Expand the Laurent series 
Expand $f(z)= \frac {z}{(z+1)(z-2)}$ in a Laurent series valid for the given annular domain: $0 \lt \lvert z+1 \rvert \lt 3$

I'm having a lot of trouble with this one.  The answer, per the back of the book, is:

$...+ \frac{3^3}{z^5} + \frac{3^2}{z^4} + \frac{3}{z^3} + \frac{1}{z^2}$

I just don't really see how they got that at all.  Some help would be greatly appreciated!
 A: Since you want powers of $z+1$ anyway, I find it much easier, even if it’s a purely mechanical device, to make the substitution $\zeta=z+1$ and $z=\zeta-1$. This gives you
$$
\frac z{(z+1)(z-2)}=\frac{\zeta-1}{\zeta(\zeta-3)}=\frac1{\zeta-3}-\frac1{\zeta(\zeta-3)}\,,
$$
at which point the problem just becomes writing out the power series for $1/(\zeta-3)$. Write this as $-\frac13\frac1{1-\zeta/3}$ to get $-1/3-\zeta/9-\zeta^2/27-\cdots=-\sum_0^\infty\zeta^n/3^{n+1}$, which you see is convergent for $|\zeta|<3$, in other words $|z+1|<3$.
The rest of it I’m sure you can do.
By the way, the book answer is clearly wrong—probably prepared by an overworked and underpaid graduate student. That series is geometric all right, with initial term $1/z^3+3/z^3$ and common ratio $9/z^2$. You’ll want $|9/z^2|<1$, in other words, $3<|z|$, so that series relates to the domain outside your stated annulus. The answer seems to be stated in terms of my $\zeta$, too, but it doesn’t seem to work out correctly.
