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Roots of unity

  1. Nov 23, 2003 #1


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    [tex]a^k =1[/tex]
    [tex]a \in \mathbb{C}[/tex]
    [tex]k \in Z^+[/tex]
    and for some k
    [tex]A = \{a|a^k = 1\}[/tex]
    [tex]|A| = k[/tex]

    edited: becasue the real numbers are a subset of the complex numbers
    Last edited: Nov 23, 2003
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  3. Nov 23, 2003 #2


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    I googled the roots of unity and it appears that there are k kth roots of unity:

    http://www.jimloy.com/algebra/roots.htm [Broken]
    Last edited by a moderator: May 1, 2017
  4. Nov 23, 2003 #3
    If k is the smallest such integer greater than 1 such that the statement is true, then yes. We can find a cyclic group isomorphic to the kth roots of unity, which takes a lot of typing, but I can provide if you want it.
  5. Nov 23, 2003 #4
    And you posted while I was typing. Oops.
  6. Nov 23, 2003 #5


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    Does this hold true for any ak = x?
  7. Nov 23, 2003 #6
    I'm not entirely sure what you're asking, so I apologise if I don't answer the question. Not every x is a root of unity. For example, the element 2 is of infinite order in C. So therefore we cannot find a k (not equal to 0) in the integers such that 2k = 1.
    The group must be of finite order to be cyclic. But even this is not enough to guarantee that the group is cyclic. However, if the order of the group is prime, the group is cyclic. This is a consequence of Lagrange's theorem.
    The fuss I am making about cyclic groups is because of the guaranteed existence of group isomorphisms to the kth roots of unity.
    Last edited: Nov 23, 2003
  8. Nov 23, 2003 #7


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    What I mean are there k kth roots of x, where x is a real number?
  9. Nov 23, 2003 #8


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    Yes, assuming that the number is not zero, there a k complex kth roots of a number. (I think this result is due to Descartes.)

    If you restrict yourself to the postive reals, then there are two roots if k is even, and one if k is odd.

    For the negative reals, there is one root on even powers, and no roots on odd powers.

    Here's an explanation.

    You're probably familiar with complex numbers written as
    with x and y real but you should be able to see that they can also be written as
    [tex]r * (cos \theta + i sin \theta)[/tex]

    if you apply Euler's forumula, you get that
    [tex]r * (cos \theta + i sin \theta)[/tex]
    is equivalent to
    [tex]r * e^{i\theta}[/tex]

    Now, let's say we have two complex numbers:
    [tex]z_1=r_1 * e^{i\theta_1}[/tex]
    [tex]z_2=r_2 * e^{i\theta_2}[/tex]
    [tex]z_1*z_2=r_1*r_2 * e^{i(\theta_1+\theta_2)}[/tex]

    Now, let's look at your question, given k, and a, how many distinct z are there such that
    I'm not going to get into the proof that there is only one solution for [tex]r_z[/tex], or that it exists, here.

    Now, since [tex]e^{i\theta}=e^{i(\theta+2n\pi)}[/tex]
    there are k solutions for [tex]\theta_z[/tex] that correspond to the solutions to
    where [tex]n[/tex] ranges over the integers from zero to k-1.
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