CVR6

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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{e^z}{z^2(z^2+9)}\,

This has poles at z=0\, of multiplicity k=2\, and at z=\pm 3i\, of multiplicity k=1\,.

For the simple roots, the residue is the numerator divided by the first derivative of the denominator when both are evaluated at the root.

\mathrm{Res}_{z=-3i} f(z) = \frac{e^z}{4z^3+18z} \Big|_{z=-3i} = \frac{e^{-3i}}{4(27i)+18(-3i)} = \frac{e^{-3i}}{54i}\,

\mathrm{Res}_{z=3i} f(z) = \frac{e^z}{4z^3+18z} \Big|_{z=3i} = \frac{e^{3i}}{ 4(27(-i))+18(3i)} = \frac{-e^{3i}}{54i}\,

For 0, 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} =\lim_{z\to 0} \frac{d}{dz} z^2 f(z) = \lim_{z\to 0} \frac{d}{dz} \frac{e^z}{z^2+9}\,

=\lim_{z\to 0} \frac{(z^2+9)e^z-e^z 2z}{(z^2+9)^2} = \frac{9}{9^2} = \frac{1}{9}\,

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=-3i} +\mathrm{Res}_{z=3i} +\mathrm{Res}_{z=0} +\mathrm{Res}_{z=\infty} = \frac{e^{-3i}}{54i} - \frac{e^{3i}}{54i} + \frac{1}{9} + \mathrm{Res}_{z=\infty}=0\,

Main Page : Complex Variables : Residues

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