Finding zeroes and poles of the Weierstrass $\wp$ and $\wp^{'}$ function associated with a lattice. I have a lattice, $\Omega$, with basis $\lbrace 2+i, 1+3i\rbrace$ and fundamental region (square) $P$ with vertices $1+2i,\ 2, \ -1+i\ $ and $-i$. I want to find the zeroes and poles of $\wp$ and $\wp^{'}$ in $P$, along with their orders; where $$\wp=\frac{1}{z^2}+\sum_{\omega\in \Omega\backslash\lbrace0\rbrace}\left(\frac{1}{\left(z-\omega\right)^2}-\frac{1}{\omega^2}\right),$$ and $$\wp^{'}=-\frac{2}{z^3}+\sum_{\omega\in \Omega\backslash\lbrace0\rbrace}\left(\frac{-2}{\left(z-\omega\right)^3}\right)=-2\sum_{\omega\in \Omega}\frac{1}{\left(z-\omega\right)^3}.$$
I'd appreciate any guidance you may have to offer. 
 A: For any lattice, the poles of $\wp'$ are triple at lattice points,
and its zeros are single at half-periods.
For a square lattice, with periods $\omega$ and $i\omega$, the poles of
$\wp$ are double at lattice points. Its zeros are also double, at
$\frac12(1+i)\omega$ modulo the lattice.
A: Using Wolfram Cloud Sandbox with
P[z_] := WeierstrassP[z ,WeierstrassInvariants[{2+I, -1+2I}/2]];
to do some calculations, I found the double poles at the square corners
 $\;(0, -1+2i, 1+3i, 2+i),\;$ clockwise and double zeros at square corners $\;(\frac32-\frac12i,\frac12+\frac32i,\frac52+\frac52i,\frac72+\frac12i)\;$ also clockwise. That is, the fundamental region is a square with a single double pole and a single double zero in it. In general, there would be two single zeros, but the order-4 symmetry combines them into one. Similarly, for $\;\wp'\;$ I found a triple pole at the origin and single zeros at $\;(-\frac12+i,\frac12+\frac32,1+\frac12i),\;$ half of the square fundamental region, and this is exactly what is to be expected in the general case.
