91Ó°ÊÓ

If a function is not analytic at z = a 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 $\underset{\left|z\right|\mathbf{→}\mathbf{∞}}{\mathbf{lim}}\mathbf{f}\mathbf{\left(}\mathbf{z}\mathbf{\right)}\mathbf{=}\mathbf{0}$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:

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

The function is given as,$f\left(z\right)=\frac{z^2-1}{z^2+1}$.

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

$$\begin{gathered} z^2+1=0 \\ z=Â\pm i \end{gathered}$$

Singularity is at$z=Â\pm i$.

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) .

Order of the pole here is 2

Since the function can differentiate up-to 2^nd derivative after that function equals to 0 .

$$\begin{gathered} f\left(z\right)=\frac{z^2-1}{z^2+1} \\ f\left(z\right)=\frac{\left({\displaystyle \frac{1}{z^2}}-1\right)}{\left(\frac{1}{z^2}+1\right)} \\ f\left(\frac{1}{z}\right)=\frac{1-z^2}{1+z^2} \end{gathered}$$

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

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

So,

$R\left(f\right(z),z→∞)=0$

.

Hence, for a function,$f\left(z\right)=\frac{z^2-1}{z^2+1}$.

$z=∞$ Is a simple pole of f(z), and residue $\left(∞\right)=0$.