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Complex number simplification

  1. Sep 23, 2010 #1
    Hello,
    I need to simplify and find Re and Im of gif.latex?\frac{1+e^{ix}}{1+e^{i2x}}.gif .

    I've multiplied by gif.latex?\frac{1+e^{-i2x}}{1+e^{-i2x}}.gif and got that Re= gif.latex?\frac{1}{2}\cdot(1+\frac{1}{cos(x)}).gif and Im= gif.latex?\frac{-1}{2}tan(x).gif .

    Is that the correct way and a correct result ?
     
  2. jcsd
  3. Sep 23, 2010 #2
    seems fair to me
     
  4. Sep 24, 2010 #3

    ehild

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    Show your work in detail. The final result is wrong. How did you get cos x in the denominator?

    ehild
     
  5. Sep 24, 2010 #4
    here's my work in details,
    http://img811.imageshack.us/img811/7444/exsol.jpg" [Broken]
     
    Last edited by a moderator: May 4, 2017
  6. Sep 24, 2010 #5

    ehild

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    It is better if you type in your work.

    I found a typo: You wrote that e-ix=cos (2x)-isin(2x).

    ehild
     
  7. Sep 24, 2010 #6
    it should have been
    [URL]http://latex.codecogs.com/gif.latex?e^{-i2x}[/URL]
    so the latter expansion is correct.
     
    Last edited by a moderator: Apr 25, 2017
  8. Sep 24, 2010 #7

    ehild

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    OK, I see now. But you should exclude x=(2k+1)pi/2 before multiplying both numerator and denominator by 1+exp(-2ix) which is 0 if x=(2k+1)pi/2.
     
  9. Sep 24, 2010 #8
    oh, forgot about that.
    thanks
     
  10. Sep 24, 2010 #9

    ehild

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    It would have been simpler to factor out exp(ix) from the denominator:

    [tex]\frac{1+e^{ix}}{1+e^{2ix}}=\frac{1+e^{ix}}{e^{ix}(e^{-ix}+e^{ix})}=\frac{e^{-ix}(1+e^{ix})}{2\cos{x}}[/tex]

    ehild
     
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