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Solving integral in complex analysis

  1. Oct 27, 2013 #1
    Hi! I'm new here, been a fan of this site for years, but only now I felt the need of registering.

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

    Use the Residue Theorem to solve the integral:

    ∫[(cos(2t)) / (5-4*cos(t)) ] dt from t=0 to t=2pi


    2. The attempt at a solution

    I did a variable change z=eit. With that:

    (1/(2i))*∫[ (z2 + z-2) / (-2z2 +5z -2)]

    Only taking the integral, I came up with:

    ∫[ (z4 +1) / (z2 * (z-1/2) * (z-2))]

    Which has 3 singularaties: 1/2 , 2 and 0
    Only 2 is not in the path considered.

    The residue in 1/2 is -17/6 and in 0 is 5/2 as they are poles of order 1 and 2 respectively.

    And with that, I have:

    1/(2i) * 2*pi*i * (-17/6 + 5/2) = - Pi/3

    Which is wrong, as the answer is actually Pi/6

    Where did I go wrong?

    Thanks in advance!
     
    Last edited: Oct 27, 2013
  2. jcsd
  3. Oct 27, 2013 #2
    Re-do the residue calculations
     
  4. Oct 27, 2013 #3

    Dick

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    And notice that (z-1/2)*(z-2) is not equal to (-2z^2+5z-2).
     
  5. Oct 28, 2013 #4
    ouch. Would have failed that test. Thanks for pointing that out Dick. Hope the op didn't spend a lot of time re-doing the residue calculations with the wrong factor.
     
  6. Oct 28, 2013 #5

    vanhees71

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    Note that to get your original integral you have to integrate
    [tex]f(z)=\frac{1}{\mathrm{i} z} \cdot \frac{z+z^{-2}}{2(5-2(z+1/z))}=\frac{\mathrm{i}(1+z^4)}{z^2(2z^2-5z+2)}[/tex]
    along the unit circle [itex]z=\exp(\mathrm{i} t)[/itex] with [itex]t \in \{0,2 \pi\}[/itex].

    Now you can use the residue theorem, which gives you immediately [itex]\pi/6[/itex]. To be honest, I've been lazy and did this with help of Mathematica :-).
     
  7. Oct 29, 2013 #6
    Thanks everyone! I feel really stupid right now, my mistake was quite dumb... I had forgotten that when you use quadratic formula, if the quadratic coefficient (a) is not 1, you'll have a*(x-z1)(x-z2). I often forget the most basic stuff.
    Thank you!
     
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