Cauchy

The result of a complex integral over a closed path is always:

$∮ f(z)⋅dz=2⋅π⋅i⋅∑_{1}^{s} Res f(z)$

where:

$f(z)=∑_{n=-∞}^{∞} a_n⋅(z-z_0)^n$ Laurent series.

$a_n=÷{1}{2⋅π⋅i}⋅∮÷{f(z)⋅dz}{(z-z_0)^{n+1}}$

$Res f(z)=a_{-1}$

$Res f(z)=÷{1}{(n-1)!}⋅\lim\limits_{z\rightarrow b} ÷{∂^{n-1}}{∂z^{n-1}}[(z-b)^n⋅f(z)]$

Author: Danny (remove the ".nospam" to send)

Last modification on: Thu, 09 May 2013 .