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Decomposition using roots of unity

  1. Apr 1, 2015 #1
    1. The problem statement, all variables and given/known data
    Decompose x5 - 1 into the product of 3 polynomials with real coefficients, using roots of unity.

    2. Relevant equations
    As far as I know, for xn = 1 for all n ∈ ℤ, there exist n distinct roots.

    3. The attempt at a solution

    So, let ω = e2πi/5. I can therefore find all the 5th roots of unity:


    ω1 = e2πi/5
    ω2 = ω2 = e4πi/5
    ω3 = ω3 = e6πi/5
    ω4 = ω4 = e8πi/5
    ω5 = ω5 = e5πi/5 = 1

    As far as I can get all the roots, I still don't quite understand how to decompose it into a product of 3 polynomials... What does it mean?
     
  2. jcsd
  3. Apr 1, 2015 #2
    I can get (x - 1)(x4 + x3 + x2 + x + 1), but then what to do?

    To decompose (x4 + x3 + x2 + x + 1) into 2 more?
     
  4. Apr 1, 2015 #3

    Dick

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    Your polynomial is ##(x-\omega_1) (x-\omega_2) (x-\omega_3) (x-\omega_4) (x-\omega_5)##. Try looking at a pair of factors corresponding to complex conjugate roots.
     
  5. Apr 1, 2015 #4
    You mean this pair are conjugates to each other?
     
  6. Apr 1, 2015 #5

    Dick

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    I mean if ##r## is a complex number and ##r^*## is its conjugate then ##(x-r) (x-r^*)## is a real polynomial. Use that.
     
  7. Apr 1, 2015 #6
    You mean I can make two pairs into the form (x−r)(x−r∗) that makes them real?
     
  8. Apr 1, 2015 #7

    Mark44

    Staff: Mentor

    The last one in your list is wrong. e5πi/5 = eπi = -1.
     
  9. Apr 1, 2015 #8

    Dick

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    I mean that if you multiply that out the coefficients of each power of x will be real. Try it. Do you see why?
     
  10. Apr 1, 2015 #9

    Ray Vickson

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    Just try it out for yourself!
     
  11. Apr 1, 2015 #10
    I can solve it now. Thanks.
     
  12. Apr 1, 2015 #11

    Sorry. My typo... It should be (e2πi/5)5, so it equals 1.
     
  13. Apr 1, 2015 #12

    Mark44

    Staff: Mentor

    I figured as much. To keep the same form as the other roots in your list, you could write it as e10πi/5
     
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