# Problem with DeMoivre's Theorem

#### 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)
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)

#### Failure

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 $$\displaystyle z$$.

Last edited:
• MuhTheKuh