# Thread: Complex - six roots of unity :S

1. ## Complex - six roots of unity :S

I managed to get to the point where there is an infinite range of values $theta + 2n pi$ (Problem & Ans Attached). I am confused how he put it in exp form.

2. Originally Posted by trojsi
I managed to get to the point where there is an infinite range of values $theta + 2n pi$ (Problem & Ans Attached). I am confused how he put it in exp form.
I think you need to be more specific as to where you are at.

You should have
$
z^6 = 1 = \cos (2k\pi) + i \sin (2k\pi)
$

By deMoivre's theorem:

$z = \cos \frac{k\pi}{3} + i \sin \frac{k\pi}{3}$

You do not have an infinite range, because you are given the conditions that $-\pi < \phi \leq \pi$
So you want to pick the values k for which the angle lies between $-\pi$ and $\pi$

These values of k are: $0, \pm 1, \pm 2 , 3$ (Note: it's 3, not -3 because your range includes $\pi$, but not $-\pi$)

To put it into exponential form as required, you'll need euler's formula:

$e^{i\theta} = \cos \theta + i \sin \theta$

3. ## Many thanks

Perfect explanation... i understood it right away. I need to brush up on the theorem .. i bet maths is only abstract when not knowing the proofs and the whole story behind them.

4. The other way is to remember that all the roots are evenly spaced around a circle. So if you can figure out the angle between them, you can keep adding this angle and find all the roots.

In your case, the real roots are easy ( $\pm 1$)...
For the complex roots...

$z^6 = 1$

$z^6 = \cos{2\pi} + i\sin{2\pi}$

$z = \cos{\frac{2\pi}{6}} + i\sin{\frac{2\pi}{6}}$

$z = \cos{\frac{\pi}{3}} + i\sin{\frac{\pi}{3}}$.

So all roots differ by an angle of $\frac{\pi}{3}$. From here we can say the roots are:

$\cos{\frac{\pi}{3}} + i\sin{\frac{\pi}{3}} = \frac{1+i\sqrt{3}}{2}$

$\cos{\frac{2\pi}{3}} + i\sin{\frac{2\pi}{3}} = \frac{-1+i\sqrt{3}}{2}$

$\cos{\frac{3\pi}{3}} + i\sin{\frac{3\pi}{3}} = -1$

$\cos{\frac{4\pi}{3}} + i\sin{\frac{4\pi}{3}} = \frac{-1-i\sqrt{3}}{2}$

$\cos{\frac{5\pi}{3}} + i\sin{\frac{5\pi}{3}} = \frac{1-i\sqrt{3}}{2}$

$\cos{\frac{6\pi}{3}} + i\sin{\frac{6\pi}{3}} = 1$.

5. ohoho lovely sir... I really needed that pattern of the root .. things are crystal clear now.

thanks