Complex Analysis – Calculating Laurent Series Expansion

Question

I have to calculate the Laurent series expansion of $$f(z) = \frac {2z−2}{(z+1)(z−2)}$$ in $1 < |z| < 2$ and $|z| > 3$.

For first annulus, I know I must manipulate the given expression to contain terms $1/z$ and $z/2$ so that some expansion is valid for $|\frac1{z}|<1$ and $|\frac z{2}|<1$, so I try doing that.

Decomposing into partial fractions,
$$f(z) = \frac 4{3} (\frac 1{z+1}) + \frac 2{3}(\frac 1{z-2})$$
$$= \frac4{3z} \frac1{1-\frac {(-1)}{z}} – \frac1{3} \frac 1{1 – \frac z{2}}$$

So can I now expand the $2$ terms by a G.P. for $|\frac1{z}|<1$ and $|\frac z{2}|<1$ to get the Laurent series?

For second case of $|\frac 3{z}|<1 $, how should I proceed?

Answer 1

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*}