Find all entire functions with $f(z) = f(\frac{1}{z})$

Question

Find all entire functions with $f(z) = f(\frac{1}{z})$, for all $z\ne 0$.

I tried to use the power series of $f$ but this did not help. I also tried to use Liouville's Theorem on $\frac{f(z)}{f(\frac{1}{z})}$ but then I have to deal with possible singularities.

Can someone help me with this problem?

Answer 1

I presume your $f$ is defined on the whole of $\mathbb C$, but the equality $f(z) = f(1/z)$ is only meant to hold for $ z \in \mathbb C \backslash \{ 0 \}$?

Well, $f$ is holomorphic, so it is bounded on $\{ z \in \mathbb C : |z | \leq 1 \}$. But since $f(z) = f(1/z)$, this means that $f$ is also bounded on $\{ | z | \geq 1 \}$. Do you see where this is going?

Answer 2

Since $f$ is entire, $f$ is defined and holomorphic on $\mathbb C$.

Let's take a sequence $(z_n)$ such that

$$\vert z_n\vert\to \infty.$$

Then $$\vert f(z_n)\vert\to \vert f(0)\vert$$

which is defined since $f$ is entire.

So $f$ is bounded and holomorphic.

Then by Liouville's theorem, $f$ is constant.

Reciprocally, all constant functions works.

Answer 3

Aside
This is one reason that modular functions are so interesting. If we require your equation only on a sub-region of the complex plane, we can get interesting things. Traditionally, instead of using the right half-plane (which is closed under $z \mapsto 1/z$) we use the upper half-plane (which is closed under $z \mapsto -1/z$). Modular functions $f$ are defined on the upper half plane, and satisfy $$ f(z) = f\left(\frac{-1}{z}\right)\quad\text{and}\quad f(z) = f(z+1) . $$