CVR4

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

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

This has poles at $z=\pm i\,$ of multiplicity $k=2\,$ .

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

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

$=\lim_{z\to -i} \frac{(z-i)^2 2z -z^2 2(z-i)}{(z-i)^4}\,$

$=\frac{(-2i)^2 2(-i) - (-i)^2 2 (-2i)}{(-2i)^4}\,$

$=\frac{8i-4i}{16} = \frac{1}{4}i\,$

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

$=\lim_{z\to i} \frac{(z+i)^2 2z -z^2 2(z+i)}{(z+i)^4}\,$

$=\frac{(2i)^2 2i - i^2 2 (2i)}{(2i)^4}\,$

$=\frac{-8i+4i}{16} = \frac{-1}{4}i\,$

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

So $\mathrm{Res}_{z=-i}+\mathrm{Res}_{z=i}+\mathrm{Res}_{z=\infty}= \frac{1}{4}i-\frac{1}{4}i+\mathrm{Res}_{z=\infty}\,$ and $\mathrm{Res}_{z=\infty}=0\,$

Main Page : Complex Variables : Residues

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