Question

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,$\frac{\mathbf{2}\mathbf{z}\mathbf{+}\mathbf{3}}{{\left(z+2\right)}^{\mathbf{2}}}$.

Short answer

For a function $f\left(z\right)=\frac{2z+3}{{\left(z+2\right)}^2}z=\infty$ is a regular point and residue $\left(\infty \right)=-2$.

Step-by-step solution

Concept of Residue at infinity

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:

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

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

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

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

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

Simplify the function

Function is given as$\frac{\mathrm{z}}{{\mathrm{z}}^2+1}$.

Singularity can be check by equating denominator equals to 0.

$\begin{array}{rcl}{\left(z+2\right)}^2& =& 0\\ z& =& -2\end{array}$

Singularity is at, $\begin{array}{rcl}z& =& -2\end{array}$.

Assuming that at$\begin{array}{rcl}\mathrm{z}& =& 0,\mathrm{f}\left(\mathrm{z}\right)\end{array}$ exists.

So, from the definition of regular point:

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

$\begin{array}{rcl}\mathrm{z}& =& \infty \end{array}$ is regular point of f(z).

Order of the pole here is 2.

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

Find the residue of the function at  ∞

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

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

localid="1664424861934" $$\begin{gathered} \begin{array}{rcl}& & \mathrm{Res}\left(\mathrm{f}\right(\mathrm{z}),\infty )=-\mathrm{Res}\left(\frac{1}{{\mathrm{z}}^2}\mathrm{f}\left(\frac{1}{\mathrm{z}}\right),0\right)\end{array}\begin{array}{rcl}& & \end{array}\begin{array}{rcl}\left(\mathrm{f}\right(\mathrm{z}),\mathrm{z}& \to & \infty )=-\mathrm{Res}\left(\frac{1}{{\mathrm{z}}^2}·\frac{\mathrm{z}(3\mathrm{z}+2)}{{\left(1+2\mathrm{z}\right)}^2},0\right)\end{array} \\ \begin{array}{rcl}\mathrm{R}\left(\mathrm{f}\right(\mathrm{z}),\mathrm{z}& \to & \infty )=-\mathrm{Res}\left(\frac{(3\mathrm{z}+2)}{\mathrm{z}{\left(1+2\mathrm{z}\right)}^2},0\right)\end{array}\begin{array}{rcl}& & \end{array}\begin{array}{rcl}\mathrm{R}\left(\mathrm{f}\right(\mathrm{z}),\mathrm{z}& \to & \infty )=-\underset{\mathrm{z}\to \infty }{\lim }\left(\frac{(3\mathrm{z}+2)}{{\left(1+2\mathrm{z}\right)}^2}\right)\end{array} \end{gathered}$$

Putting z = 0 in above equation as follows:

$\begin{array}{rcl}\mathrm{R}\left(\mathrm{f}\right(\mathrm{z}),\mathrm{z}& \to & \infty )=-2\end{array}$

Hence, for a function, $\begin{array}{rcl}f\left(z\right)& =& \frac{2z+3}{{\left(z+2\right)}^2}z=\infty \end{array}$, is a regular point and residue, $\begin{array}{rcl}\left(\infty \right)& =& -2\end{array}$.