1550251515-Classical_Complex_Analysis__Gonzalez_

Integration 445

Fig. 7.8

If z + h is another point within the same disk, we have

1

z+h

F(z+h)=<n zo J(()d(

where r consists of the horizontal side followed by the vertical side of the

rectangle with opposite vertices z 0 and z + h. Hence

1

z+h 1z

F(z+h)-F(z)=(r) f(()d(-<-ri) f(()d(

ZQ Zo

r+h

= (-y)}z J(()d( (7.10-9)

the path 'Y now consisting of the horizontal side followed by the vertical

side of the rectangle with opposite vertices z and z + h.

We may write (7.10-9) in the form

1

z+h 1z+h

F(z+h)-F(z)=('Y) z f(z)d(+h) z [f(()-f(z)]d(

1

z+h

=f(z)h+<-r) z [f(()-f(z)]d(

which gives

F(z + h) -F(z) - f(z) = ~ (-y) 1z+h [f(() - f(z)] d(

h h z

Since f is continuous at z, for any given e > 0 there is 8 > 0 such that

If(() - f(z)I < e provided that I( - zl ~ lhl < 8. Hence if we write

h = h 1 + ih 2 , we have, for lhl < 8,

I F(z + hk-F(z) - f(z)I ~ l~I e(lh1I + lh21) ~ 2e