4
$\begingroup$

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!

$\endgroup$
0
6
$\begingroup$

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 $\omega_1$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.