= Solution
A <pole> of order $m$ at $a$ means that $f(z)=h(z)/(z-a)^m$ near $a$, where $h$ is holomorphic and $h(a)\ne0$. Equivalently its <Laurent series> has a finite principal part with leading nonzero term $a_{-m}(z-a)^{-m}$. The <residue> $c$ is the coefficient $a_{-1}$. For a simple pole, $m=1$, this coefficient is $h(a)$, so
$$
\boxed{\operatorname{Res}_{z=a}f=\lim_{z\to a}(z-a)f(z).}
$$
For $\cot z=\cos z/\sin z$, the zeros of the denominator are exactly $z=n\pi$, $n\in\mathbb Z$: this follows from $e^{2iz}=1$. Each is simple, because $\sin'(n\pi)=\cos(n\pi)=(-1)^n\ne0$. Therefore
$$
\boxed{\cot z\text{ has simple poles at }n\pi,\quad\operatorname{Res}_{z=n\pi}\cot z=1.}
$$
There are no other finite <singularities>. If infinity is included, it is an accumulation point of these poles, so it is not an isolated singularity to which the usual removable/pole/essential classification could be applied.
Back to article page
