Let $(\gamma_1, \gamma_2)$ be a base for a lattice $\Gamma$ in $\mathbb C$, and $\theta$ a theta function, ie an holomorfic function such that $\theta(z+ \gamma) = \theta(z)e^{2i\pi(a_\gamma z + b_{\gamma})}$ for every $\gamma \in \Gamma$. Show that
$a_{\gamma_1}\gamma_2 - a_{\gamma_2}\gamma_1 = deg(div(\theta))$
The hint is: integrate $\theta'/\theta$ on a parallelogram of sides $\gamma_1$ and $\gamma_2$ which doesn't pass on a root, and use the equality $(\theta'/\theta)(z + \gamma_j) = (\theta'/\theta)(z) + 2i\pi a_{\gamma_j}$.
I deduced the equality from the definition of a theta function and I know that I can integrate $\theta'/\theta$ obtaining $1/2i\pi \cdot deg(div(\theta))$ thanks to the Residue Theorem, so I would like to multiply the equality by $1/2i\pi$ and integrate both sides but I can't continue.
Thanks!