1550251515-Classical_Complex_Analysis__Gonzalez_

488 Chapter^7

= J J(() d( + J hf(() d( (7.20-3)

((-z)^2 ((-z-h)((-z)^2

'Y 'Y

Since f is continuous on 'Y, there exists a constant M > 0, such that

IJ(()I :::; M for all ( E 'Y· Therefore

J

hf(()d( < lhlM L('Y) = 2lhlM L('Y)

((-z-h)((-z)^2 - (1/ 2 r)r^2 r^3

'Y

and we see that the last integral in (7.20-3) tends to zero as h ~ 0. Thus
lim [F(z + h) - F(z)]/h exists and equals J f( () d(/( ( - z)^2.
In the general case, we have, with the sa'fne notation as above,

Fn(z+h)-Fn(z)= J J(()d( -! J(()d(

((-z-h)n ((-z)n

'Y 'Y

= J cc -z r -cc -z - hr Jc

0

d(

((-z-h)n((-z)n

'Y

-j h[(( -z)n-^1 + (( -z)n-^2 (( - z - h) +. · · + (( -z - hr-^1 ]

• ((-z-h)n((-z)n J(()d(
  'Y

h

J [

=^1 + ... +^1 ' ] f.,, (1') d(

((-z-h)n((-z) ((-z-h)((-z)n

'Y

(7.20-4)

Hence

IFn(z + h) - Fn(z)J

[

1 1 1 ]

::=; lhlM (%r)nr + (%r)n-lr2 + ... + (%r)rn L('Y)

= lhlM^2 (^2 n - I) L( )

rn+l 'Y

so that IFn(z + h) - Fn(z)I ~ 0 as h ~ 0. This shows 'that Fn(z) is
continuous at z.
Now (7.20-4) can be written as

Fn(z + h) - Fn(z) = J J(() d(

h ((-z-h)n((-z)

'Y

+j J(()d( +···+] J(()d(

((-z - h)n-1(( - z)2 (( - z - h)(( -z)n

'Y 'Y