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.

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- Jan 12th 2008, 01:55 PMInstigatorA 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, 02:14 PMgalactus
Can you arrive at this:

$\displaystyle (-64)^{\frac{1}{3}}=2\sqrt{3}-2i$ - Jan 12th 2008, 02:33 PMThePerfectHacker
First $\displaystyle (-64)^{1/3} = -4$. If you let $\displaystyle \zeta = \cos \frac{2\pi}{3}+i\sin \frac{2\pi}{3}$ then the roots are: $\displaystyle -4,-4\zeta,-4\zeta^2$.

- Jan 12th 2008, 02:36 PMPlato
Written in polar form: $\displaystyle - 64i = 64\left( {\cos \left( {\frac{{ - \pi }}{2}} \right) + i\sin \left( {\frac{{ - \pi }}{2}} \right)} \right) = 64cis\left( {\frac{{ - \pi }}{2}} \right)$

So the three cube roots are:

$\displaystyle 4cis\left( {\frac{{ - \pi }}{6}} \right)\,,\,4cis\left( {\frac{{ - 5\pi }}{6}} \right)\,\& \,4cis\left( {\frac{\pi }{2}} \right)\,\,

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