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Powers of complex numbers

  1. Feb 6, 2013 #1
    1. The problem statement, all variables and given/known data

    Let z be a complex number such that z^n=(z+1)^n=1. Show that n|6 (n divides 6) and that z^3=1.

    2. Relevant equations

    n|6 → n=1,2,3,6

    3. The attempt at a solution

    The z+1, I think, is what throws me off. Considering z^n=1 by itself, for even n, z=±1 and for odd n, z=1. The (z+1) term, however, contradicts this result and leaves me right back where i started. How should I begin looking at this?
     
  2. jcsd
  3. Feb 6, 2013 #2

    tiny-tim

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    hi ptolema! :wink:

    draw the unit circle …

    obviously both z and z + 1 must lie on that circle

    and the line joining them must be horizontal and of length 1

    sooo … ? :smile:
     
  4. Feb 6, 2013 #3

    HallsofIvy

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    Surely you know better than this. z^n= 1 has real roots 1 and -1 but this is a problem about complex numbers! The fact that z^n= 1 means, as tiny-tim said, that z lies on the circle with center 0 and radius 1. And the same for z+ 1. There are two possible values for z.
     
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