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.

$\frac{\mathbf{1}\mathbf{+}\mathbf{c}\mathbf{o}\mathbf{s}\mathbf{z}}{{\mathbf{(}\mathbf{\pi }\mathbf{-}\mathbf{z}\mathbf{)}}^{\mathbf{2}}}$at$\mathit{z}\mathbf{}\mathbf{=}\mathit{\pi }$

Short answer

The residue of the function at $z=\pi$is $R\left(\pi \right)=-\frac{1}{2}$.

Step-by-step solution

Determine the formula for the higher order poles 

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

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

Determine the residue of the higher order pole:

Consider the given function as:

$f\left(z\right)=\frac{1+\cos z}{{(\pi -z)}^3}$

At$z=\pi$

The function has a pole of orderatand the residue of higher order pole is as follows:

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

 Determine the residue of the higher order function as:

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

$\begin{array}{rcl}R\left(\pi \right)& =& \frac{1}{(3-1)!}\underset{\mathrm{z}\to {\mathrm{z}}_0}{\lim }\frac{{\mathrm{d}}^2}{{\mathrm{dz}}^2}[1+\cos z]\\ & =& \frac{1}{\left(2\right)!}\underset{\mathrm{z}\to {\mathrm{z}}_0}{\lim }\frac{\mathrm{d}}{\mathrm{dz}}[-\sin z]\\ & =& \frac{1}{\left(2\right)!}\underset{\mathrm{z}\to {\mathrm{z}}_0}{\lim }\frac{\mathrm{d}}{\mathrm{dz}}\left[\cos z\right]\\ & =& -\frac{1}{2}\end{array}$

Therefore, the residue of a function at $\begin{array}{rcl}z& =& \pi \end{array}$is $\begin{array}{rcl}R\left(\pi \right)& =& -\frac{1}{2}\end{array}$.