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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.

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    $\begingroup$ you might like to look at Balarka Sen's answer here math.stackexchange.com/questions/795070/… . 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 $\endgroup$ – David Holden May 8 '15 at 11:13
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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.

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  • $\begingroup$ 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) $\endgroup$ – Ant May 8 '15 at 11:06
  • $\begingroup$ @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. $\endgroup$ – hmakholm left over Monica May 8 '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.

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  • $\begingroup$ How do you evaluate that sum? $\:$ (If I'm figuring things right, then there is no point at which it converges absolutely.) $\;\;\;\;$ $\endgroup$ – user57159 May 8 '15 at 10:29
  • $\begingroup$ thanks @Ricky for spotting the error. i've adjusted the formula $\endgroup$ – David Holden May 8 '15 at 10:51
  • $\begingroup$ 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. $\endgroup$ – Mike Housky May 8 '15 at 12:02
  • $\begingroup$ thanks for the kind thought. i appreciated the lateral thinking shown in your question. $\endgroup$ – David Holden May 8 '15 at 20:08

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