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Roots to Complex Equations

  1. Jun 11, 2008 #1
    1. The problem statement, all variables and given/known data

    Determine how many roots the equation

    [tex](z + \frac{i\sqrt{3}}{2})^{29} = \frac{1+i}{\sqrt{2}}[/tex]

    has that are in the first quadrant.

    3. The attempt at a solution

    I would like to treat the right hand side in the following way.

    [tex](z + \frac{i\sqrt{3}}{2})^{29} = \frac{1+i}{\sqrt{2}}[/tex]

    [tex]z + \frac{i\sqrt{3}}{2} = (\cos \frac{\pi}{4} + i \sin \frac{\pi}{4})^{1/29} = \cos \frac{\pi}{4 \cdot 29} + i \sin \frac{\pi}{4 \cdot 29}[/tex]

    It seems reasonable to rewrite the left hand side into

    [tex]z + \frac{i\sqrt{3}}{2} = z + i \sin \frac{\pi}{3}[/tex]

    Which give us

    [tex]z + i \sin \frac{\pi}{3} = \cos \frac{\pi}{4 \cdot 29} + i \sin \frac{\pi}{4 \cdot 29}[/tex]

    Now the real part of the LHS must match the real part of the RHS. This means that the real part of z, must be

    [tex]Re ~z~ = \cos \frac{\pi}{4 \cdot 29}[/tex]

    and that the imaginary part of x must be

    [tex]Im ~z ~= \sin \frac{\pi}{4 \cdot 29} - \sin \frac{\pi}{3}[/tex]

    From here, I am pretty much lost.
  2. jcsd
  3. Jun 11, 2008 #2


    User Avatar
    Science Advisor
    Homework Helper

    Hi Moridin! :smile:

    Yes, that all looks pretty good (and nice LaTeX, by the way)! :smile:

    Your only error is in:
    because (something)^(1/29) has 29 roots.

    (much as √(something) = (something)^(1/2) has 2 roots)

    So it should be:

    [tex]z + \frac{i\sqrt{3}}{2} = (\cos \frac{\pi}{4} + i \sin \frac{\pi}{4})^{1/29} = \cos \left(\frac{\pi}{4 \cdot 29}\,+\,\frac{2n \pi}{29}\right) + i \sin \left(\frac{\pi}{4 \cdot 29}\,+\,\frac{2n \pi}{29}\right)[/tex]

    for 0 ≤ n ≤ 28. :smile:
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