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Complex roots and de Moivre's formula

  1. Sep 11, 2011 #1
    1. The problem statement, all variables and given/known data
    Find all the roots of [itex]z^{6} = -2[/itex]


    2. Relevant equations
    [itex]z = \sqrt[n]{r} (cos(\frac{\theta}{n} + k\cdot \frac{2\pi}{n}) + isin(\frac{\theta}{n} + k\cdot \frac{2\pi}{n})[/itex] where [itex]k = 0, 1, \cdots, n - 1[/itex]


    3. The attempt at a solution
    I know that [itex]r = \sqrt{x^2 + y^2} = \sqrt{(-2)^2} = 2[/itex] but I don't understand how I find [itex]\theta[/itex]. [itex]arg z = arg -2[/itex] isn't a nice number to work with and now I'm stumped. This is the only thing holding me back since the formula is pretty straightforward so if any one could tell me how I find [itex]\theta[/itex] that would be great.

    Thanks in advance.
     
  2. jcsd
  3. Sep 11, 2011 #2

    Hootenanny

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    Welcome to Physics Forums.

    Perhaps it would be easier if you wrote the equation in exponential notation:

    [tex]r^6 e^{i6\theta} = 2 e^{i\pi}[/tex]

    Can you take it from here?
     
  4. Sep 11, 2011 #3
    Wow! Of course! Thanks for the quick help!
     
  5. Sep 11, 2011 #4

    Hootenanny

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    A pleasure :smile:
     
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