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The residue of $f(z)=z\sin\frac{1}{1-z}$ at $z=1$. Any hint is appreciated.

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up vote 2 down vote accepted

Rearranging to get everything in terms of $(z-1)$ might help:

$$z\sin\frac{1}{1-z}=-((z-1)+1)\sin\frac{1}{z-1}=-(z-1)\sin\frac{1}{z-1}-\sin \frac{1}{z-1}.$$ You could use the beginning of the power series of $\sin$.

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Thank you very much, @Jonas. I am clear. – Sam Dec 7 '12 at 3:55

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