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Math Help - Complex numbers

  1. #1
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    Complex numbers

    I am trying to solve this set of problems, any ideas:

    Let 1, omega, omega squared, ....., omega^n-1 be nth rooots of unity
    then
    (a) Show the conjugate of any nth root of unity is another root of unity, by expressing omega(bar)^j in the form omega^k for appropriate k.
    (b) Find the product of the nth roots of unity
    (c) Find the sum Summation (n-1), k=0 for omega^k of nth roots of unity. (this is a geometric series).
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  2. #2
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    Quote Originally Posted by PacManisAlive View Post
    I am trying to solve this set of problems, any ideas:

    Let 1, omega, omega squared, ....., omega^n-1 be nth rooots of unity
    Let \zeta to the generator of the n roots of unity. That means, \{1,\zeta,\zeta^2,...,\zeta^{n-1}\} contains the roots of unity.
    (a) Show the conjugate of any nth root of unity is another root of unity, by expressing omega(bar)^j in the form omega^k for appropriate k.
    Choose j=n-k.
    Then,
    \cos \left( \frac{2\pi (n-k)}{n} \right) + i \sin \left( \frac{2\pi (n-k)}{n} \right) = \cos \left( \frac{2\pi k}{n}\right) - i \sin \left( \frac{2\pi k}{n} \right)
    (b) Find the product of the nth roots of unity
    We have, (since \zeta \not = 1).
    1+\zeta + \zeta^2 + \zeta^3 + ... + \zeta^{n-1} = \frac{1-\zeta^n}{1-\zeta} = 0
    Because \zeta^n - 1=0.

    (c) Find the sum Summation (n-1), k=0 for omega^k of nth roots of unity. (this is a geometric series).
    How is that different from (b)?
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  3. #3
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    Hello, PacManisAlive!

    Here's some help . . .


    Let 1,\:\omega,\:\omega^2,\:\omega^3,\:\cdots\,\:\omeg  a^{n-1} be the n^{th} roots of unity.

    (a) Show the conjugate of any n^{th} root of unity is another root of unity.

    Let a root be: . a + bi \;=\;\cos\theta + i\sin\theta

    Then: . (a + bi)^n \;=\;(\cos\theta + i\sin\theta)^n \;=\;\cos(n\theta) + i\sin(n\theta) \;=\;1

    . . Hence: . \begin{array}{ccc}\cos(n\theta) & = & 1 \\ \sin(n\theta) & = & 0 \end{array}


    The conjugate is: . a - bi \;=\;\cos(\text{-}\theta) + i\sin(\text{-}\theta) \;=\;\cos\theta - i\sin\theta

    \text{Then: }\;(a - bi)^n \;=\;(\cos\theta - i\sin\theta)^n \;=\;\underbrace{\cos(n\theta)}_{\text{this is 1}} - i\underbrace{\sin(n\theta)}_{\text{this is 0}} \;=\;1

    Therefore, the conjugate is also an n^{th} root of unity.

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