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Finding the nth root of a complex number?

  1. Mar 14, 2015 #1
    1. The problem statement, all variables and given/known data
    Find the solutions to [itex]z^{\frac{3}{4}}=\sqrt{6}+\sqrt{2}i[/itex]

    2. Relevant equations
    de Moivre's theorem

    3. The attempt at a solution
    [itex]z^{\frac{3}{4}}=2\sqrt{2}e^{\frac{\pi i}{6}}=2\sqrt{2}e^{\frac{\pi i}{6}+2k\pi}=2\sqrt{2}e^{\frac{\pi +12k\pi}{6}i}[/itex]

    [itex]z=4e^{\frac{4}{3}{\frac{\pi +12k\pi}{6}i}=4e^{\frac{2\pi +24k\pi}{9}i}[/itex]

    For answers with a principal argument,

    [itex]-\pi<\frac{2\pi +24k\pi}{9}\leq\pi[/itex]
    [itex]-\frac{11}{24}<k\leq\frac{7}{24}[/itex]

    So the only integer value of k satisfying this is zero. So I get [itex]z=4e^{\frac{2\pi}{9}i}[/itex] as my only solution. However [itex]4e^{\frac{8\pi}{9}i}[/itex] and [itex]4e^{\frac{-4\pi}{9}i}[/itex] are also solutions...

    Can anyone see where I have lost my solutions?

    Thank you in advance :)
     
  2. jcsd
  3. Mar 14, 2015 #2

    Mentallic

    User Avatar
    Homework Helper

    Just fixing up your post since one of your latex expressions had a syntax error.

    You're missing solutions because you introduced [itex]2k\pi i[/itex] too soon.

    [tex]z^{3/4}=\sqrt{8}e^{\frac{\pi i}{6}}[/tex]

    [tex]z^3=64e^{\frac{2\pi i}{3}}=64e^{\frac{2\pi i}{3}+2k\pi i}[/tex]

    and now continue from there.

    The reason it doesn't work your way is because you begin with [itex]2k\pi i[/itex] but then you raise each side to the 4th power, hence multiplying [itex]2k\pi i[/itex] by 4, giving you [itex]8k\pi i[/itex] which is obviously missing possible solutions.
     
  4. Mar 14, 2015 #3

    Ray Vickson

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    Science Advisor
    Homework Helper

    To find all the roots of ##z^p = W##, take any one root, ##r## and then multiply it by the ##p##th roots of unity. Thus, if ##\omega \neq 1## is a root of ##\omega^p = 1##, then so are ##\omega^{-1}, \omega^{\pm 2}, \omega^{\pm 3}, \ldots##. That gives the other roots of ##W## as ##r \omega^{\pm 1}, r \omega^{\pm 2}, \ldots ##.
     
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