Q30P - 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.

$\frac{\mathbf{c}\mathbf{o}\mathbf{s}\mathbf{z}\mathbf{-}\mathbf{1}}{{\mathbf{z}}^{\mathbf{7}}}$ at z = 0

Step-by-Step Solution

Answer

The residue of the function at z = 0 is $R\left(0\right)=\frac{1}{6!}$

Step 1: Determine the formula for the residue

The Residue of a function at higher order poles is of the form as:

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

Step 2: Residue of higher order pole

Consider the given function is:

$f\left(z\right)=\frac{\cos z-1}{z^7}$

The function has a pole of order 7 at z = 0.

Step 3: Determine residue of the function:

Substitute the values in the equation for the residue of higher order pole and solve as:

$\begin{array}{rcl}R\left(0\right)& =& \frac{1}{(7-1)!}\underset{z\to 0}{\lim }\frac{d^6(\cos hz-1)}{d{z}^6}\\ & =& \frac{1}{6!}\underset{z\to 0}{\lim }\frac{d^5\left(\sin hz\right)}{d{z}^5}\\ & =& \frac{1}{6!}\underset{z\to 0}{\lim }\frac{d^4\left(\sin hz\right)}{d{z}^4}\end{array}$

Solve further as:

$\begin{array}{rcl}R\left(0\right)& =& \frac{1}{6!}\underset{z\to 0}{\lim }\frac{d^4\left(\sin hz\right)}{d{z}^4}\\ & =& \frac{1}{6!}\underset{z\to 0}{\lim }\frac{d^3\left(\sin hz\right)}{d{z}^3}\\ & =& \frac{1}{6!}\underset{z\to 0}{\lim }\frac{d^2\left(\sin hz\right)}{d{z}^2}\end{array}$

Solve further as:

$\begin{array}{rcl}R\left(0\right)& =& \frac{1}{6!}\underset{z\to 0}{\lim }\frac{d\left(\sin hz\right)}{dz}\\ & =& \frac{1}{6!}\underset{z\to 0}{\lim }\cos hz\\ & =& \frac{1}{6!}\end{array}$

Therefore, the residue of a function at z = 0 is $\begin{array}{rcl}R\left(0\right)& =& \frac{1}{6!}\end{array}$.