Spring 2018 - Problem 10

Solution.

First, let

which is an automorphism of $\C^*$ mapping $0 \mapsto 0$ and $1 \mapsto \infty$. Thus, if $f$ is as in the problem, then $g = \phi^{-1} \circ f \circ \phi$ is an injective holomorphic function everywhere except at $\phi^{-1}\p{0} = 0$ and $\phi^{-1}\p{1} = \infty$, i.e., $\func{g}{\C \setminus \set{0}}{\C \setminus \set{0}}$.

Since $g$ is injective, it follows that $g$ does not have an essential singularity at $\infty$. Indeed, by the open mapping theorem, $g\p{\set{\abs{z} < 1}}$ is open, and by Casorati-Weierstrass, $g\p{\set{\abs{z} > 1}}$ is open and dense. Thus, $g\p{\set{\abs{z} < 1}} \cap g\p{\set{\abs{z} > 1}} \neq \emptyset$, which is impossible.

Because $g$ omits $0$, we know that $\frac{1}{g}$ is holomorphic. By injectivity, $\frac{1}{g}$ any $0$ of $g$ must be simple by the argument principle, so any pole of $g$ is simple. Thus, we may write

for some $a, b, c \in \C$. Observe that

i.e., if $a, c \neq 0$, then $g$ has two non-zero roots, which is impossible, so one of them must vanish. If $a = 0$, then $g\p{z} = b + cz$, which means $b = 0$ or else $g$ still has a non-zero root. Similarly, if $c = 0$, then $g\p{z} = \frac{a}{z} + b$ and as before, we need $b = 0$ in this case. Thus, $g\p{z}$ is either $\frac{a}{z}$ or $az$ for some non-zero $a \in \C$. Hence,

In either case, $f$ is a Möbius transform, so $f$ is automatically surjective.