Problem with DeMoivre's Theorem

May 2010
19
0
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 (Happy)
 
Jul 2009
555
298
Zürich
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 \(\displaystyle \frac{1}{z}=\frac{\overline{z}}{z\overline{z}}=\frac{\overline{z}}{|z|^2}=\overline{z}\).

Thus, overall you get that

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

\(\displaystyle =\frac{1}{2}\left(\cos\theta+i\sin\theta+(\cos\theta - i\sin\theta)\right)=\frac{1}{2}\left(2\cos\theta\right)=\cos\theta\)
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 \(\displaystyle z^2=\left(e^{i\theta}\right)^2=e^{i2\theta}=\mathrm{cis}(2\theta)\).

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 \(\displaystyle z\).
 
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