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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?

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What about constant functions? – Cameron Buie Oct 29 '12 at 4:18
up vote 5 down vote accepted

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.

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The "and since $a_n \ne 0$ for some $n > 0$, then $f$ has a singularity at infinity" needs some elucidation. – Robert Israel Oct 29 '12 at 5:22
Is that more straightforward, Robert? – Cameron Buie Oct 29 '12 at 5:26

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.

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This one is short and sweet. I like this one better. Just wondering if it make any difference while you take the neighborhood of infinity open set not the closed like you did in the proof you gave? – Deepak Dec 26 '12 at 18:13

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