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Thread: Primitive roots

  1. #1
    Member kezman's Avatar
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    Primitive roots

    This problem appeared on my latest exam and I couldnt solve it.

    Knowing that w is a 36-primitive unit root solve:

    $\displaystyle w^{6}+w^{12}+3 $
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  2. #2
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    Quote Originally Posted by kezman View Post
    This problem appeared on my latest exam and I couldnt solve it.

    Knowing that w is a 36-primitive unit root solve:

    $\displaystyle w^{6}+w^{12}+3 $
    First what is a 36 primitive root unit?
    If you mean a primitive root of 36, that is wrong. 36 does not have a primitive root.

    And solve for what?
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  3. #3
    Member kezman's Avatar
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    Sorry maybe I wasnt clear.

    36-unit roots (root of unity), are all the solutions of $\displaystyle z^{36} = 1 $


    Knowing that w is a 36-Primitive root of unity I have to solve

    $\displaystyle w^{12} + w^6 + 3 $
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    Quote Originally Posted by kezman View Post
    $\displaystyle z^{36} = 1 $
    Knowing that w is a 36-Primitive root of unity I have to solve
    $\displaystyle w^{12} + w^6 + 3 $
    But there is nothing to solve
    $\displaystyle w^{12} + w^6 + 3 = ?. $
    Is that what you mean?
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  5. #5
    Member kezman's Avatar
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    Yes, sorry. The exam says Calculate $\displaystyle w^6 + w^{12} + 3 $
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  6. #6
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    In polar form we have $\displaystyle z = \cos \left( {\frac{\pi }{{18}}} \right) + i\sin \left( {\frac{\pi }{{18}}} \right).$
    Now, $\displaystyle z^{12} = \cos \left( {\frac{{2\pi }}{3}} \right) + i\sin \left( {\frac{{2\pi }}{3}} \right)\quad \& \quad z^6 = \cos \left( {\frac{\pi }{3}} \right) + i\sin \left( {\frac{\pi }{3}} \right).$

    Note that $\displaystyle \cos \left( {\frac{{2\pi }}{3}} \right) + \cos \left( {\frac{\pi }{3}} \right) = 0\quad \& \quad \sin \left( {\frac{{2\pi }}{3}} \right) + \sin \left( {\frac{\pi }{3}} \right) = \sqrt 3.$
    Now you can finish.
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