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

Short answer

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

Step-by-step solution

Determine the formula for the residue 

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

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

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.

 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 $\begin{array}{rcl}R\left(0\right)& =& \frac{1}{6!}\end{array}$.