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Nth roots of a number

  1. Oct 15, 2011 #1
    I'm slightly confused about why complex numbers are required to find all n roots of a number. Is it specifically because of the fact that you can represent complex numbers as a rotation of the plane? I understand why a number should have n roots, I'm just not sure which part of the definition of complex numbers allows this.

  2. jcsd
  3. Oct 15, 2011 #2
    Finding nth roots of a number c is the same as finding the roots of the polynomial [itex] x^n - c[/itex]. The real numbers are not algebraically closed, meaning there are nonconstant polynomials that have no roots, such as [itex] x^2 + 1 [/itex]. The complex numbers are algebraically closed, so every polynomial of degree n has n complex roots. (They are in fact the algebraic closure of the reals.) It's kind of miraculous, actually, that adjoining a single element, namely [itex] i [/itex] gives us an algebraically closed field.

    One special class of roots are roots of unity, i.e. numbers such that [itex] \zeta^n = 1[/itex] for some positive integer n. For instance, suppose we're trying to find the cube roots of 2, considering the polynomial [itex] x^3 - 2[/itex]. This has one real solution, namely the unique real number we have named [itex] \sqrt[3]{2}[/itex]. However, there are 2 more complex solutions, namely [itex] \sqrt[3]{2} \zeta[/itex] and [itex] \sqrt[3]{2} \zeta^2[/itex], where [itex] \zeta[/itex] is a primitive cube root of 1, like [itex] e^{2 \pi i/3}[/itex].
  4. Oct 16, 2011 #3
    "It's kind of miraculous, actually, that adjoining a single element, namely i gives us an algebraically closed field."

    La goutte qui fait d├ęborder le vase.
  5. Oct 16, 2011 #4
    I had to look that one up. Literally "the drop of water that makes the vase overflow." Or as we say in the U.S., the straw that broke the camel's back.
  6. Oct 16, 2011 #5
    it simply means that not all straws or drops have been created equal.
  7. Oct 17, 2011 #6
    I'm just making sure I got this. So, the reason a number can have n nth roots when using complex numbers is because of the way that i is defined? So, if one defined some other different k, would it be possible to find say n+1 nth roots? I guess my question is: does the fact that the polynomial x^n - c =0 has n roots, something that holds true independent of the way complex numbers are defined?

  8. Oct 17, 2011 #7
    Over any field (or integral domain), a polynomial of degree n can have at most n roots. So if we're dealing with the polynomial [itex]x^n -c[/itex] (which is of degree n) over the real or complex numbers (or the rationals or integers), you can never have n+1 nth roots. But maybe you could if you were instead working over a ring that had zero divisors.

    The fact that every polynomial of degree n over the complex numbers has exactly n roots is called the http://en.wikipedia.org/wiki/Fundamental_theorem_of_algebra" [Broken]. Just to be clear, this is a nontrivial result and is not immediately obvious from the way i is defined. I don't think I saw this theorem until a couple months into my complex analysis class.
    Last edited by a moderator: May 5, 2017
  9. Oct 19, 2011 #8
    Yes, I guess that was my basic question. The fundamental theorem of algebra is not a result of the way i is defined, right? So i had to be defined the way it is so that the fundamental theorem of algebra will hold....
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