91Ó°ÊÓ

If a function is not analytic at then has a singularity at z = a.

Order of pole is the highest no of derivative of the equation.

Residue at infinity:

$\mathbf{Res}\mathbf{\left(}\mathbf{f}\mathbf{\right(}\mathbf{z}\mathbf{)}\mathbf{,}\mathbf{∞}\mathbf{)}\mathbf{=}\mathbf{-}\mathbf{Res}\left(\frac{1}{z^2}f\left(\frac{1}{z}\right),0\right)$

If$\underset{\left|z\right|\mathbf{→}\mathbf{∞}}{\mathbf{l}\mathbf{i}\mathbf{m}}\mathit{f}\mathbf{\left(}\mathit{z}\mathbf{\right)}\mathbf{=}\mathbf{0}$ then the residue at infinity can be computed as:

$\mathbf{Res}\mathbf{(}\mathbf{f}\mathbf{,}\mathbf{∞}\mathbf{)}\mathbf{=}\mathbf{-}\underset{\left|z\right|\mathbf{→}\mathbf{∞}}{\mathbf{lim}}\mathbf{z}\mathbf{·}\mathbf{f}\mathbf{\left(}\mathbf{z}\mathbf{\right)}$

If$\underset{\left|z\right|\mathbf{→}\mathbf{∞}}{\mathbf{lim}}\mathbf{f}\mathbf{\left(}\mathbf{z}\mathbf{\right)}\mathbf{=}\mathbf{c}\mathbf{≠}\mathbf{0}$ then the residue at infinity is as follows:

$\mathbf{Res}\mathbf{(}\mathbf{f}\mathbf{,}\mathbf{∞}\mathbf{)}\mathbf{=}\underset{\left|z\right|\mathbf{→}\mathbf{∞}}{\mathbf{lim}}{\mathbf{z}}^{\mathbf{2}}\mathbf{·}\mathbf{f}\mathbf{\text{'}}\mathbf{\left(}\mathbf{z}\mathbf{\right)}$

The function is given as $f\left(z\right)=\ln \left(\frac{z+1}{z-1}\right)$.

Singularity can be checked by equating the denominator equals to 0 .

z = 1

z = -1

By putting z = 1 and z = -1 function will be not defined.

Singularity is at z = -1, 1 .

Assuming that at z = 0, f(z) exists.

So, from the definition of the regular point:

z = 0 Is a regular point of f(z).

$z=∞$ is a regular point of f(z)

$\begin{array}{rcl}f\left(z\right)& =& \ln \left(\frac{z+1}{z-1}\right)\\ f\left(\frac{1}{z}\right)& =& \ln \frac{{\left({\displaystyle \frac{1}{z}}+1\right)}^3}{\left({\displaystyle \frac{1}{z}}-1\right)}\\ f\left(\frac{1}{z}\right)& =& \ln \frac{1+z}{1-z}\end{array}$

Using residue formula and putting $\begin{array}{rcl}& & f\left(\frac{1}{z}\right)\end{array}$ as follows:

$\begin{array}{rcl}Res\left(f\right(z),∞)& =& -Res\left(\frac{1}{z^2}f\left(\frac{1}{z}\right),0\right)\\ \left(f\right(z),z& →& ∞)=-Res\left(\frac{1}{z^2}·\ln \frac{1+z}{1-z},0\right)\\ \left(f\right(z),z& →& ∞)=-\underset{z→0}{\lim }\frac{d}{dz}\ln \left(\frac{1+z}{1-z}\right)\\ & & \end{array}$

Differentiation of ln x is $\begin{array}{rcl}\frac{d}{dx}(\ln x)& =& \frac{1}{x}\end{array}$.

$\begin{array}{rcl}& & -\underset{z→0}{\lim }\frac{1-z}{1+z}×\frac{1-z+z-1}{{\left(1-z\right)}^2}\\ -\underset{z→0}{\lim }\frac{2}{\left(1+z\right)\left(1-z\right)}R\left(f\right(z),z& →& ∞\\ & =& -2\end{array}$

Hence, for a function,$\begin{array}{rcl}f\left(z\right)& =& \ln \left(\frac{z+1}{z-1}\right)\end{array}$ .

$\begin{array}{rcl}z& =& ∞\end{array}$ Is a regular point of f(z) , and residue$\begin{array}{rcl}\left(∞\right)& =& -2\end{array}$ .