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Thread: 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."

    $\displaystyle z^{3}=i$

    My working out:
    Magnitude:
    $\displaystyle z=1$
    Angle?
    $\displaystyle tan(\theta)=0$
    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 $\displaystyle \displaystyle\frac{\pi}{2}$

    $\displaystyle (\cos{\frac{\pi}{2}+i\sin{\frac{\pi}{2})^3$
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  3. #3
    Junior Member Cthul's Avatar
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    $\displaystyle z^{3}=cis(\frac{\pi}{2})$
    $\displaystyle z=cis(\frac{\pi}{2} \times \frac{1}{3} )$
    $\displaystyle z=cis(\frac{\pi}{6})$
    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 \displaystyle e^{i (\frac{\pi}{6} + \frac{2 \pi}{3} n)} $ for n = 0,1,2.
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  6. #6
    MHF Contributor
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    Quote Originally Posted by Cthul View Post
    $\displaystyle z^{3}=cis(\frac{\pi}{2})$
    $\displaystyle z=cis(\frac{\pi}{2} \times \frac{1}{3} )$
    $\displaystyle z=cis(\frac{\pi}{6})$
    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 $\displaystyle \displaystyle \frac{2\pi}{3}$.

    What do you think the other roots are?
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