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