# Computing the residue of $\frac{z-2}{z^2} \sin\left(\frac{1}{1-z}\right)$ for $z = 1$.

Consider the function $$f(z) = \frac{z-2}{z^2} \sin\left(\frac{1}{1-z}\right)$$

We have that $0$ is a double pole and $1$ is a single pole (essential singularity) of $f$. It is simple to compute ${\rm Res}(f,0)$. But I'm having trouble computing ${\rm Res}(f,1)$. If I didn't mess up any calculations, so far I have: $$f(z) = (1-(z-1))\left(1+\sum_{n \geq 0}(-1)^n(n+1)(z-1)^n\right)\left(\sum_{n \geq 0}\frac{(-1)^n}{(2n+1)!(z-1)^{2n+1}}\right).$$If I only had power series, I could just use some Cauchy product to do it, but I'm stuck. Wolfram Alpha doesn't compute the residue, and don't compute the Laurent expansion either. Can someone help?