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How many square roots does a complex number have?

  1. Sep 16, 2011 #1
    In general, how many square roots does a complex number have?
  2. jcsd
  3. Sep 16, 2011 #2


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    A square root u of z is the solution to the polynomial equation
    u2 = z
    (where z must be considered as a fixed number).
    It is a general theorem that this second-degree polynomial has two complex roots.

    In fact, you can write them down explicitly: if [itex]z = r e^{i\phi}[/itex] then
    [tex]u_1 = \sqrt{r} e^{i\phi / 2} = \sqrt{r} \left( \cos \frac{\phi}{2} + i \sin \frac{\phi}{2} \right)[/tex]
    [tex]u_2 = -u_1[/tex]
    both square to z.
  4. Sep 16, 2011 #3


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    Two - try looking up the Fundamental Theorem of Algebra.
  5. Sep 16, 2011 #4


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    In fact, it is easy to show that any non-zero complex number has precisely n distinct nth roots:

    Let [itex]z= re^{i\theta}[/itex] with r> 0. Then the nth roots of z are given by [itex]r^{1/n}e^{i(\theta+ 2k\pi)/n}[/itex] where [itex]r^{1/n}[/itex] is the positive real nth root of the positive real number r and k is a non-negative integer.

    For k= 0 to n-1, those are distinct because [itex]0\le 2k\pi/n< 2\pi[/itex] but when k= n, [itex]2k\pi/n= 2n\pi/n= 2\pi[/itex] and [itex]e^{i(\theta+ 2\pi)}= e^{i\theta}[/itex].
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