Let $f$ be an elliptic function with a fundamental set of periods $\omega_1, \omega_2$ (so their ratio is non-real).
Let $S$ be the sum of its poles and zeros inside a period parallelogram.
My attempt:
If $f=\prod_i (z-a_i)^{\alpha_i}$, then $\frac{f'}{f}=\sum_i \frac{\alpha_i}{z-a_i}$ and $\frac{1}{2i\pi}\int_c z\frac{f'}{f}(z)dz=\sum a_i\alpha_i=S$
$S$ is a period iff $S=n \omega_1 + m\omega_2$ for some integers $n,m$
Now, $g(z)=\frac{f'}{f}(z)$ is also elliptic so if I picture a parallelogram ABCD, then obviously $\int_{AB+CD} zg(z)dz=\int_{AB} zg(z)dz+\int_{CD} zg(z)dz$
Since ABCD is a period parallelogram, then $z_{CD}=z_{AB}+\omega_1$, and $g$ is elliptic so,
$\int_{AB+CD} zg(z)dz=\int_{AB} zg(z)dz-\int_{AB} (z+\omega_1)g(z)dz=-\omega_1\int_{AB}g(z)dz$
A similar reasoning yields a similar result for BC and $\omega_2$.
My problem is that I can't prove that $\frac{1}{2i\pi}\int_{AB}g(z)dz$ is an integer. The only things I know about $g$ is that it is elliptic, its integral over the whole parallelogram is $0$ and it counts the difference between the number of zeros and poles inside it. But this is not a closed contour so it is not so useful.