Suppose that $a_1,...,a_r$ and $b_1,...,b_r$ are the zeroes and poles in the fundamental parallelogram of an elliptic function f.
Show that $$\sum_{n=1}^{r} a_n-b_n = n \omega_1 + m \omega_2$$ for some $n,m \in \mathbb{Z}$
Solution Attempt: One can safely ignore the possibility of there being a zero or pole on the boundary by shifting the fundamental parallelogram over by a differential amount.
By the argument principle it is known that $$\int_{\partial P_0} \frac{f'(z)}{f(z)}dz = 2\pi i (N_{zero} - N_{poles}).$$
Adapting this integral to $\int_{\partial P_0} z\frac{f'(z)}{f(z)}dz$ one should receive the value of the zeroes minus the poles. However, I am not sure how to find the value of the right hand side of the equation. Arbitrary constants multiplied by the fundamental parallelogram vectors seem distinct from the previous integral. Any help or suggestions would be greatly appreciated
[although a similar question was posted on the site, their solution doesn't answer my above concern]
Let $F$ be an elliptic function and let $\beta\in\mathbb C$ such that $F$ has no zeros or poles on $\partial(\beta+\bar\Pi)$, where we define $\Pi=\{t_1w_1+t_2w_2|t_{1,2}\in[0,1)\}$ and $w_1,w_2$ are such that the lattice $\Lambda=\mathbb Zw_1\oplus\mathbb Zw_2$.
First of all we observe that the Laurent series of $F$ centered in $a$ is given by $F=c_0(z-a)^n+c_1(z-a)^{n+1}+\dots=c_0(z-a)^{n}[1+\dots]$ with $n=ord_aF$, hence
$F'/F=\frac{n}{z-a}+h(z)$ ($h$ is an holomorphic function) and $z\cdot F'/F=(a+(z-a))\cdot F'/F=\frac{an}{z-a}+\tilde h(z)\implies Res(z\cdot F'/F)=a\cdot ord_a(F)$.
Now we have $$\sum_{a\in\beta+\bar \Pi}ord_a(F)\cdot a=\frac{1}{2\pi i}\int_{\partial(\beta+\bar{\Pi})}z\cdot \frac{F'}{F}(z)dz=\frac{1}{2\pi i}\sum_{k=1}^4\int_{\gamma_k}z\cdot \frac{F'}{F}(z)dz,$$ where $\gamma_k$ is one of the four sides of the parallelogram. Taking the integral on the third side: $$\int_{\gamma_3}z\cdot \frac{F'}{F}(z)dz=-\int_{\gamma_1}(z+w_2)\underbrace{\frac{F'}{F}(z+w_2)}_{=F'(z)/F(z)}dz=-\int_{\gamma_1}z\frac{F'}{F}-w_2\int_{\gamma_1}\frac{F'}{F}\implies\\\int_{\gamma_3}+\int_{\gamma_1}z\cdot\frac{F'}{F}(z)dz=-w_2\int_{\gamma_1}\frac{F'}{F}=-w_2[\log(F(\beta+w_1)-\log(F(\beta))]\\=-w_22\pi ik, k\in\mathbb Z.$$ Same thing for $\gamma_2,\gamma_4$ therefore $\sum_{a\in\beta+\bar\Pi}ord_a(F)\cdot a\in\Lambda:=\text{span}_{\mathbb Z}\{w_1,w_2\}$.