Essential singularity of a holomorphic function of two variables

Question

I have a holomorphic function $G(z_1,z_2)$ in 2 variables, such that $G(z_1 + 1, z_2) = G(z_1,z_2)$, hence for a fixed $z_2$, $G(z_1,z_2)$ has a Laurent expansion in $e^{2\pi i z_1}$. I'm trying to show that this function doesn't have an essential singularity, so the Laurent series in $e^{2\pi i z_1}$ has only a finite negative terms. I would like to ask if anyone knows a theorem that is related to the problem? Something like if we know some properties of $G(z_1,z_2)$ then we know that it cannot have an essential singularity in the variable $z_1$ at $\infty$ or a relationship between singularities of the 2 variables.

Answer 1

Pick any entire function $f$ of one variable which is not a polynomial, say $f=\sum_{k=0}^\infty a_k z^k$. Then $f(e^{-2\pi i z_1})$ is a counterexample, as it has the form $$ f(e^{-2\pi iz_1}) = \sum_{k=0}^\infty a_k (e^{2\pi iz_1})^{-k}$$ considered as a Laurent series in $e^{2\pi iz_1}$.

Answer 2

This answer may be seen a completion of the one of vujazzman: even if $G(z_1,z_2)$ has a Laurent expansion respect to $e^{2\pi i z_1}$ with no negative term (and thus it is a Taylor expansion), it can have an essential singularity on the whole complex line $\{(z_1,z_2)\in \Bbb C^2\,:\, z_2=0\}$ as, for example of the function $$ G(z_1,z_2)=\sum_{k=0}^\infty \frac{e^{2k\pi iz_1}}{k!z_2^{2k}}=\exp\left({\frac{e^{2\pi iz_1}}{z_2^2}}\right) $$ This implies that the behavior of a holomorphic function of several variables cannot be predicted only from its properties as a holomorphic function of a single one of its variables. Also the above examples shows a deep property of holomorphic functions of several variables: they cannot have compact singularities (by Hartogs's Extension theorem).