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.

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

Short answer

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

Step-by-step solution

Determine the formula for the higher order poles 

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

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

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)

 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 $\begin{array}{rcl}R\left(2i\right)& =& \frac{3i}{32}{e}^{-2}\end{array}$.