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Homework Help: Roots of Unity Question

  1. Apr 12, 2014 #1
    1. The problem statement, all variables and given/known data
    Hello everyone,

    In this problem, I was to mark all the sixth roots of 1 in the complex plane. Then, I was to figure out what the primitive root W6 is.

    However, I am stuck by the question: "Which power of W6 is equal to 1/W6?"

    2. Relevant equations

    See Below

    3. The attempt at a solution

    In this case, W6 appears to be 1, so could I just not use the power n=-6 to do this?
    This would yield me the value cos(-2*pi) which would be equal to 1. Thus, in general, for Wn, do I simply multiply by -n to get the reciprocal?

  2. jcsd
  3. Apr 12, 2014 #2


    Staff: Mentor

    What does a power of -6 have to do with this problem? The question is asking whichy root is equal to its own reciprocal.

    When you say "W6 appears to be 1" did you do as the problem asks, and mark each of the six roots?
    Yes, but cos(2##\pi##) is also equal to 1. Why do you have the number -2 multiplying pi?
    From your question, I'm guessing that you did NOT mark the six sixth roots of 1. If you have all six of them drawn on the unit circle, it should be obvious that two of them are their own reciprocals.
  4. Apr 12, 2014 #3
    I have W3=-1 and W6=1, so these two equal their own reciprocal. Is that what the question is asking? I seem to have completely misunderstood the question.
  5. Apr 12, 2014 #4


    Staff: Mentor

    That's what I think the problem is asking for, but the problem statement isn't as clear as it should be, IMO. A clearer problem statement would be this:
    Mark the six sixth roots of 1 as points on the unit circle in the complex plane. Which of these roots is its own reciprocal?
    Answer to the second question:
    w0 = 1, (w0)6 = 1, so w0 is a sixth root of 1, and 1/w0 = 1

    w3 = -1, (w3)6 = 1, so w3 is another sixth root of 1, and 1/w3 = -1

    I'm not sure how they are numbering the roots. I am numbering them w0 (= 1), w1, ..., w5. If they are numbering them w1, w2,..., w6, it's hard to say what w6 is if I don't know what they're calling w1.
  6. Apr 12, 2014 #5


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    Homework Helper
    Gold Member

    I interpret the notation rather differently. I believe the author is writing ωn for the principal nth root of unity. The 6 roots of unity are therefore, by definition, the powers of ω6, ω6 having the least nonzero argument of them. ωn for n≠6 is irrelevant to the question.
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