Under case:

If is a bounded domain which contains origin,

If is analytic with ,

then for any .

Try to solve it by **Riemann Mapping Theorem.**

Scheme of Proof:

Think about that if is a disk. Then use

Schwarz Lemma.

**Riemann Mapping Theorem:**

If is simple connected domain of , and . Then for any point in , there exists unique function , such that:

- is holomorphic and univalent in .
- .
- .

Thus, if , then , by **Schwarz Lemma, ** with . The equality holds for .

By** Riemann Mapping Theorem:**

There exists a unique holomorphic and univalent function , such that .

Set , then is holomorphic in and satisfies that , by** Schwarz Lemma:**

, and .

Thus consider that is univalent, , for any , therefore , for any .

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