# CVR14

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

$f(z) = \frac{z^2+z-1}{z^2(z-1)}\,$

This has poles at z = 0, of multiplicity $k=2\,$ and at z = 1, of multiplicity $k=1\,$

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

$\mathrm{Res}_{z=1} =\lim_{z\to 1} (z-1)f(z) = \lim_{z\to 1} \frac{z^2+z-1}{z^2}=1\,$

$\mathrm{Res}_{z=0} =\lim_{z\to 0} \frac{d}{dz}\left[z^2 f(z)\right] = \lim_{z\to 0} \frac{d}{dz}\left[\frac{z^2+z-1}{(z-1)}\right]=\frac{(2z+1)(z-1)-(z^2+z-1)}{(z-1)^2}=-1+1=0\,$

If $f(z)\,$ is analytic except at isolated singular points, then the sum of all the residues of $f(z)\,$ equals 0.

So $\mathrm{Res}_{z=0}+\mathrm{Res}_{z=1}+\mathrm{Res}_{z=\infty}= 1+\mathrm{Res}_{z=\infty}\,$ and $\mathrm{Res}_{z=\infty}=-1\,$

##### Toolbox

 Get A Wifi Network Switcher Widget for Android