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Contour Integral around |z|=1

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

    Convert the following to an equivalent cotour integral around |z|=1 then use Cauchy's integral formula to evaluate it.

    ##\int_{0}^{2 \pi} \frac {d \theta}{13+5 \sin \theta}##

    2. Relevant equations
    let ##z=e^{i \theta}##


    3. The attempt at a solution

    ##d \theta = \frac{dz}{i z}##

    ##\displaystyle \int_{0}^{2 \pi} \frac {d \theta}{13+5 \sin \theta}=\int_{|z|=1} \frac {dz/iz}{13+5/2i(e^{i \theta} -e^{-i \theta})}\frac{2iz}{2iz}##

    ##\displaystyle =\int_{|z|=1} \frac {2dz}{26iz+5z^2-5}##

    where the denominator has the roots ##i(2.6 \pm 2.4)## using quadratic formula....so far ok?
     
  2. jcsd
  3. May 1, 2012 #2
    The integral is ok but I get different signs on the roots. Maybe double check them.
     
  4. May 2, 2012 #3
    Yes, you are right, should be ##(-2.6 \pm 2.4)i##

    continuing on

    ##|(-2.6 +2.4)i| < 1 \implies (-2.6 +2.4)i## lies inside ##|z|=1##
    ##|(-2.6 -2.4)i| > 1 \implies (-2.6 -2.4)i## lies outside ##|z|=1 \therefore##

    ##\displaystyle \int_{|z|=1} \frac {2dz}{26iz+5z^2-5}=2 \pi i f(-2.6+2.4)i=##

    ##\displaystyle 2 \pi i \frac{2}{(-2.6-2.4)i-(-2.6+2.4)i}=-\frac{\pi}{1.2}##...?
     
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