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Roots of Complex Numbers (proof)

  1. Sep 10, 2009 #1
    1. The problem statement, all variables and given/known data

    If c is any nth root of unity other than 1, then

    [tex] 1 + c + c^2 + \cdots + c^{n-1} = 0[/tex]

    3. The attempt at a solution

    This is what is done so far and I am at a dead stall for about 2 hours lol. Any ideas on what I should be thinking of next? Should I continue simplifying? I have tried to continue simplifying but it always leads to nothing relevant. Do I need to be using [tex]re^{i\theta}[/tex] ?

    Proof:
    Assume [tex]c^n = 1[/tex] and [tex]c \not = 1[/tex]. Then [tex]c^n -1 = 0[/tex]. Note that

    [tex](c-1)(1 + c + c^2 + \cdots + c^{n-1}) = (c)(1 + c + c^2 + \cdots c^{n-1}) + (-1)(1 + c + c^2 + \cdots + c^{n-1}) = c + c^2 + c^3 + \cdots + c^n - 1 - c - c^2 - \cdots - c^{n-1}[/tex]
     
  2. jcsd
  3. Sep 10, 2009 #2

    lanedance

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    Homework Helper

    my browser doesn't show tex properly, but simplify your final equation by cancelling terms & i think you're pretty much there...

    you have shown c-1 does not = 0, and you know c^n-1=0, so what does equating your start & final expressions tell you?
     
    Last edited: Sep 10, 2009
  4. Sep 10, 2009 #3
    [tex]S=1 + c + c^2 + \cdots + c^{n-1}[/tex]

    [tex]Sc=c+c^2+c^3 + \cdots + c^{n}[/tex]

    [tex]Sc-S=c^n-1[/tex]

    [tex]S(c-1)=c^n-1[/tex]

    [tex]S=\frac{c^n-1}{c-1}[/tex]

    So you got

    [tex]\frac{c^n-1}{c-1}=0[/tex]

    I think you can solve it now. :wink:
     
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