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Math Help - A problem related to complex numbers

  1. #1
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    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.
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  2. #2
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    Can you arrive at this:

    (-64)^{\frac{1}{3}}=2\sqrt{3}-2i
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  3. #3
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    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.
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    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)\,\,<br />
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