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complex analysis: laurent, residues |
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| Apr24-09, 03:15 PM | #1 |
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complex analysis: laurent, residues
This is addressed to people who know complex analysis (hope this is the right section). Here's the Laurent theorem from my book for my later reference: Suppose a function f is analytic throughout an annular domain R1<|z-z0|<R2, centered at z0, and let C denote any positively oriented simple closed contour around z0 and lying in that domain. Then, at each point in the domain, f(z) has the series representation f(z)=sum (n=0 to inf) of an(z-z0) + sum(n=1 to inf) bn/(z-z0)^n, where an=(1/ 2pi*i) integral over C of f(z)dz/(z-z0)^(n+1) and bn=(1/ 2pi i) integral over C of f(z)dz/(z-z0)^(-n+1).
So say there is a finite number of singularities of f, so they're all in a circle of some radius R. It seems to me you could select any point p in the complex plane, and then make an annulus around that point such that its inner perimeter encloses the singularities. The theorem seems to say that 2pi i *Res(f,p) is equal to the integral of a closed contour around the singularities. I know this is wrong b/c it doesn't agree with what I've seen with the Cauchy Residue Thm, where I calculated that integral by summing residues at singularities. Where is the disconnect? |
| Apr24-09, 10:14 PM | #2 |
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Good question.
b1 is not necessarily Res(f,z0) if the Laurent expansion is for R1<|z-z0|<R2 with R1>0. If the Laurent expansion is for 0<|z-z0|<R2, then b1=Res(f,z0). |
| Apr26-09, 02:17 AM | #3 |
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i don't understand what billy bob said and i don't understand what you said.
residue theorem says the integral over a closed contour of f(z)/z-z0= res(f,z0) where the residue is the coefficient of the 1/z term in the laurent expansion. a corollary of cauchy goursat theorem states that if a function is analytic in some domain and contour encloses another contour the then integrals around those contours are equal. extending that a little and you get that the integral around a contour containing other contours equals the sum of integrals around each contour. this is why the integral of a function with several singularities around a contour equals the sum of the residues. note that the laurent series is different in the neighborhood around each singularity. |
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