So the question is as follows. Prove that there is no meromorphic function $f$ such that at every $z\in \mathbb{C}$ we have $f(z)=f(z+1)$ and $f(z)=f(z+i)$ with only simple poles at the points $m+ni, m,n \in \mathbb{Z}$.
So, this looks like a holomorphic map from the torus to the Riemann sphere. Since the map is continuous, the image of the torus is compact, hence closed. On the other hand, holomorphic maps are open, hence the image is also open, so the map is surjective.
Now, if I knew that $f'(z) \neq 0$, then I would have a covering, and there are no coverings from the torus to the sphere. However, I can't make that statement.
I'm wondering, is there a proof that uses only arguments from complex analysis?
Thanks for any help!
Best Answer
Consider $R = \{z \in \mathbb{C} \mid \operatorname{Re}(z), \operatorname{Im}(z) \in \left[-\frac{1}{2}, \frac{1}{2}\right]\}$. By the periodicity of $f$,
$$\int_{\partial R}f(z) dz = 0.$$
However, by the residue theorem,
$$\int_{\partial R}f(z) dz = 2\pi i\operatorname{Res}(f, 0).$$
As $f$ has a simple pole at $0$, $\operatorname{Res}(f, 0) \neq 0$. Therefore, no such $f$ exists.