# Thread: Complex numbers

1. ## Complex numbers

2. $- i = \cos \left( {\frac{{ - \pi }}{2}} \right) + i\sin \left( {\frac{{ - \pi }}{2}} \right)$.

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?

3. 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

$\textrm{cis}\,\left(-\frac{\pi}{6}\right), \textrm{cis}\,\left(\frac{\pi}{6}\right), \textrm{cis}\,\left(-\frac{5\pi}{6}\right)$.

4. Now I'm really confused!

5. Originally Posted by .::MadMaX::.
Now I'm really confused!
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.

6. So I did good job?

Thank you all,especially you Plato!

7. Originally Posted by .::MadMaX::.
Now I'm really confused!
I was having exactly the same confusion last night. Haha. See here.