Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Help grading papers: Residue Thm. w/Rational fcns

  1. Feb 8, 2006 #1


    User Avatar
    Homework Helper

    Ok, the quick and dirty: The given Theorem: If f is analytic everywhere in the finite complex plane except for a finite number of singular points interior to a positively oriented simple closed contour C, then

    [tex]\int_{C}f(z)dz=2\pi i \mbox{Res}\left[ \frac{1}{z^2}f\left( \frac{1}{z}\right) ;z=0\right][/tex]

    The problem: If [tex]P(z)=a_0+a_1z+\cdots +a_nz^n[/tex] and [tex]Q(z)=b_0+b_1z+\cdots+b_mz^m[/tex] where [tex]m\geq n+2,a_0\neq 0, b_0\neq 0[/tex] and all the zeros of Q(z) lie interior to C, prove that


    Apply the theorem to get

    [tex]\int_{C}\frac{P(z)}{Q(z)}dz=2\pi i \mbox{Res}\left[ \frac{1}{z^2}\frac{P\left( \frac{1}{z}\right) }{Q\left( \frac{1}{z}\right) } ;z=0\right][/tex]
    now what??? Pratial fractions? That is somewhat brutal, isn't there some simple way to do this. Thanks. --Ben

    Papers due in a few hours, please hurry...
  2. jcsd
  3. Feb 8, 2006 #2


    User Avatar
    Homework Helper

    I got it...

    [tex]\frac{1}{z^2}\frac{P\left( \frac{1}{z}\right) }{Q\left( \frac{1}{z}\right) } ;z=0\right] = \frac{p(z)}{q(z)}[/tex] where deg(p) +2<= deg(q), and the constant term of q is b_m which is non-zero and hence the residue is 0.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook