# Complex numbers

• Jan 7th 2010, 12:07 PM
Complex numbers
• Jan 7th 2010, 12:38 PM
Plato
$- i = \cos \left( {\frac{{ - \pi }}{2}} \right) + i\sin \left( {\frac{{ - \pi }}{2}} \right)$.

Now find the three cube roots. One of then is
$\cos \left( {\frac{{ - \pi }}{6}} \right) + i\sin \left( {\frac{{ - \pi }}{6}} \right)$.

What are the other two?
• Jan 8th 2010, 11:25 PM
• Jan 9th 2010, 02:14 AM
Prove It
One of the cube roots is $\cos{\left(-\frac{\pi}{6}\right)} + i\sin{\left(-\frac{\pi}{6}\right)}$.

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

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

The radius is 1, so the other roots are at

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

and

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

Therefore the three roots are

$\textrm{cis}\,\left(-\frac{\pi}{6}\right), \textrm{cis}\,\left(\frac{\pi}{6}\right), \textrm{cis}\,\left(-\frac{5\pi}{6}\right)$.
• Jan 9th 2010, 10:40 AM
Now I'm really confused! (Thinking)
• Jan 9th 2010, 11:15 AM
Plato
Quote:

Now I'm really confused! (Thinking)

Why are you cnfused?
Those answers are identical as the ones you posted.
The only difference is the last set is in a more mathematical form.
That is some of us like answers to be $-\pi.
• Jan 9th 2010, 01:26 PM