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Complex number in Denominator

  1. Aug 26, 2010 #1
    1. The problem statement, all variables and given/known data
    Express (see attachment) with a real denominator


    2. Relevant equations

    Not sure if there is really relevant equations to use here.

    3. The attempt at a solution

    First I multiply the top and bottom by the exponential. That gives me e^ix/(e^ix-r^2). I think this is a good first step, but I am unsure where to go from here.
     

    Attached Files:

  2. jcsd
  3. Aug 26, 2010 #2

    Mentallic

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    To make the denominator real (called rationalizing the denominator) you need to convert it into the form a+ib and then multiply by its complex conjugate a-ib. Can you turn it into that form first? It will be easier if you convert eix into its polar form expression.
     
  4. Aug 26, 2010 #3
    If I convert the denominator as given in the problem I end up with: i r^2 sin(x)+r^2 (-cos(x))+1 I don't see how I can make a conjugate with 3 variables here? Am I missing something? I have went over this several times in attempt to make sure I hadn't made a mistake......
     
  5. Aug 26, 2010 #4

    vela

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    I take it r and δ are real. If that's the case, it's a bit simpler to simply replace i with -i to form the conjugate of the denominator, rather than converting to rectangular form. Use the fact that [itex]z+\bar{z} = 2 \textrm{Re}(z)[/itex] to simplify the cross term.
     
  6. Aug 26, 2010 #5

    Pengwuino

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    Whenever you have a problem of the form [tex]\frac{a}{b}[/tex] where 'a' and 'b' are in general, complex, you can immediately multiply top and bottom by the conjugate of b.

    [tex]\frac{a}{b} \frac{b^*}{b^*}[/tex]

    and the result in the denominator is b*b = |b|, which is always real.

    If [tex]r, \delta[/tex] are real, the only change when you conjugate is changing the sign on the i's.
     
  7. Aug 27, 2010 #6

    HallsofIvy

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    You don't have to write the complex number as a+ bi. Just multiply both numerator and denominator by the complex conjugate, [itex]1- r^2e^{i\delta}[/itex]. You get
    [tex]\frac{1- r^2e^{i\delta}}{1- r^2(e^{i\delta}+ e^{-i\delta})+ r^4}[/tex]
    [tex]= \frac{1- r^2e^{i\delta}}{1+ r^4- 2r^2 cos(\delta)}[/tex]

    (I see now that Vela had said essentially the same thing.)
     
    Last edited: Sep 1, 2010
  8. Sep 1, 2010 #7
    Thanks very much for all the help. I did get this right thanks to all your help!
     
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