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Math Help - Complex Integration II

  1. #1
    Junior Member
    Joined
    Apr 2010
    Posts
    26

    Complex Integration II

    How would I compute:

    I=\int_{2\pi }^{0}\frac{4d\theta }{2 cos\theta + 3 sin\theta + 2i}

    I have tried finding the poles by changing the values into complex values

    where z=e^{i\theta }

    2cos\theta =z+\frac{1}{z}

    2sin\theta =-i({z-\frac{1}{z}})

    I have concluded

    I=\oint \frac{\frac{4dz}{iz}}{(z+\frac{1}{z})-\frac{3i}{2}(z-\frac{1}{z})+2i}

    and have found the poles:

    z_{1}=\frac{(12-8i)(2+\sqrt{7})}{52}

    and

    z_{2}=\frac{(12-8i)(2-\sqrt{7})}{52}

    Now how would I find out which one of the poles contributes and how do I continue to finish off the rest of the integral, thanks in advance..
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  2. #2
    Super Member
    Joined
    Aug 2008
    Posts
    903
    When I just integrate it from 0 to 2pi and make all the subs I get:

    -4i\oint \frac{2i}{2i(z^2+1)+3(z^2-1)-4z}dz

    lil' of this, lil' of that and I get:

    \frac{8}{2i+3}\oint \frac{1}{z^2-\frac{4}{2i+3}z+\frac{2i-3}{2i+3}}dz

    \frac{8}{2i+3}\oint \frac{1}{(z-\alpha)(z-\beta)}dz

    and only one is in the contour (via Mathematica Abs command) giving a residue of -(3/2+i)\sqrt{17}

    then I get:

    -2\pi i\frac{8}{2i+3}(3/2+i)/\sqrt{17}\approx -6.1 i which agrees with the numerical result. Yours is the negative of that.
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