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.
 A: For every lattice $L$ with basis $\omega_1,\omega_2$, the Weierstraß $\wp$-function is an even function, and it attains each value exactly twice in a period parallelogram counting multiplicities. $p_L'$ is an odd function and attains each value exactly thrice [in a period parallelogram], counting multiplicities. Since $p_L'$ is odd and has period group $L$, the zeros of $p_L'$ are easy to find:
$$p_L'\left(\frac{\omega_1}{2}\right) = -p_L'\left(-\frac{\omega_1}{2}\right) = -p_L'\left(-\frac{\omega_1}{2} + \omega_1\right) = -p_L'\left(\frac{\omega_1}{2}\right),$$
hence $p_L'\left(\frac{\omega_1}{2}\right) = 0$. Similarly we see that
$$p_L'\left(\frac{\omega_1+\omega_2}{2}\right) = p_L'\left(\frac{\omega_2}{2}\right) = 0.$$
So either $p_L$ has a double zero in one of the zeros of $p_L'$, or $p_L$ has two distinct simple zeros in the period parallelogram.
If $z_0$ is a zero of $p_L$ in the period parallelogram $P = \{ \alpha \omega_1 + \beta\omega_2 : \alpha,\beta\in [0,1)\}$, then $z_1 = -z_0 + \omega_1+\omega_2$ is also a zero of $p_L$ in $P$, unless $\alpha = 0$ or $\beta = 0$, in which cases $z_1 = -z_0 + \omega_2$ resp. $z_1 = -z_0 + \omega_1$ will be a zero of $p_L$ in $P$. In any case, the zeros of $p_L$ in $P$ are $\{z_0,z1\}$, for if $z_1 = z_0$, then $z_0$ is also a zero of $p_L'$.
Here, with $L = \mu\mathbb{Z}[i]$, the Weierstraß $\wp$-function has the further symmetry
$$p_L(iz) = - p_L(z),$$
so if $z_0$ is a zero of $p_L$, then so is $i\cdot z_0$. This rules $\frac{\omega_k}{2}$ (here $\mu/2$ and $i\mu/2$) out as zeros, since then $p_L$ would have four zeros (counting multiplicities) in the period parallelogram. Therefore $z_2 = i\cdot z_0 + \mu$ is also a zero of $p_L$ in the period parallelogram. Hence $z_2 \in \{z_0,z_1\}$. Also, $z_3 = -i\cdot z_0 + i\mu$ is a zero of $p_L$ in the period parallelogram.
These constraints suffice to determine the zeros of $p_L$ in the extremely symmetric situation here.
A: 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.
