# Elliptic functions proof - finitely many zeros and poles

1. Feb 8, 2017

### binbagsss

1. The problem statement, all variables and given/known data
Hi

I have questions on the attached lemma and proof.

$f(z)$ is an elliptic function here, and non-consant $\Omega$ is a period lattice.
So the idea behind the proof is this is a contradiction because the function was assumed to be non-constant but by the theorem that if f is analytic in a region $R$ with zeros at a sequence of points $a_i$ that tend to $a_0$ $\in R$, then $f$ is identically zero in $R$.

Questions - mainly I don't understand where the consruction of the sequence comes from

- which of the conditions out of the three: bounded, closed, infinitely many zeros, means that a convergent sequence of zeros can be constructed? I don't understand the reasoning behind the sequence, and does it make use of the fact of the periodicity of $f(z)$?
- I'm guessing this sequence would not be possible to construct if there were only finitely many zeros for the proof to work...?
- When it argues that by continuity $f(a_0)=0$, we have that $\Omega$ has inifnitely many zeros as our assumption, but we haven't said anything about poles? We haven't disallowed poles, and this means the region is not analytic so not continous and so the limit may not neccessary exist?

2. Relevant equations
see above

3. The attempt at a solution
see above

#### Attached Files:

• ###### lemm proof.png
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2. Feb 16, 2017