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Complex n-th roots

  1. Jan 19, 2012 #1
    1. The problem statement, all variables and given/known data

    Suppose z is a nonzero complex number [itex]z=re^{i\theta}[/itex] . Show that z has exactly n distinct complex n-th roots given by [itex]r^{(1/n)}e^{i(2\pi k+\theta)/n}[/itex] for [itex]0\leq k\leq n-1[/itex].

    3. The attempt at a solution

    My attempt: [itex]z^{n}=(r\cos\theta+i\sin\theta)^{n}=r^{m}(\cos \theta+i\sin\theta)^{n}=r^{m}(\cos(n \theta)+i\sin(n \theta))=r^{m}e^{i\theta n}[/itex] ...Not sure where to go from here.
     
  2. jcsd
  3. Jan 19, 2012 #2
    Note that you took [itex]z[/itex] to the [itex]n[/itex]th power; you don't want to do that. Rather, you want to find [itex]a\in\mathbb{C}[/itex] such that [itex]a^n=z[/itex] (see the difference?). To do this, put [itex]a[/itex] in exponential form also, i.e. write [itex]a=\rho e^{i\varphi}[/itex]. Now you have to find [itex]\rho[/itex] and [itex]\varphi[/itex] such that
    [tex]\rho^n(\cos(n\varphi)+i\sin(n\varphi))=r(\cos(θ)+i\sin(\theta)).[/tex]
     
  4. Jan 19, 2012 #3
    So I take [itex]a=\rho e^{i\varphi} \rightarrow a^{n}=(\rho e^{i\varphi})^{n}=\rho^{n}(\cos(n\varphi)+i\sin(n \varphi)[/itex]. Let [itex]\rho^{n}=r so r^{1/n}=\rho[/itex]. Let [itex]n\varphi=\theta\rightarrow\varphi=\theta/n[/itex] . Then roots a have the form [itex]r^{1/n}(\cos(\frac{\theta}{n})+i\sin(\frac{\theta}{n})) \rightarrow r^{1/n}e^{i/n}[/itex]... not sure if I'm doing this the right way?
     
    Last edited: Jan 19, 2012
  5. Jan 19, 2012 #4
    That looks good (don't forget the [itex]i[/itex] in that last trigonometric expression for [itex]a[/itex]). Now you have one of the roots. To find the other [itex]n-1[/itex] of them, exploit the periodicity of the trigonometric or exponential functions. That is, either use [itex]e^{i(θ+2\pi)}=e^{iθ}[/itex] or use [itex]\sin(θ+2\pi)=\sin(θ)[/itex] and [itex]\cos(θ+2\pi)=\cos(θ)[/itex] (it really doesn't matter which you use). The key thing here is that, while [itex]\rho[/itex] must equal [itex]r^{1/n}[/itex], there is more than one angle that works; in fact, there are [itex]n[/itex] of them.
     
  6. Jan 19, 2012 #5
    Thanks for the insight -- didn't really understand the intuition behind there being [itex]n[/itex] roots before your post. Just one more thing, when I use periodicity I get:

    [itex]r^{1/n}(\cos(\frac{\theta}{n})+i\sin(\frac{\theta}{n}))=r^{1/n}(\cos(\frac{\theta}{n}+2\pi k)+i\sin(\frac{\theta}{n}+2\pi k))=r^{1/n}e^{i(\frac{\theta+2\pi kn}{n})}.[/itex]

    The exponent in that last term is [itex]i(\frac{\theta+2\pi kn}{n})[/itex] where it should be [itex]i(\frac{\theta+2\pi k}{n})[/itex]. Since n is in N and k is in Z, can I define j in Z s.t. j = kn and use that?
     
  7. Jan 19, 2012 #6
    You're right that the exponent should be [itex]i(θ+\frac{2πk}{n}).[/itex] The reason for this is that the angle in the exponent doesn't have to be a multiple of [itex]2\pi[/itex]. Rather, it is the angle times [itex]n[/itex] that should be a multiple of [itex]2\pi[/itex], so that when you raise [itex]a[/itex] to the [itex]n[/itex]th power, then you get that period which leaves your answer unchanged.
     
  8. Jan 19, 2012 #7
    One final remark: to interpret all of this geometrically, notice that the n nth roots of z are the vertices of an n-gon in the complex plane, where the modulus of each point is r1/n. This is the way I usually think of the different possible angles.
     
  9. Jan 19, 2012 #8
    Thanks A. Bahat!
     
  10. Jan 20, 2012 #9
    You're welcome! Happy to help.
     
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