Spring 2018 - Problem 11

Solution.

First, suppose $F$ maps $\partial\D$ to $\partial\D_{R_2}$. Then if we replace $F$ with $\frac{R_2}{F}$, which is holomorphic since $F\p{z} \neq 0$, we see that if $\abs{z} = 1$, then

Since $\frac{1}{z}$ is an automorphism of $\C \setminus \set{0}$, it follows that $\frac{R_2}{F}$ is still conformal. Hence, without loss of generality, we may assume that $F$ maps $\partial\D$ to $\partial\D$, and so $F$ maps $\partial\D_{R_1}$ to $\partial\D_{R_2}$.

Let

If $\abs{z} = 1$, then because $\abs{F\p{z}} = 1$ also, we get

so $G$ is continuous at $\partial\D$. By the Schwarz reflection principle, $F$ extends holomorphically to a map $\func{F}{A_{R_1}}{\D}$ such that if $\abs{z} = R_1^{-1}$, then

By induction, $F$ extends to a holomorphic map $\func{F}{\D \setminus \set{0}}{\D}$ such that if $\abs{z} = R_1^{-k}$, then $\abs{F\p{z}} = R_2^{-k}$ for infinitely many $k \in \N$. Since $F$ is bounded near $0$, we see that $0$ is a removable singularity, and because $R_1^{-k} \mapsto R_2^{-k}$, we see that $F\p{0} = 0$. Hence, the Schwarz lemma applies, so

Running the same argument with $F^{-1}$, we see that $R_1 = R_2$, by symmetry.