Differentiationand sinceI( - zl = l(zo -z) + (( -zo)I > r -^1 / 2 r = %r
it follows thatlf(zo)-f(z) _ f(()-f(z)I <e
Zo - Z (-Z387Theorem 6.36 Suppose that f is analytic in a region R and z 0 E R.
Then for any e > 0 there is a 8 > 0 such that
If((~= {(A) - f'(zo)I < e
provided that I( - zol < 8 and IA - zol < 8.
Proof By the definition of the derivative we haveI
f(zo) - f(z) -f'(zo)I < ~e,
Zo -z 2
z ER (6.22-2)
provided that lz -zol $ r, and by Theorem 6.35, there is a 8, 0 < 8 < ~r,
such thatI
f(zo) - f(z) _ f(() - f(z) I< ~e
Zo - Z (-Z 2(6.22-3)whenever lz - zol = r and I( - zol < 8. From (6.22-2) and (6.22-3) it
follows thatI
J(() - f(z) - f'(zo)I < e
(-z
For each ( such that I( - zol < 8, define
g (A)= f(()-f(A) (-A if A i= (
and
g(A) = !'(() if A= ((6.22-4)for all A with IA - zol $ 8. Then Theorem 6.34 is applicable, and we find,
using (6.22-4),IY(A) - f'(zo)I $ max I!(()-f(z) - f'(zo)I < e (6.22-5)
lz-zo l=r ( - Z
Theorem 6.37 If f is analytic in a region R, then f' is continuous in R.Proof For A= ( (6.22-5) reads If'(() - f'(zo)I < e whenever I( - zol < 8.
This expresses the continuity off' at zo. Since zo is an arbitrary point of
R, the theorem holds.