Determining Laurent Series expansion and residues Determining Laurent Series expansion and residues of $f(z)=\frac{z}{(z+1)(z+2)}$ around $z = -2$. What is the validity of the expanded region? What is $res(f, -2)$??

A related technique. Here is how you advance.
$$ f(z)=\frac{z}{(z+1)(z+2)} = \frac{(z+2)-2}{(z+1)(z+2)}$$ 
$$= \frac{1}{(z+1)} - \frac{2}{(z+2)( (z+2) - 1 ) } $$
$$ = \frac{1}{((z+2)-1)} - \frac{2}{(z+2)( (z+2) - 1 ) } $$
$$ = \frac{1}{(z+2)(1-1/(z+2))} - \frac{2}{(z+2)^2( 1-1/(z+2)  ) } $$
$$ = \frac{1}{z+2}(1-1/(z+2))^{-1} - \frac{1}{(z+2)^2}(1-1/(z+2))^{-1}. $$
Now you need to expand $(1-1/(z+2))^{-1}$ using the geometric series. For residues see here.
