Let $f$ be an entire function such that $f(z) = f(1/z)$ for all $z \in \mathbb{C} \setminus \{0\}$. Show that $f(z)$ is constant.

Question

Let $f$ be an entire function such that $f(z) = f(1/z)$ for all $z \in \mathbb{C} \setminus \{0\}$. Show that $f(z)$ is constant.

I understand Liouville's Theorem could come into play here but unsure where to apply.

Answer 1

Hint: $\lim_{z\to\infty} f(z)=\lim_{z\to\infty} f(\frac{1}{z})=f(0)$.

Answer 2

The mapping $z\mapsto\frac1z$ is a bijection between the punctured closed disk $$ \{|z|\le1,z\neq0\} $$ and what stays outer: $$ \{|z|\ge1\}. $$ Thus $$ \sup_{|z|\ge1}|f(z)|=\sup_{|z|\ge1}|f(1/z)| =\sup_{|z|\le1,z\neq0}|f(z)| $$ and the last one is clearly bounded, say by $B$. Thus $$ |f(z)|\le\max\{B,|f(0)|\}\;\;\;\;\; \forall z\in\Bbb C. $$ So you have a bounded entire function, which is constant by Liouville.