Principal part of Laurent expansion.

Question

Trying to find the principal part of the Laurent Expansion of $(\exp(z)-1)^{-2}$ about $z = 0$. I know that because there is a pole of order two at $z=0$ I need to only find the constants for $n=-1,-2$.

Answer 1

Without words:

$$(e^z-1)^2=\left(1+z+\frac{z^2}2+\ldots-1\right)^2=\left(z\left(1+\frac z2+\ldots\right)\right)^2=z^2\left(1+z+\ldots\right)\implies$$

$$\frac1{(e^z-1)^2}=\frac1{z^2(1+z+\ldots)}=\frac1{z^2}(1-z+z^2+\ldots)=\frac1{z^2}-\frac1z+1+\ldots$$