1550251515-Classical_Complex_Analysis__Gonzalez_

474 Chapter^7

have, by (7.12-2),

_1 J J(()d( = _1 J J(()d(

27l"i ( - z 27ri ( - z

C C1

= _1 j f(z)d( + _1 j f(() - f(z) d(

27l"i ( - z 21l"i ( - z

C1 C1

= nc(z)f(z) + ~ J !(() - f(z) d(

27ri ( - z

(7.17-2)

C1

But the last integral vanishes. In fact, if we let k = nc 1 (z), we have

~ j J(() -J(z) d( ~ _2._ · ~ · 2JkJ7rr = JkJ1:

27l"i ( - z 211". 7'

C1

and since E > 0 can be taken arbitrarily small, it follows that the integral

above is zero.

The reader will note that the left-hand side of (7.17-2), and the term

nc(z)f(z) on the right-hand side, are independent of r = r(1:). Hence

formula (7.17-1) holds.
Alternatively, we may prove that

J

J(()-J(z)d(=O

(-z

C1

by observing that the integrand is analytic on and within Ci, except at

the poinLz where it is undefined. However, since

f(()-f(z) =f'(z)+TJ

(-z

where 7J ---+ 0 as ( --+ z, we see that the left-hand side is locally bounded

at z, and it suffices to apply Theorem 7.21.

Next, suppose that z E Ext C*. In this case nc(z) = 0 and

J

J(()d( = 0

(-z
c
also, since the integrand is now analytic in and within C*, so that (7.17-1)
holds good for this case.

Note Formula ( 7 .17-1) is still valid if f is analytic in D = Int C* and

continuous on D (strong form of Cauchy's formula). For the particular