Zo Zo

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

1

f(z) =[q(z)f

which becomes the integrand in the integral to be evaluated when z = x. The method

developed in Exercise 7, Sec. 69, reveals that z is a pole of order 2 of /. To be

specific, we note that q is entire and recall from part (a) that q(Zo) = 0. Furthermore,

q\z) = 4z(z2 - a) and z\=a + i, as pointed out above in part (a). Consequently,

2'(z ) = 4z„ (Zq -a) = 4iz * 0. The exercise just mentioned, together with the relations

zl=a + i and 1 + a1 = A2

, also enables us to write the residue of/at Zq'.

g"(Zo) = 12z2 - 4a _ 3zp

2 - a _ 3(a + i)-a a — i __ a — /(2a2+ 3)

WW (4iz )3

16iz2z 16i(a + i)zo' a-i 16A

2Zo

As for the point -z , we observe that

q'(-z) = -q\z) and q'\-z) = q'Xz).

Since q(-z ) = and #'(-z ) = -q'do) = 4i'z 0, the point -Zq is also a pole of order

2 off. Moreover, if B2denotes the residue there,

_ <?"(-?<)) _ <?"(*(,) _ f g»frb) 1 _[^'(-^,)]

3

[<?'(z )]

3

l^'(^o)]3

= -5x

.

Thus

B.+B2= B.-B.= 2r'Im5. =—Viml 2 1 1 1

8A2/

-g + f(2fl2+3)

Zn

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

R >\z \ and CR denotes the semicircular portion of the path. The residue theorem tells

us that

R

jf(x)dx + f(z)dz = 27ri(A + B2 ),

or

I;

dx

In order to show that

Im-a + i(2a

2+3) dz

[<?(z)J2

•

lim f

dx= 0,

*^~ Jc« [?(z)]

2

we start with the observation that the polynomial q(z) can be factored into the form

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

lz±Zol>llzl-IZoll = /?-lzol and \z±Zq\>\ Izl-lz^l I = R-\z \.

This enables us to see that \q(z)\ ^ (/?-lz„l)4when z is on CR . Thus

129

1

(R-\z \)'

for such points, and we arrive at the inequality

nR

(tf-IZoO8 8 '

which tells us that the value of this integral does, indeed, tend to as R tends to <».

Consequently,

P.V.j

dx n

_J.[(*

2 -a)2+l]

2 4A2

But the integrand here is even, and

Im-q + /(2a

2 + 3)

^0

Im-a + i(2a

z+3)

Zo

= Im V2-a + i(2a

2 + 3) -jA + a - i-yjA-a

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

So, the desired result is

k2 -a)2 + l]

2=^[(2a2 +3)VA^ + .VA^],

where A = Vfl2 + 1.

SECTION 74

r cosxdx1. The problem here is to evaluate the integral — —= where a > b > 0. To do

_*.(*+

a

2X*+&)

this, we introduce the function f(z) =— —n~' whose singularities ai and bi lie

(z +a )(z +b )

inside the simple closed contour shown below, where R > a. The other singularities are, of

course, in the lower half plane.

130

According to the residue theorem,

eadx

jj? + a2)(x

2 +b2

where

and

That is,

or

B, = Res[/(zK<] = - -.v 2

*=<« (z + ai)(z +b2)

-+ J/(z)eft

£fe = 2»XA + ^)./ r.

52 =Res[/(z)e'

z

]=

Ise"dx n (e

b

_£1a

2 -b2\ b a

cos*dfr

(x2 + a

2)(x

2 + b2) a

2 -b2

( -b -a\

:=oi.2a(*

2 -a2)/

e-»

=bi2b(a

2 -b2)i-

- \Az)e*dz,

Rejf(z)ekdz.

Now, if z is a point on CR ,

l/(z)l<MR where M» =—= = = 5-fi

* (R2 -a2

)(R2 -b2

)

md\e iz

\=e~y <1. Hence

ReJc/(z)^&|<|j

Cj[

/(z)e'Vz

So it follows that

r cosjtd!*

<>MBnR =nR

(R2 -a2

)(R2 -b2

)

-» as R -» 00.

i (jc2 + a

2)(jc

2 + fc2) a

2 -&2l ft a

(a>b>0).

f COS fl.y

2. This problem is to evaluate the integral —= dx, where a > 0. The function

/(z) =——- has the singularities ±1; and so we may integrate around the simple closedz +1

contour shown below, where R > 1.

y

We start with

where

Hence

or

)-1-—dx+ jf(z)eiaidz = 2mB,

-R X +1C,

B = Res[f(z)eUK

] = ^-:*=" 1 J

z + i 2i

lOX

\—^dx = ne-a - \f{z)e

iaidz,

131

Since

we know that

and so

r cosax , _a _ r ,, . ,„T ,

J -rrr^ = ^ a -ReJ/(zy<fe,

\f{z)\<MR where L-

Re \f{z)e^dzicR

R2 -V

J

cosax

x2 + l

dx = ne~

That is,

r cosax , it—5 ox = —

<

J*2 +l 2(a>0).

4. To evaluate the integral \

XS^Xdx, we first introduce the function

I x2 +3

z2 +3 (z-zOiz-zJ

where z, = V3i. The point ztlies above the x axis, and zi lies below it. If we write

fi&P^M. where Hz) = ,

z - Zl Z-Zt

132

we see that zxis a simple pole of the function f(z)e

i2zand that the corresponding residue is

B1= (j>(zl )

= V3/exp(-2V3) _ exp(-2V3)

2V3i 2

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

.

y

Integrating f(z)e'2i

around the closed contour, we have

R ilx

jf—dx = 2mBl -if(z)e

i2idz.

Thus

j -^T^dx = Im(2^)-ImjcJ(z)e

i2idz.

Now, when z is a point on CR ,

l/(z)l<MR , where MR=— > as R -» °°;

and so, by limit (1), Sec. 74,

Consequendy, since

limf f(z)eiudz = 0.

R-*~JC,

JmjcJ(z)e

i2tdz\<\j

cnz)e

iUdz

we arrive at the result

|^<fa=^p(-2V3), or j™H.,fr = fexp(-2V3).JLJf +3 J„*+3 2

133

The integral to be evaluated isf*f^"* dx, where a > 0. We define the function

L x +4

/(z) = ~Tj7^' and» by computing the fourth roots of -4, we find that the singularities

^=V2ete/4 =l + i and z2 = V2V3 *'4 = V2V*'

V*'2 = (1 + i)i = -1 + i

both lie inside the simple closed contour shown below, where R > V2 . The other twosingularities lie below the real axis.

The residue theorem and the method of Theorem 2 in Sec. 69 for finding residues at simplepoles tell us that

R 3 tax- x e

where

and

f 7+4* +k f(z)eiKdz = 27n'^ +^

5 - Res^g* - zi

giaZ' - = g

'a(1+,)

= ggH1

t-tt z4 + 4 4z

34 4 ~ 4

o7 — Kes— = — = = =z=zlZ* + 4 4z* 4 4 4

Since

27n'(A+£2 ) = flu;. e +e

\

= i7te cos a,

we are now able to write

r x smax . r

J -^r^pfr = « "cosa-Im J /Uy*.

134

Furthermore, if z is a point on CR , then

R3

l/(z)l < MR where MR =——- -> as R -> <»;R — 4

and this means that

^LRfWehZdz

\ * \fCRmeiazdZ

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

7 x 3sinax , .—3

—

—ax = ne cosaJ x +4

-» as /? -> °°,

(a>0).

8. In order to evaluate the integralf

*, we introduce here the function

J (x2 + l)(*

2+9)

f(z)= 2 2 ^ox - Its singularities in the upper half plane are i and 3i, and we

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

Since

Res[/(z)e*]=z

'f2(z+0(r+9)

1

16e

and

Res[/(z)eiz

]= 71

-«= 3'

1 J(z +

z3e*

(zz + l)(z + 3i)

Z=3iI6e

3 »

the residue theorem tells us that

t x3eadx r /'i

,( (^ + l)(^ + 9)+Jc/^ =H-I?16<? 16e

3 '

or

r j: sinxdx n ( 9 \ t

Now if z is a point on CR , then

135

\f{z)\<sMR where MR =—- —n

* * (R2-l)(R2 -9)

as R-> oo.

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

Im\cJ(z)e*dz\ <

|\cJ(z)e*dz ->0 as /?-> oo;

and this means that

"r _jc^nxdx_ _ n_(9__\°f

xi smxdx it (9

l(x2+\){x

2+9) 8e{e

2)' °T J (x

2 + l)(x2 +9)~ I6e \e

2

The Cauchy principal value of the integral f fnx<ix

can |,e foun(j ^ih^ f^J x

2 +4* + 5

function /(z) =1

r +4z + 5and the simple closed contour shown below, where R > V5.

Using the quadratic formula to solve the equation z2+4z + 5 = 0, we find that / has

singularities at the points zl=-2 + i and z

x=-2-i. Thus f(z) = 3—— , where z^

is interior to the closed contour and zxis below the real axis.

(z-zjiz-zj

The residue theorem tells us that

where

S = Res

and so

r smxdxI—5 r = Im

J„ x

l + Ax +

2me ,Zl

5 L(zi-^i).

-Imf f(z)ekdz,

136

or

r sinxdx % . „ , f

Now, if z is a point on Q, then \elz

\=e~y < 1 and

l/(z)l<M„ where MR=

Hence

Imf /(z)^|<|f

and we may conclude that

<MRnR = — 2-» as R -» oo,

(/?-V5)2

. 7 sin;t<£t # . „VJ~—

'a 7= sm2-

J x2 + 4x + 5 e

10. To find the Cauchy principal value of the improper integral f^x

.

+ 1^cosxdx, we shall use

J x2 +4x + 5

the function /(z) =z+l z+1

r + 4z + 5 (z-zJiz-zO— , where z

x= -2 + /, and z

x= -2 - 1, and the

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

where

Thus

or

5 = Res(z + l)e*

(z-^Xz-zi)_ (Zt+l)g

fel

= (-l + Qg-2 ''

(z-zi) 2ez

r (x + Dcosjc . tt, . v r

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

R + l

\f(z)\<MR where Ms= R + l

(R-izMR-m (/?-V5)2-» as R -» oo.

Limit (1), Sec. 74, then tells us that

RcJCa

/(z)«*<fe|^|jCa/(z)e*&

137

-» as /? -> oo,

and so

_,, r (* + l)cos;t , n, . _pv - ^—: r^ =— (sin 2 -cos 2).

12. (aj Since the function /(z) = exp(i'z2) is entire, the Cauchy-Goursat theorem tells us that its

integral around the positively oriented boundary of the sector 0£r<R, 0<6<nl A

has value zero. The closed path is shown below.

A parametric representation of the horizontal line segment from the origin to the point

R is z = x (0<x<,R), and a representation for the segment from the origin to the point

Re'"* is z = re""4

(0 < r <, R). Thus

« A

jeixl

dx +jc

elll

dz-eiK'*\e-

rl

dr = <d,

or

)eixl

dx = ei

"'*]e-rl

dr-\c

ee

dz.

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

and

R 1*1jcos(x

2)dx = -j=je~

r*dr-Rej e*dzo

C*

R|

R

fsin(jc2)dbt = -H<f r2

</r-Imf e*dz.1 V2 i

Jc*

138

(b) A parametric representation for the arc CR is z = Reie(0 < < 7r / 4). Hence

it/* a/4

jce

izl

dz= \e*1'a'Rie

ied9 = iR\e-RlA* 26

eiRla"ie

ei6de.

= 1 and \eiB

\

= 1, it follows thatSince.iR

z cosZ0

, ic/4

Jce*dz<R je-*

lsia26dd.

Then, by making the substitution <j) = 2d in this last integral and referring to the form

(3), Sec. 74, of Jordan's inequality, we find that

2 2¥~4R-» as R -» oo.

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

oz

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

]cos(x2)dx =^^ and Jsin(*

2)d!* =

SECTION 77

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

- cos(ax) - cos(foc) ,

2 dx = —(p - a) (a>0,b>0),

using the indented contour shown below.

y

Applying the Cauchy-Goursat theorem to the function

139

/(*) =

we have

or

f f(z)dz+[ f(z)dz+l f(z)dz+i f(z)dz = 0,JM JC* JL1 «C.

Since ^ and have parametric representations

Ll:z = re

i0 = r(p<r£R) and - L2 :z = re" = -r(p<r^R),

we can see that

^ wr ibr & —iar —ifer

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

S ~ edr +

\

e ~ edr

f (g^+g^ ) - (e*r + e"*') , „ J cos(ar) - cos(fcr) ,

J-5 dr =

2J-5 dr.

Thus

p

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

fiz) =\z

^ |

iaz|

(iaz)2

(

(/az)3

|

N

/(a-fc)

1! 2! 3!

+ ••• (0<lzl<°°).

(ibz(ibz)\(ibz)\

1H 1 H h •••

I 1! 2! 3!

From this we see that z = is a simple pole of f{z), with restdue~Ik = i(a - b). Thus

lim f f(z)dz = -B 7ti = -i(a - b)m = %{a - b).p—»0 JCp

140

As for the limit of the value of the second integral as R -» «>, we note that if z is a point onCR , then

rt ^ \e"*\+\eM

\e-v + e-* ^1 + 1 2

<—tkR =— -*0 as

R2 R

Consequently,

It is now clear that letting p -» and /?-><» yields

^7 cos(ar)-cos(frr) J

2Jj Ldr= n(b-a)

o

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

Observe that when a - and b = 2, that result becomes

7 1 - cos(2;t),

J3 dx = n-A

But cos(2jc) = 1 - 2 sin2x, and we arrive at

I * 2

2. Let us derive the integration formula

r x (l-a)TT, ,

|(7TIF'fa =

4cos(a,/2)

(-1<a<3)'

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

_ z" exp(alogz) . _ 3;^

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

p

Branch cut

By Cauchy's residue theorem,

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

iJL

zJC

p i=i

That is,

f f(z)dz+ f f(z)dz = 2mResf(z)~ f f(z)dz- f f(z)dz.

Since

Ll:z = re

i0 =r(p<r<R) and -I2 :z = refat = -r(p<r</?),

the left-hand side of this last equation can be written

* a(lnr+i"0) R o(lnr+/)r)

« a * a * a

= 171— dr + «"*\ ,i 2 <fr = (1 + f / . dr.

J (r2

-4- 1)2 J(r2

+1)2 'J(r2 + 1)

2

Also,

Res/(z) = where </>(z) =(z + 2 '

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

reveals that

'(-\ _ -(a-l)logz0'(z) = e

a(z + i)-2z

L (z +3

142

and from this it follows that

Res f(z) = -ie-iaKllfl-a

We now have

Once we show that

limf f(z)dz = and liraf f(z)dz = 0,

p-*0 iC„ JC.

we arrive at the desired result:

„a _/i _\ JaKll -iaKll \_r ^ _ — a) _e e ^(1 - a)

Jo(^+l)2

(l-a)ff

2 l + em* eiaic —iaKll

4 eia*'2 + e

-ia*l2AcQS(a7t / 2)

The first of the above limits is shown by writing

I f(z)dz(l-p2

)

2r(1-p

2

)

np =Tip

a+l

and noting that the last term tends to as p -» since a + 1 > 0. As for the second limit,

1 _ 1

\jcmdz Ra

(tf2-l)

2 ""(R

2-l)

2 1R4 |1-

and the last term here tends to as /?-»<» since 3 - a > 0.

3. The problem here is to derive the integration formulas

"(Mxlnx i n2, . 7 VIf

-yxln* , jr . _ f Vjc , 7t

l x2 + l 62 J*2 + l V3

by integrating the function

f(r, z

mlogz _ e^logzH )

z2 +l ~

z2 + l

lzl>0,— <argz<—

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

f f(z)dz + f f(z)dz = 2mResf(z) -f f(z)dz - f f(z)dz.

Since

/W-M where =z-i z+i

the point z = i is a simple pole of f(z), with residue

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

.

The parametric representations

Ll:z = re

i0 =r(p<r<R) and -L1:z = reiK = -r(p<r<R)

can be used to write

J. and f =^.J^l^fr.

Thus

143

J^TT^ J r> + i

dr=Te-ic/^z-jc

mdz.f Vrlnr^ (

gl„/3

1Ifrlnr+in Vr ^ ;r

2

I r2 + 1 ;p p

By equating real parts on each side of this equation, we have

m^dr + cos(;r / 3)l^fdr - nsin{7t 1 3)\-^-dr =-—sin(;r / 6)

-Ref /(zVfe-Ref /(*)&;

and equating imaginary parts yields

* 3/~~ i * 3/~" 2

sin(;r / 3) f-^-f flfr + *rcos(w / 3)\——dr =— cos(;r / 6)

Jr+1 J r

2 +l 2

•Imf /(z)<fe-Imfr

/(z)<fe.

Now sin(^/3) =^, cos(^/3) = ^-, sin(7r/6) = ^-, cos(^/6) =— and it is routine toz Z 2 2

show that

limf f(z)dz = and limf /(z)<fc = 0.

p->0 Jc, ' /}_- Jc,v 7

144

Thus

That is,

3f V7lnr W3 r W . ;r2

2 * r2 +l 2 J r

2 +l 4

3 _ W3 _ n2

2l

22 ~ T*

21

22

4

Solving these simultaneous equations for /j and 72 , we arrive at the desired integration

formulas.

4. Let us use the function

v (logz)2

f, i n n I*/(z) =-2— lzl>0,--<argz<-

Z +1 V 2 2

and the contour in Exercise 2 to show that

f(lnx)

2

j n3

} Inx, n——-<& =— and -5— dtc = 0.

{x2 +l 8 JQx

2 + l

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

ff(z)dz + j

'

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

Since

/W-M whereZ-i z+i

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

Res/(z) = 0(0 =^ =(lnl + t-;r/2)

2

= _*1*=•' 2/ 2/ 8i

'

Also, the parametric representations

Ll:z = re

i0 =r{p<r<R) and -L2 :z = reiic = -r(p<r<R)

enable us to write

J,/W*"f$£* and y^,]^fdr.

P PSince

p p pthen,

p p p

Equating real parts on each side of this equation, we have

p p

and equating imaginary parts yields

r ,

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

p

It is straightforward to show that

lim f f(z)dz = and lim f f(z)dz = 0.

Hence

and

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

r dr _ %Jr2 + 1

=2'

we arrive at the desired integration formulas.

146

5. Here we evaluate the integral dx, where a>b>0. We consider theJ (x + a)(x + b)

function

.1/3 expl ^logz

(z + a)(z + b) (z + a)(z + fc)

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

z = -b are between the circles.

A parametric representation for the upper edge of the branch cut from p to R is z = rei0

(p < r < R), and so the value of the integral of/along that edge is

r exp:(ln/- + iO)

f_LL ^r = f £ ^o

(r + a)(r + b) {(r + a)(r + b)

A representation for the lower edge from p to is R is z = ren* (p<r< R). Hence the

value of the integral of/ along that edge from R to p is

*exp -(lnr + /2;r)

fL3 \ dr = -e

i2*n\

i(r + a)(r + b)dr.

According to the residue theorem, then,

where147

3=Res/(z)exp|jlog(-a)j exp|j(lna +

/;r)J gM3^-a+b a-b a-b

and

exp

B2 =Resf(z) = -

Z=-b

|log(-t) exp1(\nb + in)

i"nMb

-b + a -b + a a — b

Consequently,

• V7 ^-2«-^) .J/(^,

J/(^(\-e™)\ ^ dr = --

Now

L f(z)dz(a-p)(b-p) (a-p)(b-p)

and

/(z)&V* D 2*/?

21 n „< 2nR =— — •- >0 as

(R-a)(R-b) (R-a)(R-b) MR2

Hence

f , 2me i*/3e/a~-Mb) e™ 2^ -Mb)

o(r + fl)(r + ft)

r(1 - e''

2"3^ - *)'

<f'*/3

(e'*/3 - - *)

n(Ma-Mb) _ n{Ma~-Mb) = 2ar V^-Vfrsin(;r/3)(a-fc) ^/3~ 7~~

V3~" a-fc'

-y(«-*)

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

(V* fc^n Ma'-Mb

{(x + a)(x + b) V3 a-b(a>b>0).

148

6. (a) Let us first use the branch

z-i/2 exp^--logz

z' + l z2 + l

lzl>0,--^<argz<^

and the indented path shown below to evaluate the improper integral

P

Branch cut

Cauchy's residue theorem tells us that

f f(z)dz+i f(z)dz + f f(z)dz+l f(z)dz = 2mResf(z),M -' (-* '^2 •'Cp Z=i

or

f /(z)<fe + J f(z)dz = 2mResf(z)- f /(z)rfz-f /(z)<fe.

Since

I^-.z^re' =r(p<r<R) and -l^.z-re^ = -r(p<r<R),

we may write

dr

1)

Thus

Now the point z = i is evidently a simple pole of f(z), with residue

exp[-|logij exp -I^inl + f.5

149

z~m

Res f{z) =—.J=< z + i 2i 2i 2i 2iVV2

Furthermore,

L f(z)dz. no it Jp n-TvTT =72^ as p->0Vp(i-p

2

) 1-p2

and

\\cm<k k4r n

Finally, then, we have

-> as /?->oo.

(1

VKr2 + l) V2'

which is the same as

r dbc _ it

J-vW+1) ~V2'

To evaluate the improper integral [-j=— , we now use the branch

JQ^x(x2 + l)

-1/2 expf-^-logz

zz + l

(kl>0,0<argz<2;r)

and the simple closed contour shown in the figure below, which is similar to Fig. 99 in

Sec. 77. We stipulate that p < 1 and R > 1, so that the singularities z = ±i are between

C„ and CR .

Since a parametric representation for the upper edge of the branch cut from p to R is

z = re' (p£r< R), the value of the integral of/along that edge is

«exp1

„_..r —(lnr + i'O) „

V^(r2 + l)

A representation for the lower edge from p to is R is z = re'1" (p<r<R), and so the

value of the integral of/ along that edge from R to p is

«exp ~(\nr + iliz)

- fL 2

2-U = -e-* dr =

f_i dr.

Hence, by the residue theorem,

R R 1

where

-1/2

B[=Res/(z) =

z + i

exp --log/ exp -i-flnl-H-*2l 2

-;«/4

2/ 2/ 2i

and

.-1/2

52= Res/(z) =^

*=-' z -

1

exp4iogH)1

expIf, , .3*— lnl + i

—

2V 2-i3*r/4

-2i -2i 2i

That is,

Since

and

J„Vr(r +1) J J

In p 2k Jp

VP(i-p2

) l-p2

2;cR 2;r-»0 as oo,

151

we now find that

J

i , e —e e +e edr — it = n

J V^(r2 + l)

= n = Kco&urvrWhen x, instead of r, is used as the variable of integration here, we have the desired

result:

7 dx _ it

SECTION 78

1. Write2r dd _ r 1 dz_ r dz

]5 + 4sin0~-k J z -z~l \ iz ~^2z2 + 5iz-2'

5 + 4

where C is the positively oriented unit circle \z\= 1. The quadratic formula tells us that the

singular points of the integrand on the far right here are z = -i/2 and z - -2i. The point

z = -il 2 is a simple pole interior to C; and the point z = -2i is exterior to C. Thus

? M artier'—w-u -wfi)-^.Jo5 + 4sin0 *=--42z

2 +5iz-2j L4z + 5i_L_ 1/2 U»7 3

2. To evaluate the definite integral in question, write

f dd _ r 1 dz _ t 4izdz

Jl + sin2 0~Jc ( 7 - 7

-i\2 '

iz ~Jc z*_6z2 + r1 +

2i )

where C is the positively oriented unit circle \z\= 1. This circle is shown below.

152

Solving the equation (z2

)

2 - 6(z2) + 1 = for z

2with the aid of the quadratic formula, we

find that the zeros of the polynomial z4 - 6z

2 + 1 are the numbers z such that z2 = 3 ± 2V2

.

Those zeros are, then, z = ±^3 + 2-^2 and z = ±V3-2>/2. The first two of these zeros areexterior to the circle, and the second two are inside of it. So the singularities of theintegrand in our contour integral are

z; =V3-2V2 and z1 =-zv

indicated in the figure. This means that

where

By = Res— 4iz _ 4izt

/

*-* z4-6z

2+ l 4^-12^ zf-3 (3 -2V2) -3 2-V2

and

B2= Res

4/z _ -4/z,

Since

the desired result is

*=-*.z4-6z

2+ l -4z

1

3 + 12z1 zf-3 2V2"

i \ 2k V22^(A-H52 ) = 2^|--^j = ±|.-l| = V2^

J l + sin2

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

Jsin2" BdO =

I Jsin

2" ddd = ±J

1

r

( z-z2M 2i

In

dz

iz 22"+1

(-l)(-1)"/ Jc

-l\2«

dz

22"+ii^/c|(T) z2

"(-z

"1)iz"^

22"+1

(-l)"i ft ^ *

Now each of these last integrals has value zero except when k = n:

153

Consequently,

jcz

1

dz = 2m.

jsm2n ddd =1 (2w) !(-!)"2m = (2n)!

22»+i "

(n!)2 22fl

(«!)2

7T.

SECTION 80

5. We are given a function / that is analytic inside and on a positively oriented simple closed

contour C, and we assume that /has no zeros on C. Also, / has » zeros zt (& = l,2,...,n)

inside C, where each zk is of multiplicity mt . (See the figure below.)

The object here is to show that

To do this, we consider the kth zero and start with the fact that

f(z) = (z-zk )

m>g(z),

where g(z) is analytic and nonzero at zk . From this, it is straightforward to show that

zf'iz) _ mkz zg\z) _ mk(z-zk ) +mkzk zg\z) _ _ ,zg\z)

,mkzk— 1 — 1 -WlH 1 .

/(z) z-zk g(z) z-zt g(z) g(z) z-zk

zg'(z) zf'iz)Since the term — — here has a Taylor series representation at zk , it follows that

g(z)

has a simple pole at zk and that

Hz)

/(«)

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

154

6. (a) To detennine the number of zeros of the polynomial z6 - 5z

4 + z3 - 2z inside the circle

lzl= 1, we write

/(z) = -5z4

and g(z) = z6+z

3

-2z.

We then observe that when z is on the circle,

l/(z)l = 5 and ls(z)l<lzl6 + lzl

3+2lzl = 4.

Since \f(z)\>\g(z)\ on the circle and since /(z) has 4 zeros, counting multiplicities,inside it, the theorem in Sec. 80 tells is that the sum

Az) + g(z) = z6-5z*+z3

-2z

also has four zeros, counting multiplicities, inside the circle.

(b) Let us write the polynomial 2z4 - 2z

3 + 2z2 - 2z + 9 as the sum /(z) + g(z), where

/(z) = 9 and g(z) = 2z4-2z

3 + 2z2-2z.

Observe that when z is on the circle lzl= 1,

l/(z)l = 9 and l#(z)l<2lzl4+2lzl

3 + 2ld2+2ld = 8.

Since l/(z)l>l#(z)l on the circle and since /(z) has no zeros inside it, the sum

/(z) + g(z) = 2z4 - 2z

3 + 2z2 - 2z + 9 has no zeros there either.

7. Let C denote the circle I zl= 2

.

(a) The polynomial z4+ 3z

3 + 6 can be written as the sum of the polynomials

/(z) = 3z3

and $(z) = z4+6.

On C,

l/(z)l = 3lzl3= 24 and l^(z)l = lz

4+6l^izl

4 + 6 = 22.

Since l/(z)l>l#(z)l on C and /(z) has 3 zeros, counting multiplicities, inside C, it

follows that the original polynomial has 3 zeros, counting multiplicities, inside C.

(b) The polynomial z4 - 2z

3 + 9z2 + z - 1 can be written as the sum of the polynomials

/(z) = 9z2

and g(z) = z4-2z

3+z-l.

On C,

|/(z)| = 9lzl2= 36 and U(z)l=lz

4-2z

3 + z-ll<lzl4+2lzl

3+lzl+l = 35.

155

Since l/(z)l >\g(z)\ on C and /(z) has 2 zeros, counting multiplicities, inside C, it

follows that the original polynomial has 2 zeros, counting multiplicities, inside C.

(c) The polynomial z5 + 3z

3 + z2 + 1 can be written as the sum of the polynomials

/(z) = z5

and $(z) = 3z3+z

2 + l.

On C,

l/(z)l=lzl5 = 32 and I#(z)l = l3z

3+z

2+ll<3lzl

3+lzl

2+l = 29.

Since \f(z)\>\g(z)\ on C and /(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

P(z) = a + a,z + • • • + a^z""1 + aHz" (a. * 0),

where n > 1, has precisely n zeros, counting multiplicities. Without loss of generality, we

may take an=l since

P(z) = a„

rEa. + BLz + ... + EB=L^ +z^

Let

/(z) = zB

and g(z) = a +o1z + - + an_l

z"'1

.

Then let R be so large that

/?>l + la l + lo1

l + ---+laB_ 1l.

If z is a point on the circle C: lzl= R, we find that

lg(z)l fS la l + l^llzl + ••• + la^llzl"-1 = la l + iaJR + ••• + la^ltf-

1

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

-1+ ••• + t^.J/?"

-1= (la l + laj + - +

</?/r-1 =/2"=lzr = l/(z)l.

Since /(z) has precisely n zeros, counting multiplicities, inside C and since R can be made

arbitrarily large, the desired result follows.

156

SECTION 82

1. The singularities of the function

2s3

5-4

are the fourth roots of 4. They are readily found to be

or

V2, <j2i, -V2, and -V2*.

See the figure below, where y > -4l and R > V2 + y.

Y+iR

The function

has simple poles at the points

2?V'e =——

-

s -4

and

j = V2, jt= V2/, j, = -V2, and *3

=

V 3 o„3*.r

X Res^FW] =2 Res ^il =f 2j£

2 2 2 2

• +

= coshV2f + cos V2f.

Suppose now that s is a point on CR , and observe that

157

\s\=\y+Re'°\ <y+R = R+ y and \s\=\y+Re ie\>\y-R\ = R-y>«j2.

It follows that

\2s3

\ = 2\s\3

<2(R+yf

and

Consequendy,

This ensures that

I s4 - 41 > lid

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

lF(*)l< ^(J?+

.

y)'

->0as/?->(/?-y)

4 -4

/(/)= coshV2r + cosV2f.

2. The polynomials in the denominator of

F(s)2s -2

(s + l)(s2 + 2s + 5)

have zeros at s = -1 and s = -1 ± 2i. Let us, then, write

e"F(s) =e*(2j-2)

(j + IXj-^X^-J!)

where jt= -1 + 2/'. The points -1, su and s

{are evidently simple poles of e"F(s) with the

following residues:

i_ gJt(2^-2)

-Si)^=Resfe"F(j)]

z=-l L J

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

g"<(2^- 2) = fi - 1W<,

B3=Res[e«F(s)]=

^'^^3

~, L WJ(J, +1X51-5.)

eSl'(2Sl -2)

(5,+lX^-J!)

158

It is easy to see that

i2t

-t , -t— —e + e

f ei2'-e-

i2'

+en

' + e-i2''

= <f'(sin2f+ cos2f-l).v 2/ 2

Now let s be any point on the semicircle shown below, where y > andR > -\/5 + y.

Since

\s\=\y + Re ie\<y+R = R+y and \s\=\y+ Rei6\>\y- R\= R- y> V5,

we find that

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

and

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

\s2 +2s + 51=1 s - Sl \\s - s,\ > (\s\-\sj)

2 > \{R - 7)2 - V5 f > 0.

Thus

Im = ^=1 < . Y) + 2

Is + II \s2 + 2j + 51 [(* - y) - !][(/?

- 7)

2 - V5]

and we may conclude that

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

as R -> oo,

4. The function

159

„2 2s —a(r+a2

)2 (a>0)

has singularities at s = ±ai. So we consider the simple closed contour shown below, where

Y>0 and R>a+y.

Upon writing

F(s) = yv. 2

where 0(s) = -

(s - a/)

we see that is analytic and nonzero at s = ai. Hence sQ is a pole of order m = 2 of

F(j). Furthermore, F(s) = at points where F(s) is analytic. Consequently, s is also

a pole of order 2 of F(j); and we know from expression (2), Sec. 82, that

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

(fc1+ b

2t)] ,

where and b2 are the coefficients in the principal part

s - ai (s - ai)2

of F(s) at ai. These coefficients are readily found with the aid of the first two terms in the

Taylor series for </>(s) about s = ai:

1

(s - ai)(s) =

1

(s - ai)

160

</>(ai),0'(ai)=^)r + 7^- + - (0 <!,-«!< 2a).

It is straightforward to show that <f>(ai) = 1 / 2 and <j>\ai) = 0, and we find that ^ = and

b2=l/2. Hence

Res [e*F(sj\+ Res [e"F(s)] = 2Re ^(j'j = tcosat.

We can, then, conclude that

f{t) = tcosat(fl > 0),

provided that F(s) satisfies the desired boundedness condition. As for that condition, when

z is a point on CR ,

\z\=\y+Reie\£y+R = R+y and \z\=\y+Re

ie\>\y-R\= R- y > a;

and this means that

\z2 -a2

\<\z\2 +a2 <(R+y)2 + a

2and \z

2 +a2\>\\z\

2 -a2\>(R-y)2 -a2>0.

Hence

6. We are given

„. . sinh(xs)

which has isolated singularities at the points

-n „ _ (2n-l)^ . _ (2n-l)n., , „= °> *.= 2 ,f 311(1 J

"= ~

2*

(n = l,2,...).

This function has the property F(s) = F(s), and so

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

To find the residue at j = 0, we write

sinh(xy) _ xy + (xy)3/ 3 ! + • • • oc + jcV/6 + —

j2cosh* 5

2(l + *

2/2! + ...)" s + s

3/2 + - { 2

0<ld<-|.

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 le^Fis)] = Res F(s) = x.s=s„ L 1 s=s

As for the residues of F(s) at the singular points a; (n = 1,2, ...), we write

F(s) = - where = sinh(jcs) and = s2cosh j .

We note that

. . . . (2n-l);a _ , v „= isin^—^— 5*0 and q(sn ) = 0;

furthermore, since

q'(s) = 2j cosh j + j2sinh a,

we find that

„ . (2n-l)V. . (2w-1)tf .(2n-l)27r

2. ( n\?W = ism = -,i -J. sm\nn--

J

.(2n-l)2^V. jt , art (2n-l)

2 ^2,

' n-r smnttcos— -costt7rsin— = v — (-l)"/*0.4 V 2 2) 4

In view of Theorem 2 in Sec. 69, then, sn is a simple pole of F(s), and

?'(*.) (2n-l)J

2

Expression (4), Sec. 82, now gives us

Res He. [e"F(s)] - 2 Re(-± .-^sini^^expTifi^l]'='.

1 J'='"»

L J[;r

2(2n-l)

22 L 2 JJ

8 (-1)" . (2n-l)*ct (2n-l)nt= —?•——^—rsin- -—cos- —

.

n1(2«-l)

22 2

162

Summing all of the above residues, we arrive at the final result:

f(t) - x+— 2_,— -j sm cos- —

,

7. The function

1F(s) = -

JCOSh(51/2

)

'

where it is agreed that the branch cut of sm

does not lie along the negative real axis, has

isolated singularities at s = and when cosh(jl/2

) = 0, or at the points s„ = _(2n Jj n

in = 1,2, ...). The point s is a simple pole of F(s), as is seen by writing

.ycosh(.y1/2

) 4l + (j1/2

)

2/2! + (/

/2

)

4/4!+-] s + s

2 /2 + s3 I2A+-

and dividing this last denominator into 1. In fact, the residue is found to be 1; and

expression (3), Sec. 82, tells us that

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

As for the other singularities, we write

F(s) =^ where p(s) = 1 and q(s) = scosh(sm

).

Now

p(sn ) = 1*0 and q(sn ) = 0;

also, since

q\s) = ±sm sinh(51/2

) + cosh(*1/2

),

it is straightforward to show that

x (2n-l);r . ( n\ (2n-V)7t4 (*„) = -^-^-sinl nit -- I = ±J±-^L(-Vf±^

So each point sn is a simple pole of F(s), and

ResF(J)= pw = i.i=«l# CO K 2n-l

163

Consequently, according to expression (3), Sec. 82,

Res \e"F(s)] = e'"' Res F(s) =- • £-^-exp'=', L J

;r 2n -

1

Finally, then,

or

(2n-l)Vf

/(/) = Res [estF(s)] +£ Res [**F(j)1 ,

nZi 2n-l

(2n-l)Vf

(n = l,2,...).

Here we are given the function

. . _ coth(^y / 2) _ cosh(^y/2)

s2 + l ~(/ + l)sinh(^s/2)'

which has the property F(s) = F(J). We consider first the singularities at s = ±i. Upon

writing

\ u jl /• \ coshers/ 2)F(j) = where = -—

,

s-i (s + i)sinh(ns f 2)

we find that, since 0(i) = 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

the residue of e"F(s) is 0. The other singularities occur when nsl2 = nni

(/i = 0,±l,±2,...), or at the points s = 2ni (n = 0,±1,±2,...). To find the residues, we write

F(s) = £yl where p(s) = coshf—Jand q(s) = (s

2 + l)sinh[—q(s) \2 J \2

and note that

p(2ni) = cosh(nm) = cos(nn) = (-1)" * and q{2ni) = 0.

Furthermore, since

q\s) = (s2 +

1)|cosh^H

j+ 2,sinh(-0

we have

q\2ni) = {-An2 + l)-^cosh(n^i) = (-4n 2 + l)-cos(n^) =^ 2

/r(4w2 - 1)

(-1)"*0.

Thus

*=2 »<- q\2ni) n An2 -I(« = 0,±1,±2,...).

Expressions (3) and (4) in Sec. 82 now tell us that

Res[e"F(5)l = ResF(1y) =-

and

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

The desired function of t is, then,

ei2"'(

n An 2 -I

A_ cos 2nt

n An2 -\

*r*\ 2 4^cos2/if

(« = 1,2,...).

The function

F(V) -sinhjxs

1'2

)K

' J2sinh(^

1/2

)

(0<*<1),

where it is agreed that the branch cut of sm

does not lie along the negative real axis, has

isolated singularities at « = and when sinh(jl/2

) = 0, or at the points s = -n 2it

2

in = 1,2, .. .)• The point s = is a pole of order 2 of F(s), as is seen by first writing

sinh(x?1/2

) _ xsm+ (xs

U2)

31 3! + (xs

m)5/ 5! +• • • x + x 3

sl 6 +xV 1 120 +• •

s2smh(s

m) s

2[s

1'2 + {smf / 3! + (s

m)51 5! +•••]

~s2 + s

31 6TP 1 120 +•••

and dividing the series in the denominator into the series in the numerator. The result is

- =x— + -(x -x]s

sinh(xy1/2

) _ 1 13 1

2 • . / 1/2 \— X~T + ~T\X ~X)—h'

rsinhCs1 '2

) s2

6

165

(0<\s\< 7T2).

In view of expression (1), Sec. 82, then,

Res [e"F(s)] = Ux'-x) + xt = \x{x% - 1) + xt."°° 1 1 6 6

As for the singularities s - -n2n2(n = 1,2, ...), we write

F(s) =4t where = sinh(^1/2) and q(s) = s

2sinh(j

1/2).

q(s)

Observe that p(-n2n2)*0 and q{-n

2n2) = Q. Also, since

q'is) = 2s sinh(51/2

)+ ijj 1/2cosh(j

1/2

),

it is easy to see that q'(-n2n2

) * 0. So the points s = -n2^2(n = 1,2, ...), are simple poles

of and

>=-« 2jtj

£ (j)

2sinh(xs1/2

)

™1/2cosh(5

1/2

)

2 (-1)"+1

= —5- 5—sinn;cc (/i = 1,2,...).

.1-1 * «

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

2 (-1)B+1

Finally, since

Res le"F(s)] =4 • ^f-*""V'^nnx

f(t) = Rcj[e^F(s)] +±ReAyF(S)],

we arrive at the expression

fit) = ~x(x2-l) + xt —1

—

g " ***

sin6 jr - 3

2v(-ir',,v,,n

(n = l,2,...).

10. The function

F00=——

-

5 jsinhj

has isolated singularities at the points

s = and j„ = nm, s„ = -nm (n = 1,2,. . .).

166

jsinhj = 5| s + —s 3+---) = s

2 + —s4+-

Now

jsinhj = 5i s +6 ) 6

and division of this series into 1 reveals that

in 1

(0 <ld< oo),

(0<\s\<7t).

This shows that F(s) has a removable singularity at s . Evidently, then, es'F(s) must also

have a removable singularity there; and so

Res \es'F(s)] = 0.

s=s L J

To find the residue of F(s) at sn = nm (n = 1,2,. . .), we write

^0) =-— where p(.y) = sinhj-j and = ,y

2sinh.s

and observe that

p{nni) = -nm * 0, q(nm) = 0, and q'(nm) = n27t

2(-l)

n+l*0.

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

Res F(.) =4^ =-rrar =^ 0» = U,...).tf'(n;n) nV(-l)n+1

n;r

Since F(j) = F(s ), the points sn are also simple poles of F(s); and we may write

Res \estF(s)] + Res \e"F(sj\ = 2Re ( lT

ieinn = 2Re

Ml

(-1)"(icos/i7#-sin/Mtt)

= 2 sinnnt.nit

Hence the desired result is

/(0 = Res [e"F(sj\ +f (Res [e*F(j)l + Res [^'F(.y)l},

or

-smnm.

167

11. We consider here the function

„. . sinh(xj)F(s) =—

2 h . (0<x<l),s(s +co)coshs

where a > and cd*G)„ =— (n = 1,2,...). The singularities of F(j) are at

5 = 0, 5 = ±tui, and ^ = ±6)J (n = l,2,...).

Because the first term in the Maclaurin series for sinh(xs) is xs, it is easy to see that s = is

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

Res[«*F(j)l = 0.s=s 1 1

To find the residue of F(s) at s = coi, we write

cv x u jl v sinh(xy)F(5) =^^: where <t>(s) =- ;

',

s — coi s{s + cm)coshs

from which it follows that s = coi is simple pole and

t> rv \ ^/ -\ sinh(jcflw) /sin oxRes F(s) = </>(0)i) =——

7

coilcoi cosh(o)i) -2co coso

Since fjs) = F(J), then,

Res \e"F(s)]+ Res [e*F(5)l = 2Re|" jSi"^ iVl =

lj

lj L-2tf) cos© J

- sinfitt sin oar sin g»2—= sin G)f =

2fiJ2COSfl) G)

2cosg>

As for the residues at s = ©„/ (n = 1,2,. . .), we put F(s) in the form

F(s) = 2^- where p(j) = sinh(*.y) and q(s) = (s3 + G)

2s)coshs.

q(s)

Now p(fl)„i) = swh(x(Q„i) = i sin fi)„.x * and q(0)„i) = 0. Also, since

q'(s) = (s3 + 0)

2s)sinhs + (3s

2 + fiJ2)cosh s,

we find that

q'(o)J) = (-coli + Q)2G)J)smh(o)ni) = -Q)„ (m

2 - <a2)sin con * 0.

Hence we have a simple pole at s = ani, with residue

q'(o)ni) -con(a)

2 - G)2)sm G)n

168

Consequently,

Resfe*F(s)]+ Res \e"F(s)] = 2Re-G)n ((Q

2-G)

2

n )sin(Dn

= 2-smw.xsinco.t

0)n (a}2 -Q)2

„)smo)n

But sin o)„ = sin( n;r - -j j= (-l)"

+l, and this means that

Res[e"F(*)l + Resfg'T(j)1 =2^^>^*sin6)"'

j-*.! 1- > 6>_ a)

2-co

2

Finally,

/(f) = Res [e"F(s)] + (Res[e"F(*)] + Res [e"F(*)]J + J)fRes [e"F(*)] + Res [e"F(s)]\

That is,

_ sinffit*sin6tt[ 2y sina)„;csinfly

CO COSO) ^ 0)n © - a: