Q2P - Mathematical Methods in Physical Sciences

Question

(a) Show that if f(z)  tends to a finite limit as z  tends to infinity, then the residue of f(z)  at infinity is.

(b) Also show that if  f(z) tends to zero as z  tends to infinity, then the residue of f(z)  at infinity is  $\mathbf{-}\mathbf{lim}\mathbf{ }{\mathbf{z}}^{\mathbf{2}}\mathbf{f}\mathbf{\text{'}}\mathbf{\left(}\mathbf{z}\mathbf{\right)}$.

Step-by-Step Solution

Answer

(a) The resultant answer $\mathrm{Res}\left(\mathrm{f}\left(\mathrm{z}\right) \mathrm{at} \mathrm{z} =\infty \right)=\underset{\mathrm{z}\to \infty }{\lim }{\mathrm{z}}^2\mathrm{f}\text{'}\left(\mathrm{z}\right)$ at $z=\infty )=\underset{z\to \infty }{\lim }{z}^{2}f\text{'}(z)$  is proved.

(b) The resultant answer $\mathrm{Res}\left(\mathrm{f}\right(\mathrm{z}) \mathrm{at} \mathrm{z}=\infty )=-\underset{\mathrm{z}\to \infty }{\lim }\mathrm{zf}\left(\mathrm{z}\right)$  is proved.

Step 1: Concept used Laurent:

The Laurent's expansion of $f\left(\frac{1}{z}\right)$ , about  z = 0.

 $\mathrm{f}\left(\frac{1}{\mathrm{z}\text{'}}\right)=\sum _{\mathrm{n}=0}^{\infty }{\mathrm{b}}_{\mathrm{n}}\left(\mathrm{z}\text{'}\right)+\sum _{\mathrm{n}=1}^{\mathrm{n}}{\mathrm{a}}_{\mathrm{n}}{\left(\mathrm{z}\text{'}\right)}^{-\mathrm{n}}$

If the function  f(z) tends to a finite limit as $\mathrm{z}\to \infty$  .

So, the function   is analytic at the point  $\mathrm{z}=\infty$.

Step 2: Simplify the function

The function f(z)  tends to a finite Limit as z  tends to infinity.

Consider the equation shown below:

 $$\begin{gathered} Res(z=\infty )=\frac{-1}{2\pi i}{\int }_{c}f\left(z\right)dz \\ \Rightarrow -b_1 \end{gathered}$$

The curve  C is described in the counter clockwise direction.

 

Make an assumption that the function f(z)  has a pole of order m  at the point  $z=\infty$.

Step 3: Substitute the value

Use the Laurent's expansion of $f\left(\frac{1}{z}\right)$ , about z = 0 .

 $\mathrm{f}\left(\frac{1}{\mathrm{z}\text{'}}\right)=\sum _{\mathrm{n}=0}^{\infty }{\mathrm{b}}_{\mathrm{n}}(\mathrm{z}\text{'})+\sum _{\mathrm{n}=1}^{\mathrm{n}}{\mathrm{a}}_{\mathrm{n}}{(\mathrm{z}\text{'})}^{-\mathrm{n}}$

Substitute the value Z for  $\frac{1}{s}$.

 $\mathrm{f}\left(\mathrm{z}\right)=\sum _{\mathrm{n}=0}^{\infty }{\mathrm{b}}_{\mathrm{n}}{\mathrm{z}}^{-\mathrm{n}}+\sum _{\mathrm{n}=1}^{\infty }{\mathrm{a}}_{\mathrm{n}}{\mathrm{z}}^{\mathrm{n}}$                                                                                          …… (1)

Since, the function f(z)  tends to a finite limit as  $z\to \infty$

So, the function   is analytic at the point $z=\infty$ .

Step 4: Substitute the value further

Consider that, $a_n=0$  .

For  $\forall \mathrm{n}$ as  $\mathrm{I}\le \mathrm{n}\le \mathrm{m}$.

 

Obtain from equation (2) as follows:

 

 $\begin{array}{rcl}f\left(z\right)& =& \sum _{n=0}^{\infty }{b}_n{z}^{-n}\\ & =& b_0+\frac{b_1}{z}+\frac{b_2}{z}+...\\ f\text{'}\left(z\right)& =& \frac{-b_1}{z^2}-\frac{2b_2}{z^3}+...\\ z^{2}f\text{'}\left(z\right)& =& -b_1-\frac{2b_2}{z}+...\end{array}$

 

Determine the limits as follows:

 $\begin{array}{rcl}\underset{\mathrm{z}\to \infty }{\lim }{\mathrm{z}}^2\mathrm{f}\left(\mathrm{z}\right)& =& -{\mathrm{b}}_1-0-0...\\ & =& -{\mathrm{b}}_1\\ & =& \mathrm{Res}\left(\mathrm{z}=\infty \right)\end{array}$

 

Hence, it is shown that $\begin{array}{rcl}\underset{\mathrm{z}\to \infty }{\lim }{\mathrm{z}}^2\mathrm{f}\left(\mathrm{z}\right)& =& \mathrm{Res}\left(\mathrm{z}=\infty \right)\end{array}$ .