solutions chapter5.pdf

7/27/2019 solutions chapter5.pdf

1/33

Section 5.1 Cauchys Residue Theorem 101

Solutions to Exercises 5.1

1. Write

f(z) =1 + z

z=

1

z+ 1.

f has one simpe pole at z0 = 0. The Laurent series expansion of f(z) at z0 = 0 is already given.The residue at 0 is the coefficient of 1z in the Laurent series a

1. Thus a

1 = Res (f, 0) = 1.

5. We have one pole of order 3 at z0 = 3i.Laurent series off around z0:

z 1z+ 3i

3=

1

(z+ 3i)3[(z+ 3i) + (3i 1)]3

=1

(z+ 3i)3

(z+ 3i)3 + 3(z+ 3i)2(1 3i) + 3(z+ 3i)(1 3i) + (3i 1)3= 1 + 3

(1 3i)z+ 3i

+ 3(1 3i)(z+ 3i)2

+(3i 1)3(z+ 3i)3

Thus a1 = 3(1 3i) = Res (3i).

9. Write f(z) = csc(z) z+ 1z 1 =

1sin z

z+ 1z 1 .

Simple poles at the integers, z= k, z= 1. For k = 1,

Res(f, k) = limzk

(z k) 1sin z

z+ 1

z 1= lim

zkz+ 1

z 1 limzk(z k)sin z

=k + 1

k 1 limzk1

cos z(LHospitals rule)

=k + 1

k 11

cos k=

(1)k

k + 1

k 1 .

At z0 = 1, we have a pole of order 2. To simplify the computation of the residue, lets rewrite f(z)as follows:

1

sin z

z+ 1

z 1 =1

sin z

(z 1) + 2z 1

=1

sin z+

2

(z 1)sin zWe have

Res(f, 1) = Res(1

sin z, 1) + Res (

2

(z 1)sin z, 1);

Res(1

sin z, 1) = lim

z1(z 1) 1

sin z= 1

(Use lHospitals rule.);

Res( 2(z 1)sin z, 1) = limz1

ddz

2(z 1)sin z

= 2 limz1

sin z (z 1) cos z(sin z)2

= 2 limz1

cos z cos z+ (z 1) sin z2 sin zcos z

= 0


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102 Chapter 5 Residue Theory

So Res(f, 1) = 1 .13. The easiest way to compuet the integral is to apply Cauchys generalized formula with

f(z) =z2 + 3z 1

z2 3 ,

which is analytic inside and on C1(0). HenceC1(0)

z2 + 3z 1z(z2 3) dz= 2if(0) = 2i(

1

3) =

2i

3.

Note that from this value, we conclude that

Res(z2 + 3z 1

z(z2 3) , 0) =1

3

because the integral is equal to 2i times the residue at 0.

17. The function

f(z) =1

z(z 1)(z 2) (z 10)has simple poles at 0 and 1 inside C3

2

(0). We have

Res(f(z), 0) =1

(0 1)(0 2) (0 10) =1

10!

Res(f(z), 1) =1

1 (1 2) (1 10) = 1

9!.

Hence C3

2(0)

dz

z(z 1)(z 2) (z 10) = 2i (Res(f(z), 0) + Res (f(z), 1))

= 2i

1

10! 1

9!

= 18

10!i

21. The function

f(z) =ez

2

z6

has a pole of order 6 at 0. To compute the residue at 0, we find the coefficient a1 in the Laurentserie expansion about 0. We have

1

z6ez

2

=1

z6

n=0

(z2)n

n!.

It is clear that this expansion has no terms with odd powers ofz, positive or negative. Hence a1 = 0and so

C1(0)

ez2

z6dz = 2i Res (0) = 0.

25. Same approacha s in Exercise 21:

sin z

z6=

1

z6

k=0

(1)k z2k+1

(2k + 1)!

=1

z6

z z

3

3!+

z5

5!


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Section 5.1 Cauchys Residue Theorem 103

Coefficient of 1z : a1 =15! , so

C1(0)

sin z

z6dz = 2i Res (0) =

2i

5!.

29. (a) The Order of a pole of csc(z) = 1sin z is the order of the zero of

1

csc(z)= sin z.

Since the zeros of sin z occur at the integers and are all simple zeros (see Example 1, Section 4.6),it follows that csc z has simple poles at the integers.(b) For an integer k,

Res

csc z, k

= limzk

(z k)csc z = limzk

z ksin z

= limzk

1

cos z(lHospitals rule)

=(1)k

.

(c) Suppose that f is analytic at an integer k. Apply Proposition 1(iii), then

Res

f(z)csc(z), k

=(1)k

f(k).

33. A Laurent series converges absolutely in its annulus of convergence. Thus to multiply twoLaurent series, we can use Cauchy products and sum the terms in any order. Write

f(z)g(z) =

n=

an(z z0)n

n=bn(z z0)n

=

n=

cn

(z

z0

)n,

where cn s obtained by collecting all the terms in (z z0)n, after expanding the product. Thus

cn =

j=

ajbnj;

in particular

c1 =

j=ajb1j,

and hence

Res f(z)g(z), z0 =

j=aj bj1.

37. (a) We have, Exercise 35(a), Section 4.5,

J0(z) =1

2i

C1(0)

ez2

1

)

d

.


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Thus 0

J0(t)est dt =

1

2i

0

C1(0)

et

s 12 ( 1 )

d

dt

(b) For on C1(0), we have

1

= = 2i Im (),

which is 0 if Im() = 0 (i.e., = 1) or is purely imaginary. In any case, for all C1(0), and allreal s > 0 and t, we have ets 12 ( 1 )

= ets e t2 ( 1 ) = ets.So, by the inequality on integrals (Th.2, Sec. 3.2), 12i

C1(0)

et

s12 ( 1 )

d

22 maxC1(0)ets 12 ( 1 ) 1

= max

C1(0)

et

s12 ( 1 )

(|| = 1)

ets

Thus the iterated integral in (a) is absolutely convergent because 12i

0

C1(0)

et

s12 ( 1 )

d

dt

0

C1(0)et

s12 ( 1 )

d

dt

0

ets dt =1

s<

(c) Interchange the order of integration, and evaluate the integral in t, and get

0

J0(t)est dt =

1

2i C1(0)

0

ets12 ( 1 ) dt

d

=1

2i

C1(0)

1s 1

2 ( 1)ets 12 ( 1 )

0

d

=1

i

C1(0)

1

2 + 2s+ 1 d

because, as t , ets 12 ( 1 ) = ets 0.

(d) We evaluate the integral using the residue theorem. We have simple poles at

= s

s2

+ 11 = s s2 + 1.

Only s s2 + 1 is inside C0(1). To see this, note that because s > 0 and

s2 + 1 > 1, we haves +

s2 + 1 > 1. Also

s 0, 0

J0(t)est dt =

2i

i

1

2

s2 + 1=

1s2 + 1

.

(e) Repeat the above steps making the appropriate changes.Step (a): Use the integral representation ofJn and get

In =

0

Jn(t)est dt =

1

2i

0

C1(0)

et

s12 ( 1 )

d

n+1dt

Step (b) is exactly like (b) because, for on C1(0), we have |n+1

| = || = 1.Step (c): As in (c), we obtain

In =1

i

C1(0)

1

(2 + 2s+ 1)n d.

Let = 1 = . Then d = 12 d. As runs through C1(0) in the positive direction, runs throughC1(0) in the negative direction. Hence

In =1

i

C1(0)

12 d( 12 + 2s 1 + 1) 1n

=1

i

C1(0)

nd

2 + 2s 1 .

Step (d): We evaluate the integral using the residue theorem. We have simple poles at

=s s2 + 1

1 = s

s2 + 1.

Only s + s2 + 1 is inside C0(1). (Just arge as we did in (d).) By Proposition 1(ii)

Res

n

2 + 2s 1 , s +

s2 + 1

=

n

2 + 2s

=s+s2+1

=

s2 + 1 s

n

2

s2 + 1

Thus, for s > 0, 0

Jn(t)est dt =

1s2 + 1

s2 + 1 s

n.


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1. Let z = ei, dz = ieid, d = iz dz, cos =z+1/z

2 . Then20

d

2 cos =

C1(0)

iz dz2 (z+1/z)2

= iC1(0)

dz2z z2

2 1

2

= i

C1(0)

dz

z22 + 2z 12= 2

j

Res

z

2

2+ 2z 1

2, zj

,

where the sum of the residues extends over all the poles of 12z z22 12

inside the unit disk. We have

z2

2+ 2z 1

2= 0 z2 4z+ 1 = 0.

The roots are z = 2 3, and only z1 = 2 3 is inside C1(0). We compute the residue usingProposition 1(ii), Sec. 5.1:

Res

z

2

2+ 2z 1

2, z1

=

1

z1 + 2 =1

3.

Hence 20

d

2 cos =2

3.


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Section 5.2 Definite Integrals of Trigonometric Functions 107

5. Let z = ei, dz= ieid, d = i dzz , cos =z+1/z

2 , and cos2 =e2i+e2i

2 =z2+ 1

z2

2 . Then20

cos2

5 + 4 cos d = i

C1(0)

12 (z

2 + 1z2

5 + 4 (z+1/z)2

dz

z

= i2 C1(0)

z4 + 1

5z2 + 2z3 + 2z

dz

z

= i2

C1(0)

z4 + 1

z2(2z2 + 5z+ 2)dz

= i2

2i

j

Res

z4 + 1

z2(2z2 + 5z+ 2), zj

=

j

Res

z4 + 1

z2(2z2 + 5z+ 2), zj

,

where the sum of the residues extends over all the poles of z4+1

z2(2z2+5z+2)inside the unit disk. We

have a pole of order 2 at 0 and possible more poles at the roots of 2 z2 + 5z+ 2. Lets compute theresidue at 0.

Res

z4 + 1

z2(2z2 + 5z+ 2), 0

= limz0

d

dz

z4 + 1

(2z2 + 5z+ 2)

=4z3(2z2 + 5z+ 2) (z4 + 1)(4z+ 5)

(2z2 + 5z+ 2)2

z=0

= 54

.

For the nonzero poles, solve2z2 + 5z+ 2 = 0.

The roots are z = 534 . Only z1 = 12 is inside C1(0). We compute the residue using Proposition1(ii), Sec. 5.1:

Res z4 + 1

z2(2z2 + 5z+ 2), z1 =

z41 + 1

z2

1

1d

dz(2z2 + 5z+ 2)z1

=( 12 )

4 + 114 (4(12 ) + 5)

=17

12.

Hence 20

cos2

5 + 4 cos d =

17

12 5

4

=

6.


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9. . Let z = ei, dz= ieid, d = i dzz

, cos = z+1/z2 , sin =z1/z

2i. Then2

0

d

7 + 2 cos + 3 sin = i

C1(0)

dz

z(7 + (z+ 1/z) + 32i z2 32i

= i

C1(0)

dz

(1 32

i)z2 + 7z+ (1 + 32

i)

= 2

j

Res 1

(1 32

i)z2 + 7z+ (1 + 32

i), zj

,

where the sum of the residues extends over all the poles of 1(132 i)z2+7z+(1+ 32 i)

inside the unit disk.

Solve

(1 32

i)z2 + 7z+ (1 +3

2i) = 0.

Youll find

z =7

49 4(1 3

2i)(1 + 3

2i)

2(1 32 i)=

7

49 4(1 + 94

)

2(1 32 i)

= 7 362 3i = 7 62 3i

=13

2 3i or1

2 3i=

13(2 + 3i)13

= (2 + 3i) or (2 + 3i)13

.

We have

|2 + 3i| =

13 > 1 and

(2 + 3i)13 =

13

13< 1.

So only z1 =(2+3i)

13 is inside C1(0). We compute the residue using Proposition 1(ii), Sec. 5.1:

Res(z1) =

1

2(1 32 i)z1 + 7=

1

2(1 32 i)(2+3i)13 + 7=

1

6.

Hence 20

d

7 + 2 cos + 3 sin = 2

1

6=

3.


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Section 5.2 Definite Integrals of Trigonometric Functions 109

13. The solution will vary a little from what is in the text. Note the trick based on periodicity.Step 1: Double angle formula

a + b cos2 = a + b

1 + cos 2

2

=

2a + b + b cos2

2,

so1

a + b cos2 =

2

2a + b + b cos2.

Step 2. Change variables in the integral: t = 2, dt = 2d. Then

I =

20

d

a + b cos2 =

40

dt

2a + b + b cos t.

The function f(t) = 12a+b+b cos t is 2-periodic. Hence its integral over intervals of length 2 areequal. So

I =

20

dt

2a + b + b cos t+

42

dt

2a + b + b cos t= 2

20

dt

2a + b + b cos t.

Step 4. Now use the method of Section 5.2 to evaluate the last integral. Let z = eit, dz = ieitdt,

dt = i dzz , cos t =z+1/z

2 . Then

2

20

dt

2a + b + b cos t= 4i

C1(0)

dz

bz2 + (4a + 2b)z+ b

= 8

j

Res

1

bz2 + (4a + 2b)z+ b, zj

,

where the sum of the residues extends over all the poles of 1bz2+(4a+2b)z+b inside the unit disk. Solve

bz2 + (4a + 2b)z+ b = 0,

and get

z =

(4a + 2b)

(4a + 2b)2

4b2

2b=

(2a + b)

2a(a + b)b

= z1 =(2a + b) + 2

a(a + b)

bor z2 =

(2a + b) 2

a(a + b)

b.

It is not hard to prove that |z1| < 1 and |z2| > 1. Indeed, for z2, we have

|z2| = 2a + bb

+2

a(a + b)

b= 1 +

2a

b+

2

a(a + b)

b> 1

because a, b are > 0. Now the product of the roots of a quadratic equation z2 + z+ = 0 ( = 0)is always equal to . Applying this in our case, we find that z1 z2 = 1, and since |z2| > 1, we musthave |z1| < 1.

We compute the residue at z1 using Proposition 1(ii), Sec. 5.1:

Res 1bz2 + (4a + 2b)z+ b

, z1 = 1

2bz1 + (4a + 2b)

=1

(2a + b)2 + 4

a(a + b) + (4a + 2b)

=1

4

a(a + b).


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Hence 20

d

a + cos2 = 2

1

4

a(a + b)=

2a(a + b)

.


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Section 5.3 Improper Integrals Involving Rational and Exponential Functions 111


1. Use the same contour as Example 1. Several steps in the solution are very similar to thosein Example 1; in particular, Steps 1, 2, and 4. Following the notation of Example 1, we haveIR = I[R, R] + IR . Also, limR IR = 0, and

limR

I[

R, R] = I =

dx

x4

+ 1

.

These assertions are proved in Example 1 and will not be repeated here. So all we need to do isevaluate IR for large values of R and then let R . We have

IR = 2i

j

Res

1

z4 + 1, zj

,

where the sum ranges over all the residues of 1z4+1 in the upper half-plane. The function1

z4+1 have

four (simple) poles. These are the roots of z4 + 1 = 0 or z4 = 1. Using the result of Example 1,we find the roots to be

z1 =1 + i

2, z2 =

1 + i

2, z3 =

1 i

2, z4 =

1 i

2.

In exponential form,

z1 = ei4 , z2 = e

i 34 , z3 = ei 54 , z4 = e

i 74 .

Only z1 and z2 are in the upper half-plane, and so inside R for large R > 0. Using Prop.1(ii),Section 5.1,

Res

1

z4 + 1

,z1

=

1d

dz z4 + 1

z=z1

=1

4z31=

1

4ei

34 ;

Res

1

z4 + 1

,z2

=

1d

dz z4 + 1z=z2

=1

4z32=

1

4ei

94 =

1

4ei

4 .

So

IR = 2i1

4

ei

34 + ei

4

=

i

2

cos

3

4 i sin 3

4+ cos

4 i sin

4

=

i

2(i

2) =

2

.

Letting R , we obtain I = 2

.


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5. The integral converges absolutely, as in Step 1 of Example 1. We will reason as in this example,and omit some of the details. Here IR = I[R, R] + IR ; limR IR = 0, and

limR

I[R, R] = I =

dx

(x2 + 1)3= .

All we need to do is evaluate IR for large values of R and then let R

. The function f(z) =

1(z2+1)3 has poles of order 3 at z1 = i and z2 = i, but only z1 is in the upper half-plane. You canevaluate the integral IR using the residue theorem; however, an equal good and perhaps faster wayin this case is to use Cauchys generalized integral formula (Sec. 3.6). Write

IR =

R

1

(z2 + 1)3dz =

R

1

[(z+ i)(z i)]3 dz =

R

1

(z+ i)3dz

(z i)3 .

Let g(z) = 1(z+i)3 . According to Theorem 2, Sec. 3.6,

IR = 2ig(i)

2!.

Compute:

g(z) = 3(z+ i)4, g(z) = 12(z+ i)5,so

g(i) = 12(2i)5 =12

25(i).

Finally,

IR = 12

25=

3

8.

Letting R , we obtain I = 38 .9. Proceed as in Example 3, as follows. Let x = et, dx = et dt. Then

0

dx

x3

+ 1

=

et

e3t

+ 1

dt =

ex

e3x

+ 1

dx,

where in the last integral we reverted to the variable x for convenience. Use a contour as in Fig.6.Refer to Example 3 for notation. As in Example 3, limR |I2| = 0 and limR |I4| = 0. Let usrepeat the proof for I4 here. For I4, z is on 4, so z = R + iy, where 0 y 23 . Hence

|I4| =

4

ez

e3z + 1dz

l(I4) M,where l(I4) =

23

= length of vertical side 4, and M is the maximum value of eze3z+1 on 4. For z

on 4,

ez

e3z + 1 = eR+iy

e3(R+iy) + 1 eR

1

e3R

,

which tends to 0 as R . To justify the last inequality, note that by the reverse triangle inequality:

e3(R+iy) + 1 1 e3R=1 e3iy

= 1 e3R,


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so1e3(R+iy) + 1 11 e3R ,

as desired.

As in Example 3, I1 I as R . For I3, we have z = x + 2i3 , x varies from R to R,dz = dx, so

I3 =

RR

ex+2i3

e3x+2i + 1dx

= e 2i3RR

ex

e3x + 1dx = e 2i3 I1.

Now

e3z + 1 = 0 e3z = 1 = ei 3z = i + 2ki, (k = 0, 1, 2, . . .) z = i

3(2k + 1).

Only one root, z = i3 , is inside R, and so the functionez

e3z+1 has one pole at z =i3 inside R.

Hence (see Example 3 for a justification)

1 e 2i3

I = 2i Res

ez

e3z + 1,

i

3

= 2iei3

ddz (e

3z + 1)

z= i3

= 2iei3

3e3i3

= 2iei3

3 = 2iei3

3.

Solving for I,

I = 23

iei3 1

1 e 2i3 =2

3i 1

ei3 ei3

,

where we have multiplied numerator and denominator by ei3 . Now e

i3 ei3 = 2i sin 3 . So

I =

3sin 3=

3

32

=2

3

3.

13. Use the same reasoning as in Example 3. Let x = et, dx = et dt. Then

I =

0

x

x3 + 1dx =

et2

e3t + 1etdt =

e3x2

e3x + 1dx.

Use a contour as in Fig.6. Refer to Example 3 for notation: limR |I2| = 0, limR |I4| = 0, andlimR |I1| = I, the desired integral. For I3, we have z = x + 2i3 , x varies from R to R, dz = dx,so

I3 =

RR

e32 x+

322i3

e3x+2i + 1dx =

RR

e32xei

e3x + 1dx =

RR

e32x

e3x + 1dx = I1.


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Same poles as in Exercise 9. So

2I = 2i Res

e32z

e3z + 1,

i

3

= 2ie32i3

ddz (e

3z + 1)z= i3= 2i

ei2

3ei=

2i(i)

3 =2

3;

I =

3

17. As in Example 4: Let 2x = et, 2dx = et dt. Then

I =

0

ln(2x)

x2 + 4dx =

1

2

te2t

4+ 4

etdt = 2

xex

e2x + 16dx.

Use a rectangular contour as in Fig.6, whose vertical sides have length . Refer to Example 3 for

notation: limR |I2| = 0, limR |I4| = 0, and limR |I1| = I, the desired integral. For I3,z = x + i, x varies from R to R, dz= dx, so

I3 = 2

RR

(x + i)ex+i

e2x+2i + 16dx

= 2RR

(x + i)ex(1)e2x + 16

dx

= 2

RR

xex

e2x + 16dx +

=BRi 2i

RR

ex

e2x + 16dx = I1 + BRi,

where BR is a real constant, because the integrand is real-valued. We have

IR = I1 + I2 + I3 + I4 = 2I1 + I2 + I4 + iBR.

Letting R , we getlim

RIR = 2I + iB,

where

B = limR

BR =

ex

e2x + 16dx.

At the same time

IR = 2i Res 2zez

e2z + 16, z0

,

where z0 is the root ofe2z + 16 = 0 that lies inside R (there is only one root, as you will see):

e2z + 16 = 0 e2z = 16ei = eln(16)+i 2z= 4 ln2 + i + 2ki z = 2ln2 + i

2(2k + 1).


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Only z = 2 ln2 + i 2 is inside the contour. So

IR = 2i Res

2zez

e2z + 16, 2 ln2 + i

2

=2(2 ln2 + i 2 )e

2 ln 2+i2

2e2(2 ln2+i2 )

=

4 (4ln 2 + i)

Thus

2I+ 2iB = ln2 + i2

4.

Taking real and imaginary parts, we find

I = ln 2

2and B =

8.

This gives the value of the desired integral I and also of the integral

8

= B =

ex

e2x

+ 16

dx.

21. We use the contour R in Figure 9 and follow the hint. Write IR = I1 + I2 + I3. On 1, z = x,dz = dx,

1

1

z3 + 1dz =

R0

1

x3 + 1dx I =

0

1

x3 + 1dx, as R .

On 3, z= e23 ix, dz = e

23 idx, x varies from R to 0:

3

1

z3 + 1dz = e

23 i

0R

1

e323 ix3 + 1

dx

=

e23 i

R

0

1

x

3

+ 1

dx

e23 iI as R

.

For I2, we have z = Reit, 0 t 23 :

2

1

z3 + 1dz

l(2) maxz on 2 1z3 + 1

2

3R 1

R3 1 0as R .

Now

I1 + I2 + I3 = IR = 2i Res

1

z3 + 1, z0

,

where z0 = ei3 is the only pole of 1z3+1 inside R. By Proposition 1(ii), Sec. 5.1,

Res

1

z3 + 1, ei

3

=

1

3ei23

=ei

23

3.

So

I1 + I2 + I3 = 2iei

23

3.


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Letting R , we get

I e 23 iI = 2i ei 23

3 (1 e 23 i)I = 2i e

i 23

3

I = 2i3

ei23

(1 e 23 i) e

3 i

e3 i

I =2i

3

e

i

(e3 i e3 i) =

3sin 3

I = 23

3.


1. Use a contour as in Fig. 2 and proceed exactly as in Example 1, replacing s by 4. No need torepeat the details.

5. The degree of the denominator is 2 more than the degree of the numerator; so we can use thecontour in Fig. 2 and proceed as in Example 1.Step 1: The integral is absolutely convergent.

x2 cos2x(x2 + 1)2 x2

(x2 + 1)2 1

x2 + 1 ,

becausex2

(x2 + 1)2=

(x2 + 1) 1(x2 + 1)2

=1

x2 + 1 1

(x2 + 1)2 1

x2 + 1.

Since

1x2+1 dx < (you can actually compute the integral = ), we conclude that our integral

is absolutely convergent.Step 2:

x2 cos2x

(x2 + 1)2dx =

x2 cos2x

(x2 + 1)2dx + i

x2 sin2x

(x2 + 1)2dx =

x2e2ix

(x2 + 1)2dx

because

x2 sin2x

(x2 + 1)2 dx = 0,

being the integral of an odd function over a symmetric interval.Step 3: Let R and R be as in Fig. 2. We will show that

R

z2e2iz

(z2 + 1)2dz 0 as R .

|IR | =

R

z2e2iz

(z2 + 1)2dz

l(R) maxz on R z2e2iz(z2 + 1)2

= R M.As in (7) in Sec. 5.4, for z on R, e2iz e2R sin 1.So z2e2iz(z2 + 1)2 z2(z2 + 1)2 z2 + 1 1(z2 + 1)2

1z2 + 1

+ 1(z2 + 1)2

1

R2 1 +1

(R2 1)2 .


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18/33


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20/33


25. (a) For w > 0, consider

J =

C

eiwz

e2z 1 dz,

where C is the indented contour in Figure 14. The integrand is analytic inside and on C. So

(1) 0 = J = Ceiwz

e2z

1

dz = I1 + I2 + I3 + I4 + I6 + I6,

where Ij is the integral over the jth component of C, starting with the line segement [, R] andmoving around C counterclockwise. As 0+ and R , I1 I, the desired integral.

For I3, z = x + i, where x varies from R to :

I3 =

R

eiw(x+i)

e2(x+i) 1 dx = ew

R

eiwx

e2x 1 dx.

As 0+ and R , I3 ew I.For I2, z = R + iy, where y varies from 0 to 1:

|I3| 1 maxz=R+iy

0

y

1

eiwz

e2z 1

;

eiwze2z 1 =

eiw(R+iy)e2(R+iy) 1

=

eiwRe2R 1 0, as R .

So I2 0, as R .For I5, z = iy, where y varies from 1 to :

I5 =

1

eiw(iy)

e2(iy) 1 i dy= i1

ewy

e2iy 1 dy.

For I4, the integral over the quarter circe from (, + i) to (0, i i), we apply Corollary 2, tocompute the limit

lim0

4

eiwz

e2z 1dz = i

2Res

eiwz

e2z 1 , i

= i 2

eiw(i)

2e2(i)= i e

w

4.

For I6, the integral over the quarter circe from (0 , i) to (, 0), we apply Corollary 2, to computethe limit

lim0

6

eiwz

e2z 1 dz =

2Res

eiwz

e2z 1 , 0

= 2

i eiw(0)

2e2(0)= i

4.

Plug these findings in (1) and take the limit as R then as 0, and get

lim0

eiwx

e2x 1 dx ew

eiwx

e2x 1 dx i1

ewy

e2iy 1 dy

i ew

4 i

4= 0


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Section 5.4 Improper Integrals of Products of Rational and Trigonometric Functions 121

or

lim0

eiwx

e2x 1 dx ew

eiwx

e2x 1 dx i1

ewy

e2iy 1 dy

= i(ew

4+

1

4).

(Note: The limits of each individual improper integral does not exist. But the limit of the sum,as shown above does exist. So, we must work wth the limit of the three terms together.) Takeimaginary parts on both sides:

(2) lim0

Im

eiwx

e2x 1 dx ew

eiwx

e2x 1 dx i1

ewy

e2iy 1 dy

=ew + 1

4.

Now

Im

eiwx

e2x 1dx ew

eiwx

e2x 1 dx i1

ewy

e2iy 1 dy

=

sin wx

e2x 1(1 ew) dx +

1

Im

(i)ewye2iy 1

dy;(3)

Im (i)ewy

e2iy 1 = Re (i)ewy

e2iy 1= Re

ewy

e2iy 1

= ewy Re

1

e2iy 1

= ewy Re

eiy

eiy eiy

= ewy Re

cos y i sin y2i sin y

=

ewy

2.

Plugging this in (3) and using (2), we get

lim0

sin wx

e2x

1

(1 ew) dx +

1

ewy

2dy

=

ew + 14

.

Evaluate the second integral and get

lim0

sin wx

e2x 1 (1 ew) dx +

ew(1) ew2(w)

=

ew + 14

.

lim0

sin wx

e2x 1 (1 ew) dx + lim

0ew(1) ew

2(w) =ew + 1

4.

(1 ew) lim0

sin wx

e2x 1 dx +1 ew

2w=

ew + 14

.

So

(1 ew) lim0

sin wxe2x 1 dx =

ew + 14

1 ew2w

;

lim0

sin wx

e2x 1 dx =1

1 ew

ew + 14

1 ew

2w

0

sin wx

e2x 1 dx =12w

+1

4

ew + 1

ew 1 .


22/33


(b) From (10), Section 4.4 (recall B0 = 1):

zcoth z =

n=0

22nB2n

(2n)!z2n, |z| <

z

2coth

z

2=

n=022n

B2n(2n)!

(z

2)2n, |z

2| <

z

2

ez2 + e

z2

ez2 e z2 =

n=0

B2n(2n)!

z2n, |z| < 2

1

2

ez2 + e

z2

ez2 e z2 =

B0z

+

n=1

B2n(2n)!

z2n1, |z| < 2

1

2

ez2 + e

z2

ez2 e z2

B0z

=

n=1

B2n(2n)!

z2n1, |z| < 2

1

4

ez + 1

ez 1 1

2z=

1

2

n=1

B2n(2n)!

z2n1, |z| < 2

(c) Replace sin wx in the integral in (a) by its Taylor series

sin wx =

k=1

(1)k1 w2k1x2k1

(2k 1)! ,

ue (b), and interchange order of integration, and get0

1

e2x 1

n=0

(1)n (wx)2n+1

(2n + 1)!dx =

1

2

n=1

B2n(2n)!

w2n1

n=0

(1)n(2n + 1)!

w2n+1

0

x2n+1

e2x 1 dx =1

2

n=1

B2n(2n)!

w2n1

n=1

(

1)n1

(2n 1)! w2n

1

0

x2n1

e2x 1 dx =1

2

n=1

B2n

(2n)!w2n

1

.

Comparing the coefficients ofw, we obtain0

x2n1

e2x 1 dx =(1)n1

4nB2n (n = 1, 2, . . .).


23/33

Section 5.5 Advanced Integrals by Residues 123


1. As in Example 1, take w > 0 (the integral is an even fnction of w),

12

ex2

2 cos wxdx =12

ex2

2 eiwx dx (because

ex2

2 sin wxdx = 0)

=1

2

e

12 (xiw)2 12 w2 dx

= e12w

2 12

=J

e12 (xiw)2 dx =

e12 w

2

2

J.

To evaluate J, consider the integral

I =

R

e12 (ziw)2 dz,

where R is a rectangualr contour as in Fig. 1, with length of the vertical sides equal to w. ByCauchys theorem, I = 0 for all R.

Let Ij denote the integral on j (see Example 1). Using the estimate in Example 1, we see that

I2 and I4 tend to 0 as R . On 3, z = z+ iw, where x varies from R ro R:RR

e12 (x+iwiw)2 dx =

RR

e12 x

2

dx,

and this tends to 2 as R , by (1), Sec. 5.4. Since I = 0 for all R, it follows that

J = limR

I1 = limR

I3 =

2.

Consequently,12

ex2

2 cos wxdx = ew2

2

for all w > 0. Since the integral is even inw, the formula holds for w < 0. For w = 0 the formulafollows from (1), Sec. 5.5.


24/33


5. . (a) Let f(x) = cos2x, then f(x) = 2sin2x and f(x) = 4cos2x. If x (0, 4 ), thenf(x) < 0 and the graph concaves down. So any chord joining two points on the graph ofy = cos 2xabove the interval (0,

4) lies under the graph of y = cos 2x. take the two points on the graph, (0, 1)

and ( 4 , 0). The equation of the line joining them is y = 4 x + 1. Since it is under the graph ofy= cos 2x for x (0, 4 ), we obtain

4

x + 1

cos2x for 0

x

4

(b) Let Ij denote the integral ofez2 over the path j in Fig. 13. Since ez

2

is entre, by Cauchystheorem, I1 + I2 + I3 = 0.(c) For I1, z = x,

I1 =

R0

ex2

dx

0

ex2

dx =

2, as R ,

by (1), Sec. 5.5. On 2, z = Rei z2 = R2(cos 2 + i sin2),ez2 = eR2(cos 2+i sin 2) = eR2 cos2 eR2(1 4 ).

Parametrize the integral I2 and estimate:

|I2| =

4

0

eR2e2iRieid

R

4

0

eR2e2ieid R40

eR2(1 4

)d

ReR2

4

0

eR2 4

d

= ReR2 1

R2

4eR

2 4

4

0

=

4ReR

2

eR2 1

,

which tends to 0 as R .(d) On 3, z = xei

4 , where x varies from R to 0, dz = ei

4 dx. So

I3 = ei 4R

0

ex2ei

2 dx = ei 4

R0

ex2idx

= ei 4R

0

(cos x2 i sin x2)dx.

As R , I3 converges toei 4

0

(cos x2 i sin x2)dx.

(e) Let R in the sum I1 + I2 + I3 = 0 and get

2 ei

4

0 (cos x

2 i sin x2)dx = 0;ei4

0(cos x2 i sin x2)dx =

2;

0 (cos x2 i sin x2)dx =

2 ei 4 ;

0(cos x2 i sin x2)dx =

2

2

2 i

22

.

The desired result follows upon taking real and imaginary parts.


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27/33

Section 5.5 Advanced Integrals by Residues 127

(d) If |a| < b, take t = bx and w = a, then

P.V.

eax

sinh bxdx =

btan

a

2b.

(e) Replace a by a in (d) and get

P.V.

eax

sinh bx dx = b tan a2b .

Subtract from (d) and divide by 2:

sinh ax

sinh bxdx =

btan

a

2b(b > |a|).

Note that the integral is convergent so there is no need to use the principal value.

17. We use a contour like th eone in Fig. 17: Let1 denote the small circular path around 0 (negative direction);2 the line segment from r to R, above the x-axis (positive direction);3 the large circular path around 0 (positive direction);

4 the line segment from R to r, below the x-axis (negative direction). We integrate the function

f(z) =

z

z2 + z+ 1

on the contour , where

z = e12 log0 z (branch cut along positive x-axis). By the residue theorem,

f(z) dz = 2i

j

Res(f, zj ) 4

j=1

Ij = 2i

j

Res(f, zj ),

where the sum is over all the residues of f in the region inside . The poles of f in this region areat the roots ofz2 + z+ 1 = 0 or

z = 1 32 ; z1 = 12 + i

32 , z2 = 12 i

32 .

We have

|z1| =

(12 )2 + (

3

2 )2 = 1, z1 = ei

23

log0(z1) = ln |z1| + i arg 0(z1) = 0 + i 23 = i 23Res (z1) =

z1

2z1+1= e

12log0(z1)

2z1+1

Res (z1) =ei3

2ei23 +1

.

Similarly,

|z2| = (12 )2 + (32 )2 = 1, z2 = ei 43log0(z2) = ln |z2| + i arg 0(z2) = 0 + i 43 = i 43

Res (z2) =

z22z2+1

= e12log0(z2)

2z2+1

Res (z2) =ei

23

2ei43 +1

.


28/33


29/33


30/33


31/33

Section 5.7 The Counting Theorem and Rouche Theorem 131


1. The roots of the polynomial are easy to find using the quadratic formula:

z2 + 2z+ 2 = 0 z = 1 i.

Thus no roots are in the first quadrant. This, of course, does not answer the exercise. We must

arrive at this answer using the method of Example 1, with the help of the argument principle.First, we must argue that f has no roots of the positive x axis. This is clear, because if x > 0then x2 + x + 2 > 2 and so it cannot possibly be equal to 0. Second, we must argue that there areno roots on the upper imaginary axis. f(iy) = y2 + 2y+ 2 = 2 y2 + 2y. If y = 0, f(0) = 2 = 0.If y > 0, then Im(f(y)) = 2y= 0. In all cases, f(iy) = 0 ify 0.

The number of zeros of the polynomial f(z) = z2 + 2z+ 2 is equal to the number of times theimage ofR wraps around the origin, where R is the circular path in the first quadrant, in Fig. 4.This path consists of the interval [0, R], the circular arc R, and the interval on the imaginary axisfrom iR to 0. To find the image on R, we consider the image of each component separately.

Since f(x) is real for real x, we conclude that the image of the interval [0, R] is also an interval,and it is easy to see that this interval is [2 , R2 + 2R + 2]. So its initial point is w0 = 2 and itsterminal point is w1 = R2 + 2R + 2.

The image of the arc R starts at the point w1 = R2+2R+2 and ends at f(iR) =

R2+2iR+2 =

w2, which is the image of the terminal point ofR. We have Im(w2)) = 2R and Re(w2) = 2R2 < 0if R is very large. Hence the point f(w2) is in the second quadrant. Also, for very large R, and|z| = R, the mapping z f(z) is approximately like z z2. So f(z) takes R and maps itapproximately to the semi-circle (the map w = z2 doubles the angles), with initial point w1 andterminal point w2.

We now come to the third part of the image ofR. We know that it starts at w2 and end at w0.As this image path go from w2 to w0, does it wrap around zero or not? To answer this question, weconsider f(iy) = 2 R2 + 2iy. Since Im(f(iy) > 0 if y > 0, we conclude that the image point of iyremains in the upper half-plane as it moves from w2 to w0. Consequently, the image curve does notwrap around 0; and hence the polynomial has no roots in the first quadrant-as expected.

5. Follow the steps in the solution of Exercise 1, but here the roots of z4 + 8z2 + 16z+ 20 = 0 arenot so easy to find, so we will not give them.

Argue that f has no roots of the positive x axis. This is clear, because if x > 0 thenx4 + 8x2 + 16z+ 20 > 20 and so it cannot possibly be equal to 0. Second, we must argue that thereare no roots on the upper imaginary axis. f(iy) = y4 8y2 + 16iy+ 20 = y4 8y2 + 20 8iy. Ify= 0, f(0) = 20 = 0. If y > 0, then Im(f(y)) = 8y= 0. In all cases, f(iy) = 0 ify 0.

The number of zeros of the polynomial f(z) = z4 + 8z2 + 16z+ 20 is equal to the number oftimes the image of R wraps around the origin, where R is the circular path in the first quadrant,in Fig. 4. This path consists of the interval [0, R], the circular arc R, and the interval on theimaginary axis from iR to 0. To find the image on R, we consider the image of each componentseparately.

Since f(x) is real for real x, we conclude that the image of the interval [0, R] is also an interval,and it is easy to see that this interval is [20 , R4 + 8R2 + 16R + 20]. So its initial point is w0 = 2and its terminal point is w1 = R4 + 8R2 + 16R + 20.

The image of the arc R starts at the point w1 = R4 + 8R2 + 16R + 20 on the real axis and ends

at f(iR) = R4 8R2 + 20 + 16iR = w2, which is the image of the terminal point of R. We haveIm (w2)) = 16R > 0 and Re(w2) = R4 8R2 + 20 > 0 if R is very large. Hence the point f(w2) isin the first quadrant. Also, for very large R, and |z| = R, the mapping z f(z) is approximatelylike z z2. So f(z) takes R and maps it approximately to a circle (the map w = z4 multipliesangles by 4), with initial point w1 and terminal point w2. So far, the image of [0, R] and R wrapsone around the origin.


32/33


We now come to the third part of the image of R. We know that it starts at w2 and end atw0. As this image path go from w2 to w0, does it close the loop around 0 or does it unwrap it? Toanswer this question, we consider f(iy) = R4 8R2 +20+16iy. Since Im(f(iy) = 16y > 0 ify > 0,we conclude that the image point of iy remains in the upper half-plane as it moves from w2 to w0.Consequently, the image curve wraps around 0; and hence the polynomial has one root in the firstquadrant.

9. Apply Rouches theorem with f(z) = 11, g(z) = 7z3

+ 3z2

. On |z| = 1, |f(z)| = 11 and|g(z)| 7 + 3 = 10. Since |f| > |g| on |z| = 1, we conclude that N(f) = N(f + g) inside C1(0).Since N(f) = 0 we conclude that the polynomial 7z3 + 3z2 + 11 has no roots in the unit disk.

13. Apply Rouches theorem with f(z) = 3z, g(z) = ez . On |z| = 1, |f(z)| = 3 and|g(z)| =

ecos t+i sin t = ecos t e.Since |f| > |g| on |z| = 1, we conclude that N(f) = N(f + g) inside C1(0). Since N(f) = 1 weconclude that the function ez 3z has one root in the unit disk.17. Apply Rouches theorem with f(z) = 5, g(z) = z5 + 3z. On |z| = 1, |f(z)| = 5 and |g(z)| 4.Since |f| > |g| on |z| = 1, we conclude that N(f) = N(f + g) inside C1(0). Since N(f) = 0 weconclude that the function z5 + 3z+ 5 has no roots in the unit disk. Hence 1z5+3z+5 is analytic inthe unit disk and by Cauchys theorem

C1(0)

dz

z5 + 3z+ 5= 0.

21. (a) Let g(z) = z and f(z) = zn 1. Then

1

2i

CR(0)

znzn1

zn 1 dz =1

2i

CR(0)

g(z)f(z)f(z)

dz=

nj=1

g(zj ),

where zj are the roots of f(z). These are precisely the n roots of unity. Hencen

j=1 g(zj ) =nj=1 zj = S. Let =

1z

, d = 1z2

dz, as z runs through CR(0) in the positive direction, runsthrough C1/R(0) in the negative direction. So

1

2i

CR(0)

znzn1

zn 1 dz =1

2i

C1

R(0)

n

n( 1n 1)d

2 =n

2i

C1

R(0)

d

(1 n).

(b) Evaluate the second integral in (a) using Cauchys generalized integral formula and concludethat

n

2i

C1

R(0)

d

(1 n) =d

d

1

1 n

=0

=nn1

(1 n)2

=0

= 0.

Using (a), we find S = 0.A different way to evaluate S is as follows: From (a),

S =n

2i CR(0)zn

zn

1

dz

=n

2i

CR(0)

zn 1zn 1 dz+

n

2i

CR(0)

1

zn 1 dz

=n

2i

CR(0)

1 dz+ n2i

CR(0)

1

zn 1 dz= 0 + 0 = 0.


33/33

Section 5.7 The Counting Theorem and Rouche Theorem 133

The first integral is 0 by Cauchys theorem. The second integral is zero because CR(0) contains allthe roots ofp(z) = zn 1 (see Exercise 38, Sec. 3.4).25. Suppose that f is not identically 0 in , and let Br (z0) be a disk such that f(z) = 0 forall z on Cr(z0). Since |f| is continuous, it attains its maximum and minimum on Cr(z0). So thereis a point z1 on Cr(z0) such that |f(z1)| = m = min |f| on Cr(z0). Since |f(z1)| = 0, we find thatm > 0.

Apply uniform convergence to get an index N such that n > N implies that |fn f| < m |f|on Cr(z0). Now, |f| m on Cr(z0), so we can apply Rouches theorem and conclude that N(f) =N(fn f + f) = N(fn) for all n N. That is, the number of zeros of fn inside Cr(z0) is equal tothe number of zeros of f inside Cr(z0).

29. Follow the steps outlined in Exercise 28. Let denote the open unit disk and let fn(z) =z5 + z4 + 6z2 + 3z+ 1 1 + 1n . Then fn(z) converge uniformly to p(z) = z

5 + z4 + 6z2 + 3z+ 11 on anyclosed disk in . Since p(z) is clearly not identically zero, it follows from Hurwitzs theorem thatN(fn) = N(p) in |z| < 1 for all sufficiently large n. But, an application of Rouches theorem showsthat fn(z) has no zeros in N1(0) or on C1(0). (Take f(z) = 11 +

1n and g(z) = z

5 + z+6z2 + 3z.)Hence p(z) has no roots in N1(0).

Note: Hurwitzs theorem does not apply to p(z) on the boundary ofN1(0). On this boundary,p(z) may have roots. Consider another simpler example: p(z) = z2 + 1, fn(z) = z

2 + 1 + 1n . Then

fn converges to p uniformly on any closed set. Clearly, p(z) has two roots, i, on the boundary ofthe unit disk, but fn(z) has no roots inside or on the boundary of the unit disk.

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