5.1 Laurent expansion

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

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

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

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

$$\sin{z} = z - \frac{z^3}{3!} - \frac{z^5}{5!} + \cdots +$$

$$\cos{z} = 1 - \frac{z^2}{2!} + \frac{z^4}{4!} - \cdots +$$

$$\frac{1}{1-z} = 1 + z + z^2 + z^3 + z^4 + \cdots + ( , \vert z\vert < 1)$$

(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

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

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

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

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

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

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

Note that $\vert\frac{z}{2}\vert < 1$. Then we can use Taylor expansion.

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

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

Therefore,

$$\frac{1}{z^2 - 3z + 2} = \frac{-1}{z-1} + \frac{1}{z-2}$$

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

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

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

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

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.

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

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

Note that since $\vert z\vert > 1$, we write $\frac{1}{z-1}$ by $\frac{-1}{z-1} = \frac{-1}{z}\frac{1}{1-\frac{1}{z}}$.

$$\frac{-1}{z-1} = \frac{-1}{z}\frac{1}{1-\frac{1}{z}} = -\frac{1}{z}(1 + \frac{1}{z} + (\frac{1}{z})^2 + + \cdots)$$

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

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

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

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

Note that $\vert\frac{z}{2}\vert < 1$. Then we can use Taylor expansion.

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

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

Therefore,

$$\frac{1}{z^2 - 3z + 2} = \frac{-1}{z-1} + \frac{1}{z-2}$$

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

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

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

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.

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

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

Note that since $\vert z\vert > 2$, we write $\frac{1}{z-1}$ by $\frac{-1}{z-1} = \frac{-1}{z}\frac{1}{1-\frac{1}{z}}$.

$$\frac{-1}{z-1} = \frac{-1}{z}\frac{1}{1-\frac{1}{z}} = -\frac{1}{z}(1 + \frac{1}{z} + (\frac{1}{z})^2 + + \cdots)$$

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

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

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

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

Here $\vert\frac{2}{z}\vert < 1$. Then it can be Taylor expanded. Thus

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

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

Therefore,

$$\frac{1}{z^2 - 3z + 2} = \frac{-1}{z-1} + \frac{1}{z-2}$$

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

$$= -\sum_{n=0}^{\infty}\frac{2^{n} -1}{z^{n+1}}$$

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

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

Therefore,

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

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

$$= \frac{1}{z^{3}} - \frac{1}{z^2} + \frac{1}{z} - 1 + z - z^2 + \cdots$$

Note that the singularity 0 is the 3rd pole.

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

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

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

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

Therefore,

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

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

$$= -[\frac{1}{t} + 3 + 6t + 10t^2 + \cdots]$$

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

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.

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

Thus,

$$\frac{e^{z^2}}{z^{3}} = \frac{1}{z^{3}}e^{z^2}$$

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

$$= \frac{1}{z^{3}} + \frac{1}{z} + \frac{z}{2} + \frac{z^3}{3!} + \cdots$$

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

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

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

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.

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

Therefore,

$$\frac{\sin{z}}{z - \pi} = \frac{-\sin{t}}{t}$$

$$= -\frac{1}{t}(t - \frac{t^3}{3!} + \frac{t^5}{5!} - \frac{t^7}{7!} + \cdots)$$

$$= -1 + \frac{t^2}{3!} - \frac{t^4}{5!} + \cdots$$

$$= -1 + \frac{(z-\pi)^2}{3!} - \frac{(z-\pi)^4}{5!} + \cdots$$

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