5.1 Laurent expansion

1. Laurent expansion around $z=0$ is expressed by

$\displaystyle \sum_{n=1}^{\infty}\frac{a_{n}}{z^{n}} + \sum_{0}^{\infty}{a_{n}z^{n}}$

To find Laurent expansion, it is useful to know the following Taylor expansion..

$\displaystyle e^{z}$ $\displaystyle =$ $\displaystyle 1 + z + \frac{z^2}{2!} + \frac{z^3}{3!} + \cdots +$  
$\displaystyle \sin{z}$ $\displaystyle =$ $\displaystyle z - \frac{z^3}{3!} - \frac{z^5}{5!} + \cdots +$  
$\displaystyle \cos{z}$ $\displaystyle =$ $\displaystyle 1 - \frac{z^2}{2!} + \frac{z^4}{4!} - \cdots +$  
$\displaystyle \frac{1}{1-z}$ $\displaystyle =$ $\displaystyle 1 + z + z^2 + z^3 + z^4 + \cdots +   ($here$\displaystyle , \vert z\vert < 1)$  

(a) Since $\vert z\vert < 1$, $\frac{1}{1-z}$ can be Taylor expanded. First we expand $\frac{1}{z^2 - 3z + 2}$ by using partial fraction expansion. Then

$\displaystyle \frac{1}{z^2 - 3z + 2}$ $\displaystyle =$ $\displaystyle \frac{1}{(z-1)(z-2)} = \frac{A}{z-1} + \frac{B}{z-2}$  
  $\displaystyle =$ $\displaystyle \frac{-1}{z-1} + \frac{1}{z-2}$  

Note that $\frac{-1}{z-1} = \frac{-1}{1-z}$. Then

$\displaystyle \frac{-1}{z-1} = \frac{-1}{1-z} = -(1 + z + z^2 + z^3 + z^4 + \cdots) = -\sum_{n=0}^{\infty}z^{n}$

Next since $\vert z\vert < 1$, we write $\frac{1}{z-2}$ by

$\displaystyle \frac{1}{z-2} = \frac{-1}{2}\frac{1}{1 - \frac{z}{2}}$

Note that $\vert\frac{z}{2}\vert < 1$. Then we can use Taylor expansion.
$\displaystyle \frac{1}{z-2}$ $\displaystyle =$ $\displaystyle \frac{-1}{2}\frac{1}{1 - \frac{z}{2}}$  
  $\displaystyle =$ $\displaystyle -\frac{1}{2}(1 + \frac{z}{2} + (\frac{z}{2})^2 + (\frac{z}{2})^3 + \cdots + = \sum_{n=0}^{\infty}(\frac{z}{2})^{n}$  

Therefore,
$\displaystyle \frac{1}{z^2 - 3z + 2}$ $\displaystyle =$ $\displaystyle \frac{-1}{z-1} + \frac{1}{z-2}$  
  $\displaystyle =$ $\displaystyle -\sum_{n=0}^{\infty}z^{n} + \sum_{n=0}^{\infty}(\frac{z}{2})^{n} = \sum_{n=0}^{\infty}(z^{n} - (\frac{z}{2})^{n})$  
  $\displaystyle =$ $\displaystyle \sum_{n=0}^{\infty}(1 - \frac{1}{2^{n}}z^{n})$  

(b) Since $1 < \vert z\vert < 2$, $\frac{1}{1-z}$ can not be Taylor expanded. But if we write

$\displaystyle \frac{1}{1-z} = \frac{-1}{z}\frac{1}{1 - \frac{1}{z}}$

Then $\vert\frac{1}{z}\vert < 1$. Thus we can use Taylor expansion. We expand $\frac{1}{z^2 - 3z + 2}$ using partial fraction expansion.
$\displaystyle \frac{1}{z^2 - 3z + 2}$ $\displaystyle =$ $\displaystyle \frac{1}{(z-1)(z-2)} = \frac{A}{z-1} + \frac{B}{z-2}$  
  $\displaystyle =$ $\displaystyle \frac{-1}{z-1} + \frac{1}{z-2}$  

Note that since $\vert z\vert > 1$, we write $\frac{1}{z-1}$ by $\frac{-1}{z-1} = \frac{-1}{z}\frac{1}{1-\frac{1}{z}}$.
$\displaystyle \frac{-1}{z-1}$ $\displaystyle =$ $\displaystyle \frac{-1}{z}\frac{1}{1-\frac{1}{z}} = -\frac{1}{z}(1 + \frac{1}{z} + (\frac{1}{z})^2 + + \cdots)$  
  $\displaystyle =$ $\displaystyle -\frac{1}{z}\sum_{n=0}^{\infty}(\frac{1}{z})^{n}$  
  $\displaystyle =$ $\displaystyle -\sum_{n=0}^{\infty}(\frac{1}{z})^{n+1}$  

Next since $\vert z\vert < 2$, we write $\frac{1}{z-2}$ by

$\displaystyle \frac{1}{z-2} = \frac{-1}{2}\frac{1}{1 - \frac{z}{2}}$

Note that $\vert\frac{z}{2}\vert < 1$. Then we can use Taylor expansion.
$\displaystyle \frac{1}{z-2}$ $\displaystyle =$ $\displaystyle \frac{-1}{2}\frac{1}{1 - \frac{z}{2}}$  
  $\displaystyle =$ $\displaystyle -\frac{1}{2}(1 + \frac{z}{2} + (\frac{z}{2})^2 + (\frac{z}{2})^3 + \cdots + = -\sum_{n=0}^{\infty}(\frac{z}{2})^{n}$  

Therefore,
$\displaystyle \frac{1}{z^2 - 3z + 2}$ $\displaystyle =$ $\displaystyle \frac{-1}{z-1} + \frac{1}{z-2}$  
  $\displaystyle =$ $\displaystyle -\sum_{n=0}^{\infty}(\frac{1}{z})^{n+1} - \sum_{n=0}^{\infty}(\frac{z}{2})^{n}$  

(c) Since $\vert z\vert > 2$, $\frac{1}{1-z}$ can not be Taylor expanded. But if we write

$\displaystyle \frac{1}{1-z} = \frac{-1}{z}\frac{1}{1 - \frac{1}{z}}$

Then $\vert\frac{1}{z}\vert < 1$. Thus it can be Taylor expanded. We expand $\frac{1}{z^2 - 3z + 2}$ using partial fraction expansion.
$\displaystyle \frac{1}{z^2 - 3z + 2}$ $\displaystyle =$ $\displaystyle \frac{1}{(z-1)(z-2)} = \frac{A}{z-1} + \frac{B}{z-2}$  
  $\displaystyle =$ $\displaystyle \frac{-1}{z-1} + \frac{1}{z-2}$  

Note that since $\vert z\vert > 2$, we write $\frac{1}{z-1}$ by $\frac{-1}{z-1} = \frac{-1}{z}\frac{1}{1-\frac{1}{z}}$.
$\displaystyle \frac{-1}{z-1}$ $\displaystyle =$ $\displaystyle \frac{-1}{z}\frac{1}{1-\frac{1}{z}} = -\frac{1}{z}(1 + \frac{1}{z} + (\frac{1}{z})^2 + + \cdots)$  
  $\displaystyle =$ $\displaystyle -\frac{1}{z}\sum_{n=0}^{\infty}(\frac{1}{z})^{n}$  
  $\displaystyle =$ $\displaystyle -\sum_{n=0}^{\infty}(\frac{1}{z})^{n+1}$  

Next since $\vert z\vert > 2$, we write $\frac{1}{z-2}$ by

$\displaystyle \frac{1}{z-2} = \frac{1}{z}\frac{1}{1 - \frac{2}{z}}$

Here $\vert\frac{2}{z}\vert < 1$. Then it can be Taylor expanded. Thus
$\displaystyle \frac{1}{z-2}$ $\displaystyle =$ $\displaystyle \frac{1}{z}\frac{1}{1 - \frac{2}{z}}$  
  $\displaystyle =$ $\displaystyle \frac{1}{z}(1 + \frac{2}{z} + (\frac{2}{z})^2 + (\frac{2}{z})^3 + \cdots + = \sum_{n=0}^{\infty}\frac{2}{z^{n+1}}$  

Therefore,
$\displaystyle \frac{1}{z^2 - 3z + 2}$ $\displaystyle =$ $\displaystyle \frac{-1}{z-1} + \frac{1}{z-2}$  
  $\displaystyle =$ $\displaystyle -\sum_{n=0}^{\infty}(\frac{1}{z})^{n+1} + \sum_{n=0}^{\infty}\frac{2}{z^{n+1}}$  
  $\displaystyle =$ $\displaystyle -\sum_{n=0}^{\infty}\frac{2^{n} -1}{z^{n+1}}$  

2.

(a) Since it is a Laurent expansion with $z=0$, we do not do anything with $\frac{1}{z^3}$. Thus we expand $\frac{1}{z+1}$ using Taylor expansion. Then

$\displaystyle \frac{1}{z+1} = \frac{1}{1 - (-z)} = 1 +(-z) + (-z)^2 + (-z)^3 + \cdots = \sum_{n=0}^{\infty}(-z)^{n}$

Therefore,
$\displaystyle \frac{1}{z^{3}(z+1)}$ $\displaystyle =$ $\displaystyle \frac{1}{z^{3}}\frac{1}{z+1}$  
  $\displaystyle =$ $\displaystyle \frac{1}{z^{3}}(1 +(-z) + (-z)^2 + (-z)^3 + \cdots)$  
  $\displaystyle =$ $\displaystyle \frac{1}{z^{3}} - \frac{1}{z^2} + \frac{1}{z} - 1 + z - z^2 + \cdots$  

Note that the singularity 0 is the 3rd pole.

(b) For Laurent expansion of $z=-1$, let $t = z+1$. Then

$\displaystyle \frac{1}{z^{3}(z+1)} = \frac{1}{(t-1)^{3}t}$

and it is a Laurent expansion with $t=0$. Now we have nothing to do with $\frac{1}{t}$. Thus, we expand $\frac{1}{(t-1)^3}$ using Taylor expansion. Then

$\displaystyle \frac{1}{(t-1)^3} = (-\frac{1}{1-t})^{3} = -(1 + t + t^2 + \cdots)^{3} = -(1 + 3t + 6t^2 + 10t^3 + \cdots)$

Therefore,
$\displaystyle \frac{1}{z^{3}(z+1)}$ $\displaystyle =$ $\displaystyle \frac{1}{(t-1)^{3}t}$  
  $\displaystyle =$ $\displaystyle -\frac{1}{t}(1 + 3t + 6t^2 + 10t^3 + \cdots)$  
  $\displaystyle =$ $\displaystyle -[\frac{1}{t} + 3 + 6t + 10t^2 + \cdots]$  
  $\displaystyle =$ $\displaystyle -[\frac{z+1} + 3 + 6(z+1) + 10(z+1)^2 + \cdots]$  

Note that the singularity $z=-1$ is the 1st pole.

(c) For a Laurent expansion with $z=0$, we have nothing to do with $\frac{1}{z^3}$. Then expand $e^{z^2}$ using Taylor expansion.

$\displaystyle e^{z^2} = 1 + z^2 + \frac{(z^2)^2}{2!} + \frac{(z^2)^{3}}{3!} + \cdots $

Thus,
$\displaystyle \frac{e^{z^2}}{z^{3}}$ $\displaystyle =$ $\displaystyle \frac{1}{z^{3}}e^{z^2}$  
  $\displaystyle =$ $\displaystyle \frac{1}{z^{3}}(1 + z^2 + \frac{(z^2)^2}{2!} + \frac{(z^2)^{3}}{3!} + \cdots)$  
  $\displaystyle =$ $\displaystyle \frac{1}{z^{3}} + \frac{1}{z} + \frac{z}{2} + \frac{z^3}{3!} + \cdots$  

Note that the singularity $z=0$ is the 3rd pole.

(d) For a Laurent expansion with $z=\pi$, let $t = z-\pi$. Then

$\displaystyle \frac{\sin{z}}{z - \pi} = \frac{\sin(t+\pi)}{t} = \frac{-\sin{t}}{t}$

and it is a Laurent expansion with $t=0$.

Note that we have nothing to do with $\frac{1}{t}$. Thus expand $-\sin{t}$ using Taylor expansion.

$\displaystyle -\sin{t} = -(t - \frac{t^3}{3!} + \frac{t^5}{5!} - \frac{t^7}{7!} + \cdots$

Therefore,
$\displaystyle \frac{\sin{z}}{z - \pi}$ $\displaystyle =$ $\displaystyle \frac{-\sin{t}}{t}$  
  $\displaystyle =$ $\displaystyle -\frac{1}{t}(t - \frac{t^3}{3!} + \frac{t^5}{5!} - \frac{t^7}{7!} + \cdots)$  
  $\displaystyle =$ $\displaystyle -1 + \frac{t^2}{3!} - \frac{t^4}{5!} + \cdots$  
  $\displaystyle =$ $\displaystyle -1 + \frac{(z-\pi)^2}{3!} - \frac{(z-\pi)^4}{5!} + \cdots$  

Note that the singularity $z=\pi$ is removal singularity.