 Solution

A pole of order $m$ at $a$ means that $f(z)=h(z)/(z-a{)}^m$ near $a$, where $h$ is holomorphic and $h(a)\ne 0$. 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 $h(a)$, soFor $\cot z=\cos z/\sin z$, the zeros of the denominator are exactly $z=n\pi$, $n\in \mathbb{Z}$: this follows from $e^{2iz}=1$. Each is simple, because ${\sin }^{\prime }(n\pi )=\cos (n\pi )=(-1{)}^n\ne 0$. ThereforeThere 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.