Q32P - Mathematical Methods in Physical Sciences

Question

Find the residues of the following functions at the indicated points. Try to select the easiest of the methods outlined above. Check your results by computer.

$\frac{{\mathbf{e}}^{\mathbf{iz}}}{{\mathbf{(}{\mathbf{z}}^{\mathbf{2}}\mathbf{+}\mathbf{4}\mathbf{)}}^{\mathbf{2}}}\mathbf{ }\mathbf{at}\mathbf{ }\mathbf{z}\mathbf{=}\mathbf{2}\mathbf{i}$

Step-by-Step Solution

Answer

The residue of the function at $z = 2i$ is $R( 2i)=\frac{3i}{32}{e}^{-2}$

Step 1: Determine the formula for the higher order poles

Residue of a function at higher order poles is given as follows:

$\mathbf{\text{R[f(z)]=}}\frac{\mathbf{1}}{\mathbf{(}\mathbf{n}\mathbf{-}\mathbf{1}\mathbf{)}\mathbf{!}}\underset{\mathbf{z}\mathbf{\to }{\mathbf{z}}_{\mathbf{k}}}{\mathbf{lim}}\frac{{\mathbf{d}}^{\mathbf{n}\mathbf{-}\mathbf{1}}}{{\mathbf{dz}}^{\mathbf{n}\mathbf{-}\mathbf{1}}}{\mathbf{(}\mathbf{z}\mathbf{-}{\mathbf{z}}_{\mathbf{k}}\mathbf{)}}^{\mathbf{n}}\mathbf{f}\mathbf{\left(}\mathbf{z}\mathbf{\right)}$

Step 2: Determine the residue of the higher order pole:

Consider the given function as:

 $f\left(z\right)=\frac{e^{iz}}{{(z^2+4)}^2}$

Resolve the function as:

$\begin{array}{rcl}f\left(z\right)& =& \frac{e^{iz}}{{(z^2+4)}^2}\\ & =& \frac{e^{iz}}{{(z^2+2i)}^2{(z-2i)}^2}\end{array}$

At z = 2i

 

The function has a pole of order n = 2 at z = 2i and the residue of higher order pole is given by:

$R\left(z_0\right)=\frac{1}{(n-1)!}\underset{z\to z_0}{\lim }\frac{d^{n-1}}{d{z}^{n-1}}(z-z_0)f\left(z\right)$                                    ……. (1)

Step 3: Determine the residue of the higher order function as:

Substitute the values in equation (1) and solve as:

$\begin{array}{rcl}R\left(2i\right)& =& \frac{1}{(n-1)!}\underset{z\to 0}{\lim }\frac{d}{dz}\left(\frac{e^{iz}}{{(z+2i)}^2}\right)\\ & =& {\left[\frac{i{e}^{iz}{(z+2i)}^2-2e^{iz}(z+2i)}{{(z+2i)}^4}\right]}_{z=2i}\\ & =& {\left[\frac{i{e}^{iz}(z+2i)-2e^{iz}}{{(z+2i)}^3}\right]}_{z=2i}\end{array}$

 

Solve further as:

$\begin{array}{rcl}R\left(2i\right)& =& {\left[\frac{i{e}^{iz}(z+2i)-2e^{iz}}{{(z+2i)}^3}\right]}_{z=2i}\\ & =& \left[\frac{i{e}^{iz}\left(4i\right)-e^{-2}\times 2}{{\left(4i\right)}^3}\right]\\ & =& \frac{3i}{32}{e}^{-2}\end{array}$

 

Therefore, the residue of a function at z = 2i is $\begin{array}{rcl}R\left(2i\right)& =& \frac{3i}{32}{e}^{-2}\end{array}$.