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Math Help - DeMoivre's Theorem

  1. #1
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    DeMoivre's Theorem

    I am using DeMoivre's Theorem to find the 6th power of the complex number (2+3i). I start by putting the complex number in trig form, but am having trouble with the angle. I have r=sqrt(13), and tan0=3/2. The solution guide has an angle for this tan, but I can not determine how to get it. Also the angle they get is then used as a reference angle. I thought I could use tan(3/2) and then use that value with arctan to find the angle but it is not working.
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  2. #2
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    Hello, dashreeve!

    Use DeMoivre's Theorem to find: . (2 + 3i)^6

    I start by putting the complex number in trig form, but am having trouble with the angle.
    I have: . r\:=\:\sqrt{13},\;\;\tan\theta \:=\:\frac{3}{2} . . . . Good!
    We have: . z \;=\;2 + 3i \;=\;r(\cos\theta + i\sin\theta)

    . . And: . \tan\theta \,=\,\frac{3}{2}\quad\Rightarrow\quad \sin\theta \,=\,\frac{3}{\sqrt{13}} \quad\Rightarrow\quad \cos\theta \,=\,\frac{2}{\sqrt{13}} .[1]

    . . Then: . z^6 \;=\;(\sqrt{13})^6\,\left(\cos\theta + i\sin\theta\right)^6 \;=\;13^3\,(\cos6\theta + i\sin6\theta)


    We can use DeMoivre's Theorem to derive the following identities:

    . . \begin{array}{ccc}\cos6\theta &=& \cos^6\theta - 15\cos^4\theta\sin^2\theta + 15\cos^2\theta\sin^4\theta - \sin^6\theta \\ \sin6\theta &=& 6\cos^5\theta\sin\theta -20\cos^3\theta\sin^3\theta + 6\cos\theta\sin^5\theta \end{array}


    Hence: . z^6 \;=\;13^3\,\bigg[\left(\cos^6\theta - 15\cos^4\theta\sin^2\theta + 15\cos^2\theta\sin^4\theta - \sin^6\theta\right)
    . . . . . . . . . . . . . .  + i\left(6\cos^5\theta\sin\theta -20\cos^3\theta\sin^3\theta + 6\cos\theta\sin^5\theta\right)\bigg]


    Then substitute the values from [1].

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  3. #3
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    Thanks for the help!!
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