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Adding primitive roots of unity

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

    I was trying to figure out whether or not ##\zeta_5 + \zeta_5^2## and ##\zeta_5^2 + \zeta_5^3## were complex (where ##\zeta_5## is the fifth primitive root of unity).

    2. Relevant equations



    3. The attempt at a solution

    ##\zeta_5 + \zeta_5^2 = \cos(2\pi/5) + i\sin(2\pi/5) + (\cos(2\pi/5) + i\sin(2\pi/5))^2 = \cos(2\pi/5) + i\sin(2\pi/5) + \cos(4\pi/5) + i\sin(4\pi/5)##.

    Since ##i\sin(2\pi/5)## and ##i\sin(4\pi/5)## do not cancel out each other, ##\zeta_5 + \zeta_5^2## must be complex, right?

    ##\zeta_5^2 + \zeta_5^3 = (\cos(2\pi/5) + i\sin(2\pi/5))^2 + (\cos(2\pi/5) + i\sin(2\pi/5))^3 = \cos(4\pi/5) + i\sin(4\pi/5) + \cos(6\pi/5) + i\sin(6\pi/5)##

    But again, the complex numbers don't cancel out each other, right?
     
  2. jcsd
  3. Aug 6, 2013 #2

    tiny-tim

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    draw an argand diagram!! :rolleyes:

    (ie mark them out as vectors on a unit circle)
     
  4. Aug 6, 2013 #3

    haruspex

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    ##\bar{\zeta_n}\zeta_n = 1##, so ##\bar{\zeta_n} = 1/\zeta_n##. Suppose ##\zeta_n^r+\zeta_n^s## is real. What can you deduce about r+s?
     
  5. Aug 7, 2013 #4

    When I draw a diagram, I need to go three times the angle of ##\zeta_5## for ##\zeta_5+\zeta_5^2##, and so I end up in quadrant three, right? So it is not real.

    But for ##\zeta_5^2 + \zeta_5^3## or ##\zeta_5+\zeta_5^4## we need to go 2+3=1+4=5 times the angle of ##\zeta_5##, and so we will be going all the way and end up back in 1, right? So it must be real.

    Do you think that's correct?
     
    Last edited: Aug 7, 2013
  6. Aug 7, 2013 #5

    tiny-tim

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    Hi Artusartos! :smile:

    I'm not sure what you're doing there (looks more like multiplication than addition :confused:).

    Draw the vectors OB and OC (for ##\zeta_5^2## and ##\zeta_5^3##), and then use the parallelogram law to add them …

    what do you see? :wink:
     
  7. Aug 7, 2013 #6
    Well, that looks like it's in the second quaderant, right?
     
  8. Aug 7, 2013 #7

    tiny-tim

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    what's in the second quadrant? :confused:

    do you know the parallelogram law?

    you should be seeing two vectors OB and OC, and a parallelogram with BOC on three sides​
     
  9. Aug 7, 2013 #8
    Oh I'm sorry. I was looking at ##\zeta_5## instead of ##\zeta_5^2## and ##\zeta_5^2## instead of ##\zeta_5^3##...

    So if I look at ##\zeta_5^2 + \zeta_5^3##, then it lies on the x-axis and so the sine is zero and it must be real, right?
     
  10. Aug 7, 2013 #9

    tiny-tim

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    yup! :biggrin:

    (though it worries me that you add "the sine is zero" …

    i see what you mean, but it really has nothing to do with the problem, does it? :wink:)​
     
  11. Aug 7, 2013 #10
    Thanks a lot. :smile:

    But I'm not sure why it doesn't have to do with this problem. We want sine to be zero, because it is the complex term in ##\cos(2\pi/n) + i\sin(2\pi/n)##, right?
     
  12. Aug 7, 2013 #11

    tiny-tim

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    First, you need an "r" there: ##r[\cos(2\pi/n) + i\sin(2\pi/n)]##.

    Second, it's much simpler to say that a real number is a complex number x + iy with y = 0 …

    and the x + iy form is perfect for addition (of two complex numbers), while the re (polar) form is pretty much useless! :wink:
     
  13. Aug 7, 2013 #12

    haruspex

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    Seems you didn't like my hint, so I'll take it a bit further:

    If ##\zeta_n^r+\zeta_n^s## is real then ##\zeta_n^r+\zeta_n^s = \bar{\zeta_n^r}+\bar{\zeta_n^s} = 1/\zeta_n^r+1/\zeta_n^s##. Multiplying out, ##\zeta_n^{2r+s}+\zeta_n^{r+2s} = \zeta_n^r+\zeta_n^s = \zeta_n^{r+s}(\zeta_n^r+\zeta_n^s)##. Can you take it from there?
     
  14. Aug 7, 2013 #13
    Thanks. So if ##\zeta_n^r+\zeta_n^s## is real, then ##r+s## must be a multiple of ##n##.

    So now I should prove that ##\zeta_n^r+\zeta_n^s = \bar{\zeta_n^r}+\bar{\zeta_n^s}##, right?

    But shouldn't we be proving the converse instead? That if ##r+s## is a multiple of ##n##, then ##\zeta_n^r+\zeta_n^s## must be real?
     
  15. Aug 7, 2013 #14

    haruspex

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    There is another solution to the equation.
    Not sure what you mean. If the sum is real then ##\zeta_n^r+\zeta_n^s = \overline{\zeta_n^r+\zeta_n^s}##, from which the above follows quickly.
    It depends. You asked whether a particular sum of powers is real. We've shown that if it is then a certain relationship holds. If the powers do not satisfy that relationship then the sum is not real. If the powers do satisfy it then, yes, there's a bit more work to do, but it's basically running the same argument backwards.
     
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