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Complex Analysis Integral

  1. Sep 29, 2009 #1
    1. The problem statement, all variables and given/known data
    Compute the integral

    [tex] \oint_{|z|=30}\frac{dz}{z^9+30z+1} [/tex]


    2. Relevant equations
    Residue theorem for a regular closed curve [tex]C[/tex]

    [tex]\onit_C f(z)dz=2\pi i\sum_k\textrm{Res}(f,z_k)[/tex]

    [tex]z_k[/tex] a singularity of [tex]f[/tex] inside [tex]C[/tex]


    3. The attempt at a solution
    I'd rather not compute the integral numerically. I know that the polynomial in the denominator has a root close to [tex]-1/30[/tex]. By Rouche's theorem I know that all the roots of [tex]f(z)=z^9+30z+1[/tex] lie inside the contour of integration and that they are close to [tex]|z|=1[/tex] also there are 9 distinct roots. I also tried building a comparison with the integral
    [tex]\oint_C\frac{1}{z^9}dz=0[/tex]
    But I did not have any luck in computing the integral
     
  2. jcsd
  3. Sep 29, 2009 #2

    Dick

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    If you know all of the roots are inside of |z|=30 then you know that the integral around |z|=30 is equal the integral around |z|=R for R>30. Let R->infinity.
     
  4. Sep 29, 2009 #3

    Hurkyl

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    If you just take that formula and grind things out symbolically, I think you can compute it without ever actually computing a specific value for any of the roots -- symmetric functions of the roots of a polynomial are pretty nice like that.

    Have you thought about making a transformation to move the known singularities outside of the contour of integration?

    Or, since all the singularities are on the inside, what about just pushing the contour of integration off to infinity?
     
  5. Sep 29, 2009 #4

    Hurkyl

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    To elaborate on the geometric picture here... there's no meaningful difference between the "inside" and the "outside" of your contour -- you can use whichever one you like. dz has a pole at infinity, though, so you have to be careful of that, and many usual formulas are only applicable for finite points, so you have to find the right version that works for infinity (or make a coordinate change).
     
  6. Sep 29, 2009 #5
    I tried letting the radius of the contour go to infinity but I got [tex]0[/tex] as my final answer. Numerical methods seem to point to the conclusion that the integral is non-zero, so I don't think this is correct. I know that [tex]\int_{C_{30}}f=\int_{C_R}f\quad R>30[/tex] but is it true in the limit as [tex]R\to\infty[/tex]?
     
  7. Sep 29, 2009 #6

    Dick

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    Sure. You know the integral over C_R is actually independent of R. But you also have an estimate telling you it's arbitrarily small for large R. That means it's zero.
     
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