Fall 2016 - Problem 11

Solution.

1. By continuity, we have $\abs{f\p{z}} \geq \frac{1}{2}\abs{f\p{0}}$ if $\abs{z} < \delta$, for some $\delta > 0$. Let $M$ be an upper bound of $g$ so that

   $$\abs{f\p{1/z}} \leq \frac{2M}{\abs{f\p{0}}}$$

   near $0$. But this implies that $f\p{z}$ is bounded as $\abs{z} \to \infty$, so by Liouville's theorem, $f$ must be constant.

2. If $f$ is identically zero, then the claim is true with $a = 0$. Otherwise, there exists some $n \in \N$ such that $h\p{z} = \frac{f\p{z}}{z^n}$ is entire with $h\p{0} \neq 0$. Thus,

   $$g\p{z} = \frac{f\p{z}}{z^n} \cdot z^nf\p{1/z} = h\p{z} h\p{1/z}.$$

   By (1), it follows that $h$ is constant, i.e., there exists $a \in \C$ so that

   $$h\p{z} = a \implies f\p{z} = az^n,$$

   as required.