Complex integration and Laurent Series

Question

i was asked to calculate the following integral: $$\int_{|z|<1}\frac{1}{z^2\sinh(z)}\: dz$$ I was suggested to calculate the Laurent Series of the integrand function, but how does one determine such series? So far i've just done some manipulations, such $$\frac{1}{z^2\sinh(z)}=\frac{1}{z^2}\frac{2}{e^{z}-e^{-z}}$$ but i'm far away from finding a series. Can someone help me finding such series? Thanks!

Answer 1

We have that

$$z^2\sinh z = z^3\left(1+ \frac{z^2}{6} + \cdots\right)$$

Since we are finding a residue for a pole located at $0$, we can use geometric series to conclude that

$$\frac{1}{z^2\sinh z} = z^{-3}\left(1 - \frac{z^2}{6} - \cdots\right) = \frac{1}{z^3} - \frac{1}{6z}+\cdots$$

which means the residue is $-\frac{1}{6}$.