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Math Help - Roots of a complex number

  1. #1
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    Roots of a complex number

    Find all the roots in rectangular coordinates, exhibit them as vertices of certain squares, and point out which is the principal root.

    z_0=(-8-8i \sqrt3)^{1/4}=z^{1/4}

    The given answer is \pm (\sqrt 3-i), \pm (1+ i\sqrt3), but I just do not see at all how they arrive at that answer. Here's what I got:

    r=\sqrt(64+192)=\sqrt(256)=16

    tan(\theta)=\frac{-8\sqrt3}{-8}=\sqrt3...\theta= \frac{\pi }{3}

    z=16e^{i(\frac{\pi}{3})}

    z_0={16e^{i(\frac{\pi}{3})}}^{1/4}=\pm2e^{(\frac{\pi}{12}+\frac{2\pi k}{4})}=\pm2e^{(\frac{\pi}{12}+\frac{\pi k}{2})},    k=\pm1, \pm2,...

    Switching back to rectangular coordinates gives:

    \pm2\{cos(\frac{\pi}{12}+k\frac{\pi}{2})+isin(\fra  c{\pi}{12}+k\frac{\pi}{2})\}

    This is clearly not correct, and my eyes are beginning to glaze over from looking at this problem all day. What am I doing wrong?
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  2. #2
    Senior Member
    Joined
    Mar 2009
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    First, the radius should always be positive. Therefore 16^{1/4}=2.
    Second, the angle you should have gotten is in the third quadrant, since both a and b are negative. This gives an original angle of \frac{4\pi}{3}.
    This should help get you the correct answer.
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