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Math Help - De Moivre's Theorem

  1. #1
    Super Member Quacky's Avatar
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    De Moivre's Theorem

    I have absolutely no idea where this question should go, so please feel free to move it to the correct subforum.

    a)If z = cos\theta + i sin\theta, use de Moivre's theorem to show that \displaystyle z^n + \frac{1}{z^n} = 2cos (n\theta)

    b) Express (z^2 + \frac{1}{z^2})^3 In terms of Cos{6\theta} and Cos{2\theta}

    Part a) I have done correctly, but part b) Is confusing me. My working so far:
    (z^2 + \frac{1}{z^2})^3

    = [2Cos(2\theta)]^3

    =2^3(Cos(2\theta))^3

    =8(Cos(2\theta))^3

    Then I hit a wall. At first glance, I try to say ' =8Cos6\theta' but that doesn't fit the question. I see that I have to use the theorem in some way, but I have no idea on the correct approach from here.
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  2. #2
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    Isn't 8\cos^32\theta already expressed in terms of \cos{2\theta}?
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  3. #3
    Super Member Quacky's Avatar
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    Yes, but the question means without the exponent involved.
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  4. #4
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    Quote Originally Posted by Quacky View Post
    Yes, but the question means without the exponent involved.
    Why do you think that?
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  5. #5
    Super Member Quacky's Avatar
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    Quote Originally Posted by Plato View Post
    Why do you think that?
    Experience with the textbook and past questions, mainly. I doubt the question would include the part about Cos6\theta otherwise?
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     \displaystyle \cos^3{t} = \cos{t}\cos^2{t} = \cos{t}\left(\frac{1+\cos{2t}}{2}\right) = \frac{\cos{t}+\cos{t}\cos{2t}}{2} = \frac{\cos{t}+\frac{\cos(2t-t)+\cos(2t+t)}{2}}{2}

     \displaystyle = \frac{\cos{t}+\frac{\cos{t}+\cos{3t}}{2}}{2} = \frac{\frac{2\cos{t}+\cos{t}+\cos{3t}}{2}}{2}  =  \frac{2\cos{t}+\cos{t}+\cos{3t}}{4} = \frac{3\cos{t}+\cos{3t}}{4}.

    \therefore ~ \displaystyle \cos^3{t} = \frac{3\cos{t}+\cos{3t}}{4}.

    Put t = 2\theta and multiply by 8, you get what you want.
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  7. #7
    Super Member Quacky's Avatar
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    Thanks for the help. That's exactly what I needed.
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    Quote Originally Posted by Quacky View Post
    I have absolutely no idea where this question should go, so please feel free to move it to the correct subforum.

    a)If z = cos\theta + i sin\theta, use de Moivre's theorem to show that \displaystyle z^n + \frac{1}{z^n} = 2cos (n\theta)

    b) Express (z^2 + \frac{1}{z^2})^3 In terms of Cos{6\theta} and Cos{2\theta}

    Part a) I have done correctly, but part b) Is confusing me. My working so far:

    (z^2 + \frac{1}{z^2})^3

    = [2Cos(2\theta)]^3

    =2^3(Cos(2\theta))^3

    =8(Cos(2\theta))^3

    Then I hit a wall. At first glance, I try to say ' =8Cos6\theta' but that doesn't fit the question. I see that I have to use the theorem in some way, but I have no idea on the correct approach from here.
    (b)

    Using the Binomial Expansion,

    \displaystyle\left(z^2+\frac{1}{z^2}\right)^3=\bin  om{3}{0}\left(z^2\right)^3+\binom{3}{1}\left(z^2\r  ight)^2\frac{1}{z^2}+\binom{3}{2}z^2\left(\frac{1}  {z^2}\right)^2+\binom{3}{3}\left(\frac{1}{z^2}\rig  ht)^3

    =z^6+3z^2+3\left(\frac{1}{z^2}\right)+\frac{1}{z^6  }

    =\left(z^6+\frac{1}{z^6}\right)+3\left(z^2+\frac{1  }{z^2}\right)

    =2Cos6\theta+3\left(2Cos2\theta}\right)

    =2Cos6\theta+6Cos2\theta
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