Consider the following theorem: suppose $f:\mathbb C \to \mathbb C_{\infty}$ is a non-constant doubly periodic meromorphic function. (So, $f(z) = f(z+ \omega_1) = f(z+\omega_2)$ for all $z \in \mathbb C,$ where $\omega_1,\omega_2$ are linearly independent.) Then degree of $f$ is $\geq 2.$
One of the proofs I see says that we can choose a parallelogram $P$ with vertices $w, w+ \omega_1, w+ \omega_2, w + \omega_1 + \omega_2$ so that $\partial P$ does not contain any singularities of $f.$ Then
$$ \sum_{p \text{ is a pole in }P} (\text{residue of $f$ at $p$}) = \frac{1}{2\pi i} \int_{\partial P} f(z) dz =0. $$ From this, we conclude that $f$ cannot have exactly one pole in $P.$
However, why does the integral above equal zero?
The integral above equals $0$ not by any clever observation on the residues, but rather on the contour integral itself.
Notice that if $z$ is in the straight line segment $\ell_1:w\to w+\omega_1$ then $z+\omega_2$ is in the segment $\ell_3:w+\omega_2\to w+\omega_1+\omega_2$ but $f(z)=f(z+\omega_2)$. Likewise, if $z$ is in the segment $\ell_4:w\to w+\omega_2$ then $z+\omega_1$ is in the segment $\ell_2:w+\omega_1\to w+\omega_1+\omega_2$ but $f(z)=f(z+\omega_1)$ again.
Then the integral around $\partial P$ consists of four line segments where each opposite pair has equal integral by the comments above, but due to the orientation of a contour integral (clockwise or counter-clockwise doesn't matter here) the integral reduces to: $$\oint_{\ell_1+\ell_2-\ell_3-\ell_4}f=\oint_{(\ell_1-\ell_3)}f+\oint_{(\ell_2-\ell_4)}f=0$$
To help a visualisation, I have described a clockwise-oriented integral over $\partial P$, a parallelogram where $w$ is the bottom left corner and $w+\omega_1+\omega_2$ the top right corner, $w+\omega_1$ the top left corner and $w+\omega_2$ the bottom right corner. The actual orientation or placements of the $\omega_{1,2}$ is irrelevant however and the same argument works to any formation of $\partial P$.