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, $\frac{\mathbf{z}}{{\mathbf{z}}^{\mathbf{2}}\mathbf{+}\mathbf{1}}$.

Short answer

The resultant answer $z=\infty$ is regular point of, f(z) . Order of the pole here is 2 and .$\mathrm{z}=\infty \mathrm{Re}{\left|\mathrm{f}\left(\mathrm{z}\right)\right|}_{\mathrm{z}=\infty }=-1$

Step-by-step solution

Concept of Residue at infinity:

If a function is not analytic at z = a then f(z) has a singularity at z = a.

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

Residue at infinity:

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

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

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

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

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

Simplify the function:

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

Singularity can be check by equating denominator equals to 0 .

$\begin{array}{rcl}{z}^2+1& =& 0\\ z^2& =& -1\\ z& =& \pm i\end{array}$

Singularity is at $\begin{array}{rcl}& & \pm i\end{array}$.

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

So, from the definition of regular point:

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

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

Order of the pole here is 2.

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

Find the residue of the function at ∞ :

$Re{\left|f\left(z\right)\right|}_{z=\infty }=-Re{\left|g\left(z\right)\right|}_{z=0}$ …… (1)

Where,.$g\left(z\right)=\frac{1}{z^2}f\left(\frac{1}{z}\right)$

Finding g(z) first as follows:

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

Since,

$\begin{array}{rcl}\mathrm{Re}\left|\mathrm{g}\left(\mathrm{z}\right)\right|& =& \mathrm{co}\mathrm{efficient}\mathrm{of}\frac{1}{\mathrm{z}}\\ & =& 1\\ \mathrm{Re}{\left|\mathrm{f}\left(\mathrm{z}\right)\right|}_{\mathrm{z}=\infty }& =& -1\end{array}$

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

Order of the pole here is 2 and $\begin{array}{rcl}Re{\left|f\left(z\right)\right|}_{z=\infty }& =& -1\end{array}$.