Laurent Series Expansion for $f(z)=\dfrac{z+2}{(z+1)(z-2)}$ in $\{1<|z|<2\}$ and $\{2<|z|<\infty\}$

Question

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?

Answer 1

For $\;|z|>2\;$ :

$$-\frac13\frac1{1+z}+\frac43\frac1{z-2}=-\frac1{3z}\frac1{1+\frac1z}+\frac4{3z}\frac1{1-\frac2z}=$$

$$=-\frac1{3z}\left[\left(1-\frac1z+\frac1{z^2}-\frac1{z^3}+\ldots+\right)-4\left(1+\frac2z+\frac{2^2}{z^2}+\ldots\right)\right]=$$

$$=-\frac1{3z}\left(-3-\frac9z-\frac{15}{z^2}-\ldots\right)=\frac1z+\frac3{z^2}+\frac5{z^3}+\ldots$$

When $\;1<|z|<2$ :

$$-\frac13\frac1{1+z}+\frac43\frac1{z-2}=-\frac1{3z}\frac1{1+\frac1z}-\frac23\frac1{1-\frac z2}=$$

$$=-\frac13\left[\frac1z\left(1-\frac1z+\frac1{z^2}-\frac1{z^3}+\ldots\right)+2\left(1+\frac z2+\frac{z^2}{2^2}+\ldots\right)\right]=$$

$$=-\frac13\left(\ldots-\frac1{z^4}+\frac1{z^3}-\frac1{z^2}+\frac1z+2+z+\frac{z^2}2+\ldots\right)=\sum_{n=-\infty}^\infty a_nz^n$$

with

$$ a_n=\begin{cases}\frac{(-1)^n}{3}&,\;\;\;n<0\\{}\\ -\frac1{3\cdot2^{n-1}}&,\;\;n\ge 0\end{cases}$$

and thus I agree with you and not with the book.