# Functional equation of elliptic function with exactly two poles of order one

This is exercise I.2.2 of "Elliptische Funktionen und Modulformen" by Koecher and Krieg.

Let $f(z)$ be an elliptic function that has exactly two poles of order one at $a$ and $b$ in a fundamental parallelogram $P$ and no other poles there. Show that $f(z) = f(a+b-z)$.

What I have done:

We have $\sum_{c\in P} \mbox{res}_c f = 0$, that is $\mbox{res}_a f + \mbox{res}_b f = 0$, so $\mbox{res}_a f = -\mbox{res}_b f$.

The assertion is equivalent to

$$h(z):= f(z) - f(a+b-z) \equiv 0$$

and $h$ is of course an elliptic function. It suffices to show that $h$ is holomorphic (since it is then forced to be constant) and takes the value $0$ at some point. We have

$$\mbox{res}_a h(z) = \mbox{res}_a f(z) - \mbox{res}_a f(a+b-z) = \mbox{res}_a f(z) - \mbox{res}_{-(a-(a+b))} f(z) = \mbox{res}_a f - \mbox{res}_b f = 2\mbox{res}_a f \not=0$$

so we get a contradiction.

Where did I commit an error? Thank you for your insights.

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## 1 Answer

$res_af(a+b-z)=res_{-b} f(-z)=res_b f(z)$. The sign of the residue changes when going from $z$ to $-z$.

Cheers,

Rofler

Edit: Fixed sign error

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Of course, thank you very much. Time and again, sign errors are fun. You made one, too. :) –  Gregor Bruns Sep 5 '12 at 10:04