TITLE: Suppose $f$ is entire and $\lim_{z\to\infty}f(z)=\infty$. Show that $f(\mathbb{C})=\mathbb{C}$
QUESTION [5 upvotes]: Suppose $f$ is entire and $\lim_{z\to\infty}f(z)=\infty$. Show that $f(\mathbb{C})=\mathbb{C}$.
First of all I don't really understand this question. I know $z\to\infty$ means $|z|\to\infty$, but what does $f(z)\to\infty$ means? Does it mean $|f(z)|\to \infty$? Also, I just learnt about the one point compactification $\infty$ to the complex plane. So the reason we write $z\to\infty$ instead of $|z|\to \infty$ is because we are referring to $\infty$ as a point in the extended complex plane $\bar{\mathbb{C}}$? So $f(z)\to\infty$ is also referring to $\infty$ in $\bar{\mathbb{C}}$?

REPLY [2 votes]: Let $g(z):=f(1/z)$ for $z \ne 0$. Then $g$ has at $0$ an isolated singularity. Since $g(z) \to \infty$ as $z \to 0$, $g$ has a pole at $0$.
Let $f(z)=\sum_{n=0}^{\infty}a_nz^n$, then the Laurent expansion of $g$ at $0$ is  given by
$g(z)=\sum_{n=0}^{\infty}a_n \frac{1}{z^n}$, which shows that there is $m$ such that $a_n=0$ for $n>m$.
Hence $f$ is a non- constant poynomial. Now use the  fundamental theorem of algebra.