Show that $\int_{\partial P}z\frac {f'(z)} {f(z)} dz $ is on the lattice $\Lambda$

Question

Problem: Let $f(z)$ be a meromorphic function on the complex torus $\mathbb C/\Lambda$ that as a function on $\mathbb C$ has no zeros and no poles on $\partial P$, the boundary of the fundamental parallelogram $P$. Show that

\begin{equation} \frac{1}{2 \pi i}\int_{\partial P}z\frac {f'(z)} {f(z)} dz \in \Lambda \, . \end{equation}

Thoughts: By the Residue Theorem \begin{equation} \frac{1}{2 \pi i}\int_{\partial P}z\frac {f'(z)} {f(z)} dz = \sum_{z_0 \in \text{ Int }P} v_{z_0}(f)z_0. \end{equation} I don't know why the latter is on the lattice.

Thanks!

Answer 1

Consider the contributions of two opposite sides of the fundamental parallelogram to the integral:

$$ \frac{1}{2 \pi i} \int_{a}^{a+\omega_1} z\frac {f'(z)} {f(z)} dz +\frac{1}{2 \pi i} \int_{a + \omega_1 + \omega_2}^{a+\omega_2} z\frac {f'(z)} {f(z)} dz \\ = \frac{1}{2 \pi i}\int_{a}^{a+\omega_1} z\frac {f'(z)} {f(z)} dz - \frac{1}{2 \pi i}\int_{a + \omega_2}^{a + \omega_1+\omega_2} z\frac {f'(z)} {f(z)} dz \\ = \frac{1}{2 \pi i}\int_{a}^{a+\omega_1} z\frac {f'(z)} {f(z)} dz - \frac{1}{2 \pi i}\int_{a}^{a + \omega_1} (z + \omega_2)\frac {f'(z)} {f(z)} dz \\ $$ because $f$ is $\omega_2$-periodic. This expression simplifies to $$ \frac{-\omega_2}{2 \pi i} \int_{a}^{a+\omega_1} \frac {f'(z)} {f(z)} dz = - k \omega_2 $$ where $$ k = \frac{1}{2 \pi i}\int_{a}^{a+\omega_1} \frac {f'(z)} {f(z)} dz $$ is the winding number of a closed curve (the image of the segment $[a, a+\omega_1]$ under $f$) with respect to $z=0$, and therefore an integer (compare Integrate logarithmic derivative of a periodic function).

With the same argument, the contributions of the remaining two sides of the parallelogram add up to an integer multiple of $\omega_1$.