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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

{\frac  {z^{{2n}}}{(1+z)^{n}}},n\in {\mathbb  {Z^{+}}}\,

This has a pole at z=-1\, of multiplicity k=n\,.

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}}f(z)={\frac  {1}{(n-1)!}}\lim _{{z\to -1}}{\frac  {d^{{n-1}}}{{dz}^{{n-1}}}}(z+1)^{n}f(z)={\frac  {1}{(n-1)!}}\lim _{{z\to -1}}{\frac  {d^{{n-1}}}{{dz}^{{n-1}}}}z^{{2n}}\,

The derivative can be computed like this:

{\frac  {d^{{n-1}}}{{dz}^{{n-1}}}}z^{{2n}}=(2n)(2n-1)(2n-2)\cdot \cdot \cdot (2n-n)z^{{2n-(n-1)}}\,

multiplicand numerador and denominador for (n-1)!\,

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


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

={\frac  {(2n)!}{[(n-1)!]^{2}}}\lim _{{z\to -1}}z^{{n+1}}={\frac  {(2n)!}{[(n-1)!]^{2}}}\cos(n+1)\pi \,

If f(z)\, is analytic except at isolated singular points, then the sum of all the residues of f(z)\, equals 0.

Therefore {\mathrm  {Res}}_{{z=\infty }}f(z)={\frac  {(2n)!}{[(n-1)!]^{2}}}\cos(n+1)\pi \,

Main Page : Complex Variables : Residues