Q3P - Mathematical Methods in Physical Sciences

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

Step-by-Step Solution

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 1: 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:

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

Step 2: 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 .

Step 3: 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  $\begin{array}{rcl}Re{\left|f\left(z\right)\right|}_{z=\infty }& =& -1\end{array}$.