Show that $\int_{\partial P}z\frac {f'(z)} {f(z)} dz $ is on the lattice $\Lambda$ 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!
 A: 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 $\omega_1$.
