<s>[INST] Find the Laurent series at the given point and its residue 
a) $\displaystyle \frac{\sin(z)}{(z-\pi)^2}$ at $z_0=\pi$
b) $\displaystyle \frac{1}{1-\cos(z)}$ at $z_0=0$

I'm having trouble understanding Laurent series, please help!
 [/INST] A related problem. For the second one, you can advance as
$$ \frac{1}{1-\cos(z)} = \frac{1}{\frac{z^2}{2!}-\frac{z^4}{2!}+\frac{z^6}{6!}-\ldots }= \frac{1}{\frac{z^2}{2!} \left(1-\frac{2! z^2}{4!}+\frac{2! z^4}{6!}-\ldots \right) } $$
$$ =\frac{2}{z^2} \left(1-\left(\frac{2! z^2}{4!}-\frac{2! z^4}{6!}+\ldots\right)  \right)^{-1} =\frac{2}{z^2}( 1-t )^{-1}, $$
where $t=\left(\frac{2! z^2}{4!}-\frac{2! z^4}{6!}+\ldots\right).$
$$\implies \frac{2}{z^2}( 1-t )^{-1}=\frac{2}{z^2}( 1+t+t^2+t^3+\ldots )$$
$$ =\frac{2}{z^2} \left(1+\left(\frac{2! z^2}{4!}-\frac{2! z^4}{6!}+\ldots\right)+\left(\frac{2! z^2}{4!}-\frac{2! z^4}{6!}+\ldots\right)^2 + \dots \right)  $$
$$ =\frac{2}{z^2} + \frac{1}{6} + \frac{z^2}{120}+\ldots\,. $$
Clearly, the coefficient of $z^{-1}$ is $0$. Note that, we have used the identity
$$ (1-x)^{-1}=1+x+x^2+x^3+\ldots\,. $$
Different approach: You can use another technique to find the residue without using the Laurent series. See here for details.
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