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Complex_Analysis_Prelim_Solutions.pdf page 15
Let $f(z)$ be an entire holomorphic function. Suppose that $f(z)=f(z+1)$ and $|f(z)| \leq e^{|z|}$ for all $z \in \mathbb{C}$. Prove that $f$ is constant.
Observe that the function
\[
g(z)=\frac{1}{\sin (\pi z)}
\]
also satisfies $g(z+1)=g(z)$, and $g$ has a simple pole at each integer. Therefore
\[
h(z)=\frac{f(z)-f(0)}{\sin (\pi z)}
\]
is an entire, periodic function with period 1 . We claim that $h$ is identically 0 . To show this, we will first show that $h$ is bounded on $S=\{x+i y \in \mathbb{C}: 0 \leq x \leq 1\}$. For any $x+i y \in S$,
\[
\begin{aligned}
|h(x+i y)| &=\left|\frac{f(x+i y)-f(0)}{\sin (\pi x+i y)}\right| \\
& \leq \frac{2 e^{|x+i y|}+|f(0)|}{| e^{-\pi y} e^{i \pi x}-e^{\pi y} e^{-i \pi x }|} \ll \frac{e^{|y|}}{e^{|\pi y|}} .
\end{aligned}
\]
It follows that $|h(z)| \rightarrow 0$ uniformly as $|y| \rightarrow \infty$, so $h$ is bounded on $S$. Since $h(z+1)=h(z)$, we have that $h$ is bounded on $\mathbb{C}$, and so by Liouville, $h$ is constant. Furthermore, since $|h(x+i y)| \rightarrow 0$ as $y \rightarrow \infty$, we conclude that $h \equiv 0$ in $\mathbb{C}$. Then $f(z)-f(0) \equiv 0$ on $\mathbb{C}$, so $f$ is constant.
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