I'm trying to get the Laurent Series expansion of the function stated in the title in the stated regions. My approach is as follows:
We can first break up $f(z)$ using partial fractions
$$\frac{z+2}{(z+1)(z-2)} = \frac{-\frac{1}{3}}{z+1}+\frac{\frac{4}{3}}{z-2}$$
For the first region, $\{1<|z|<2\}$, we can rewrite the two fractions above as
$$ \begin{align*} \frac{-\frac{1}{3}}{z+1}+\frac{\frac{4}{3}}{z-2}&= \left(-\frac{1}{3z}\frac{1}{1+\dfrac{1}{z}}\right) -\left(\frac{2}{3}\frac{1}{1-\dfrac{z}{2}}\right) \\ &=\left[-\frac{1}{3z}\sum_{n=0}^{\infty}(-1)^n\left(\frac{1}{z}\right)^n\right]-\left[\frac{2}{3}\sum_{n=0}^{\infty}\left(\frac{z}{2}\right)^n\right]\\ &=\sum_{-\infty}^{\infty}a_nz^n, \begin{cases} \dfrac{(-1)^n}{3},& n<0\\ \dfrac{-1}{3\cdot2^{n-1}},& n\geq0 \end{cases} \end{align*} $$
Now for the region $\{2<|z|<\infty\}$, I just divide the top and bottom of that second fraction, $\left(\dfrac{2}{3}\dfrac{1}{1-\dfrac{z}{2}}\right)$, by $z$, which would give me an expansion valid in the whole region.
The problem with all of this is that the book I'm using gives me a different answer than what I got. The answer in the book for the first region is
$$\sum_{-\infty}^{\infty}a_nz^n, \begin{cases} \dfrac{(-1)^n}{3},& n<0\\ \dfrac{-4}{3^{n+2}},& n\geq0 \end{cases}$$
Which doesn't seem to be quite what I got.
Have I made the correct approach? Could you nudge me in the right direction?