Let $f$ be a non-constant enrire function on $\mathbb{C}$ such that $f(z+i)=f(z)$ for all $z$. Let $U$ be an open subset of $C$ and $z_0\in U$. Let $g:U\setminus\{z_0\}\longrightarrow\mathbb{C}$ be holomorphic. If $z_0$ is a non-removable singularity of $g$, prove that $z_0$ is an essential singularity if $f\circ g$.
How to prove?
In the case that $g$ has a pole of order $m$ at $z_0$, let $g(z)=\sum_{k=-m}^\infty b_k (z-z_0)^k$. since $f$ is periodic hence not a polynomial, $f(z)=\sum_{n=0}^\infty a_n z^n$. Hence \begin{eqnarray} f(g(z))&=&\sum_{n=0}^\infty a_n(g(z))^n\\ &=&\sum_{n=0}^\infty a_n(\sum_{k=-m}^\infty b_k(z-z_0)^k)^n\\ &=&\sum_{n=0}^\infty\sum_{k_1,\cdots,k_n\geq -m} a_nb_{k_1}\cdots b_{k_n}(z-z_0)^{k_1+\cdots +k_n}\\ &=&\sum_{p=-\infty}^\infty(\sum_{\sum_{j=1}^n k_j=p,n\in\mathbb{N},k_j\geq -m}a_nb_{k_1}\cdots b_{k_n})(z-z_0)^p. \end{eqnarray} How to prove that there are infinite many terms such that the coefficients $\sum_{\sum_{j=1}^n k_j=p,n\in\mathbb{N},k_j\geq -m}a_nb_{k_1}\cdots b_{k_n}$ are nonzero?
Dear Professor, I am still confused. Since $z_0$ is a pole, for any $M>0$ there exists $\epsilon>0$ such that $g(D^*(z_0,\epsilon))\subseteq \{|z|>M\}$ (they may not be equal). $f(\{|z|>M\})$ is dense in $\mathbb{C}$. How to obtain $f(g(D^*(z_0,\epsilon))$ dense in $\mathbb{C}$?
The Taylor series for $f$ must be an infinite series because the periodicity property guarantees that $f$ is not a polynomial. Thus, we are done if the singularity of $g$ at $z_0$ is a pole of order $>0$. Entire functions that are not constant may not omit two points in the image from the complex plane. (This is Picard's theorem. Its weakening to only say "has dense image" instead of "may not omit two points" is also acceptable.)
The image of every deleted neighborhood of $z_0$ under $f\circ g$ is then dense in $C$. This is equivalent to the singularity being essential actually, even though Casorati-Weierstrass is only stated usually in one direction. Think about why.
It seems like I only used the fact that $f$ is not polynomial and yet entire. Is something wrong with this proof? It has been years since I last did complex analysis.