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 $\frac{z^2+5}{2}$.

Singularity is at z = 0.

Here, z = 0 is regular point of f(z) , f(z) has a simple pole at z = 0 , f(z) has a simple pole at $z=∞$, $z=∞$ is regular point of f(z) .

Order of the pole here is 2 .

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

$$\begin{gathered} f\left(z\right)=\frac{\left(z^2+5\right)}{z} \\ f\left(\frac{1}{z}\right)=\frac{{\displaystyle \frac{1}{x^2}}+5}{{\displaystyle \frac{1}{z}}} \\ f\left(\frac{1}{z}\right)=\frac{5z^2+1}{z} \end{gathered}$$

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

$$\begin{gathered} Res\left(f\right(z),∞)=-Res\left(\frac{1}{z^2}f\left(\frac{1}{z}\right),0\right) \\ \left(f\right(z),z→∞)=-Res\left(\frac{1}{z^2}·\frac{5z^2+1}{z},0\right) \\ \left(f\right(z),z→∞)=-Res\left(\frac{5z^2+1}{z^3},0\right) \\ R\left(f\right(z),z→∞)=-\underset{z→0}{\lim }\frac{1}{2!}·\frac{d^2}{d{z}^2}\left(5z^2+1\right) \end{gathered}$$

Therefore,

$\begin{array}{rcl}R\left(f\right(z),z& →& ∞)=-\underset{z→0}{\lim }\frac{1}{2!}·\frac{d^2}{d{z}^2}\left(5z^2+1\right)\\ & =& -\underset{z→0}{\lim }\frac{1}{2}×10\\ & =& -5\end{array}$

.

Hence, for a function $\begin{array}{rcl}f\left(z\right)& =& \frac{z^2+5}{z}\end{array}$. f(z) has a simple pole at $\begin{array}{rcl}z& =& ∞\end{array}$ and residue $\begin{array}{rcl}\left(∞\right)& =& -5\end{array}$.