Use the Laurent Theorem to evaluate an integral

Question

$I = \oint_C z^4 \sinh(3/z)\,dz$

Along the contour $C(0,1)$. How do I approach this? If I convert $\sinh(3/z)$ to a Maclaurin series I get a never pole at z = $0$. Any help is appreciated

Answer 1

HINT:

The Laurent series for $\sinh(3/z)$ is given by

$$\sinh(1/z)=\sum_{n=0}^\infty \frac{3^{2n+1}}{(2n+1)!\,z^{2n+1}}$$

Now, multiply by $z^4$ and find the coefficient on the term $z^{-1}$.