Argument Principle for the elliptic function

Suppose that $a_1, \ldots, a_r$ and $b_1, \ldots, b_r$ are the zeros and poles, respectively, in the fundamental parallelogram of an elliptic function $f$. Show that $$a_1+\cdots +a_r-b_1-\cdots -b_r=n\omega_1+m\omega_2$$ for some integers $n,m.$

I need some help understanding this problem. One of my ideas is to take the Taylor expansion near the zeros (and not around the poles since the function if not defined at the pole). So we have $f(z)=\sum_{n=0}^{\infty} \frac{f^{(n)}(a_i)}{n!}(z-a_i)^n$ for some fixed zero $a_i$. Then I compute the derivative $f'(z)=\sum_{n=0}^{\infty} \frac{f^{(n)}(a_i)}{(n-1)!}(z-a_i)^{n-1}.$

I don't see where to go from here so feel free to provide any hints or let me know if I said something wrong!

Suppose I am near a zero WLOG $a_1$ of order $k_1$, then can I just say that $f(z)=\frac{f^{k_1}(a_1)}{k_1!}(z-a_1)^{k_1}+o(z-a_1)^{k_1+1}$ or do I need to account for the terms less that the $k_1$th term as well?

Solution 1:

$f$ is $u\Bbb{Z}+v\Bbb{Z}$ periodic. Take $c$ such that $f$ has no pole and zero on the boundary of $P=c+[0,1]u+[0,1]v$.

Then $$\int_{\partial P} z \frac{f'(z)}{f(z)}dz = -v\int_0^1 \frac{f'(c+ut)}{f(c+ut )}d(ut)- u\int_0^1 \frac{f'(c+vt)}{f(c+vt)}d(vt)$$ $$= - v \log f(c+ut)|_0^1- u \log f(c+vt)|_0^1 =v 2i\pi n + u2i\pi m$$ Next, apply the residue theorem to the LHS.