1550251515-Classical_Complex_Analysis__Gonzalez_

Integration 483

For ( E C 1 and .>. sufficiently large, we have

A 2A

= IClm+lll - z/(I < .>.m+i

since ICI ;::: .>.for ( E C1 and lz/(I < % for.>. large enough, i.e., for.>. > 2lzl.

Hence

J

f ( () d( < ~ (4.>. - 2b) =^2 A (4 - 2! ) ~ 0

' - z - A m+l Am A

C1

as .>. ~ oo. Thus letting .>. ~ oo in (7.19-2), we obtain (7.19-^0 1).

If we put ( = b + iy in (7.19-1), the formula can be written in the form

J(z) = - ]_ J+oo J(b: iy) dy

27r -oo b + iy - z

(7.19-3)

In fact, since the integrand is a continuous function of y and

I

f(b+iy) I A'

b + iy - Z < IYim+l

for IYI > y 0 > 0, the improper integral converges to a value that is the

same as its principal value.

Exercises 7.3

1. If C: z = i + 3eit, 0 :::=; t :::=; 27r, evaluate each of the following integrals.
   (a) J 2z2 -z+3 dz (b) J e-z dz
   z+l z
   c c

(c) J cos27rz dz (d) J sin2z. dz

3z -1 2z - i

c c

( e) J ( z + 1) dz ( f) J sin z dz

(z2 - 4)(z + 4) z^2 - 2z

c c

2. If C: z = 2eit, 0 :::=; t:::; 47r, evaluate each of the following:

(a) j ~dz (b) j z

3

+ z :-

3

dz

z4 -1 z - i

c c

J

ez cos z J ez-^1 dz

(c) (z - i7r)(z - %i7r) dz (d) z(z^2 -1)

c c