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Math Help - Factoring z^5-1

  1. #1
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    Factoring z^5-1

    I have be asked to factorize z^5-1


    Attempt at a solution:

    z-1 is a factor because 1^5-1=0

    So (z-1)(z^4+z^3+z^2+z+1)

    I'm not sure how to factor (z^4+z^3+z^2+z+1)

    Help please
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  2. #2
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    Re: Factoring z^5-1

    Quote Originally Posted by ImConfused View Post
    I have be asked to factorize z^5-1

    This amounts to finding the five fifth roots of one.

    Let's agree that r\exp(i\theta)=r(\cos(\theta)=i\sin(\theta).

    Then \rho _k  = \exp \left( {\frac{{2k\pi }}{5}i} \right),\;k = 0,1,2,3,4 are the roots.

    Thus z^5  - 1 = \prod\limits_{k = 0}^4 {\left( {z - \rho _k } \right)} .
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  3. #3
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    Re: Factoring z^5-1

    You might want to locate the five roots on the unit circle. Remember DeMoivre's theorem? That might help you get a more intuitive grasp of it.
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