Problem with DeMoivre's Theorem

**MuhTheKuh** · May 18th 2010, 09:27 AM

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

**Failure** · May 18th 2010, 10:41 AM

Originally Posted by MuhTheKuh

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 $z$ .

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