Q:

If $g(z) = f(1/z)$ and $g$ has a pole at $0$ and $f$ entire.Then there is a $z_0$ such that $f(z_0) = 0$.

I figured I can somehow use Liouville's theorem. I don't really know what the method is for solving or where to begin.
Suppose that $f: \mathbb{C} \to \mathbb{C}$ is an entire function. Let $g(z) = f(\frac{1}{z})$. Prove that if $g$ has a pole at $0$ then there is a $z_{0}$ in $\mathbb{C}$ such that $f(z_{0}) = 0$. Can assume $g$ is holomorphic on $\mathbb{C} \setminus \{0\}$.

A:

Since $f$ is entire, there is a power series $\sum_{n=0}^\infty a_nz^n$ centered at $0$ such that$$(\forall z\in\Bbb C):f(z)=\sum_{n=0}^\infty a_nz^n$$and therefore$$(\forall z\in\Bbb C\setminus\{0\}):g(z)=a_0+\frac{a_1}z+\frac{a_2}{z^2}+\cdots$$So, since $g$ has a pole at $0$, there is some $N\in\Bbb N$ such that $n>N\implies a_n=0$. So, $f$ is a polynomial function and therefore you can use the Fundamental Theorem of Algebra to prove that it has a zero.

