The Weierstrass Zeta Function $\zeta$ is a meromorphic function satisfying
- $\zeta(z+\omega_1) = \zeta(z) + \eta_1$
- $\zeta(z+\omega_2) = \zeta(z) + \eta_2$
- The singularities of $\zeta$ are poles of residue 1 at the points $m\omega_1+n\omega_2$ for $m,n \in \mathbb{Z}$
Here $\omega_1, \omega_2, \eta_1, \eta_2$ are complex constants with $\frac{\omega_2}{\omega_1}$ not real. Use Cauchy's Residue Theorem to prove Legendre's relation $\omega_2\eta_1 - \omega_1\eta_2 = \pm 2\pi i$ and express the sign in terms of $\omega_1$ and $\omega_2$.
Take a parallelogram with vertex at the origin with sides $\omega_1$ and $\omega_2$ and shift it appropriately so the center is 0. I get that Cauchy's Residue Theorem gives us that the integral around the parallelogram is $2\pi i$.
Here's where I get lost. Take the case $\Im(\frac{\omega_2}{\omega_1}) > 0$. Then integrate along the sides. How is this done? Parametrize somehow? Don't seem to be able to understand the abstraction here.
By the pseudo periodicity of the Weierstrass zeta functions, the integrals of $\zeta(z)\,dz$ over two opposite sides of the parallelogram almost cancel, but don't.
Remember that $\zeta(z+\omega_j)=\zeta(z)+\eta_j$.