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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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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.2012-12-07