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?
Best Answer
$\sin \xi = \xi -\frac {\xi^{3}} {3!}+\frac {\xi^{5}} {5!}...$. 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 $f$.