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Expressing geometrically the nth roots of a complex number on a circle

  1. Nov 5, 2012 #1
    1. The problem statement, all variables and given/known data

    Let [itex]z \in \mathbb{C}[/itex]. Prove that [itex]z^{1/n}[/itex] can be expressed geometrically as [itex]n[/itex] equally spaced points on the circle [itex]x^2 + y^2 = |z|^2[/itex], where [itex]|z|=|a+bi|=\sqrt{a^2 + b^2}[/itex], the modulus of [itex]z[/itex].

    2. Relevant equations

    //

    3. The attempt at a solution

    My problem is that I am confused with the question.

    I understand that

    [itex]z^{1/n} = |z|^{1/n} (\cos(\frac{\theta}{n} + \frac{2k\pi}{n}) + \sin(\frac{\theta}{n} + \frac{2k\pi}{n}) i)[/itex]

    for [itex]k=0,1,\ldots,n-1[/itex], where [itex]\theta=\mathrm{Arg}(z)[/itex].

    Yet, my confusion is this: Each above complex number we get for each [itex]k[/itex] will produce a complex number with modulus [itex]|z|^{1/n}[/itex]. Because the given circle is of radius [itex]|z|[/itex], I cannot see how any of these complex numbers can lie on it for a arbitrary complex number [itex]z[/itex].

    Where is my confusion lying, or is there a mistake in this question? I had asked the question giver, but they explained that the modulus didn't matter for plotting on such circle, which didn't make sense...

    Thanks you.
     
  2. jcsd
  3. Nov 5, 2012 #2

    Dick

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    Science Advisor
    Homework Helper

    I agree with you. E.g. If you put z=2 and n=2 then the values of 2^(1/2) are sqrt(2) and -sqrt(2). They lie on the circle of radius sqrt(2) not the circle of radius 2. The question is bad.
     
  4. Nov 5, 2012 #3
    Yeah, that's what I am thinking. Thanks for the reply
     
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