I want to find the residues of the integral
$F = \int_{-\infty}^{\infty} \dfrac{1}{x+(a-ib)} \dfrac{1}{\exp(-x/c)-1} dx$
I know that $x=-(a-ib)$ is a simple pole which contributes a non-zero residue.
How can I use the Laurent series expansion of $\dfrac{1}{\exp(-x/c)-1}$ to determine if $x=2\pi icn$, $n = 0, \pm 1, \pm 2, ...$ are also simple poles which give non-zero residues?
(The Laurent expansion at $x=0$ is $-\dfrac{c}{x} - \dfrac{1}{2} - \dfrac{x}{12c} + \dfrac{x^3}{720c^3} + O(5)$ )
Consider $\,x:=2\pi icn+y\;$ with $n\in \mathbb{Z}\,$ then $$\frac{1}{\exp(-x/c)-1}=\frac{1}{\exp(-2\pi in-y/c)-1}=\frac{1}{\exp(-y/c)-1}$$ Since for $\,x\to 2\pi icn\;$ we have $\,y\to 0\;$ your expansion may be used with $y$ instead of $x$.
Replacing $\,y\,$ by $\;(x-2\pi icn)\;$ there will give you the Laurent expansion as $\,x\to 2\pi icn$.
The residues of $f(x):=\dfrac{1}{x+(a-ib)} \dfrac{1}{\exp(-x/c)-1}$ at $\,x= 2\pi icn\,$ will then be given by $\;\operatorname{Res}(f(x),2\pi icn)=\dfrac{1}{2\pi icn+(a-ib)}(-c)\ \ $ (at least if $\,a\neq 0\,$ or $\,2\pi cn\neq b$).
Hoping this clarified more (else ask questions!).