How to grasp the relationship between Laurent series depending on the region they are developed at

Question

in trying to understand how to set up the Laurent series for a fractional expression, I have the given function

\begin{equation} \frac{1}{(z-2)}-\frac{1}{(z-1)} \end{equation}

The Laurent series of each of the two fractions are evaluated over the three intervals:

a) On the ring $|z|<1$ the series are:

\begin{equation} \frac{1}{(z-1)}=-\frac{1}{(1-z)}=\sum_{n=0}^{\infty}(-1)^nz^n \end{equation}

\begin{equation} \frac{1}{(z-2)}=-\frac{1}{2}\frac{1}{(1-z/2)}=\sum_{n=0}^{\infty}(-1)^n\frac{z^n}{2^{n+1}} \end{equation}

b) On the annulus $1<|z|<2$

\begin{equation} \frac{1}{(z-1)}=\frac{1}{z}\frac{1}{(1-1/z)}=\sum_{n=0}^{\infty}\frac{1}{z^{n+1}} \end{equation}

\begin{equation} \frac{1}{(z-2)}=-\frac{1}{2}\frac{1}{(1-z/2)}=\sum_{n=0}^{\infty}(-1)^n\frac{z^n}{2^{n+1}} \end{equation}

c) On the open region $|z|>2$

\begin{equation} \frac{1}{(z-1)}=\frac{1}{z}\frac{1}{(1-1/z)}=\sum_{n=0}^{\infty}\frac{1}{z^{n+1}} \end{equation}

\begin{equation} \frac{1}{(z-2)}=\frac{1}{z}\frac{1}{(1-2/z)}=\sum_{n=0}^{\infty}\frac{2^n}{z^{n+1}} \end{equation}

If one looks closely, whenever the fractions are evaluated over an open disk, a factor of $\frac{1}{z}$ is extracted from the fraction, so that one obtains the form $\sum_{n=0}^\infty \frac{a^n}{z^{n+1}}$. Then whenever the fractions are evaluated over a closed disk, one seeks a solution in the form of $\sum_{n=0}^\infty \frac{z^n}{a^{n(+1)}}$, with alternating sign. When one considers an annulus squeezed between the two other regions, then the fraction with the pole close to the upper bound is evaluated as for the closed ring, while the fraction with the pole whose value is closer to the lower bound, is rather "inverted" such that one seeks a solution by extracting the factor $\frac{1}{z}$ from the denominator. There is an evident pattern here, but can it be generalized in some understandable formula or explanation?

Thanks

Answer 1

The function \begin{align*} &f:\mathbb{C}\setminus\{1,2\}\to\mathbb{C}\\ &f(z)=\frac{1}{z-2}-\frac{1}{z-1} \end{align*} has simple poles at $z=1$ and $z=2$. Let's assume we want a Laurent expansion of $f$ at $z=0$. We distinguish three regions of convergence \begin{align*} \color{blue}{D_1:}&\color{blue}{\quad 0\leq |z|<1}\\ \color{blue}{D_2:}&\quad \color{blue}{1<|z|<2}\\ \color{blue}{D_3:}&\quad \color{blue}{ |z|>2} \end{align*}

We recall if a function $g$ admits a power series expansion \begin{align*} g(z)=\sum_{n=1}^{\infty}a_nz^n \end{align*} with positive convergence radius $R$ the series converges for $|z|<R$ and diverges for $|z|>R$ whereas no general statement at the boundary of the disc can be given. This implies that \begin{align*} g\left(\frac{1}{z}\right)=\sum_{n=1}^{\infty}a_n\frac{1}{z^n} \end{align*} converges for $\left|\frac{1}{z}\right|<R$ i.e. converges for $|z|>R$. With this in mind we can analyse the regions $D_1, D_2$ and $D_3$.

• 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.

Region $D_1: 0\leq |z|<1$

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

Region $D_2: 1<|z|<2$

We obtain \begin{align*} \color{blue}{f(z)}&=\frac{1}{z-2}-\frac{1}{z-1}\\ &=-\frac{1}{2}\cdot\frac{1}{1-\frac{z}{2}}-\frac{1}{z}\cdot\frac{1}{1-\frac{1}{z}}\\ &=-\frac{1}{2}\sum_{n=0}^{\infty}\frac{1}{2^n}z^n-\sum_{n=0}^{\infty}\frac{1}{z^{n+1}}\\ &\,\,\color{blue}{=-\sum_{n=0}^{\infty}\frac{1}{2^{n+1}}z^n-\sum_{n=1}^{\infty}\frac{1}{z^n}} \end{align*}

Region $D_3: |z|>2$

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