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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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f has a pole of order m at a iff a is a zero of 1/f of multiplicity m. – user938272 May 7 '11 at 13:28
At $z=0$ it is easy (follow the suggestion of user938272). At $z=1$ there is something more nasty happening (check out this – Fabian May 7 '11 at 13:33
For essential singularity, the only way I am aware of finding the residue at the singularity is from its Laurent expansion or if you manage to integrate the function by other means over a closed contour containing the singularity. – user17762 May 7 '11 at 17:03
How do we calculate the Laurent expansion for such a strange function? – Ash May 7 '11 at 17:22
up vote 4 down vote accepted

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 $$ \begin{align} \text{Res}_{z=1}\left[\frac{z-2}{z^2}\sin\left(\frac{1}{1-z}\right) \right] &= \sum_{k=0}^\infty \frac{(-1)^k}{(2k+1)!} (-1)^{2k} (4k+1) \\ &=\sum_{k=0}^\infty \frac{(-1)^k [2 (2k +1 ) -1]}{(2k+1)!} \\ &= 2\cos(1) - \sin(1) \approx 0.24 \end{align}$$

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