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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^{3}-z^{5}}}\,

z^{3}-z^{5}=z^{3}(1-z^{2})\, has roots at z=0,\pm 1\, with multiplicities k=3,1,1\,.

Use the formula {\mathrm  {Res}}_{{z=z_{0}^{k}}}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^{3}}}f(z)={\frac  {1}{(3-1)!}}\lim _{{z\to 0}}{\frac  {d^{2}}{{dz}^{2}}}\left[(z-0)^{3}f(z)\right]={\frac  {1}{2}}\lim _{{z\to 0}}{\frac  {d^{2}}{{dz}^{2}}}{\frac  {1}{1-z^{2}}}\,

={\frac  {1}{2}}\lim _{{z\to 0}}{\frac  {d}{dz}}\left[{\frac  {2z}{(1-z^{2})^{2}}}\right]={\frac  {1}{2}}\lim _{{z\to 0}}{\frac  {(1-z^{2})^{2}2-2z2(1-z^{2})(-2z)}{(1-z^{2})^{4}}}={\frac  {1}{2}}{\frac  {2}{1}}=1\,

{\mathrm  {Res}}_{{z=-1}}f(z)=\lim _{{z\to -1}}(z+1)f(z)=\lim _{{z\to -1}}{\frac  {1}{z^{3}(1-z)}}={\frac  {-1}{2}}\,

{\mathrm  {Res}}_{{z=1}}f(z)=\lim _{{z\to 1}}(z-1)f(z)=-\lim _{{z\to 1}}{\frac  {1}{z^{3}(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