CVR13

From Exampleproblems

Find the residues of $f(z)\,$ at all its isolated singular points and at infinity (if infinity is not a limit point of singular points), where $f(z)\,$ is given by

$f(z) = \frac{\cos z}{z^2(z-\pi)^3}\,$

This function has poles at $z=0, \pi\,$ with multiplicities $k=2,3\,$ .

Use the formula $\mathrm{Res}_{z=z_0} f(z) = \frac{1}{(k-1)!}\lim_{z\to z_0} \frac{d^{k-1}}{{dz}^{k-1}}\left[(z-z_0)^k f(z)\right]\,$

$\mathrm{Res}_{z=0} f(z)=\lim_{z\to 0} \frac{d}{dz}\left[z^2f(z)\right]=\lim_{z\to 0}\frac{d}{dz}\frac{\cos z}{(z-\pi)^3}\,$

$=\lim_{z\to 0} \left[ \frac{(z-\pi)^3(-\sin z)-(\cos z) 3 (z-\pi)^2}{(z-\pi)^6} \right] = \frac{-3\pi^2}{\pi^6} = \frac{-3}{\pi^4}\,$

$\mathrm{Res}_{z=\pi} f(z)=\frac{1}{2}\lim_{z\to \pi} \frac{d^2}{{dz}^2}\left[(z-\pi)^3f(z)\right]=\frac{1}{2}\lim_{z\to \pi}\frac{d^2}{{dz}^2}\frac{\cos z}{z^2}\,$

$=\frac{1}{2} \lim_{z\to \pi} \frac{d}{dz} \frac{z^2 (-\sin z) - 2z\cos z}{z^4} = ... = \frac{1}{2}\frac{\pi^2-6}{\pi^4}\,$

Main Page : Complex Variables : Residues

Retrieved from "http://www.exampleproblems.com/wiki/index.php/CVR13"

Views

Personal tools

Create an account or log in

Search

Toolbox

Argan Oil
Natural Skin Care
Organic Skin Care