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.
$res_af(a+b-z)=res_{-b} f(-z)=res_b f(z)$. The sign of the residue changes when going from $z$ to $-z$.
Cheers,
Rofler
Edit: Fixed sign error