Let $f:\mathbb C\rightarrow \mathbb C$ be an entire function such that the function $g(z)$ given by $g(z)=f(\frac{1}{z})$ has a pole at 0. Prove or disprove $f$ is onto.
I think $f$ is an onto function.
My attempt: Since $f$ is an entire function $f$ has a power series representation about 0 given by $\sum_{m=0}^\infty a_m z^m$. Then $g(z)=\sum_{m=0}^\infty \frac {a_m}{z^m}$. Since $g$ has pole at 0 then $g$ is of the form $g(z)=a_0+\frac{a_1}{z}+\frac{a_2}{z^2}+...+\frac{a_n}{z^n}$ for some fixed $n\in \mathbb N$. Then $f(z)=a_0+a_1z+a_2z^2+....+a_nz^n$. Since $f$ is a polynomial it follows from Fundamental Theorem of Algebra that $f$ is onto.
Correct, with only a small quibble: since $f$ is a non-constant polynomial, ...