help with meromorphic doubly periodic function

Question

Given a non constant meromorphic doubly periodic function $f$ with real-independent periodes $f(z+\omega_1)=f(z+\omega_2)=f(z)$. Consider $F:=\{\lambda_1\omega_1+\lambda_2 \omega_2, \lambda_1, \lambda_2 \in [0,1]\}$. Prove that $f$ has the same number of poles and zeros on $F$ with multiplicities counted.

I tried basically everything (Laurent-series etc.) but what I'm missing is that I need to prove the hint: $f$ has no poles and zeroes on the boundary of $F$.

I know that $\partial F$ is a contour and for the rest of the exercises I can use the Argument Principle but I really don't know how to prove the hint.

Answer 1

Use the Argument principle. Integration of $f'/f$ on the contour $\partial F$ is possible, and the result is

• the number of zeros minus the number of poles, multiplicities counted, on the one side, and
• zero, on the other side, because contour integrals on opposite sides of the fundamental parallelogram $F$ cancel each other, same values, opposite direction of the contour paths.