<s>[INST] Expand $ z^2 \sin \frac {1} {z-1}$ in a Laurent series in the neighbourhood of the point $z=1$ Expand the function $ z^2 \sin \frac {1} {z-1}$ in a Laurent series in the neighbourhood of the point $z=1$.
With the substitution $w=z-1$,
$ z^2 \sin\left(\frac{1}{z-1}\right) = (w+1)^2\sin\left(\frac{1}{w}\right) = (w^2+2w+1) (\frac{1}{w} - \frac{1}{3!w^3} + \frac{1}{5!w^5} - \frac{1}{7!w^7} + \cdot \cdot \cdot$).
When I multiply, I get three series and I try to get the next series but I get stuck.
$w+2 + \sum_{k=1}^{\infty} [\frac{(-1)^{k-1} [\frac{1}{(2k-1)!}-\frac{1}{(2k+1)!}]}{w^{2k-1}} + \frac{2(-1)^{k}}{(2k+1)! w^{2k}}]$, $\quad$
para $0< \mid  w \mid <\infty$.
 [/INST] You just have to be careful in rearranging terms. We have
$$ w^2\sin\frac{1}{w} = \sum_{n\geq 0}\frac{(-1)^n}{(2n+1)! w^{2n-1}}\qquad \sin\frac{1}{w} = \sum_{n\geq 0}\frac{(-1)^n}{(2n+1)! w^{2n+1}} $$
$$ 2w\sin\frac{1}{w} = 2\sum_{n\geq 0}\frac{(-1)^n}{(2n+1)! w^{2n}} $$
hence
$$ (w+1)^2\sin\frac{1}{w} = w +2\sum_{n\geq 0}\frac{(-1)^n}{(2n+1)!w^{2n}}+\sum_{n\geq 0}\left(\frac{1}{(2n+1)!}-\frac{1}{(2n+3)!}\right)\frac{(-1)^n}{w^{2n+1}} $$
where the coefficient of $\frac{1}{w^m}$ depends on the parity of $m$.
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