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,$\frac{\mathbf{1}\mathbf{+}\mathbf{z}}{\mathbf{1}\mathbf{-}\mathbf{z}}$

Short answer

For a function $f\left(z\right)=\frac{1+z}{1-z}=\infty$ is a simple pole of f(z) and residue $\left(\infty \right)=2$.

Step-by-step solution

Concept of Residue at infinity

If a function is not analytic at z = athen has a singularity at z = a .

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

Residue at infinity:

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

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

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

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

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

Simplify the function

The function is given as,$f\left(z\right)=\frac{1+z}{1-z}$.

Singularity can be checked by equating the denominator equals to 0 .

1 - z = 0

z = 1

Singularity is at z = 1 .

So, from the definition of the regular point:

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

$z=\infty$ Is a regular point of f(z).

The order of the pole here is 1

Since a function can differentiate up-to derivative after that function is equal to 0 .

Find the residue of the function at  ∞

Finding the residue of the function is as follows:

$$\begin{gathered} f\left(z\right)=\frac{1+z}{1-z} \\ f\left(\frac{1}{z}\right)=\frac{\left(1+{\displaystyle \frac{1}{z}}\right)}{\left(1-{\displaystyle \frac{1}{z}}\right)} \\ f\left(\frac{1}{z}\right)=\frac{z+1}{z-1} \end{gathered}$$

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

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

Simplify further as follows:

$$\begin{gathered} -\underset{z\to 0}{\lim }\frac{1}{1!}\left(\frac{z-1-z-1}{{\left(z-1\right)}^2}\right) \\ -\underset{z\to 0}{\lim }\left(\frac{-2}{{\left(z-1\right)}^2}\right) \\ R\left(f\right(z),z\to \infty )=2 \end{gathered}$$

Hence, for a function,$f\left(z\right)=\frac{1+z}{1-z}$.

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