Q2P - Mathematical Methods in Physical Sciences

Question

Show that equation (4.4) can be written as (4.5). Then expand each of the fractions in the parenthesis in (4.5) in powers of z and in powers of $\frac{\mathbf{1}}{\mathbf{z}}$ [see equation (4.7) ] and combine the series to obtain (4.6), (4.8), and (4.2). For each of the following functions find the first few terms of each of the Laurent series about the origin, that is, one series for each annular ring between singular points. Find the residue of each function at the origin. (Warning: To find the residue, you must use the Laurent series which converges near the origin.) Hints: See Problem 2. Use partial fractions as in equations (4.5) and (4.7). Expand a term $\frac{\mathbf{1}}{\mathbf{(}\mathbf{z}\mathbf{-}\mathbf{\alpha }\mathbf{)}}$ in powers of z to get a series convergent for $\left|\mathbf{z}\right|\mathbf{<}\mathbf{\alpha }$, and in powers of $\frac{\mathbf{1}}{\mathbf{z}}$  to get a series convergent for $\left|\mathbf{z}\right|>\mathbf{\alpha }$.

Step-by-Step Solution

Answer

The result is  $\left[f\left(z\right)=1+\frac{z}{2}+\frac{z}{4}+...+{\left(\frac{z}{2}\right)}^n+...+\frac{2}{z}+4\left(\frac{1}{z^2}-\frac{1}{z^3}+...+\frac{{\left(-1\right)}^n}{z^n}+...\right)\right]$

Step 1 : Introduction

The Laurent series of a complex function f(z) is a representation of that function as a power series which includes terms of negative degree. It may be used to express complex functions in cases where a Taylor series expansion cannot be applied.

Step 2: Solution

Consider a function:

 $f\left(z\right)=\frac{12}{z\left(2-z\right)\left(1+z\right)}$

  

The above function has three singular points, at z=0, z=2 and z=-1.

Thus, there are two circles $C_1 \& C_2$ about $z_0=0$ and three Laurent series about $z_0=0$, one series valid in each of the three regions $R_1\left(0<\left|z\right|<1\right),R_2\left(1\left|z\right|<2\right)\&R_3\left(\left|z\right|>2\right)$.

To find these series first write f(z) in the below shown from by the use if partial fraction:

$f\left(z\right)=\frac{12}{z\left(2-z\right)\left(1+z\right)}$ 

 

Then,

$\begin{array}{c}\frac{1}{\left(2-z\right)\left(1+z\right)}=+\frac{A}{2-z}+\frac{B}{1+z}\\ =\frac{\left(1+z\right)A+\left(2-z\right)B}{\left(2-z\right)\left(1+z\right)}\end{array}$ 

 

Put z=2

Then, 

 1=3A 

Or,

A=1/3 

put z=-1,

Then,

1=3B  

Or,

B=1/3 

Hence,

$f\left(z\right)=\frac{12}{z\left(2-z\right)\left(1+z\right)}=\frac{12}{z}\times \frac{1}{3}\left(\frac{1}{1+z}+\frac{1}{2-z}\right)$ 

Thus the required function is:

$f\left(z\right)=\frac{4}{z}\left(\frac{1}{1+z}+\frac{1}{2-z}\right)$  

 

Now for $0<\left|z\right|<1$ here expand each of the fraction;

$\begin{array}{l}f\left(z\right) =\frac{4}{z}\left[{\left(1+z\right)}^{-1}-\frac{1}{2}{\left(1-\frac{z}{2}\right)}^{-1}\right]\\ =\frac{4}{z}\left[\left(1-z+z^2-z^3+...\right)+\frac{1}{2}\left(1+\frac{z}{2}+\frac{z^2}{4}+\frac{z^3}{8}+...\right)\right]\end{array}$  

 

This implies;

 $\begin{array}{l}f\left(z\right)=\left(\frac{4}{z}-4+4z-4z^2+...\right)+\frac{4}{z}\left(\frac{1}{2}+\frac{z}{4}+\frac{z^2}{8}+\frac{z^3}{16}...\right)\\ =-3+\frac{9z}{2}+\frac{15z^2}{4}+\frac{33z^3}{8}+...\frac{6}{z}\end{array}$

  

This is the Laurent series for f(z) which is valid in the region $0<\left|z\right|<1$.

 

To obtain the series valid in the region  $\left|z\right|=2$.

Write it as;

$\begin{array}{l}f\left(z\right)=\frac{4}{z}\left(\frac{1}{1+z}+\frac{1}{2-z}\right)\\ =\frac{4}{z}\left[\frac{1}{z}{\left(1+\frac{1}{z}\right)}^{-1}-\frac{1}{z}{\left(1-\frac{2}{z}\right)}^{-1}\right]\\ =\frac{4}{z}\left[\frac{1}{z}\left(1-\frac{1}{z}+\frac{1}{z^2}+\frac{1}{z^3}+...\right)-\frac{1}{z}\left(1+\frac{2}{z}+\frac{4}{z^2}+\frac{8}{z^3}+...\right)\right]\\ =\left(\frac{4}{z^2}-\frac{4}{z^3}+\frac{4}{z^4}+\frac{4}{z^5}\right)+...-\left(\frac{4}{z^2}-\frac{8}{z^3}+\frac{16}{z^4}+\frac{32}{z^5}+...\right)\end{array}$  

This implies;

$\begin{array}{l}f\left(z\right)=\frac{-12}{z^3}-\frac{-12}{z^4}+\frac{-36}{z^5}+\frac{60}{z^6}-...\\ =\frac{-12}{z^3}\left(1+\frac{1}{z}+\frac{3}{z^2}+\frac{5}{z^3}+...\right)\end{array}$  

Finally, to obtain the series valid in the region  $1<\left|z\right|<2$,

write it as;

$\begin{array}{l}f\left(z\right)=\frac{4}{z^2}{\left(1+\frac{1}{z}\right)}^{-1}+\frac{4}{z}.\frac{1}{2}{\left(1-\frac{z}{2}\right)}^{-1}\\ =\left(\frac{4}{z^2}-\frac{4}{z^3}+\frac{4}{z^4}+\frac{4}{z^5}+...\right)+\frac{2}{z}\left(1+\frac{z}{2}+\frac{z^2}{4}+\frac{z^3}{8}+...\right)\\ =\left(\frac{4}{z^2}-\frac{4}{z^3}+\frac{4}{z^4}+\frac{4}{z^5}+...\right)+\left(\frac{2}{z}+1+\frac{z}{2}+\frac{z^2}{4}+...\right)\end{array}$

This implies

$f\left(z\right)=1+\frac{z}{2}+\frac{z}{4}+...+{\left(\frac{z}{2}\right)}^n+...+\frac{2}{z}+4\left(\frac{1}{z^2}-\frac{1}{z^3}+...+\frac{{\left(-1\right)}^n}{z^n}+...\right)$ 

Hence; the required result is;

$\left[f\left(z\right)=1+\frac{z}{2}+\frac{z}{4}+...+{\left(\frac{z}{2}\right)}^n+...+\frac{2}{z}+4\left(\frac{1}{z^2}-\frac{1}{z^3}+...+\frac{{\left(-1\right)}^n}{z^n}+...\right)\right]$