Laurent series in various regions I have the question: "Find the Laurent series which represents the function
$$ 
f(z) = (z^2 - 1)/(z + 2)(z + 3)\
$$
in the regions 
(i) $\mid z\mid < 2\ $
(ii) $ 2 < \mid z\mid < 3\ $
(iii) $\mid z\mid > 3\ $"
I can rewrite this function into partial fractions as:
$$
f(z) = 5/(z+2) - 10/(z+3)
$$
I understand that for part (ii) this is the region within the annulus. 
I also know that I can  write:
$$
5/(z+2) = 5/2 \sum (-1)^n (z/2)^n $$ for $ \mid z\mid < 2\ $
$$
10/(z+3) = 10/3 \sum (-1)^n (z/3)^n $$ for $ \mid z\mid < 3\ $
I also know how to find the respective sums for $ \mid z\mid > 2\ $ and $ \mid z\mid > 3\ $ 
But I'm not at all sure how to combine these to get the answers for the whole function $f$, for parts (i), (ii) and (iii).
Any help would be much appreciated

(i) $|z| < 2$
$$\frac5{z+2}-\frac{10}{z+3} = \sum_{n=0}^\infty (-1)^n \left(\frac52\left(\frac{z}2\right)^n-\frac{10}3\left(\frac{z}3\right)^n\right)$$
(ii) $2<|z|<3$
\begin{align*}
  & \frac{5}{z+2} - \frac{10}{z+3} \\
  =& \frac{5}{z} \frac{1}{1+\frac{2}{z}} - \frac{10}{3} \frac{1}{1+\frac{z}{3}} \\
  =& -\frac{10}{3} \sum_{n = 0}^\infty (-1)^n \left( \frac{z}{3} \right)^n + \frac{5}{z} \sum_{n = 0}^\infty (-1)^n \left( \frac{2}{z} \right)^n \\
  =& \sum_{n = 0}^\infty (-1)^{n + 1} \frac{10}{3} \left( \frac{z}{3} \right)^n + \sum_{n = 1}^\infty (-1)^{n - 1} 5 \left( \frac{2}{z} \right)^n
\end{align*}
(iii) $|z|>3$
\begin{align*}
  & \frac{5}{z+2} - \frac{10}{z+3} \\
  =& \frac{5}{z} \frac{1}{1+\frac{2}{z}} - \frac{10}{z} \frac{1}{1+\frac{3}{z}} \\
  =& \frac{1}{z} \sum_{n = 0}^\infty (-1)^n \left( 5 \left( \frac{2}{z} \right)^n - 10 \left( \frac{3}{z} \right)^n \right) \\
  =& \sum_{n = 1}^\infty (-1)^{n - 1} \left( 5 \left( \frac{2}{z} \right)^n - 10 \left( \frac{3}{z} \right)^n \right)
\end{align*}
