I am having quite the trouble combining the coefficient of the Laurent series of $z^2sin(\frac{1}{1-z})$ at $0<\lvert z-1 \rvert<\infty$.
At first, it seems pretty elementary since you can set $w=\frac{1}{z-1}$ and expand at infinity in z, which is 0 in $w$. Therefore, $$z^2sin(\frac{1}{z-1})=(\frac{1}{w^2}+\frac{2}{w}+1)sin(w)$$ $$=(\frac{1}{w^2}+\frac{2}{w}+1)\sum_{n=0}^\infty\frac{(-1)^nw^{2n+1}}{(2n+1)!}$$ $$=\sum_{n=0}^\infty\frac{(-1)^nw^{2n-1}}{(2n+1)!}+2\sum_{n=0}^\infty\frac{(-1)^nw^{2n}}{(2n+1)!}+\sum_{n=0}^\infty\frac{(-1)^nw^{2n+1}}{(2n+1)!}$$
But now, I am having troubles adjusting the sums so that the coefficients coincide into one series. The problem I have right now is the change of indices. I believe it is possible to do it for the 1st and 3rd term so that it becomes one. However, I have difficulties incorporating the term contributing to even powers of the $w$. Is it the case that Laurent series is able to be represented as 2 or more sums. Have I done something wrong??
Looking forward to any hints. Thanks in advance.
$$z^2\sin\frac{1}{1-z}=-((z-1)^2+2(z-1)+1)\sin\frac{1}{z-1}$$ $$sin(\frac{1}{z-1})=\sum_{n=0}^{\infty}(-1)^n\frac{1}{(z-1)^{2n+1}(2n+1)!}$$ $$\to(z-1)^2sin(\frac{1}{z-1})=z-1+\sum_{n=0}^{\infty}(-1)^{n+1}\frac{1}{(z-1)^{2n+1}(2n+3)!}$$ $$2(z-1)sin\frac{1}{z-1}=\sum_{n=0}^{\infty}(-1)^n\frac{2}{(z-1)^{2n}(2n+1)!}$$ therefore $$z^2sin\frac{1}{1-z}=1-z-\sum_{n=0}^{\infty}(-1)^n(\frac{1}{(2n+1)!}-\frac{1}{(2n+3)!})\frac{1}{(z-1)^{2n+1}}+\sum_{n=0}^{\infty}(-1)^n\frac{2}{(z-1)^{2n}(2n+1)!}$$