# Complex roots of polynomials

I am looking for a necessary and sufficient condition to determine whether a polynomial f(x) = a_nx^n+\ldots a_1x+a_0 = 0 has complex roots or not? Any idea?

### Re: Complex roots of polynomials

Yes! I also think so. But I thought that you knew that. The descriminant! However I gave you a particular case which it is very important in some applications. See, for instance, the Cayley-Hamilton Equation and it associate Quadratic (positive and semi-positive defined) forms and the resultant hyper-ellipsoid representation.
perucho

### Re: Complex roots of polynomials

Dear bchui,
Sincerely I don't know if, in general, there exist such a necessary and sufficient condition. However, consider any nxn'' non singular symmetryc matrix with real coefficents and find out the n'' eigenvalues through its associate characteristic equation. (could be have roots multiplicity)
Hope that helps,
peruchin

### Re: Complex roots of polynomials

I would want say: with real entries'', sorry.

### Re: Complex roots of polynomials

I think maybe he meant whether the roots are real or
complex. In that case, looking at things like
Descartes' signs theorem and Sturm's theorem
would help.

### Re: Complex roots of polynomials

Look:
(1) A quadratic equation f(x)=a_2x^2+a_1x+a_0=0 has 0 complex roots
<=> \Delta(f)=a_1^2-4 a_1 a_0 < 0
(2) A cubic equation f(x)=a_3 x^3+a_2 x^2+_a_1 x+a_0 = 0 has 0
compex roots <=> \Delta (f) =(q/2)^2+(p/3)^3 > 0
(3) A quartic equation f(x)=a_4x^4+a_3 x^3+a_2x^2+a_1x+a_0 =0
has 0 or 2 complex roots <=> \Delta (f) < 0

In general
(1) Discriminant(f) = 0 <=> repeated roots exists
(2) Discriminant(f) < 0 <=> number of pairs of
complex roots = 0,2,4,...
(3) Discriminant(f) > 0 <=> number of pairs of
complex roots = 1,3,5,...
The problem is: Do we have a computable index I(f) based on the coefficients a_i only, so that I(f) can tell us whether the number of complex roots is 0 or not (not only even numbers)

### Re: Complex roots of polynomials

Now I understand the question. The answer should be "yes", it is
just a matter of finding the right invariant for the job. I will
think about is some more and get back to you when I at least
sort out the quartic case.

### Re: Complex roots of polynomials

For the quartic case f(x)=a_4x^4+a_3x^3+a_2x^2+a_1x+a_0=0
The sign of the determinant \Detla (f) alone could only tells whether the number of pairs of complex roots is even or odd.
To check for whether there is no complex root, one method is to check for the minimum value of the curve y=f(x) and that involves solving a cubic equation. Considering \Delta(f) may not be the right route and I wonder if there are other method.
Sufficient conditions do exist. Necessary conditions do exists.
"If and only if" is a problem.

The problem is simple but difficult!

### Re: Complex roots of polynomials

Please see the fundamental theorem of algebra!

### Re: Complex roots of polynomials

I guess my problem is actually equivalent to the unsolvability of polynomials of degree higher than 4 in terms of root forms of its coefficients