Fix a $\tau \in \mathbb{C}$ with $Im(\tau)>0$ , and define $\theta(z)= \sum_{n=-\infty}^{n=\infty}{e^{\pi i(n^{2 }\tau + 2nz)}}$ . I have shown that $\theta(z)$ is analytic on all of $\mathbb{C}$.
How to show $\frac{1}{2\pi i}\int_{\gamma} \frac{\theta'(z)}{\theta(z)} dz =1 $ ? where
$\gamma = $ the fundamental parallelogram with vertex $0, 1, 1+\tau , \tau$ . I am taking ${1,\tau}$ as a basis of the lattice.
Actually this integration gives the number of zeroes of $\theta(z)$. Two observations: $\theta (z+m)=\theta(z) \forall m \in \mathbb{Z}$ and $\theta (z+m\tau)=e^{-\pi i(m^2\tau + 2mz) }\theta(z)\forall m \in\mathbb{Z}$.
I guess this two observation will be helpful for doing this integration.
For reference See Rick Miranda's Book on algebraic curves and Riemann surface page 34.