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:

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

Function is given as,$f\left(z\right)=\frac{4z^3+2z+3}{z^2}z$.

Singularity can be check by equating denominator equals to 0 .

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

Singularity is at$z=0$.

So, from the definition of regular point:

f(z) Has pole of order 2 at z = 0.

f(z) Has a simple pole at $z=∞$

Order of the pole here is 2 .

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

$$\begin{gathered} f\left(z\right)=\frac{4z^3+2z+3}{z^2} \\ f\left(z\right)=\frac{4{\left({\displaystyle \frac{1}{z}}\right)}^3+2\left({\displaystyle \frac{1}{z}}\right)+3}{{\left({\displaystyle \frac{1}{z}}\right)}^2} \\ f\left(\frac{1}{z}\right)=\frac{4+2z^2+3z^3}{z} \end{gathered}$$

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

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

Therefore, $R\left(f\right(z),z→∞)=-2$.

Hence, for a function,$f\left(z\right)=\frac{4z^3+2z+3}{z^2}$.

$z=∞$ Is a simple pole of f(z) , and residue$\left(∞\right)=-2$ .