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

This has poles at z=0, of multiplicity k=2\, and at z=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  {z^{2}+z-1}{z^{2}}}=1\,

{\mathrm  {Res}}_{{z=0}}=\lim _{{z\to 0}}{\frac  {d}{dz}}\left[z^{2}f(z)\right]=\lim _{{z\to 0}}{\frac  {d}{dz}}\left[{\frac  {z^{2}+z-1}{(z-1)}}\right]={\frac  {(2z+1)(z-1)-(z^{2}+z-1)}{(z-1)^{2}}}=-1+1=0\,

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

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