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Math Help - 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:

    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:

    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  z^{36} = 1


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

     w^{12} + w^6 + 3
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    Quote Originally Posted by kezman View Post
     z^{36} = 1
    Knowing that w is a 36-Primitive root of unity I have to solve
     w^{12} + w^6 + 3
    But there is nothing to solve
     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  w^6 + w^{12} + 3
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  6. #6
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    In polar form we have z = \cos \left( {\frac{\pi }{{18}}} \right) + i\sin \left( {\frac{\pi }{{18}}} \right).
    Now, 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 \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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