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Math Help - Roots of Unity

  1. #1
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    Roots of Unity

    Suppose w^3 = 1 and k is a positive integer

    i) Find two possible values of 1 + w^k + w^2k

    I have no idea where to begin? Please give me some hints.
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  2. #2
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    Quote Originally Posted by Lukybear View Post
    Suppose w^3 = 1 and k is a positive integer

    i) Find two possible values of 1 + w^k + w^2k

    I have no idea where to begin? Please give me some hints.
    Suppose k is a multiple of 3. Then 2k is also a multiple of 3. So 1 + w^k + w^(2k) = 3.

    Suppose k gives remainder 1 when divided by 3. Then 2k gives remainder 2 when divided by 3.

    Suppose k gives remainder 2 when divided by 3. Then 2k gives remainder 1 when divided by 3.

    In either of the two latter cases you get 1 + w + w^2.
    Last edited by undefined; August 28th 2010 at 09:22 PM. Reason: typo
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  3. #3
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    w^3 = 1

    w^3 - 1 = 0

    w^3 - 1^3 = 0

    (w - 1)(w^2 + w + 1) = 0

    So w - 1 = 0 or w^2 + w + 1 = 0

    w = 1 or

    w^2 + w + \left(\frac{1}{2}\right)^2 - \left(\frac{1}{2}\right)^2 + 1 = 0

    \left(w + \frac{1}{2}\right)^2 + \frac{3}{4} = 0

    \left(w + \frac{1}{2}\right)^2 = -\frac{3}{4}

    w + \frac{1}{2} = \pm\frac{\sqrt{3}i}{2}

    w = \frac{-1 \pm \sqrt{3}i}{2}.


    So the three solutions are

    w = 1, w = \frac{-1 - \sqrt{3}i}{2}, w = \frac{-1 + \sqrt{3}i}{2}.
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    Quote Originally Posted by undefined View Post
    Suppose k is a multiple of 3. Then 2k is also a multiple of 3. So 1 + w^k + w^(2k) = 3.

    Suppose k gives remainder 1 when divided by 3. Then 2k gives remainder 2 when divided by 3.

    Suppose k gives remainder 2 when divided by 3. Then 2k gives remainder 1 when divided by 3.

    In either of the two latter cases you get 1 + w + w^2.
    Thanks for quick reply. Just got a question with the concept of this. If k gives a remainder of 4, that is equivalent to k giving a remainder of 1 right? Also, how was it acquired that if k gives remainder of 1, then 2k gives a remainder of 3 and vice versa.

    Thanks really much.
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  5. #5
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    Quote Originally Posted by Lukybear View Post
    Thanks for quick reply. Just got a question with the concept of this. If k gives a remainder of 4, that is equivalent to k giving a remainder of 1 right? Also, how was it acquired that if k gives remainder of 1, then 2k gives a remainder of 3 and vice versa.

    Thanks really much.
    Dealing with positive integers, it is not possible to have a remainder that is greater than (or equal to) the divisor! But 4 and 1 are congruent mod 3. I used modular arithmetic to get that 1 * 2 congruent to 2 mod 3, and 2 * 2 congruent to 1 mod 3.
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    Quote Originally Posted by Lukybear View Post
    Thanks for quick reply. Just got a question with the concept of this. If k gives a remainder of 4, that is equivalent to k giving a remainder of 1 right?
    k can't "give a remainder of 4" when divided by 3. The "remainder" when dividing by the integer n must be less than n.

    If you get a "remainder of 4" then you have divided incorrectly- increase your divisor by 1 and you will get a remainder of 1. Also, how was it acquired that if k gives remainder of 1, then 2k gives a remainder of 3 and vice versa.

    Thanks really much.
    If "k, divided by 3, gives a remainder of 1" we can write k= 3n+ 1 for some integer n. Then 2k= 2(3n+1)= 6n+ 2= 3(2n)+ 2.

    If "2, divided by 3, gives a remaider of 2" we can write k= 3m+ 2 for some integer m. Then 2k= 2(3m+2)= 6m+ 4= 6m+ 3+ 1= 3(6m+1)+ 1.
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