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Math Help - find the primitive root

  1. #1
    Member kezman's Avatar
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    find the primitive root

    Let w, a 20th units primitive root.

    Prove that
    w + w^3 + w^7 + w^9

    is a pure imaginary complex.

    Im not sure how to "attack" this problem.
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  2. #2
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    Quote Originally Posted by kezman
    Let w, a 20th units primitive root.

    Prove that
    w + w^3 + w^7 + w^9

    is a pure imaginary complex.

    Im not sure how to "attack" this problem.
    That is not true, 1 is a 20th root of unity yet,
    1+1^3+1^7+1^9 is not a pure imaginary.
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  3. #3
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    I think by primitive root, he means a generator for the multiplicative group of 20th roots of unity, i.e. a complex number of the form:

    exp(I*k*Pi/10) where 0 < k < 20 and k is relatively prime to 20

    expand your expression out using Euler's formula, and you get that the real part looks like

    cos(k Pi/10) + cos(k 9 Pi/10) + cos(k 3 Pi/10) + cos(k 7 Pi/10)

    And you should be able to show that, so long as k is relatively prime to 20,

    cos(k Pi/10) + cos(k 9Pi/10) = 0 and cos(k 3Pi/10) + cos(k 7Pi/10) = 0

    showing that the real part of your expression is zero.

    I'm almost sure there's a slicker way to do it, explicitly using the relative primeness of k and 20, but my trig is a bit rusty. So at the least you could just boil it down to cases. That is, prove it for the cases
    k = 1, 3, 7, 9, 11, 13, 17, 19, which are all the possible primitive roots. k = 1 for instance, you would write

    cos(Pi/10) + cos(9Pi/10) = cos(Pi - 9Pi/10) + cos(9Pi/10) = -cos(9Pi/10) + cos(9Pi/10) = 0

    using the identity cos(Pi-x) = -cos(x).

    hope that helps. Try to find a slicker way before you resort to the cases method. And by the way, I verified the identities for all these k's on maple, so the theorem is true.
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  4. #4
    Member kezman's Avatar
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    yes I want to avoid the cases method. Its to complicated and long.
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  5. #5
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    Alright then...

    If a and b are two real numbers, then the following condition is sufficient to guarantee that cos(a) = -cos(b):

    a = Pi - b + n2Pi

    where n2Pi is any integer multiple of 2Pi. Or, rewritten,

    a + b = Pi + n2Pi

    So in other words, if a + b is an ODD multiple of 2Pi, then cos(a) = -cos(b).

    So in our case, let a = k*Pi/10, b = k*9*Pi/10, and we have

    a + b = k(Pi/10+9*Pi/10) = k*10*Pi/10 = k*Pi

    Note that k must be odd since it is assumed that k is relatively prime to 20. This shows then that

    cos(k*Pi/10) + cos(k*9*Pi/10) = 0

    whenever k is odd (but more importantly for us, when k is relatively prime to 20). Similarly you can show that

    cos(k*3*Pi/10) + cos(k*7*Pi/10) = 0

    which proves the theorem.
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  6. #6
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    Consider taking the complex conjugate of the expression. For any root of unity \zeta, the conjugate \bar\zeta satsifies \zeta\bar\zeta = 1, so \bar{w^i} = w^{20-i}. Furthermore, w^{10} = -1. Conjugating w + w^3 + w^7 + w^9 gives w^{19} + w^{17} + w^{13} + w^{11} = -w - w^3 - w^7 - w^9 which is the negative of the original expression. But a quantity which is negated by complex conjugation is a pure imaginary.
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  7. #7
    Member kezman's Avatar
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    yes thats a good idea thansk.
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