I want to compute the laurent series of
i) $(z-3)\sin(\frac{1}{z+1}); (z\ne -1)$ at $z_{0}=-1 $.
ii) $\displaystyle{\frac{z}{(z-1)(z-2)}}; ( z\ne 1,2) $ at $z_{0}=1$
and classify their singularities.
A Laurent series has the form: $\displaystyle{f(z)= \sum_{n=1}^{\infty}c_{n}\frac{1}{(z-z_{0})^{n}}}+\sum_{n=0}^{\infty} b_{n}(z-z_{0})^{n}$
my attempts :
i) $\displaystyle{\sin(z):= \sum_{n=0}^{\infty} \frac{(-1)^{n}(x+1)^{2n+1}}{(2n+1)!}}$ at $z_{0}=-1 $ so for $x=\frac{1}{z+1}$ it follows :
$\displaystyle{(z-3)(\sin(x)) = (z-3) \sum_{n=0}^{\infty} \frac{(-1)^{n}(\frac{1}{1+z}+1)^{2n+1}}{(2n+1)!}}$ then I get stuck .
ii) partial fraction decomposition leads to: $\displaystyle{\frac{2}{z-2}-\frac{1}{z-1}}$ I can write these as taylor series . It is : $\displaystyle{\frac{1}{1-x} = \sum_{n=0}^{\infty}(x-1)^{n}}$
So it follows: $\frac{2}{z-2}-\frac{1}{z-1} = 2\sum_{n=0}^{\infty}(z-4)^{n} + \sum_{n=0}^{\infty}(z-2)^{n}$
then I get stuck
What is the right path?