How to find Laurent expansion

Question

I have been presented with the function $g(z) = \frac{2z}{z^2 + z^3}$ and asked to find the Laurent expansion around the point $z=0$.

I split the function into partial fractions to obtain $g(z) = \frac{2}{z} - \frac{2}{1+z}$, but do not know where to go from here.

Answer 1

So you've obtained $g(x) = \frac{2}{z} - \frac{2}{1+z}$. The first part of it looks to be in the proper form already. How can you change something of the form $\frac{1}{1+z}$ into powers of $z$? You should be thinking of geometric sum.

$$\frac{1}{1+z} = \sum_{n=0}^\infty (-1)^n z^n$$

Answer 2

$$\frac{2z}{z^2+z^3}=\frac {2z}{ z^2}\cdot \frac 1{1+z}=\frac 2 z\cdot\sum_{k=0}^\infty(-1)^kz^k=\sum_{k=-1}^\infty-2(-1)^kz^k$$

No need for fraction decomposition.

Answer 3

The function $$g(z)=\frac{2}{z}-\frac{2}{1+z}$$ has two simple poles at $0$ and $-1$.

We observe the fraction $\frac{2}{z}$ is already the principal part of the Laurent expansion at $z=0$. We can keep the focus on the other fraction.

Since we want to find a Laurent expansion with center $0$, we look at the other pole $-1$ and have to distuinguish two regions.

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

• The first region $ |z|<1$ is a disc with center $0$, radius $1$ and the pole $-1$ at the boundary of the disc. In the interior of this disc the fraction with pole $-1$ admits a representation as power series at $z=0$.

• The second region $1<|z|$ containing all points outside the disc with center $0$ and radius $1$ 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}\frac{1}{1+\frac{z}{a}} =\frac{1}{a}\sum_{n=0}^{\infty}\left(-\frac{1}{a}\right)^nz^n\\ &=-\sum_{n=0}^{\infty}\left(-\frac{1}{a}\right)^{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}\frac{1}{1+\frac{a}{z}}=\frac{1}{z}\sum_{n=0}^{\infty}(-a)^n\frac{1}{z^n}\\ &=\sum_{n=1}^{\infty}(-a)^{n-1}\frac{1}{z^n}\\ \end{align*}

We can now obtain the Laurent expansion of $g(x)$ at $z=0$ for both regions

• Region 1: $|z|<1$

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

• Region 2: $1<|z|$

\begin{align*} g(z)&=\frac{2}{z}-2\sum_{n=1}^{\infty}(-1)^{n-1}\frac{1}{z^n}=2\sum_{n=2}^{\infty}(-1)^{n}\frac{1}{z^n}\\ \end{align*}