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

This has poles at z=\pm i\, of multiplicity k=2\,.

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=-i}}=\lim _{{z\to -i}}{\frac  {d}{dz}}(z+i)^{2}f(z)=\lim _{{z\to -i}}{\frac  {d}{dz}}{\frac  {z^{2}}{(z-i)^{2}}}\,

=\lim _{{z\to -i}}{\frac  {(z-i)^{2}2z-z^{2}2(z-i)}{(z-i)^{4}}}\,

={\frac  {(-2i)^{2}2(-i)-(-i)^{2}2(-2i)}{(-2i)^{4}}}\,

={\frac  {8i-4i}{16}}={\frac  {1}{4}}i\,

{\mathrm  {Res}}_{{z=i}}=\lim _{{z\to i}}{\frac  {d}{dz}}(z-i)^{2}f(z)=\lim _{{z\to i}}{\frac  {d}{dz}}{\frac  {z^{2}}{(z+i)^{2}}}\,

=\lim _{{z\to i}}{\frac  {(z+i)^{2}2z-z^{2}2(z+i)}{(z+i)^{4}}}\,

={\frac  {(2i)^{2}2i-i^{2}2(2i)}{(2i)^{4}}}\,

={\frac  {-8i+4i}{16}}={\frac  {-1}{4}}i\,

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=-i}}+{\mathrm  {Res}}_{{z=i}}+{\mathrm  {Res}}_{{z=\infty }}={\frac  {1}{4}}i-{\frac  {1}{4}}i+{\mathrm  {Res}}_{{z=\infty }}\, and {\mathrm  {Res}}_{{z=\infty }}=0\,

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