How to find the residue of
$\displaystyle \frac{sinh(z)}{cosh(z)-1}$
If you don't want to use long division, you could use L'Hospital's rule.
$\displaystyle \lim_{z \to 0} \ (z-0) \ \frac{\sinh z}{\cosh z - 1} = \lim_{z \to 0} \frac{\sinh z + z \cosh z}{\sinh z} $
$\displaystyle = \lim_{z \to 0} \frac{\cosh z + \cosh z + z \sinh{z}}{\cosh z} = \frac{1+1+0}{1} = 2 $