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Math Help - Problem reg De'Moivre's theorem

  1. #1
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    Question De'Moivre's theorem

    Hi,

    Use De’Moivre’s theorem to solve the following equations:

    a. X^(3) - 1 = 10


    b. (x + 1)^(5) + x^(2) = 0




    Thanks in advance,
    Lalit Chugh
    Last edited by MathGuru; June 19th 2006 at 11:58 AM.
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  2. #2
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    Quote Originally Posted by lalitchugh
    Hi,

    Use De’Moivre’s theorem to solve the following equations:

    a. X^(3) - 1 = 10
    Thus,
    x^3=11
    Express in polar form thus,
    x^3=11(\cos 2\pi k+i\sin 2\pi k)
    Thus,
    x=11^{1/3}\left( \cos \frac{2\pi k}{3}+i\sin \frac{2\pi k}{3} \right) for k=0,1,2
    Thus,
    x=11^{1/3}\left( \cos \frac{2\pi \cdot 0}{3}+i\sin \frac{2\pi \cdot}{3} \right)=11^{1/3}
    x=11^{1/3}\left( \cos \frac{2\pi \cdot 1}{3}+i\sin \frac{2\pi \cdot 1}{3} \right)=11^{1/3}\left(-\frac{1}{2}+i\frac{\sqrt{3}}{2}\right)
    x=11^{1/3}\left( \cos \frac{2\pi \cdot 2}{3}+i\sin \frac{2\pi \cdot 2}{3} \right)=11^{1/3}\left( -\frac{1}{2}-i\frac{\sqrt{3}}{2}\right)
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  3. #3
    Super Member malaygoel's Avatar
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    Quote Originally Posted by ThePerfectHacker
    Thus,
    x^3=11
    Express in polar form thus,
    x^3=11(\cos 2\pi k+i\sin 2\pi k)
    Thus,
    x=11^{1/3}\left( \cos \frac{2\pi k}{3}+i\sin \frac{2\pi k}{3} \right) for k=0,1,2
    I think we could take any value of k, why only 0,1 and 2, why not 3,4 and 5?
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    Quote Originally Posted by malaygoel
    I think we could take any value of k, why only 0,1 and 2, why not 3,4 and 5?
    Because given,
    x^n=1
    We can then show that if,
    \zeta\in \left\{ \left{ \cos\frac{2\pi k}{n}+i\sin \frac{2\pi k}{n} \right| 0\leq k\leq n-1\right\}
    Then,
    \zeta^n=1
    Furthermore, all elements are distinct.
    Thus, by the pigeonhole principle we have exactly n elements in this set. But since a polynomial can have a most n solution we see that all of the solution of, x^n=1 are found in this set.

    I like to think this is as elegant as it gets.

    Yes you can take other values for k, but I like to follow this reasoning. Because, I am certain that all the solutions are found by starting at k=0 and going to k=n-1.
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  5. #5
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    Question Problem reg De'Moivre's theorem

    Hi,

    Many thanks for the solution.

    May i request you to provide solution for 2nd question on De'Moivre's thoerem as well.

    I will be thankful to you. It will help me in undertstanding the steps.


    Best Regards,
    Lalit Chugh
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  6. #6
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    latichugh do not beg!
    -=USER WARNED=-
    ---------------------
    Your question was not answered perhaps because it has nothing with de Moiver's theorem.
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