Poles of Gamma Function

Theorem
The gamma function $\Gamma :\C \to \C$ is analytic throughout the complex plane except at $\left\{{0, -1, -2, -3, \ldots}\right\}$ where it has simple poles.

Proof
First we examine the location of the poles, if any.

We examine the Weierstrass form of the Gamma function:


 * $\displaystyle \frac 1 {\Gamma(z)} = ze^{\gamma z} \prod_{n=1}^\infty \left({\left({1 + \frac z n}\right) e^{\frac{-z} n} }\right)$

The terms of the product clearly do not tend to zero, and so the product is only zero if one of the terms is zero; this occurs when $1 + \dfrac z n = 0$, which occurs at $z \in \left\{{-1,-2, \dots }\right\}$.

We also have the expression outside the product to consider; since the exponential function is never 0, this expression is zero whenever $z=0$.

Hence $\displaystyle \frac 1 {\Gamma(z)} = 0$ when, and only when, $z \in \left\{{0, -1, -2, -3, \ldots}\right\}$.

We turn now to examine possible analytic properties.