CVR2

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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}}zf(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