# A problem related to complex numbers

• Jan 12th 2008, 02:55 PM
Instigator
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.
• Jan 12th 2008, 03:14 PM
galactus
Can you arrive at this:

$(-64)^{\frac{1}{3}}=2\sqrt{3}-2i$
• Jan 12th 2008, 03:33 PM
ThePerfectHacker
First $(-64)^{1/3} = -4$. If you let $\zeta = \cos \frac{2\pi}{3}+i\sin \frac{2\pi}{3}$ then the roots are: $-4,-4\zeta,-4\zeta^2$.
• Jan 12th 2008, 03:36 PM
Plato
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)\,\,
$