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A contour integral with Laurent Series?

  1. Apr 20, 2012 #1
    1. The problem statement, all variables and given/known data

    1. Evaluate
    [itex]\int_{c_{2}(0)} f(z)dz = \int_{c_{2}(0)} \frac{z^{m}}{1+z^{3}}dz[/itex]
    Where [itex]c_{2}(0)[/itex] is the circle of radius 2 centered at the origin with positive orientation (ccw).

    I have done the question myself and compared it with the solution. However, I don't think I am understanding the use of the Laurent Series expansion right.

    2. Relevant equations



    3. The attempt at a solution
    1. If m >= 0, then the integrand has 3 simple poles at z = -1, e^i(pi/3) and e^-i(pi/3).
    If m < 0, then the integrand has the above poles and poles of order |m| at z = 0. The circular contour contains all such poles.

    From here, what I did was simply apply the Residue Theorem.
    For m >= 0
    [itex]\int_{c_{2}(0)} \frac{z^{m}}{1+z^{3}}dz = 2i\pi(Res(-1) + Res(e^{\frac{i\pi}{3}}) + Res(e^{\frac{-i\pi}{3}}))[/itex]
    For m < = 0
    [itex]\int_{c_{2}(0)} \frac{z^{m}}{1+z^{3}}dz = 2i\pi(Res(-1) + Res(e^{\frac{i\pi}{3}}) + Res(e^{\frac{-i\pi}{3}}) + Res(0))[/itex]
    The first 3 residues are easy to find because they are simple poles. The mth pole at zero is harder but still okay.

    However, the solution does it differently. The first write the integrand as a Laurent Series about z = 0.
    [itex]f(z) = \sum_{n = 0}^{\infty}z^{m}(-z^{3})^{n} = \sum_{n = 0}^{\infty}(-1)^{n}z^{m+3n}[/itex]
    Then for 1/z to appear, m + 3n = -1 hence m = -1 - 3n. Thus:
    [itex]Res(0) = (-1)^{n}[/itex] when [itex] m = -1 - 3n, n≥ 0[/itex]

    I can see why they would do that so that the residue at z = 0 would be easier to find. However, why is the summation for the Laurent Series from n = 0 and not n = -∞? Technically, isn't this just a power series?
     
    Last edited: Apr 20, 2012
  2. jcsd
  3. Apr 20, 2012 #2
    no takers?
     
  4. Apr 20, 2012 #3

    Dick

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    1/(1+z^3) is analytic at z=0. You can expand it in a power series there. No need for negative powers. Then multiply that by z^(m).
     
  5. Apr 20, 2012 #4
    But isn't that only assuming m >= 0?
     
  6. Apr 20, 2012 #5

    Dick

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    I'm saying you can expand 1/(1+z^3) without any negative powers. That's why n=0 to infinity. Sure, after you multiply by x^(m) some of the powers of z may become negative. But the sum is still n=0 to infinity. I'm not sure what's bothering you about this.
     
  7. Apr 20, 2012 #6
    Exactly, there may be negative powers depending on your choice of m, but there certainly shouldn't be infinitely many negative terms as was suggested in the OP.

    When m<0 there is clearly a pole at z=0, not an essential singularity.
     
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