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Thread: 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 $\displaystyle \theta$ is pronounced "theta" and that you have got the wrong sign in the above formula. It should be $\displaystyle \cos\theta =\frac{1}{2}\left(z{\color{red}+}\frac{1}{z}\right )$.

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

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

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

    Thus, overall you get that

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

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

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

    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 $\displaystyle \mathrm{cis}\theta=e^{i \theta}$, it is easy to see that if $\displaystyle z=\mathrm{cis}\theta$, it follows that $\displaystyle z^2=\left(e^{i\theta}\right)^2=e^{i2\theta}=\mathr m{cis}(2\theta)$.

    Thus,

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