Calculating Laurent series expansion

The function \begin{align*} f(z)&=\frac{2z-2}{(z+1)(z+2)}\\ &= \frac{4}{3}\left( \frac{1}{z+1}\right) + \frac{2}{3}\left(\frac{1}{z-2}\right)\\ \end{align*} is to expand around the center $z=0$ in $1<|z|<2$ and $|z|>3$.

Since there are simple poles at $z=-1$ and $z=2$ we have to distinguish three regions of convergence \begin{align*} D_1:&\quad 0\leq |z|<1\\ D_2:&\quad 1<|z|<2\\ D_3:&\quad |z|>2 \end{align*}

• The first region $D_1$ is a disc with center $z=0$, radius $1$ and the pole at $z=-1$ at the boundary of the disc. It admits for both fractions a representation as power series.

• The region $D_2$ is an annulus containing all points outside the closure of $D_1$ and the closure of $D_3$. It admits for the fraction with pole at $z=-1$ a representation as principal part of a Laurent series and for the fraction with pole at $z=2$ a power series.

• The region $D_3$ contains all points outside the disc with center $z=0$ and radius $2$. It admits for both fractions a representation as principal part of a Laurent series.

Since we are interested in an expansion for $1<|z|<2$ we consider the expansion in $D_2$.

Expansion in $D_2$:

\begin{align*} f(z)&=\frac{4}{3}\left(\frac{1}{z+1}\right) + \frac{2}{3}\left(\frac{1}{z-2}\right)\\ &=\frac{4}{3z}\left(\frac{1}{1+\frac{1}{z}}\right)-\frac{1}{3}\left(\frac{1}{1-\frac{1}{2}z}\right)\\ &=\frac{4}{3}\sum_{n=0}^\infty\frac{(-1)^{n}}{z^{n+1}}+\frac{1}{3}\sum_{n=0}^\infty\frac{1}{2^n}z^n\\ &=\frac{4}{3}\sum_{n=1}^\infty\frac{(-1)^{n-1}}{z^{n}}+\frac{1}{3}\sum_{n=0}^\infty\frac{1}{2^n}z^n\\ \end{align*}

Since we are interested in an expansion for $|z|>3$ we consider the expansion in $D_3$.

Expansion in $D_3$:

\begin{align*} f(z)&=\frac{4}{3}\left(\frac{1}{z+1}\right) + \frac{2}{3}\left(\frac{1}{z-2}\right)\\ &=\frac{4}{3}\sum_{n=1}^\infty\frac{(-1)^{n-1}}{z^{n}}+\frac{2}{3z}\left(\frac{1}{1-\frac{2}{z}}\right)\\ &=\frac{4}{3}\sum_{n=1}^\infty\frac{(-1)^{n-1}}{z^{n}}+\frac{1}{3}\sum_{n=0}^\infty\frac{2^{n+1}}{z^{n+1}}\\ &=\frac{4}{3}\sum_{n=1}^\infty\frac{(-1)^{n-1}}{z^{n}}+\frac{1}{3}\sum_{n=1}^\infty\frac{2^n}{z^{n}}\\ &=\frac{1}{3}\sum_{n=1}^\infty\left(2^n-4(-1)^{n}\right)\frac{1}{z^{n}}\\ \end{align*}

The trick when solving Laurent Series is to use the following geometric serie: $$\frac{1}{1-w} = \sum _0^{\infty} (w)^n \text{ for } | w | < 1$$ and let $w = z-z_0$, where $z_0$ is the point of expansion. But since you sometimes would like to have a series which is valid outside instead of inside the circle |z| = 1 we can insert $\frac{1}{w} = w$ in the geometric series above $$\frac{1}{1-\frac{1}{w}} = \sum _0^{\infty} (\frac{1}{w})^n \text{ for } |{w}| > 1$$ this is great since this series is valid outside of the circle |z| = 1. Also the regions for the series can be alternated by chaning the formula a bit, let $a$ and $b$ be constants, then:

1)$$\qquad \frac{1}{1-a \cdot w} = \sum _0^{\infty} (a \cdot w)^n \text{ for } |{w}| > \frac{1}{|a|}$$ 2)$$\qquad \frac{1}{1-\frac{b}{w}} = \sum _0^{\infty} (\frac{b}{w})^n \text{ for } |{w}| > |b| $$ hence, we are able to create series for whatever region you would like in the complex plan. This is the principal idea behind solving these kind of problems and everything is explained in this video, with some examples which might help you solve your examples: https://www.youtube.com/watch?v=RC15R-ktnUI