$\int_{|z| = 3}\frac{\exp(-z)}{z^2}dz$ around the circle in the positive sense

Question

If I integrate $$ \int_{|z| = 3}\frac{\exp(-z)}{z^2}dz $$ around the circle $|z| = 3$ in the positive sense, what do I actually do with the fact that $|z| = 3$ to solve this problem using residues? I know the answer is $-2\pi i$, but I am unsure all the gaps are filled in for me. The biggest gap being how to use the following fact: $|z| = 3$.

Answer 1

Since the pole is of second order, you could in principle calculate it using the limit-of-derivative approach. But I would use the series expansion around $0$: $$ \frac{1}{z^2}\exp(-z) = \frac{1}{z^2} (1-z+z^2/2-\dots) = \frac{1}{z^2}- \frac{1}{z}+1/2-\dots $$ which shows what the residue is. The residue method amounts to saying that only the power ${-1}$ contributes to the integral, because it's the one integer power of $z$ that does not have a single-valued antiderivative. The series shows that the integral is the same as if you integrated $- \frac{1}{z}$ only; the latter is easy enough to do.