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Math Help - De'Moivre's theorem in polar form. Stuck

  1. #1
    Junior Member Cthul's Avatar
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    De'Moivre's theorem in polar form. Stuck

    "Use De Moivre's theorem to solve the following equation, in polar form."

    z^{3}=i

    My working out:
    Magnitude:
    z=1
    Angle?
    tan(\theta)=0
    I'm not sure what to do because it's 0 isn't it?
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  2. #2
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    Angle is \displaystyle\frac{\pi}{2}

    (\cos{\frac{\pi}{2}+i\sin{\frac{\pi}{2})^3
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  3. #3
    Junior Member Cthul's Avatar
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    z^{3}=cis(\frac{\pi}{2})
    z=cis(\frac{\pi}{2} \times \frac{1}{3} )
    z=cis(\frac{\pi}{6})
    Ah, I see. Thanks.
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  4. #4
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    Misread.
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  5. #5
    Super Member Random Variable's Avatar
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     \displaystyle e^{i (\frac{\pi}{6} + \frac{2 \pi}{3} n)} for n = 0,1,2.
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  6. #6
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    Quote Originally Posted by Cthul View Post
    z^{3}=cis(\frac{\pi}{2})
    z=cis(\frac{\pi}{2} \times \frac{1}{3} )
    z=cis(\frac{\pi}{6})
    Ah, I see. Thanks.
    You have to remember that there are 3 cube roots, and they are evenly spaced around a circle. So they are separated by an angle of \displaystyle \frac{2\pi}{3}.

    What do you think the other roots are?
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