Budan's theorem and real part of the roots Budan's theorem gives an upper bound for the number of real roots of a real polynomial in a given interval $(a,b)$. This bound is not sharp (see the example in Wikipedia).
My question is the following: let us suppose that Budan's theorem tells us "there are $0$ or $2$ roots in the interval $(a,b)$" (or more generally "there are $0$, $2$, ... $2n$ roots"). Let us suppose also that there are actually no roots in the interval. Is it true that there are $2$ roots whose real part lies in the interval $(a,b)$?
Thank you!
 A: If this were so, real parts of non-real roots will form a subset of
the real critical points which is absurd. So almost any random example is a counterexample.
Take 
$$f(x)=(1+z^2)(1-x+x^2),$$
there are no real roots, and the the real parts of the complex roots are known exactly: they are $0$ and $1/2$.
Now $$f'(x)=4x^3-3x^2+4z-1,$$
and this has a root at approximately $c=0.28839$ by Maple, but you can locate it
with less accuracy by hand. Then $f''$ has no roots and $f'''$ has
a single root at $1/4$. These roots $c$ and $1/4$ are the only Fourier critical points (the points where
Budan estimate changes). So on the interval $(1/8,3/8)$ Boudan says "at most four roots", while in fact there is no root with real part in this interval.
A: You could take
$$ f(x) = x^4 + 1$$
as a counterexample. Budan's theorem would tell you there's at most 4 roots in the interval $(-\epsilon, \epsilon)$ for any positive $\epsilon$. But $f(x)$ doesn't have any roots with real part 0.

Also $x^3 + 1$ works for the same reason. But degree 3 is minimal -- something quadratic like $x^2 + 1$ gives you the wrong intuition.
