Complex - six roots of unity :S

**trojsi** · June 1st 2010, 03:40 PM

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.

**Gusbob** · June 1st 2010, 04:48 PM

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
$<br /> z^6 = 1 = \cos (2k\pi) + i \sin (2k\pi)<br />$

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:

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

**trojsi** · June 1st 2010, 05:07 PM

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.

**Prove It** · June 1st 2010, 05:07 PM

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:

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

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

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

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

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

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

**trojsi** · June 2nd 2010, 02:31 AM

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

thanks

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Complex - six roots of unity :S

Many thanks

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