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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.

TravisJ
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Mike Housky
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    you might like to look at Balarka Sen's answer here http://math.stackexchange.com/questions/795070/questions-aobut-weierstrasss-elliptic-functions?rq=1 . if you are unfamiliar with elliptic functions it may seem obscure at first. but it is a very good very condensed introduction to the relevant technicalities and way of thinking – David Holden May 08 '15 at 11:13
  • Perhaps of interest to you ; a meromorphic function with 2 invariants exists :

    $f(x) = \frac{(x^2 - x + 1)^3}{x^2(x-1)^2}$ satisfies the properties of $f(x)=f(1-x)=f(\frac{1}{x})$.

     – mick Apr 03 '24 at 22:48

2 Answers2

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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.

hmakholm left over Monica
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  • why it's important to have non-parallel periods? Wouldn't this function be contained in a fundamental square, or something? (I know there are counterexamples, like $e^z$, I would like to know why the above proof does not work in this case) – Ant May 08 '15 at 11:06
  • @Ant: If the two periods you start out with are parallel (e.g. for $\sin z$ both $4\pi$ and $6\pi$ are periods) then the fundamental parallelogram collapses to a line segment, and the identical line copies of that line segment don't tile the entire complex plane, because neither of the periods take it away from the line it is a segment of. – hmakholm left over Monica May 08 '15 at 18:28
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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.

David Holden
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  • How do you evaluate that sum? $:$ (If I'm figuring things right, then there is no point at which it converges absolutely.) $;;;;$ –  May 08 '15 at 10:29
  • thanks @Ricky for spotting the error. i've adjusted the formula – David Holden May 08 '15 at 10:51
  • Thank you for the answer and for that function. The site doesn't allow me to accept both answers (I tried :^) so I went with chronology. – Mike Housky May 08 '15 at 12:02
  • thanks for the kind thought. i appreciated the lateral thinking shown in your question. – David Holden May 08 '15 at 20:08