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Find the Laurent series in the region $0 < \lvert z-1 \rvert < \infty$ of function $$ f(z) = (z+1)\sin{\frac{1}{(z-1)^2}}. $$

As far as I understand, this is the same as finding the Laurent series at $z=\infty$, but I do not know how to proceed with this. When we had rational functions, then we could manipulate it and use the geometric series formula, but what to do when there is a sine involved? Can I use the Taylor series for $\sin{z}$ somehow for this?

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$\sin \xi = \xi -\frac {\xi^{3}} {3!}+\frac {\xi^{5}} {5!}...$. Just put $\xi =\frac 1{(z-1)^{2}}$, write $z+1$ as $(z-1) +2$ and collect the coefficients of powers of $z-1$. You will get the Laurent series for $f$.

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  • $\begingroup$ So I would get $$ \sin{\frac{1}{(z-1)^2}} = \sum_{n=0}^\infty \frac{(-1)^n}{(2n+1)! \cdot (z-1)^{4n+2}} $$ And then I would multiply this by $(z-1)+2$? $\endgroup$ Oct 31 '19 at 8:52
  • $\begingroup$ @MarkusPunnar Yes, that is right. $\endgroup$ Oct 31 '19 at 8:57

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