1550251515-Classical_Complex_Analysis__Gonzalez_

Sequences, Series, and Special Functions

2. lf(z)I ~ 1 for lzl < 1

3. J(c) = 0, where 0 < lei < 1

Then

IJ(z)I ~I; ~c: I

1

for lzl < 1 and IJ'(c)I ~ 1 - lcl 2

Equality holds iff f(z) = a(z - c)/(1 - cz), where lal = 1.

585

(8.13-7)

Proof Let z' = g(z) = (z -c)/(1-cz). This function is analytic in lzl < 1

since its pole, namely, z 0 = 1/c, lies outside the unit circle. The inverse

function is given by

z=g -le z ') =--z' + c

1 +cz'

and it is easily verified that the unit disk lzl < 1 is mapped by g onto

lz'I < 1 (Exercises 5.1, problem 7)..

Now, if we define

F(z') = J(z) = J(g-^1 (z')) = f ( ;~~;,)

we have

1. F(z') is analytic in lz'I < 1

2. IF(z')I = IJ(z)I ~ 1 for lz'I < 1
   3. F(O) = J(c) = 0

Hence, by Schwarz's lemma,

IF(z')I ~ lz'I for lz'I <^1 and IF'(O)I ~^1

which implies that

IJ(z) ~ I z -_c I

1-cz

for lzl < 1

and since

F'(z') = f'(z) dz = J'(z) l - lcl2

dz' (1 + cz')2

we have

IF'(O)I = lf'(c)l(l - lcl^2 ) ~ 1

or