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 Exercise [09.02] 
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Joined: 12 Mar 2008, 10:57
Posts: 69
Location: India
Post Exercise [09.02]
\oint{\frac{F(z)}{z^{n+1}}}dz
=\oint{\frac{\sum{\alpha_{r}z^{r}}}{z^{n+1}}}dz(Using the laurent expansion)
=\oint{\frac{\alpha_{n}z^{n}}{z^{n+1}}}dz(For every other value of r the integral is zero courtesy [7.1])
=\alpha_{n}\oint{\frac{1}{z}}dz=2\pi\iota\alpha_{n}
Hence,
\alpha_{n}=\frac{1}{2\pi\iota}\oint{\frac{F(z)}{z^{n+1}}}dz


23 May 2008, 10:10

Joined: 12 Nov 2008, 02:55
Posts: 12
Location: Japan
Post Re: Exercise [09.02]
Sameed Zahoor
Is it necessary to provide a separate derivation of the alpha (sub-n)?
Thanks
Tim


24 Nov 2008, 02:37
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