Fall 2017 - Problem 12

Solution.

Let $B = \set{z \in \C \mid g'\p{z} = 0}$. Since $g$ is non-constant and entire, $B$ must be at most countable (it can have only finitely many zeroes on any compact ball). Thus, given $z \in \C \setminus h^{-1}\p{B}$, we see that $g'\p{h\p{z}} \neq 0$, so $g$ is biholomorphic on a neighborhood $U$ of $h\p{z}$. Consequently,

on $h^{-1}\p{U}$, so $h$ is holomorphic at $z$ and so holomorphic on all of $\C \setminus h^{-1}\p{B}$.

Let $z \in B$ so that $g'\p{h\p{z}} = 0$. Notice that $f\p{z} = g\p{h\p{z}}$, and because $f$ is holomorphic near $z$ and $g$ near $h\p{z}$, it follows that $h$ must be bounded near $z$. To complete the proof, we simply need to show that $z$ is isolated in $B$.

Suppose otherwise, and that $\set{z_n}_n \subseteq B$ with $z_n \neq z_0$ converges to $z_0$. We claim that $h\p{z_n} = h\p{z_0}$ for at most finitely many $n$. Otherwise, we get a subsequence $h\p{z_{n_k}} = h\p{z_0}$ and

which implies that the zeroes of $f\p{z} - f\p{z_0}$ accumulate, which is impossible as $f$ is non-constant. Thus, by reindexing, we may assume that $h\p{z_n} \neq h\p{z_0}$ for all $n$. But this implies that $g'\p{h\p{z_n}} = 0$ and $g'\p{h\p{z_0}} = 0$, which means that the zeroes of $g'$ have an accumulation point, so $g'$ is constant. By assumption this means $g' = g'\p{h\p{z_0}} = 0$, which implies $g$ is constant, a contradiction. Hence, no accumulation point could have existed to begin with, so $B$ is an isolated. By Riemann's removable singularity theorem, $h$ extends to an entire function, as required.