Image of $|z|>1$ is a dense set

Question

Let f be a non constant entire function which is NOT a polynomial. Then show that image of {z such that |z|>1} is a dense set.

I know that $\infty$ is not a pole because then f would be a polynomial. So f is having an essential singularity at $\infty$. I know the heavy machinery (Casorati-Weierstrass and Picard) but I can't seem to put them here in this context. How do I draw conclusion about that given set?

Answer 1

Define $g(z):=f (1/z)$ for $z \ne 0, U:=\{w \in \mathbb C: 0<|w|<1\}$ and $V:=\{z \in \mathbb C: |z|>1\}.$

Since $f$ is not constant and is not a polynomial, $g$ has an essential singularity in $0$. Casorati-Weierstraß says:

$$ \overline{g(U)}= \mathbb C.$$

We have $g(U)=f(V),$ thus $$ \overline{f(V)}= \mathbb C.$$