Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question

1 Answer 1

up vote 2 down vote accepted

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

Cheers,

Rofler

Edit: Fixed sign error

share|improve this answer
    
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

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.