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Math Help - Problem with DeMoivre's Theorem

  1. #1
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    Problem with DeMoivre's Theorem

    I am having a little struggle with two questions about deMoivre's theorem and I did not find anything helpful in my books, so I was wondering if anyone here knew it.
    First of all I assume that for those particalur questions
    cos(pheta)= 1/2(z-1/z)
    and
    sin(pheta= 1/2i(z-1/z)
    are given formulas? (if anyone knows?)
    and finally I do not understand how to get from the step where I worked out all z's in my brackets to an answer that makes sense.




    Thanks for your time and help
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  2. #2
    Super Member Failure's Avatar
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    Quote Originally Posted by MuhTheKuh View Post
    I am having a little struggle with two questions about deMoivre's theorem and I did not find anything helpful in my books, so I was wondering if anyone here knew it.
    First of all I assume that for those particalur questions
    cos(pheta)= 1/2(z -1/z)
    Please note that \theta is pronounced "theta" and that you have got the wrong sign in the above formula. It should be \cos\theta =\frac{1}{2}\left(z{\color{red}+}\frac{1}{z}\right  ).

    This formula is valid if, z=\mathrm{cis}\theta. To see this you need to know that

    (1) \mathrm{cis}\theta =\cos\theta+i\sin\theta, by definition of \mathrm{cis}\theta

    and (2) that \frac{1}{z}=\frac{\overline{z}}{z\overline{z}}=\fr  ac{\overline{z}}{|z|^2}=\overline{z}.

    Thus, overall you get that

    \frac{1}{2}\left(z+\frac{1}{z}\right)=\frac{1}{2}\  left(\mathrm{cis}\theta +\overline{\mathrm{cis}\theta}\right)

    =\frac{1}{2}\left(\cos\theta+i\sin\theta+(\cos\the  ta - i\sin\theta)\right)=\frac{1}{2}\left(2\cos\theta\r  ight)=\cos\theta
    what was to be shown.

    Similarly for \sin\theta =\frac{1}{2i}\left(z{\color{red}-}\frac{1}{z}\right)

    and
    sin(pheta= 1/2i(z-1/z)
    are given formulas? (if anyone knows?)
    and finally I do not understand how to get from the step where I worked out all z's in my brackets to an answer that makes sense.
    If you know that \mathrm{cis}\theta=e^{i \theta}, it is easy to see that if z=\mathrm{cis}\theta, it follows that z^2=\left(e^{i\theta}\right)^2=e^{i2\theta}=\mathr  m{cis}(2\theta).

    Thus,

    \frac{1}{4}\left(z^2+\frac{1}{z^2}+2\right)=\frac{  1}{2}\left({\color{blue}\frac{1}{2}\left(z^2+\frac  {1}{z^2}\right)}\right)+\frac{1}{2}{\color{red}=}\  frac{1}{2}{\color{blue}\cos(2\theta)}+\frac{1}{2}=  \frac{1}{2}\left(\cos(2\theta)+1\right)
    Where the equivalence that I have marked red follows from the above formula \cos\theta =\frac{1}{2}\left(z+\frac{1}{z}\right), if you subtitute 2\theta for \theta and z^2 for z.
    Last edited by Failure; May 18th 2010 at 11:34 PM.
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