How to expand $f(z)$ as a Laurent Series

Question

Given

$$\frac{z}{(z-1)(z+2i)}$$

expand $f(z)$ in the following regions: $|z|<1$, $1<|z|<2$, $|z|>2$

I'm preparing for an exam and Laurent Series are a weakness of mine. I would love advice regarding interpretation, understanding, and solution of the problem.

Answer 1

Here is a somewhat detailed description of the Laurent expansion of $f(z)$ around $z=0$.

The function

\begin{align*} f(z)&=\frac{z}{(z-1)(z+2i)} =\frac{1-2i}{5}\cdot\frac{1}{z-1}+\frac{4+2i}{5}\cdot\frac{1}{z+2i}\\ \end{align*} has two simple poles at $1$ and $-2i$.

Since we want to find a Laurent expansion with center $0$, we look at the poles $1$ and $-2i$ and see they determine three regions.

\begin{align*} |z|<1,\qquad\quad 1<|z|<2,\qquad\quad 2<|z| \end{align*}

• The first region $ |z|<1$ is a disc with center $0$, radius $1$ and the pole $-2i$ at the boundary of the disc. In the interior of this disc all two fractions with poles $0$ and $-2i$ admit a representation as power series at $z=0$.

• The second region $1<|z|<2$ is the annulus with center $0$, inner radius $1$ and outer radius $2$. Here we have a representation of the fraction with pole $1$ as principal part of a Laurent series at $z=0$, while the fraction with pole at $-2i$ admits a representation as power series.

• The third region $|z|>2$ containing all points outside the disc with center $0$ and radius $2$ admits for all fractions a representation as principal part of a Laurent series at $z=0$.

A power series expansion of $\frac{1}{z+a}$ at $z=0$ is using geometric series expansion \begin{align*} \frac{1}{z+a}&=\frac{1}{a}\cdot\frac{1}{1+\frac{z}{a}} =\frac{1}{a}\sum_{n=0}^{\infty}\left(-\frac{z}{a}\right)^n\\ &=\sum_{n=0}^{\infty}\frac{(-1)^n}{a^{n+1}}z^n \end{align*} The principal part of $\frac{1}{z+a}$ at $z=0$ is \begin{align*} \frac{1}{z+a}&=\frac{1}{z}\cdot\frac{1}{1+\frac{a}{z}} =\frac{1}{z}\sum_{n=0}^{\infty}\left(-\frac{a}{z}\right)^n =\sum_{n=0}^{\infty}(-a)^n\frac{1}{z^{n+1}}\\ &=\sum_{n=1}^{\infty}\left(-a\right)^{n-1}\frac{1}{z^n} \end{align*}

We can now obtain the Laurent expansion of $f(z)$ at $z=0$ for all three regions. Here we consider the second region

• Region 2: $1<|z|<2$

\begin{align*} f(z)&=\frac{1-2i}{5}\frac{1}{z-1}+\frac{4+2i}{5}\frac{1}{z+2i}\\ &=\frac{1-2i}{5}\sum_{n=1}^{\infty}\frac{1}{z^n} +\frac{4+2i}{5}\sum_{n=0}^{\infty}\frac{(-1)^n}{(2i)^{n+1}}z^{n}\\ &=\frac{1-2i}{5}\sum_{n=1}^{\infty}\frac{1}{z^n} +\frac{4+2i}{5}\cdot\frac{1}{2i}\sum_{n=0}^{\infty}\left(-\frac{1}{2i}\right)^nz^n\\ &=\frac{1-2i}{5}\left(\sum_{n=1}^{\infty}\frac{1}{z^n} +\sum_{n=0}^{\infty}\left(\frac{i}{2}\right)^nz^n\right)\\ \end{align*}

The other regions can be calculated similarly.