Can the Laurent series be found in region where there is singularity?

Question

In class my teacher showed that we can use the expansion of $\frac{1}{1-z}$ to find an expansion for $\frac{1}{(z+1)(z+2)}$ in the regions $|z|<1$, $1<|z|<2$ and $|z|>2$

My question is can this method be used to find the expansion in the regions $|z|>1$ and $|z|<2$, if not then what method can be used?

Answer 1

The function

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

Since we want to find a Laurent expansion with center $0$, we look at the poles $-1$ and $-2$ 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 poles $-1$ at the boundary of the disc. In the interior of this disc all two fractions with poles $-1$ and $-2$ 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 poles $-1$ as principal part of a Laurent series at $z=0$, while the fraction with pole at $-2$ 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 \begin{align*} \frac{1}{z+a}&=\frac{1}{a}\cdot\frac{1}{1+\frac{z}{a}}\\ &=\sum_{n=0}^{\infty}\frac{1}{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}\frac{a^n}{(-z)^n} =-\sum_{n=0}^{\infty}\frac{a^n}{(-z)^{n+1}}\\ &=-\sum_{n=1}^{\infty}\frac{a^{n-1}}{(-z)^n} \end{align*}

We can now obtain the Laurent expansion of $f(x)$ at $z=0$ for all three regions

• Region 2: $1<|z|<2$

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

The Laurent expansion for the other regions can be calculated similarly.