Find out whether infinity is a regular point, an essential singularity, or a pole (and if a pole, of what order) for each of the following functions. Find the residue of each function at infinity $\frac{{\mathbf{z}}^{\mathbf{2}}\mathbf{+}\mathbf{2}}{\mathbf{3}{\mathbf{z}}^{\mathbf{2}}}$

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

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

Residue at infinity:

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

If $\underset{\left|z\right|\mathbf{\to }\mathbf{\infty }}{\mathbf{lim}}\mathbf{f}\mathbf{\left(}\mathbf{z}\mathbf{\right)}\mathbf{=}\mathbf{0}$then the residue at infinity can be computed as:

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

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

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

The function is given as,$f\left(z\right)=\frac{z^2+2}{3z^2}$.

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

$$\begin{gathered} 3z^2=0 \\ z=0 \end{gathered}$$

Singularity is at z = 0 .

So, from the definition of the regular point:

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

$z=\infty$ 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 is equal to 0.

$$\begin{gathered} f\left(z\right)=\frac{z^2+2}{3z^2} \\ f\left(\frac{1}{z}\right)=\frac{\left({\displaystyle \frac{1}{x^2}}+2\right)}{{\displaystyle \frac{3}{x^2}}} \\ f\left(\frac{1}{z}\right)=\frac{1+2x^2}{3} \end{gathered}$$

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

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

Therefore simplify as follows:

$$\begin{gathered} -Res\left(\frac{1}{z^2}·\frac{1+2z^2}{3},0\right) \\ -Res\left(\frac{1+2z^2}{3z^2},0\right) \\ R\left(f\right(z),z\to \infty )=-\underset{z\to 0}{\lim }\frac{1}{2!}\frac{d}{d{z}^2}\left(\frac{1+2z^2}{3z^2}\right) \\ R\left(f\right(z),z\to \infty )=-4\times 0=0 \end{gathered}$$

Hence, For a function,$f\left(z\right)=\frac{z^2+2}{3z^2}$.

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