Question on the Koebe-Bieberbach Theorem

Question

Assume $f$ is injective and that $f(0) = 0$ and $f'(0) = 1$.

The theorem states that $\exists r >0$ such that $D_r(0) \subset f(\mathbb D)$ and, at best, $r=1/4$

($\mathbb D$ is the unit disc).

The idea is to show that for a function:

$$h(z) = \frac1z + \sum_{n=0}^{\infty}c_nz^n$$

,which is injective and analytic for $0<|z|<1$, then:

$$\sum_{n=1}^{\infty}n|c_n|^2 \le 1$$

Anyone have any idea how to prove this? I'm thinking of parameterizing $h$ in polar co-ordinates and seperating the function in to a bounded part and an unbounded part, but not sure where to go from there.

Answer 1

The proof is based on Green's formula for the area of a set $G$, which is $\int_{\partial G}x\,dy$.

1. Apply this formula with $G$ being the domain bounded by the curve $h(\{|z|=r\})$.
2. Express the integral in terms of coefficients $c_n$, by using the power series for $h$.
3. Note that area cannot be negative; this gives an inequality for coefficients.
4. As $r\to 1$, this inequality becomes what you want.

For details, see Area Theorem on Wikipedia.