Laurent series around certain points 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? 

You're series expansion of $\sin(z)$ isn't quite right. It should be $\sum_{n=0}^\infty \frac{(-1)^n z^{2n+1}}{(2n+1)!}$. Keep in mind that the expansion of $z-3$ about $z=-1$ is $z-3=-4+(z+1)$ Therefore, 
$$ (z-3)\sin\left(\frac{1}{z+1}\right) = \left(-4+(z+1)\right)\sum_{n=0}^\infty \frac{(-1)^n }{(2n+1)! (z+1)^{2n+1}}$$
This is
$$ (z-3)\sin\left(\frac{1}{z+1}\right) = \sum_{n=0}^\infty \frac{4(-1)^{n+1} }{(2n+1)! (z+1)^{2n+1}} + \sum_{n=0}^\infty \frac{(-1)^n }{(2n+1)! (z+1)^{2n}}$$ 
For the second problem, you only want to expand $2/(z-2)$ not both.
$$\frac{z}{(z-1)(z-2)} = \frac{2}{z-2}-\frac{1}{z-1} = \frac{2}{(z-1)-1} - \frac{1}{z-1}$$
$$\frac{z}{(z-1)(z-2)} = -2\sum_{n=0}^\infty (z-1)^n - \frac{1}{z-1}$$
