Spring 2017 - Problem 7

Solution.

Denote $A\p{r, R} = \set{r \leq \abs{z} \leq R}$ and so $\func{f}{A\p{1,R}}{A\p{1,S}}$ is a conformal mapping. $f$ maps $\partial A\p{1, R}$ to $\partial A\p{1, S}$. If $f$ maps $\abs{z} = 1$ to $\abs{z} = S$, then we can post-compose $f$ with $\frac{S}{z}$. This inversion maps $\abs{z} = S$ to $\abs{z} = 1$, and so without loss of generality, we can assume that $f$ maps $\abs{z} = 1$ to itself.

Consider

which is holomorphic via the Cauchy-Riemann equations and because $f$ never vanishes. Observe that if $\abs{z} = 1$, then $1/\conj{z} = z$ and so

Thus, $f$ and $g$ are continuous and agree at $\abs{z} = 1$ and are holomorphic on their respective domains, they extend by the Schwarz reflection principle to a single holomorphic function $A\p{R^{-1}, R} \to A\p{S^{-1}, S}$, and we will just denote this extension as $f$.

Notice that $Sf\p{\frac{z}{R}}$ is a holomorphic function $A\p{1, R^2} \to A\p{1, S^2}$, so we may repeat this argument. By induction, we obtain a holomorphic extension of $f$ to the interior of $A\p{R^{1-2^n}, R} \to A\p{S^{1-2^n}, S}$ for all $n \geq 1$. Thus, $f$ extends to a holomorphic function $B\p{0, R} \setminus \set{0} \to B\p{0, S} \setminus \set{0}$. Since $\abs{f\p{z}} \leq S$ on the entire punctured disk, $f$ extends holomorphically to a holomorphic function on $B\p{0, R}$. From our extension, we know that

so $f\p{0} = 0$, so $f$ is a holomorphic function $B\p{0, R} \to B\p{0, S}$. By applying the Schwarz lemma to $\frac{1}{S}f\p{Rz}$, we have

Setting $z = R^{1-2^n}$,

Since $f$ is a continuous bijection from the compact space $A\p{1, R}$ to the Hausdorff space $A\p{1, S}$, it is a homeomorphism. On the interior, $f$ is an injective holomorphic function, so by the inverse function theorem, $f$ is a biholomorphism on the interiors. Hence, we may apply this argument to $\func{f^{-1}}{A\p{1, S}}{A\p{1, R}}$ to see that $S \leq R$ by symmetry. Thus, $S = R$, which completes the proof.