How can I determine what the order of the pole on the following function is:
$$\ f(z)= \frac{e^{bz}}{z\sinh(az)}$$
From the Laurent series, I found that the residue would be b/a or -b/a, however, I am confused whether this is a simple pole or a pole of second order.
(Effectively answered in comments, post made to clear queue).
The numerator is nonzero, prove the denominator has a root of degree $2$ by your favorite method. Alternatively,
$$f(z)= \frac{e^{bz}}{z\sinh(az)} = \frac{2e^{bz}}{z}\left(e^z-e^{-z}\right)^{-1} = \frac{e^{bz}}{z}\sum_{k = -\infty}^\infty \frac{(-1)^kz}{(\pi k)^2+z^2}$$
The sum on the right will only yield a negative power of $z$ when $k=0,$ so the order is $-2.$