Given an holomorphic 1-periodic complex-valued function $f$ on the upper half-plane and $$q(z)=e^{2\pi i z}.$$
Is there an holomorphic function $$F:\{z\in\mathbb{C}\setminus\{0\}:|z|<1\}\rightarrow\mathbb{C}$$ with $$F(q(z))=f(z)?$$
Why do we need the '1-periodic'?
Thank you very much
It's clear that we need to assume $f$ has period $1$, since $F(q(z+1))=F(q(z))$.
Conversely, say $f$ has period $1$. Say $$D'=\{z\in\mathbb{C}\setminus\{0\}:|z|<1\}$$and let $\Pi^+$ be the upper half-plane.
Now, for every $w\in D'$ there exists $z\in\Pi^+$ with $w=q(z)$. So we can attempt to define $F$ by $$F(w)=f(z)\quad(w=q(z)).$$We need to show $F$ is well-defined, which means we need to show that if $q(z)=q(z')$ then $f(z)=f(z')$. This follows easily from the fact that $f$ has period $1$. (And then the fact that $F$ is holomorphic follows from the fact that there exists a logarithm defined in a neighborhood of any point of $D'$.)