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Math Help - Complex Numbers with multiple powers

  1. #1
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    Complex Numbers with multiple powers

    Hello,

    Im stuck on the following question

    Consider the equation z^n = 1, for the four cases N = 2, 3, 4 and 5. What are the solutions to this equation in each case? Guess (and if possible prove) how many solutions there are for arbitrary N.

    Im just unsure of what its asking, for Z^2 = -1 surely Z will just be +/- i but for the other powers i dont really know what the solutions would be

    Any help most appreciated.
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  2. #2
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    This gives too much away but…
    Let \zeta  = \exp \left( {\frac{{2\pi i}}{n}} \right) = \cos \left( {\frac{{2\pi }}{n}} \right) + i\,\sin \left( {\frac{{2\pi }}{n}} \right).

    Now, each power \zeta^k for k=0,1,\cdots,n-1 is a different complex number.

    But (\zeta^k)^n=1, so how many are there?
    Last edited by Plato; October 16th 2010 at 05:34 AM. Reason: fix copy/paste typo
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  3. #3
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    Quote Originally Posted by Plato View Post
    This gives too much away but…
    Let \zeta  = \exp \left( {\frac{{2\pi i}}{n}} \right) = \cos \left( {\frac{{2\pi i}}{n}} \right) + i\,\sin \left( {\frac{{2\pi i}}{n}} \right).
    Typo: there should be no i inside the trig functions.
    \zeta  = \exp \left( {\frac{{2\pi i}}{n}} \right) = \cos \left( {\frac{{2\pi}}{n}} \right) + i\,\sin \left( {\frac{{2\pi}}{n}} \right).



    Now, each power \zeta^k for k=0,1,\cdots,n-1 is a different complex number.

    But (\zeta^k)^n=1, so how many are there?
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