When does a real polynomial have a pair of complex conjugate roots? Suppose we have a polynomial function $f(z)=a_0+a_1z+a_2z^2+...+z^n$ with each $a_i$ between 0 and 1. Is there a method to determine if $f$ has a pair of complex conjugate roots?
There are many results on radius of roots, but I never see similar facts concerning the question here.
 A: Let $f$ be the polynomial in question and let $g$ be $f$ divided by the gcd of $f$ and $f'$. Use Sturm's theorem http://en.wikipedia.org/wiki/Sturm_theorem to compute the number of real roots of $g$ between $-n-1$ and $n+1$. The result is equal to $\deg g$ if and only if all roots of $g$ (and also of $f$) are real.
A: We can assume $f$ has no multiple root (if the gcd of $f$ and $f'$ is not constant, divide by this gcd). Let $n$ be the degree of $f$. Compute
$$\frac{f(X)f'(Y)-f(Y)f'(X)}{X-Y} = \sum_{i,j=i}^{n}a_{i,j}\; X^{i-1}\; Y^{j-1}\;.$$
Then $f$ has all roots real iff the symmetric matrix $(a_{i,j})_{i,j=1,\ldots,n}$ is positive definite. This can be checked for instance by computing the principal minors of this matrix and verifying whether they are all positive.
There are several methods for computing the number of real roots using signature of quadratic form : see for instance this note (in french).
A: It's not clear to me what sort of answer you want - a practical method or a definitive algorithm? Also, I'm not sure how to make use of the assumption on the coefficients. 
One thing to observe is that if $f'(z)$ has a complex root, then so does $f(z)$ by the Gauss-Lucas Theorem. Taking the discrminant of $f^{(n-2)}(z)$, we get a quick criterion for there to be a complex root: 
$$(n-1) a_{n-1}^2 < 2n a_{n-2}.$$
