584 ChaptersGeometric interpretation. If we let w = f(z), we see from lwl :::; lzl :::;
r < 1 that, under the assumptions of the Schwarz lemma, the disk lzl :::; r
of the z-plane is mapped by f into the disk lwl :::; r of the w-plane. Also,
it follows that a point z 0 on the circle lzl = r can map on lwl = r only
if w = az with ial = 1, i.e., only if the mapping consists of a rotation
about the origin. ·
Remark Clearly, Schwarz's lemma also applies to conjugate analytic
functions.
Corollary 8.18 Suppose that:1. f is analytic in the open disk lzl < R
2. lf(z)I :::; M for lzl < R
- f(O) = 0
Then we have the estimates
for lzl < R and IJ'(O)I:::; ~ (8.13-6)
The equality in (8.13-6) occurs iff f(z) = (M/R)az, where lal = 1.Proof Letting F(z) = f(Rz)/M, this case is easily reduced to the preceding
theorem. In fact, we have1. F is analytic in Rlzl < R or lzl < 1
- IF(z)I :::; lf(Rz)l/M :::; 1 for lzl < 1
- F(O) = 0
Hence IF'(z)I :::; lzl for lzl < 1 and IF'(O)I :::; 1, with equality occurring iff
F(z) = az, where lal = 1. If we let z' = Rz, the inequality IF(z)I :::; lzl
becomes lf(z')l/M :::; lz'l/R, or lf(z')I :::; Mlz'l/R for lz'I < R. Also,
since F'(z) = Rf'(Rz)/M, we have F'(O) = Rf'(O)/M, so that IF'(O)I :::; 1
implies lf'(O)I :::; M/R. The case F(z) = az yields f(z')/M = az'/R or
f(z') = Maz'/R.
Example Consider the function f(z) = sinz in lzl < 2. Since
eiz _ e-iz e-Y + eY
I sin z I = I
2
i I :::;
2= cosh y < cosh 2
for -2 < y < 2, we have
I sin zl :::; (% cosh 2)lzl
for lzl < 2.
Corollary 8.19 Suppose that: