Problem:
Classify the singularity of the function $$f(z) = e^{−1/z}$$ at $z = 0$ in the complex plane, and write the first three terms of the Laurent series expansion of $f(z)$ about $z = 0$.
I pull out my definition of Laurent Series:
$$f(z) = \sum_{-\infty}^{+\infty} a_n(z-z_0)^n$$
$$a_n = \frac{1}{2\pi i} \oint \frac{f(z') dz'}{(z' - z_0)^{n+1}}$$
Finding $a_{-1}$ seems like a reasonable thing to try first:
$$a_{-1} = \frac{1}{2\pi i} \oint e^{-1/z} dz$$
After about half an hour of fruitless u-substitution, I realized it wasn't solvable in terms of elementary functions, so I was on the wrong track. Not knowing what else to do, I looked at the answer, which tells me:
At $z=0$ there is an essential singularity: the Laurent series can be written as
$$f(z) = 1 + \sum_{n=1}^{\infty} \frac{1}{n!} \frac{1}{z^n} = 1 + \sum_{n=-1}^{-\infty} \frac{z^n}{n!}$$
I don't understand at all how this result was arrived at. It looks roughly like the Taylor expansion, but a) Taylor expansions aren't valid around the singular point (that's why we're using a Laurent series in the first place) and b) I'm sure it needs a $(-1)^n$ in there somewhere, because we're dealing with powers of a negative, so the terms should have alternating signs.
Can someone explain simply how this answer works, and why we don't need to compute the integrals for $a_{-n}$?