Complex analysis: laurent, residues

In summary, the Laurent theorem states that for an analytic function f in an annular domain centered at z0, f can be represented by a series consisting of an infinite sum of coefficients an and bn. These coefficients can be calculated using the Cauchy Residue Theorem and the integral of a closed contour around the singularities. However, the value of bn may not necessarily be equal to the residue of f at z0, depending on the radius of the annular domain. It is important to note that the Laurent series can be different in the neighborhood of each singularity.
  • #1
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?
 
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  • #2
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).
 
  • #3
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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