# Suppose $f$ is entire and for all $z \in \mathbb{C}$, $f(z)=f(\frac{1}{z})$. Prove that $f$ is constant.

Suppose $f$ is entire and for all $z \in \mathbb{C}$, $f(z)=f(\frac{1}{z})$. Prove that $f$ is constant.

I want to prove $f$ is bounded. Then by using Liouville theorem, $f$ is constant. But I can't proceed. Can anyone give some hint?

• What can you say about $\lim\limits_{z\to\infty} f(z)$? – Daniel Fischer Nov 17 '13 at 14:41
• @Daniel Fischer:The limit exists due to $f(z)=f(\frac{1}{z})$? – Idonknow Nov 17 '13 at 18:24
• Right. So $\infty$ is a removable singularity. Hence $f$ is bounded, whence constant. – Daniel Fischer Nov 17 '13 at 18:29
• So if $f$ has a removable singularity at $z_0$ and $f$ is entire, then $f$ is bounded? – Idonknow Nov 17 '13 at 19:22
• Not any $z_0$. But if an entire function has a removable singularity in $\infty$, that means it can be extended to a holomorphic function on the entire Riemann sphere. The sphere is compact, hence the image id compact, hence bounded. Another way to look at it, if $\infty$ is a removable singularity, then $f$ is bounded on $\{z : \lvert z\rvert > R\}$ for some $R$. The complement of that is compact, hence $f$ is bounded there too, hence $f$ is bounded. – Daniel Fischer Nov 17 '13 at 19:27

Note : If $f$ is entire, then, in particular it is bounded on closed balls around the origin. Hence, if $D = \{z : |z| \leq 1\}$, then $f$ is bounded on $D$ by some $M > 0$.
Hence, if $|z| \geq 1$, then $1/z \in D$, and hence $$|f(z)| = |f(1/z)| \leq M$$ And hence, $|f(z)| \leq M$ for all $z\in \mathbb{C}$. Hence, Louiville applies.