Laurent expansions of $\frac{1}{z-1}$

Question

I want to calculate Laurent expansion of $\frac{1}{z-1}$ thtat are valid in the annuli

$\begin{align} (a) & \;\;1<|z|<3\\ (b) & \;\;0<|z-3|<2 \end{align}$

For part $(a)$ since $|z|>1$ we use the trick $$\frac{1}{z-1} = \frac{1}{z(1-\frac{1}{z})}$$ which can be turned into a geometric series.

For part $(b)$ why did the book used $$\frac{1}{z-1} = \frac{1}{2(1+\frac{z-3}{2})}.$$ I know $|\frac{z-3}{2}| < 1$ but why can't the the method in part $(a)$ be applied here, since we also have $|z| > 1$.

The original problem was $\frac{1}{(z-1)(z-3)}$ I think this might effect the two different calculations because $z$ can not equal to $3$ in the expansion? And suppose we do not have the factor $(z-3)$ on the bottom, could the method in $(a)$ applied to $(b)$?

Answer 1

Note: The regions of convergence are somewhat different than you might think. Here I provide a complete Laurent expansion of $f$ around $z=0$. From this you should be able to check your results.

The function

\begin{align*} f(z)&=\frac{1}{(z-1)(z-3)}\\ &=-\frac{1}{2}\frac{1}{z-1}+\frac{1}{2}\frac{1}{z-3}\\ \end{align*} has two simple poles at $1$ and at $3$.

Since we want to find a Laurent expansion with center $0$, we look at the poles $1$ and $3$ and see they determine three regions.

\begin{align*} |z|<1,\qquad\quad 1<|z|<3,\qquad\quad 3<|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 all two fractions with poles $1$ and $3$ admit a representation as power series at $z=0$.

• The second region $1<|z|<3$ is the annulus with center $0$, inner radius $1$ and outer radius $3$. 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 $3 admits a representation as power series.

• The third region $|z|>3$ containing all points outside the disc with center $0$ and radius $3$ 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}} =\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=1}^{\infty}(-a)^{n-1}\frac{1}{z^n} \end{align*}

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

• Region 1: $\ |z|<1$

\begin{align*} f(z)&=\frac{1}{2}\sum_{n=0}^{\infty}z^n -\frac{1}{2}\sum_{n=0}^{\infty}\frac{1}{3^n}z^n\\ &=\frac{1}{2}\sum_{n=0}^{\infty}\left(1-\frac{1}{3^n}\right)z^n \end{align*}

• Region 2: $\ 1<|z|<3$

\begin{align*} f(z)&=-\frac{1}{2}\sum_{n=1}^{\infty}\frac{1}{z^n} -\frac{1}{2}\sum_{n=0}^{\infty}\frac{1}{3^n}z^n \end{align*}

• Region 3: $\ 3<|z|$

\begin{align*} f(z)&=-\frac{1}{2}\sum_{n=1}^{\infty}\frac{1}{z^n} +\frac{1}{2}\sum_{n=1}^{\infty}3^{n-1}\frac{1}{z^n}\\ &=\frac{1}{2}\sum_{n=1}^{\infty}\left(3^{n-1}-1\right)\frac{1}{z^n}\\ \end{align*}

---

Note: According to OPs comment let's assume a slightly different situation:

Challenge: Find the Laurent series expansion of $$f(z)=\frac{1}{(z-1)(z-3)}$$ at $z_0=3$ in the region $$0<|z-3|<2$$

We can immediately conclude from the representation of $f$ that $1$ and $3$ are simple poles.

Note, that in case of isolated singularities the radius of convergence is the distance from the center $z_0$ to the nearest singularity ($\neq z_0$).

Since $z_1=1$ is the nearest singularity to $z_0=3$ we have a radius of convergence $R=2$ and since $z_0$ is already a singularity we have a punctured disc as region of convergence \begin{align*} 0<|z-3|<2 \end{align*}

When looking at \begin{align*} f(z)=-\frac{1}{2}\frac{1}{z-1}+\frac{1}{2}\frac{1}{z-3} \end{align*} we observe the term $\frac{1}{2}\frac{1}{z-3}$ is already the principal part of the Laurent expansion at $z_0=3$ and since $\frac{1}{2}\frac{1}{z-1}$ is analytic at $z_0=3$ we can expand it as power series.

\begin{align*} f(z)&=-\frac{1}{2}\frac{1}{z-1}+\frac{1}{2}\frac{1}{z-3}\\ &=\frac{1}{2}\frac{1}{z-3}-\frac{1}{2}\frac{1}{(z-3)+2}\\ &=\frac{1}{2}\frac{1}{z-3}-\frac{1}{4}\frac{1}{1+\frac{z-3}{2}}\\ &=\frac{1}{2}\frac{1}{z-3}-\frac{1}{4}\sum_{n=0}^\infty \left(-\frac{1}{2}\right)^n(z-3)^n\\ &=\sum_{n=-1}^\infty \frac{(-1)^{n+1}}{2^{n+2}}(z-3)^n \end{align*}

Answer 2

For rational functions, the idea is essentially always the same: transform the given annulus boundary conditions $r < \lvert z - a \rvert < R$ into the two equivalent conditions $\lvert R^{-1} (z - a) \rvert < 1$ and $\lvert r(z - a)^{-1} \rvert < 1$, and then typically use partial fraction decomposition to separate your function into terms involving $1/(1-R^{-1}(z - a))$ and $1/(1-r(z - a)^{-1})$ to express them using convergent geometric series. A typical example of application can be found here, for instance.