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- Jan 7th 2010, 11:07 AM.::MadMaX::.Complex numbers
- Jan 7th 2010, 11:38 AMPlato
$\displaystyle - i = \cos \left( {\frac{{ - \pi }}{2}} \right) + i\sin \left( {\frac{{ - \pi }}{2}} \right)$.

Now find the three cube roots. One of then is

$\displaystyle \cos \left( {\frac{{ - \pi }}{6}} \right) + i\sin \left( {\frac{{ - \pi }}{6}} \right)$.

What are the other two? - Jan 8th 2010, 10:25 PM.::MadMaX::.
- Jan 9th 2010, 01:14 AMProve It
One of the cube roots is $\displaystyle \cos{\left(-\frac{\pi}{6}\right)} + i\sin{\left(-\frac{\pi}{6}\right)}$.

All of the cube roots are evenly spaced around the unit circle.

So they are all separated by an angle of $\displaystyle \frac{2\pi}{3}$.

The radius is 1, so the other roots are at

$\displaystyle -\frac{\pi}{6} + \frac{2\pi}{3} = \frac{\pi}{2}$

and

$\displaystyle \frac{\pi}{2} + \frac{2\pi}{3} = \frac{7\pi}{6} = -\frac{5\pi}{6}$.

Therefore the three roots are

$\displaystyle \textrm{cis}\,\left(-\frac{\pi}{6}\right), \textrm{cis}\,\left(\frac{\pi}{6}\right), \textrm{cis}\,\left(-\frac{5\pi}{6}\right)$. - Jan 9th 2010, 09:40 AM.::MadMaX::.
Now I'm really confused! (Thinking)

- Jan 9th 2010, 10:15 AMPlato
- Jan 9th 2010, 12:26 PM.::MadMaX::.
So I did good job? :D :D :D

Thank you all,especially you Plato! :) - Jan 9th 2010, 04:32 PMCaptcha
I was having exactly the same confusion last night. Haha. See here.