Let $f:C→P1$ be such that $f(z+1)=f(z+i)=f(z)$ for all z∈C. Let
$Γ=\{m+ni:m,n∈Z\}$. Show that if $f$ is holomorphic on $C∖Γ$, and $z⋅f(z)$ is bounded in a neighbourhood of $z=0$, then $f$ is constant.
I am stuck in the part how a condition at zero determines other point of the fundamental region. More specifically, how it prevents pole at other points on the boundary of the fundamental region (I assume that $f$ is holomorphic in the interior)
Did you prove that $f$ is holomorphic on $\mathbb{C}\backslash\Gamma$?
If you did, then it's easier: it can only have poles in the points of the form $m + in$, and its values in all such points are equal: $f(z) = f(z + 1) = f(z + i)$.
Now, if it doesn't have poles in such points, it means that it has no poles at all, so it is bounded on $\mathbb{C}$, and, by Liouville, it is constant.
The function duplicates itself for every square of the form $z = x + iy: n \leq x\leq n+1, m\leq y\leq m+1$. So, if it is bounded on one such square, it is bounded on all complex plane. And the only way it could be unbounded on a compact, is to have a pole.