# Finding zeros of the elliptic function $f(z) = \sum_{\omega \in \Omega} \frac{1}{(z - \omega)^3}$

Here, the function has periods $\omega_{1}$ and $\omega_{2}$ and $\Omega = \{m\omega_{1} + n\omega_{2}: m,n \in \mathbb{Z}\}$. I'm pretty sure $f(z) = \sum_{\omega \in \Omega} \frac{1}{(z - \omega)^3}$ is elliptic, but it seems to be a known fact that the zeros in its fundamental parallelogram are given by $\frac{\omega_{1}}{2}$, $\frac{\omega_{2}}{2}$, and $\frac{\omega_{1} + \omega_{2}}{2}$. (For example, it was mentioned in this question: Zeroes of derivative of Weierstrass's elliptic function)

How can we prove this? I'm guessing it's quite simple -- is there something obvious I'm overlooking?

• Note that $-\frac12\wp'(z)$ is your function and hence elliptic Commented Feb 10, 2018 at 13:28

This is both an odd function and $\Omega$-periodic. Then $$f(\omega_1/2)=f(-\omega_1/2)$$ as $f$ has period $\omega_1$. Also $$f(\omega_1/2)=-f(-\omega_1/2)$$ as $f$ is odd. This works also for $\omega_2/2$ and $(\omega_1+\omega_2/2)/$. As $f$ has just a triple pole in a fundamental region, it has exactly three zeroes (up to multiplicity) there, so these are all the zeroes $f$ has.