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Given the complex function $f(z) = \frac{z-2}{z^2}\sin(\frac{1}{1-z})$, how can we calculate the residue at the essential singularity at $z = 1$?

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Let us define $\xi=z-1$ (such that we can calculate the residue at $\xi=0$). The function is given by $\frac{1-\xi}{(1+\xi)^2} \sin(\xi^{-1}).$ For the residue, we need to obtain the coefficient in front of $\xi^{-1}$ in the Laurent expansion around $\xi=0$. We have the Laurent expansions $\sin(\xi^{-1}) = \sum_{k=0}^\infty \frac{(-1)^k}{(2k+1)!\xi^{2k+1}}$ and $\frac{1-\xi}{(1+\xi)^2} = \sum_{k=0}^\infty (-1)^k (2k+1) \xi^k$ valid for $0<|\xi|<1$.

As the original function is the product of these functions, the residue (= coefficient in front of $\xi^{-1}$) is given by $ \begin{align} \text{Res}_{z=1}\left[\frac{z-2}{z^2}\sin\left(\frac{1}{1-z}\right) \right] &= \sum_{k=0}^\infty \frac{(-1)^k}{(2k+1)!} (-1)^{2k} (4k+1) \\ &=\sum_{k=0}^\infty \frac{(-1)^k [2 (2k +1 ) -1]}{(2k+1)!} \\ &= 2\cos(1) - \sin(1) \approx 0.24 \end{align}$