Why is $\int\limits_{u}^{u+\omega_j} f'(z)/f(z) dz \in 2 \pi i \mathbb{Z}$?

159 Views Asked by At

Because $f(u) = f(u+\omega_j)$ for $j \in \{1,2\}$ it applies $$\int\limits_{u}^{u+\omega_j}\frac{f'(z)}{f(z)}dz \quad \in 2\pi i \mathbb{Z} \quad \text{for} \quad j=1,2, $$

Hello, I write my thesis at the moment and I really don't understand why the last integral here is in $2\pi i \mathbb{Z}$. I would really appreciate your help!

The function $f$ is elliptic, so $\omega_1, \omega_2 \in \mathbb{C}$ are both periods. $u$ is also complex. My thought was that it could be connected with the fact that the antiderivative of $\int\limits f'/f$ being $log(f(z))$ ?