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Roots of Unity Proof

  1. Feb 19, 2012 #1
    Hey everyone! I would really appreciate some help with this problem. I have been racking my brain for hours now, and nothing seems to work/convince me.

    1. The problem statement, all variables and given/known data
    Show that Un [itex]\subseteq[/itex] U2n for every positive integer, n.


    2. Relevant equations
    [1] Un = {z ε ℂ, zn = 1}
    [2] Un = {cos([itex]\frac{2m\pi}{n}[/itex]) + i sin([itex]\frac{2m\pi}{n}[/itex])}


    3. The attempt at a solution
    First I started out by comparing the two sets using the first equation:
    (i)zn = 1

    (ii)z2n = 1
    (zn)2 = 1
    zn = [itex]\sqrt{1}[/itex]
    zn = [itex]\pm[/itex]1
    But I was not sure if that was enough to show one is a subset of the other

    So, then I tried using the second formula
    (i) [itex]\Theta[/itex]n = [itex]\frac{2m\pi}{n}[/itex]

    (ii) [itex]\Theta[/itex]2n = [itex]\frac{m\pi}{n}[/itex]

    I hoped I could somehow deduce that given the above theta values, one must be a subset of the other

    But unfortunately, I am not sure if I am going about this proof in the right manner. I would really love any guidance you could give me. Thank you in advance!
     
  2. jcsd
  3. Feb 19, 2012 #2
    Is pretty much as straightforward as you've described. Pick z in Un = {z : zn = 1}. Then you want to show z is in U2n.
     
  4. Feb 19, 2012 #3

    HallsofIvy

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    Staff Emeritus
    Science Advisor

    You are going at it backwards. You want to start with [itex]z^n= 1[/itex] and show that [itex]z^{2n}= 1[/itex].
     
  5. Feb 19, 2012 #4
    So, I let z be an element of the set Un, where zn = 1.
    Then, by squaring both sides I get:
    (zn)2 = 12
    z2n = 1

    Therefore, z must also be an element of the set U2n.

    Is this correct?
     
  6. Feb 20, 2012 #5
    Yep.
     
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