Removable Singularity at Infinity implies Constant Function

Theorem

Let $f : \C \to \C$ be an entire function.

Let $f$ have an removable singularity at $\infty$.

Then $f$ is constant.

Proof

We are given that $f$ has a removable singularity at $\infty$.

By Riemann Removable Singularities Theorem, $f$ must be bounded in a neighborhood of $\infty$.

That is, there exists a real number $M > 0$ such that:

$\forall z \in \set {z : \cmod z > r}: \cmod {\map f z} \le M$

for some real $r \ge 0$.

However, by Continuous Function on Compact Space is Bounded, $f$ is also bounded on $\set {z: \cmod z \le r}$.

As $\set {z: \cmod z > r} \cup \set {z: \cmod z \le r} = \C$, $f$ is therefore bounded on $\C$.

Therefore by Liouville's Theorem, $f$ is constant.

$\blacksquare$