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.

$\frac{{\mathbf{e}}^{\mathbf{3}\mathbf{z}}\mathbf{-}\mathbf{3}\mathbf{z}\mathbf{-}\mathbf{1}}{{\mathbf{z}}^{\mathbf{4}}}$at z = 0

Short answer

The residue of the function at z = 0 is $R\left(0\right)=\frac{9}{2}$

Step-by-step solution

Determine the formula for the higher order poles 

Residue of a function at higher order poles is given by

$\mathit{R}\mathbf{\left[}\mathit{f}\mathbf{\right(}\mathit{z}\mathbf{\left)}\mathbf{\right]}\mathbf{=}\frac{\mathbf{1}}{\mathbf{(}\mathbf{n}\mathbf{-}\mathbf{1}\mathbf{)}\mathbf{!}}\underset{\mathbf{z}\mathbf{\to }{\mathbf{z}}_{\mathbf{k}}}{\mathbf{l}\mathbf{i}\mathbf{m}}\frac{{\mathbf{d}}^{\mathbf{n}\mathbf{-}\mathbf{1}}}{\mathbf{d}{\mathbf{z}}^{\mathbf{n}\mathbf{-}\mathbf{1}}}{\mathbf{(}\mathit{z}\mathbf{-}{\mathit{z}}_{\mathbf{k}}\mathbf{)}}^{\mathbf{n}}\mathit{f}\mathbf{\left(}\mathit{z}\mathbf{\right)}$

Determine the residue of the higher order pole:

Consider the given function as:

$f\left(z\right)=\frac{e^{3z}-3z-1}{z^4}$

The function has a pole of order n = 4 at z = 0 and the residue of higher order pole is given by:

$R\left(z_0\right)=\frac{1}{(\mathrm{n}-1)!}\underset{\mathrm{z}\to {\mathrm{z}}_0}{\lim }\frac{{\mathrm{d}}^{\mathrm{n}-1}}{{\mathrm{dz}}^{\mathrm{n}-1}}(z-z_0)f\left(z\right)$ ……. (1)

 Determine the residue of the higher order function as:

Substitute the values equation (1) and solve as:

$\begin{array}{rcl}R\left(0\right)& =& \frac{1}{(4-1)!}\underset{z\to 0}{\lim }\frac{d^3(e^{3z}-3z+1)}{d{z}^3}\\ & =& \frac{1}{3!}\underset{z\to 0}{\lim }\frac{d^3(3e^{3z}-3)}{d{z}^3}\\ & =& \frac{1}{3!}\underset{z\to 0}{\lim }\frac{d\left(9e^{3z}\right)}{dz}\end{array}$

Solve further as:

$\begin{array}{rcl}R\left(0\right)& =& \frac{1}{3!}\underset{z\to 0}{\lim }\frac{d\left(9e^{3z}\right)}{dz}\\ & =& \frac{1}{3!}\underset{z\to 0}{\lim }\left(27e^{3z}\right)\\ & =& \frac{9}{2}\end{array}$

Therefore, the residue of a function at z = 0 is$\begin{array}{rcl}R\left(0\right)& =& \frac{9}{2}\end{array}$