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Math Help - Roots of complex numbers

  1. #1
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    Roots of complex numbers

    I am just wanting to see if the book may have made a typo. When asking for the three cubed roots of z=-2+2i, we get sqrt(8) for r. the next step shows 6th root of 8 *((cos 135 + 360k)/(3) + i sin (135 + 360k)/(3)). I understand how to get everything here except for the 6??? Shouldn't nth root of r be 3rd root of sqrt (8)????
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  2. #2
    Super Member wingless's Avatar
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    Yes, and 3rd root of \sqrt{8} is 6th root of 8.

    \sqrt[3]{\sqrt{8}}

    (8^{\frac{1}{2}})^{\frac{1}{3}}

    8^{\frac{1}{2}\cdot \frac{1}{3}} = 8^{\frac{1}{6}} = \sqrt[6]{8}
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  3. #3
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    I see now, thank you for the help.
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