Singularity at infinity of a function entire How to prove that every non-constant entire function $\,\,f:\mathbb{C}\rightarrow\mathbb{C}\,\,$ has a singularity at infinity?
What type of singularity must this be?
 A: Just to give an example.
Clearly $f$ is unbounded near $\infty$ so the singularity cannot be removable. 
If $f(z)=az+b$ where $a\neq 0$ , then $f$ has a pole at $\infty$.
If $f(z)=e^z$, then $f$ has an essential singularity at $\infty$. 
A: Technically, $f$ has a singularity at $\infty$ by virtue of not being defined there.  What is really meant is that a non-constant entire function $f$ has a non-removable singularity at $\infty$, and this follows directly from Liouville's theorem: if the singularity at $\infty$ was removable, $f$ would be bounded in a neighbourhood of $\infty$, say $\{z: |z| \ge r\}$, and since $f$ is also bounded on $\{z: |z| \le r\}$ (because a continuous function is bounded on a compact set) $f$ would be bounded on $\mathbb C$, therefore constant by Liouville's theorem.
A: Given a function $f(z)$, we say that $f$ has a singularity (of a given type) at infinity if and only if $f(\frac1z)$ has a singularity (of said type) at $0$.
By Liouville's Theorem, we know that every bounded entire function is constant. Conversely, it's clear that every constant function (with a non-infinite constant) is bounded and entire.
Suppose $f$ is a non-constant entire function, so that $$f(z)=\sum_{n=0}^\infty a_nz^n,$$ where the radius of convergence of this power series is infinite, and there is at least one $n\geq 1$ such that $a_n\neq0$.
Since $$f\left(\frac1z\right)=\sum_{n=0}^\infty\frac{a_n}{z^n}=a_0+\sum_{n=1}^\infty\frac{a_n}{z^n},$$ and since $a_n\neq 0$ for some $n>0$, then $f$ has a singularity at infinity--that is, the Laurent expansion of $f(\frac1z)$ about $z=0$ has a non-zero singular part. If there are only finitely-many non-zero $a_n$'s, then $f$ has a pole at infinity. Otherwise, $f$ has an essential singularity at infinity.
