Reciprocal of entire function

Question

In Liouville theorem exercise we often define a new function $g(z)=\frac{1}{f(z)}$, where $f(z)$ is an entire function.

Why is $g(z)$ entire? there can be some $z$'s such that $f(z)=0$.

Answer 1

In this context $g(z)=\frac{1}{f(z)}$ is not always entire when $f(z)$ is entire. Take $f(z)=z$ as your entire function. Then $g(z)=\frac{1}{f(z)}=\frac{1}{z}$ is not entire, because it is not defined in $z=0$. Maybe you meant to say that $g(z)=\frac{1}{f(z)}$ is entire iff $f(z)$ is entire and does not have roots.