Finding zeroes of a complex function over a lattice Question:
Let $L = \mu\mathbb{Z}[i]$ be a lattice in $\mathbb{C}$, where $\mathbb{Z}[i] = \{n+mi:n,m\in\mathbb{Z}\}$ and $\mu \in \mathbb{R}_{+}$. 
Let $$\mathrm{G}_k = \displaystyle\sum_{\omega \in L}_{\omega \not= 0} \dfrac{1}{\omega^k} $$
be the corresponding Eisenstein series for $k \in \mathbb{Z}_{>0}$ and let 
$$ p_L(z) = \dfrac{1}{z^2} + \displaystyle\sum_{\omega \in L}_{\omega \not= 0} \dfrac{1}{(z-\omega)^2} - \dfrac{1}{\omega^2}  $$ be its Weierstrass $p$-function.
Find the zeros of $p_L$.

Answer: It's easy to show that $L=iL$ and $L$ is closed under complex conjugation.
Now $L=iL \Rightarrow \left(\omega \in L \iff i\omega \in L\right) \Rightarrow \displaystyle\sum_{\omega \in L}_{\omega \not= 0} \dfrac{1}{\omega^2} = 0 $
$L$ closed under complex conjugation $\Rightarrow \left(\omega \in L \iff \bar{\omega} \in L\right) \Rightarrow p_L(x) \in \mathbb{R}, \forall x \in \mathbb{R}, x \notin L$
So $p_L$ reduces to 
$$ p_L(x) = \dfrac{1}{x^2} + \displaystyle\sum_{\omega \in L}_{\omega \not= 0} \dfrac{1}{(x-\omega)^2}$$.
How would I find the zeros of $p_L$ from here?
I've tried resolving the sum by splitting into the different $\omega$s ($i\omega$,$\bar{\omega}$,$-\omega$) but that doesn't seem to go anywhere.

For an explicit formula of the zeroes of the Weierstrass $\wp$-function $P_L(z)$ see the article of Eichler and Zagier here. We have $\tau=i$ for $\mathbb{Z}[i]$. 
Theorem: The zeroes of $P_L(z)$ are given by
$$
z=m+\frac{1}{2}+ni\pm \Biggl(\frac{\log(5+2\sqrt{6})}{2\pi i}+144\pi i\sqrt{6} \int_i^{i\infty}(t-i)\frac{\Delta(t)}{E_6(t)^{\frac{3}{2}}}dt\Biggr),
$$
where $m,n\in \mathbb{Z}$, $E_6(t)$ and $\Delta(t)$ denote the normalized Eisenstein series of weight $6$ and unique normalised cusp form of weight $12$ on $SL_2(\mathbb{Z})$, respectively.
The authors give two proofs, one using modular forms, the other one elliptic integrals.
