Can a multiply-periodic complex function be analytic?

Question

It's possible to construct complex periodic functions with two periods in different directions, such as $f(z) = \cos x + i \sin 2y$. That has periods $2\pi$ and $\pi i$. It's also not analytic.

It's been a long time since complex variables, and that was self-study, so I'm very likely under-thinking this, but...Is there any analytic function with two linearly-independent periods?

I don't consider constant functions as properly periodic, since there's no minimum period...but I'm not sure if that attitude is mainstream.

Answer 1

A continuous complex function with two non-parallel periods would be globally bounded (because every value is the same as the value somewhere in a fundamental parallelogram, which is compact).

By Liouville's theorem this means that it is either constant or non-analytic.

If you allow poles, a doubly-periodic function is possible; such functions are known as elliptic functions, and there's quite a bit of theory about them.

Answer 2

try a simple function of a type introduced by Weierstrass:

$$ f(z)=\frac1{z^2}+\sum_{(m,n) \in \mathbb{Z}\times \mathbb{Z} \setminus\{(0,0)\}} \left( \frac1{(z+m +in)^2}-\frac1{(m +in)^2} \right) $$ this has poles on the lattice of Gaussian integers, but is otherwise well-behaved, and evidently has periods 1 and $i$.

if a doubly periodic function had no poles it would have to be constant, since the periodicity would force boundedness, and a bounded entire function is constant.