# Math Help - A problem related to complex numbers

1. ## A problem related to complex numbers

Hey guys, I have a problem concerning complex numbers and I could very much use your help. Please help me with it. I need to find the cube roots of -64i in the form of a+bi with a and b being real. Thank you for your help in advance.

2. Can you arrive at this:

$(-64)^{\frac{1}{3}}=2\sqrt{3}-2i$

3. First $(-64)^{1/3} = -4$. If you let $\zeta = \cos \frac{2\pi}{3}+i\sin \frac{2\pi}{3}$ then the roots are: $-4,-4\zeta,-4\zeta^2$.

4. Written in polar form: $- 64i = 64\left( {\cos \left( {\frac{{ - \pi }}{2}} \right) + i\sin \left( {\frac{{ - \pi }}{2}} \right)} \right) = 64cis\left( {\frac{{ - \pi }}{2}} \right)$

So the three cube roots are:
$4cis\left( {\frac{{ - \pi }}{6}} \right)\,,\,4cis\left( {\frac{{ - 5\pi }}{6}} \right)\,\& \,4cis\left( {\frac{\pi }{2}} \right)\,\,
$