Calculate the Laurent series of $\displaystyle\frac{e^z}{z^2}$ when $0<|z|<\infty$.
I know how the terms $a_n$ and $b_n$ of the Laurent series are defined in terms of a closed integral, but I don't think it is really useful to apply them in any situation. What should I do here?
$$\frac{e^z}{z^2}=\sum_{n=-2}^\infty \frac{z^n}{(n+2)!}$$ And yes, it is usually the other way around: most common usage is the Residue theorem where you need $a_{-1}$ to find $\oint f(z)dz$.
(Particularly for this case it immediately yields $$\oint\frac{e^z}{z^2}dz=2\pi i$$ around $0$ as $a_{-1}=1$ in this case.)