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Sum of the e's

  1. Mar 24, 2007 #1
    1. The problem statement, all variables and given/known data
    hi,
    I'm working on constructible things again and in one of the proofs our prof threw out this identity and I just don't know where it came from:
    [tex]1+e^{\frac{2\pi}{7}i}+e^{\frac{4\pi}{7}i}+e^{\frac{6\pi}{7}i}+e^{\frac{8\pi}{7}i}+e^{\frac{10\pi}{7}i}+e^{\frac{12\pi}{7}i}=\frac{e^(\frac{12\pi}{7}i)^7-1}{e^{\frac{12\pi}{7}i}-1}[/tex]
    HOW did he get that?


    Edit: I can't tell if the final term is supposed to be 2pi of 12 pi. I dunno.
    2. Relevant equations



    3. The attempt at a solution
     
    Last edited: Mar 24, 2007
  2. jcsd
  3. Mar 24, 2007 #2

    Curious3141

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    Homework Helper

    Hint : geometric progression.

    The exponents on the Right Hand Side should be 2pi, not 12 pi.
     
  4. Mar 24, 2007 #3
    [tex]x^n-1=(x-1)(1+x+x^2+x^3+x^4+x^5+...x^{n-1})[/tex]

    for n=natural number.
     
  5. Mar 24, 2007 #4
    Thanks. I see it now.
    CC
     
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