CVR2

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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{1}{z(1-z^2)}\,

This has poles at z=0, \pm 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{1}{z(1-z)} = \frac{-1}{2} \,

\mathrm{Res}_{z=0} =\lim_{z\to 0} z f(z) = \lim_{z\to 0} \frac{1}{(1-z^2)} = 1 \,

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

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=1}+\mathrm{Res}_{z=\infty}=1-\frac{1}{2}-\frac{1}{2}+\mathrm{Res}_{z=\infty}=0\, and \mathrm{Res}_{z=\infty}=0\,


Main Page : Complex Variables : Residues

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