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Plotting the roots of unity on the complex plane

  1. Apr 25, 2012 #1
    1. The problem statement, all variables and given/known data
    Find the 6th complex roots of √3 + i.

    2. Relevant equations
    z^6=2(cos(π/6)+isin(π/6))
    r^6=2, r=2^1/6
    6θ=π/6+2kπ, θ=π/36+kπ/3

    3. The attempt at a solution
    When k=0, z = 2^1/6(cos(π/36)+isin(π/36)),
    When k=1, z = 2^1/6(cos(13π/36)+isin(13π/36)),
    When k=2, z = 2^1/6(cos(25π/36)+isin(25π/36)),
    When k=3, z = 2^1/6(cos(37π/36)+isin(37π/36)),
    When k=4, z = 2^1/6(cos(49π/36)+isin(49π/36)),
    When k=5, z = 2^1/6(cos(61π/36)+isin(61π/36)).
    I just want help with plotting these roots on the complex plane. So, I am just wondering, are all roots of unity on the complex plane the same, regardless of the equation? By this, I mean, are the position of the roots the same, regardless of what the equation is?
     
  2. jcsd
  3. Apr 25, 2012 #2

    LCKurtz

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    These aren't the roots of unity. They are the roots of ##\sqrt 3 +i##. I'm not sure what you mean by the position being the same. What they are is distributed evenly around the circle. In this example, each one is rotated ##\frac \pi 3## from the next, so by the time you do that 6 times, you are all the way around.
     
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