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Symmetry group for roots

  1. Jan 7, 2010 #1
    my question is , given the Group G of symmetries for the equation

    [tex] x^{4} + a^{2}=0 [/tex]

    for some 'a' Real valued i see this equation is invariant under the changes

    [tex] x \rightarrow -x [/tex]

    [tex] x \rightarrow ix [/tex]

    [tex] x \rightarrow -ix [/tex]

    [tex] x \rightarrow -x [/tex]

    [tex] x \rightarrow i^{1/2}x [/tex]

    [tex] x \rightarrow (-i)^{1/2}x [/tex]

    under this symmetries we can see that we ONLY can have imaginary roots, since from the symmetries above any complex number solution to [tex] x^{4} + a^{2}=0 [/tex] should have an argument [tex] 4\phi = 2\pi [/tex] this is deduced from the base that [tex] x^{4} + a^{2}[/tex] is a real function for real 'x' , of course this example is TRIVIAL to prove to be true , but how about a more important case, could we deduce from my idea that ALL the roots of the function [tex] x^{-1}sinh(x)=0 [/tex] are ALL imaginary numbers ?

    given any Polynomial K(x) with the following properties

    * K(x) have ONLY pure imaginary roots (A)

    * degre of K(x) is a multiple of '4' (B)

    could we proof by any REDUCIBILITY theorem (over Real numbers) that the irreducible factors of K(x) over the field R are or will be of the form (the best possible chance) [tex] x^{4} + (a_i)^{2} [/tex] for some a_i ??

    Another question is are conditions (A) and (B) equivalent ??
     
  2. jcsd
  3. Jan 8, 2010 #2
    A few places where you're not quite right:

    1) This equation is not invariant under the transformations [tex]x \mapsto \pm i^{ \frac{1}{2} } x[/tex] since [tex]( \pm i^{\frac{1}{2}} x )^4 = i^2 x^4 = -x[/tex], not x.

    2) Do you mean that x^4 + a^2 = 0 as only pure imaginary roots? Because that's not true for a = 2, so your two *-ed conditions are certainly not equivalent.

    3) No, even under those assumptions the irreducible factors do not have to be of that form. Consider for example (x-i)^4, or if you want one that doesn't split, (x^2 + 1)^2.
     
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