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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  {\cos z}{z^{2}(z-\pi )^{3}}}\,

This function has poles at z=0,\pi \, with multiplicities k=2,3\,.

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}}f(z)=\lim _{{z\to 0}}{\frac  {d}{dz}}\left[z^{2}f(z)\right]=\lim _{{z\to 0}}{\frac  {d}{dz}}{\frac  {\cos z}{(z-\pi )^{3}}}\,

=\lim _{{z\to 0}}\left[{\frac  {(z-\pi )^{3}(-\sin z)-(\cos z)3(z-\pi )^{2}}{(z-\pi )^{6}}}\right]={\frac  {-3\pi ^{2}}{\pi ^{6}}}={\frac  {-3}{\pi ^{4}}}\,

{\mathrm  {Res}}_{{z=\pi }}f(z)={\frac  {1}{2}}\lim _{{z\to \pi }}{\frac  {d^{2}}{{dz}^{2}}}\left[(z-\pi )^{3}f(z)\right]={\frac  {1}{2}}\lim _{{z\to \pi }}{\frac  {d^{2}}{{dz}^{2}}}{\frac  {\cos z}{z^{2}}}\,

={\frac  {1}{2}}\lim _{{z\to \pi }}{\frac  {d}{dz}}{\frac  {z^{2}(-\sin z)-2z\cos z}{z^{4}}}=...={\frac  {1}{2}}{\frac  {\pi ^{2}-6}{\pi ^{4}}}\,

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