# Symmetry group for roots

1. Jan 7, 2010

### zetafunction

my question is , given the Group G of symmetries for the equation

$$x^{4} + a^{2}=0$$

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

$$x \rightarrow -x$$

$$x \rightarrow ix$$

$$x \rightarrow -ix$$

$$x \rightarrow -x$$

$$x \rightarrow i^{1/2}x$$

$$x \rightarrow (-i)^{1/2}x$$

under this symmetries we can see that we ONLY can have imaginary roots, since from the symmetries above any complex number solution to $$x^{4} + a^{2}=0$$ should have an argument $$4\phi = 2\pi$$ this is deduced from the base that $$x^{4} + a^{2}$$ 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 $$x^{-1}sinh(x)=0$$ 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) $$x^{4} + (a_i)^{2}$$ for some a_i ??

Another question is are conditions (A) and (B) equivalent ??

2. Jan 8, 2010

### rochfor1

A few places where you're not quite right:

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