CVR3

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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 \isin \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-2))z^{2n-(n-1)}\,

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

Now,

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

= {2n \choose n-1} \lim_{z\to -1} z^{n+1} = {2n \choose n-1} \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)={2n \choose n-1} \cos n\pi\,


Main Page : Complex Variables : Residues

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