Polynomial roots and convexity
First, a counterexample to your conjecture. Let $\Pi = x^4+x^3+4x^2+4x = x(x+1)(x^2+4)$, so $P = 4x^3+3x^2+8x+4$. The critical values of $\omega$ are $1.06638, 3.89455 + 2.87687i, 3.89455 - 2.87687i$, and by inspection (using Mathematica) we see that for each of these values of $\omega$, $\mbox{Conv}(\Pi_\omega)$ contains a neighborhood of $0$.
Now for a calculus on convex sets. Every convex set is the intersection of a set of halfplanes. Call a halfplane in this collection essential if removing all of the halfplanes in an open set of halfplanes (in the halfplane topology) containing it from our set of halfplanes makes the intersection of the halfplanes in our set bigger. We wish to find a characterization of the essential halfplanes of $\mbox{Hull}(P)$.
First of all, I claim that any essential halfplane of $\mbox{Hull}(P)$ occurs as an essential halfplane of $\mbox{Conv}(\Pi_\omega)$ for some $\omega$. This follows from continuity - for any open set around our essential halfplane there is some $\omega$, take the limit of a subsequence of these $\omega$s...
Now, suppose that the halfplane $\mbox{Re}(x) \le 0$ occurs as an essential halfplane of some $\mbox{Conv}(\Pi_\omega)$, i.e. there are at least two roots of $\Pi_\omega$ with real part $0$, and the rest of the roots have negative real part. If the number of roots on the line $\mbox{Re}(x) = 0$ (counted with multiplicity) is two, then by holomorphicity we can always find a direction to move $\omega$ so that either both roots move to the left, or both roots stay on the line $\mbox{Re}(x) = 0$ and move towards eachother. If we can ever make both roots move to the left, then clearly the halfplane $\mbox{Re}(x) \le 0$ is not an essential halfplane of $\mbox{Hull}(P)$, otherwise we keep pushing the roots towards eachother until either they run into eachother or until a third root hits the line $\mbox{Re}(x) = 0$. In any case, we see that if a halfplane is essential for $\mbox{Hull}(P)$, then there is some $\omega$ such that the halfplane is essential for $\mbox{Conv}(\Pi_\omega)$ and such that at least three roots (counted with multiplicity) of $\Pi_\omega$ are on the boundary of the halfplane, or two of the roots are equal and $\Pi_\omega$ has no other roots.
So if we let $L$ be the set of $\omega$s such that three of the roots of $\Pi_\omega$ lie on a line, we get that $\mbox{Hull}(P) = \cap_{\omega \in L} \mbox{Conv}(\Pi_\omega)$ if $\deg P \ge 2$.
Edit: Actually, I think there is a problem with this. It's conceivable that two roots are on the line $\mbox{Re}(x) = 0$ and have derivatives (with respect to $\omega$) pointed in opposite directions, such that we can't simply push them towards eachother. For instance, the map from one root to the other root could, locally, look like the fractional linear transform sending the left halfplane to a circle contained in the right halfplane and tangent to the line $\mbox{Re}(x) = 0$ at the other root. So, we may need to enlarge the set $L$ to contain also those $\omega$s for which the ratio of the derivatives of two of the roots (with respect to $\omega$) is a negative real number.
Edit 2: It turns out that this isn't an issue. Call the two roots on the line $\mbox{Re}(x) = 0$ $r_1$ and $r_2$. Suppose that locally, $r_1(\epsilon) = \epsilon$, $r_2(\epsilon) = i - m\epsilon + a\epsilon^k + O(\epsilon^{k+1})$, $a \ne 0$, $m > 0$. Note that if we had $r_2(\epsilon) = i-m\epsilon$, then the intersection of the halfplanes corresponding to $r_1(\epsilon), r_2(\epsilon)$ and $r_1(-\epsilon), r_2(-\epsilon)$ would be contained in the halfplane $\mbox{Re}(x) \le 0$, and the intersection of their boundaries would be located at $i/(m+1)$. Now if $k$ is even, then the correction term shifts the intersection of the boundaries by $a\epsilon^k/(m+1) + O(\epsilon^{k+1})$, so if we choose $\epsilon$ small such that $a\epsilon^k$ is real and negative, then we see that the halfplane $\mbox{Re}(x) \le 0$ is not essential. If $k$ is odd, then if we choose $\epsilon$ small such that $\mbox{Re}(\epsilon) < 0$ and $a\epsilon^k$ is a positive real times $i$, then $r_2(\epsilon)$ is shifted up and $r_2(-\epsilon)$ is shifted down, so the intersection of the boundaries will be shifted to the left (draw a picture), so again the halfplane $\mbox{Re}(x) \le 0$ is not essential.
I. First I want to share some computer experiments of H.H. Rugh. The following image supports the positive answer of the
QUESTION: is this hexagon equal to $Hull(X^3−1)$?
triangle (as a new user I was not allowed to use image tags).
A scilab program testing this problem can be found at roots-dancing. In particular an example $z^6-3z^3+z$ similar to the starting example of Zeb, showing that $Hull(P)$ can be strictly smaller then $\bigcap_vConv(f_v)$ for $v$ ranging the critical values of $f$ (shown in blue polygons). See smaller.
II. Here is a proof (communication of Rugh) of the same statement of Zeb (with $f=\Pi$):
If we let L be the set of $v$'s such that three of the roots of $f_v$ lie on a line, we get $Hull(P)=\bigcap_{v\in L}Conv(f_v)$ if $deg f≥3$.
The underlying idea is very similar to that of Zeb.
Statements: Let $f$ be a polynomial of degree at least 3. Assume that $a_0$ and $b_0$ are two distinct simple roots of $f(z)-v_0$. Then for $v$ in a small neighborhood of $v_0$, there are two simple roots $a(v)$, $b(v)$ of $f(z)-v$ with $a(v_0)=a_0$ and $b(v_0)=b_0$. In this case,
If for some fixed complex number $t\neq 0,1$, the holomorphic function $v\mapsto t a(v) + (1-t)b(v)$ is a constant $c$, then $c$ is a critical point of $f$ and $f$ has a rotational symmetry about $c$.
If the segment $[a_0, b_0]$ is a boundary edge of the polygon $Conv(f_{v_0})$ and no other point in the line through $a_0, b_0$ is mapped to $v_0$, then
2.1 for $v$ sufficiently close to $v_0$, the segment $[a(v),b(v)]$ is a boundary edge of the polygon $Conv(f_{v})$, and for any $t\in ]0,1[$, the map $v\mapsto ta(v)+(1-t) b(v)$ is an open mapping.
2.2. The line through $a_0, b_0$ is outside $\bigcap_v Conv(f_{v})$.
Proof. 1. Replacing $f(z)$ by $f(z-c)$ if necessary, we may assume $c=0$. Note that $a\mapsto b(f(a))$ is defined and holomorphic in a neighborhood of $a_0$, satisfying that $b(f(a_0))=b_0$ and $f(b(f(a)))=f(a)$. It follows that $ta+(1-t)b(f(a))\equiv 0$ so $b(f(a))=\dfrac{ta}{t-1} $. Therefore $f(a)=f(\dfrac {ta}{t-1})$ in a neighborhood of $a$, thus in the entire complex plane. Comparing the coefficients we conclude that $\dfrac t{t-1}$ is a root of unity and $f'(0)=\dfrac{t}{t-1} f'(0)$. Using $\dfrac t{t-1}\ne 1$ we get $f'(0)=0$. An example is $z^6+3z^4-5z^2$.
2.1. The condition means that all the other points in $f^{-1}(v_0)$ are contained in one of the open half planes delimited by the line through $a_0, b_0$. This is clearly an open condition.
Now if $ ta(v)+(1-t) b(v)\equiv c$, by Point 1 $c$ must be a critical point and a center of symmetry and $Conv(f_{v_0})$ would have been symmetric with respect to $c$. This is not possible by our assumption that all the other points of $f^{-1}(v_0)$ (and there is at least one due to the assumption on the degree of $f$ and on the simplicity of $a_0, b_0$ as roots of $f_{v_0}$) are on one side of the line through $a_0, b_0$.
2.2. We may look at the open set $W=\bigcup_v Outer(f_v)$ where $Outer(f_v)$ is the complement of $Conv(f_v)$. We may assume $a_0, b_0$ are on the imaginary axis and all the other points of $f^{-1}(v_0)$ are on the left half plane. We know already $i{\mathbb R}-[a_0, b_0]\subset Outer(f_{v_0})\subset W$.
Fix some $t\in [0,1]$ and let $z_0=ta_0+(1-t)b_0$. We may assume $z_0=0$. Now $v\mapsto z(v)=t a(v)+(1-t) b(v)$ is open. Choose a path $v(s)$ such that $z(v(s))$ is negative real. Then the segment $[a(v(s)), b(v(s))]$ passes through $z(v(s))$ and remains almost vertical for sufficiently small $s$, so has $z_0$ on its right side. By 2.1 we may conclude $z_0\in Outer(f_{v(s)})\subset W$. qed.
III. Finally I want to share some numerical experiments (with the help of Jos Leys) illustrating a refinement (communication by Thurston) of Gauss-Lucas property.
Consider a polynomial $f$ as a branched covering of the complex plane. Denote by ${\cal C}$ the convex hull of the critical points of $f$. It is called {\em the critical convex} of $f$. The following statements are equivalent: (1) For any $v\in \mathbb C$, we have $Conv(f_v)\supset {\cal C}$.
(2) The map $f$ is surjective on any closed half plane $H$ intersecting ${\cal C}$.
(1)$\Longrightarrow$(2). Assume $f(H)\not\ni v$. Then $f^{-1}(v)$ is contained in $\mathbb C-H$ which is an open half plane, in particular convex. Then $Conv(f|_v)$ is also contained in $\mathbb C - H$. So $Conv(f|_v)\not\supset {\cal C}$, contradicting (1).
(2)$\Longrightarrow$(1). Assume that $Conv(f|_v)$ does not contain the entire set ${\cal C}$. Then there is a closed half plane $H$ intersecting ${\cal C}$ but disjoint from $Conv(f|_v)$. Then $f(H)\not\ni v$, that is, $f$ is not surjective on $H$, contradicting (2).
Now the refinement (I'll leave the proof to Thurston if somebody requires) is that if one takes a supporting line $L$ of $\cal C$ there is a region on the outer half space of $L$ on which $f$ is a bijection onto $\mathbb C$ (injective in the interior and bijective on the union of the interior with half of the boundary arc). In fact this region is bounded by two geodesic rays of the conformal metric $|f'(z)|\cdot |dz|$ tangent to $L$ at a critical point (if the critical point is simple, otherwise the region is even smaller). This means that we use the euclidean length in the range to measure tangent vectors in the domain. Geodesics in this metric are the pullbacks by $f$ of straight lines.
The movie bijectivity is made by Jos Leys to illustrate this result.
This problem has been considered before:
The notes to chapter 4 of
Rahman/Schmeißer: Analytic Theory of Polynomials, Oxford University Press, 2002
mention this problem, state that conv(P) is a proper subset of hull(P) in general, and give two references:
1) J. L. Walsh: The location of Critical Points of Analytic and Harmonic Functions, AMS Colloquium Publications Volume 34, 1950, p. 72
2) E. Chamberlin and J. Wolfe: Note on a converse of Lucas's theorem, Proceedings of the American Mathematical Society 5, 1954, pp. 203 - 205
I had a look at both of them, the paper of Chamberlin and Wolfe can be obtained online via the AMS for free. The relevant paragraph in Walsh's book is 3.5.1 (starts on page 71). I state the 4 theorems given there for convenience:
1) conv(P) = hull(P) if P is of degree 1 or 2
2) conv(P) = hull(P) if P is of degree 3 and its zeros are collinear
3) There exists a P of degree 3, such that conv(P) is a proper subset of hull(P); example $P(z) = z^3 + 1$.
4) There exists a P of degree 4 with real zeros, such that conv(P) is a proper subset of hull(P); example $P(z) = (z^2 - 1)^2$.
Chamberlin and Wolfe prove, that
1) Any point on the boundary of hull(P) is on the boundary of conv($\Pi_\omega$) for some $\omega$.
2) If a side of any conv($\Pi_\omega$) contains a point of hull(P) and only two zeros of $\Pi_\omega$ , counting multiplicities, then P is of degree 1. (Here a side is equal to conv($\Pi_\omega$), if the zeros of $\Pi_\omega$ are collinear.)
3) Vertices of conv(P) need not lie on the boundary of hull(P), example $\Pi(z) = z^2 (z + 1)(z^2 - 2az + 1 + a^2)$, where a is positive and sufficiently small.
4) Even if P is cubic, hull(P) need not be determined by its primitives with multiple zeros, example P(z) = $4z^3 + 9/2 z^2 + 2z + 3/2$.
While I have corrected what I considered a minor typo in the example $\Pi(z) = z^2 (z + 1)(z^2 - 2az + 1 + a^2)$, I did no thorough proofreading.
Walsh gives no special references to further work in section 3.5.1 and the only reference provided by Chamberlin and Wolfe is to Walsh's book cited above.