Zeroes of meromorphic function on the Complex Tori 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.
 A: For $a,b\in \mathbb{C}$ and $t\in [0,1]$, let $\gamma_{a,b}(t)\colon=a(1-t)+bt$. For ease of typing, define
$$\gamma_1\colon=\gamma_{0,1}, \ \gamma_2\colon=\gamma_{1,1+\tau},\ \gamma_3\colon=\gamma_{1+\tau,\tau}\ \text{ and }\ \gamma_4\colon =\gamma_{\tau,0}.$$
Then
$$\frac{1}{2\pi i}\int_{\gamma}\frac{\theta'(z)}{\theta(z)}dz= \frac{1}{2\pi i}\sum_{j=1}^{4}\int_{\gamma_j}\frac{\theta'(z)}{\theta(z)}dz.$$
Since $\theta$ is 1-periodic, so is the derivative $\theta'$. Thus, the meromorphic function $z\mapsto\frac{\theta'(z)}{\theta(z)}$ is 1-periodic and we have
$$\int_{\gamma_2}\frac{\theta'(z)}{\theta(z)}dz=-\int_{\gamma_4}\frac{\theta'(z)}{\theta(z)}dz.$$
Thus
$$\frac{1}{2\pi i}\int_{\gamma}\frac{\theta'(z)}{\theta(z)}dz= \frac{1}{2\pi i}\Big(\int_{\gamma_1}\frac{\theta'(z)}{\theta(z)}dz +\int_{\gamma_3}\frac{\theta'(z)}{\theta(z)} dz\Big)= \frac{1}{2\pi i}\Big(\int_{\gamma_1}\frac{\theta'(z)}{\theta(z)}dz +\int_{\gamma_1^{-1}}\frac{\theta'(z+\tau)}{\theta(z+\tau)} dz\Big).$$ The last step follows from $\gamma_3(t)=\gamma_1^{-1}(t)+\tau$ , where $\gamma_1^{-1}(t)=\gamma_1(1-t)$ is the inverse curve of $\gamma_1$.
Consequently
$$\frac{1}{2\pi i}\int_{\gamma}\frac{\theta'(z)}{\theta(z)}dz= \frac{1}{2\pi i}\Big(\int_{\gamma_1}\frac{\theta'(z)}{\theta(z)}-\frac{\theta'(z+\tau)}{\theta(z+\tau)} dz\Big).$$
Finally, verify that, for all $z\in \mathbb{C}$ with $\theta(z)\neq 0$, we have $$\frac{\theta'(z)}{\theta(z)}-\frac{\theta'(z+\tau)}{\theta(z+\tau)}=2\pi i.$$ To see this, differentiate $\theta(z+\tau)=e^{-\pi i (\tau - 2z)}\theta(z)$ using the chain and the product rule.
In total, $$\frac{1}{2\pi i}\int_{\gamma}\frac{\theta'(z)}{\theta(z)}dz=\int_{\gamma_1}1dz=1.$$

A priori, we do not know that $\theta$ does not have a zero on the trace of $\gamma$. So technically, we cannot apply the argument principle as naively as we did. However, by the Identity Theorem, the holomorphic functions $\theta$ and $\theta'$ have only a finitely many zeroes in a compact set. Thus, if the trace of $\gamma$ contained a zero of $\theta$ or $\theta'$, then we could translate the parallelogram $\operatorname{Im}(\gamma)$ a tiny bit such that the boundary of the translated parallelogram would not contain any zeroes of $\theta$ and $\theta'$. Integrating $z\mapsto \frac{1}{2\pi i}\frac{\theta'(z)}{\theta(z)}$ along the boundary of such a translated parallelogram also gives $1$. By the argument principle, we thus know that $\theta$ has only one zero in the interior of the translated parallelogram. One verifies that $\frac{\tau +1}{2}$ is a zero of $\theta$. Thus, the Jacobi theta function has no zeroes in $\operatorname{Im}(\gamma)$.
