The following describes an extension of the Descartes Rule of Signs to polynomials with complex coefficients.

First, I need to define the notion of a "sweep"... Given a complex polynomial p(z) := c_{0} z^{m0} + c_{1} z^{m1} + ... + c_{n} z^{mn} with non-zero c_{i} and strictly increasing m_{i}, let {α_{i}} be the coefficients' "non-decreasing argument sequence": α_{i} = arg c_{i} (mod 2π), and α_{0} ≤ α_{1} ≤ ... ≤ α_{n}, with each α_{i+1}-α_{i} taken as small as possible; then α_{n}-α_{0} computes the "angular sweep" --which I'll denote **sweep(p)**-- of a needle with one end anchored at the origin of the Complex Plane that begins pointing at c_{0} and then spins counter-clockwise to point at each c_{i} in turn ("stalling in place" when consecutive coefficients have equal arguments).[*]

With this, we have:

**The Descartes Rule of Sweeps.** The number of positive real roots of p is at most $\lfloor \frac{1}{\pi} sweep(p)\rfloor$.

(The Descartes Rule of Signs represents a special case: each sign change in a polynomial's real coefficient sequence contributes π to the sweep, so that $\frac{1}{\pi}sweep(p)$ exactly counts those sign changes.)

Now, I can prove the Rule of Sweeps using the Descartes Rule of Signs itself, but that approach sheds no light on *why* the Rule of Signs works (which is what I'm really after). The result seems to be one clever contour away from falling directly out of the Cauchy Argument Principle --and it even seems curiously appropriate that the formula provides a kind of half-winding number from the coefficient sequence-- but apparently I'm not sufficiently clever. :(

Suggestions?

[*] Note: One can also compute a sweep of the coefficients taken in reverse order (or, equivalently, spinning the needle clockwise). Differently-directed sweeps are usually not equal, so we can optimize the bound in the Rule of Sweeps by taking "the" sweep as the minimum of the two directed sweeps.