A couple of years back, I came up with the following question, to which I have no
answer to this day. I asked a few people about this, most of my teachers and some
friends, but noone had ever heard of the question before, and noone knew the answer.

I hope this is an original question, but seeing how natural it is, I doubt this is
the first time someone has asked it.

First, some motivation. Take $P$ any non zero complex polynomial. It is an easy and
classical exercise to show that the roots of its derivative $P'$ lie in the convex
hull of its own roots (I know this as the Gauss-Lucas property). To show this, you
simply write
$P=a\cdot\prod_{i=1}^{r}(X-\alpha_i)^{m_i}$ where the
$\alpha_i~(i=1,\dots,r)$ are the different roots of $P$, and $m_i$ the corresponding
multiplicities, and you evaluate $\frac{P'}{P}=\sum_i \frac{m_i}{X-\alpha_i}$ on a
root $\beta$ of $P'$ which is not also a root of $P$. You'll end up with an
expression of $\beta$ as a convex
combination of $\alpha_1,\dots,\alpha_r$. It is worth mentioning that all the convex coefficients are $>0$, so the new root cannot lie on the edge of the convex hull of $P$'s roots, unless (perhaps) it is a root shared with $P$.

Now fix $P$ a certain non zero complex polynomial, and consider $\Pi$ it's primitive
(antiderivative) that vanishes at $0:~\Pi(0)=0$ and $\Pi'=P$. For each complex
$\omega$, write $\Pi_{\omega}=\Pi-\omega$ so that you get all the primitives of $P$.
Also, define (for any non zero polynomial $Q$) $\mathrm{Conv}(Q)$, to be the convex
hull of $Q$'s roots.

$\mathrm{QUESTION}$: describe
$\mathrm{Hull}(P)=\bigcap_{\omega\in\mathbb{C}}\mathrm{Conv}(\Pi_{\omega})$.

By the above quoted property, it is clear that $\mathrm{Hull}(P)$ is a convex
compact subset of the complex plane that contains
$\mathrm{Conv}(P)$, but I strongly suspect that it is in general larger.

Here are some easy remarks:

$1)$ replacing $P$ (resp. $\Pi$) by $\lambda P$ (resp. $\lambda \Pi$) will not
change the result, and considering $P(aX+b)$ will change
$\mathrm{Hull}(P)$ accordingly. Hence we don't lose any generality in supposing both
$P$ and $\Pi$ be unitary. The fact that $\Pi$ is no longer a primitive of $P$ is of
no consequence.

$2)$ the intersection defining $\mathrm{Hull}(P)$ can be taken for $\omega$ ranging
in a compact subset of $\mathbb{C}$ (depending of course on $P$), because as
$|\omega|\rightarrow\infty$, the roots of
$\Pi_{\omega}$ will tend to become close to the $(\deg (P)+1)$-th roots of $\omega$,
and for large enough $\omega$ always enclose, say,
$\mathrm{Conv}(\Pi)$.

$3)$ $\mathrm{Hull}(P)$ can be explicitely calculated in the following cases:
$P=X^n$, $P$ of degree $1$ or $2$. For degree $2$ polynomials, $1)$ implies that one
only need to calculate the folloing 2 cases: $P=X^2$ and $P=X(X-1)$. One gets the
one point set containing only $0$, which happens to equal $\mathrm{Conv}(X^2)$, and
$[0,1]=\mathrm{Conv}(X(X-1))$ respectively.

Also if $\Pi$ is a real polynomial of odd degree $n+1$, that has all its roots real
and simple, say
$\lambda_1<\mu_1<\lambda_2<\dots<\mu_n<\lambda_{n+1}$, where I have also placed
$P$'s roots $\mu_1,\dots,\mu_n$, and if you further assume that
$\Pi(\mu_{2j})\leq\Pi(\mu_n)\leq\Pi(\mu_1)\leq\Pi(\mu_{2j+1})$ for all suitable $j$
(a
condition that is best understood when drawing a picture), then it is clear that
$\mathrm{Hull}(P)=\mathrm{Conv}(P)=[\mu_1,\mu_n]$: just vary $\omega$ between
$[\Pi(\mu_n),\Pi(\mu_1)]$, the resulting polynomial $\Pi_{\omega}$ is always split
over the real numbers and you get 

$[\mu_1,\mu_n]=\mathrm{Conv}(P)\subset\mathrm{Hull}(P)\subset
\mathrm{Conv}(\Pi_{\Pi(\mu_1)})\cap
\mathrm{Conv}(\Pi_{\Pi(\mu_n)})=$
$[\mu_1,\dots]\cap
[\dots,\mu_n]=[\mu_1,\mu_n]$

$4)$ The equation $\Pi_{\omega}(z)=\Pi(z)-\omega=0$ defines a Riemann surface, but I
don't see how that could be of any use.

But trying to compute $\mathrm{Hull}(P)$ for the next most simple
polynomial $P=X^3-1$ has proven a challenge for me, and I can only conjecture what
it might be, and thus obtain a general conjecture.

In trying to compute $\mathrm{Hull}(X^3-1)$, which requires factorizing degree 4
polynomials, I naturally tried to look for the simplest values of $\omega$, namely
the $\omega$ that allow for easy factorisation of $\Pi_{\omega}=X^4-4X-\omega$, and
those are the $\omega$ that produce a double root. All that remains to be done
afterwards is to factor a polynomial of degree $2$. Also, the problem is symmetric,
and you can focus on the case where 1 is the double root (i.e. $\omega=-3$).
Plugging the result in the intersection, you obtain the following superset of
$\mathrm{Hull}(X^3-1)$: a hexagon that is the intersection of 3 similar isocele
triangles with their main vertex located on the three third roots of unity $1,j,j^2$

$\mathrm{QUESTION}$: is this hexagon equal to $\mathrm{Hull}(X^3-1)$?

How does the convex hull of the roots of $\Pi_{\omega}$ vary as $\omega$ varies?
When $\omega_0$ is such that all roots of $\Pi_{\omega_0}$ are simple, then a simple
application of the inverse function theorem shows that the roots of $\Pi_{\omega}$
with $\omega$ in a small neighborhood of $\omega_0$ vary holomorphically $\sim$
linearly in $\omega-\omega_0$: $z(\omega)-z(\omega_0)\sim \omega-\omega_0$. If however $\omega_0$ is such that $\Pi_{\omega_0}$
has a multiple root $z_0$ of multiplicity $m>1$, then a small variation of $\omega$
about $\omega_0$ will have the effect that the multiple root $z_0$ will split into
$m$ distinct roots of $\Pi_{\omega}$ that will spread out roughly as
$z_0+c(\omega-\omega_0)^{\frac{1}{m}}$, where $c$ is some non zero coefficient. This
means that for small variations, these roots will move at much higher velocities
than the simple roots, and they will do the major contribution to the variation of
$\mathrm{Conv}(\Pi_{\omega})$, also, they spread evenly out, and (at least if the
multiplicity is greater or equal to $3$) they will tend to increase the convex hull
around $z_0$. Thus it seems not too unreasonable to conjecture that the convex hull
$\mathrm{Conv}(\Pi_{\omega})$ has what one can only describe as
$critical~points$ at the $\omega_0$ that produce roots with
multiplicities. I'm fairly certain there is a sort of calculus on convex sets that
would allow one to make the above statement precise, regardless of wether it's true
or not.

$\mathrm{QUESTION\Conjecture}$: is it true that
$\mathrm{Hull}(P)=\bigcap_{\omega\in\mathrm{MR}}\mathrm{Conv}(\Pi_{\omega})$, where
$\mathrm{MR}$ is the set of all $\omega_0$ such that
$\Pi_{\omega_0}$ has a multiple root, i.e. the set of all $\Pi(\alpha_i)$ where the
$\alpha_i$ are the roots of $P$?

All previous examples of calculations agree with this, and I have tried as best I can to justify this guess heuristically.

Are you aware of a solution? Is this a classical problem? Is anybody brave enough to
make a computer program that would compute some intersections of convex hulls
obtained from the roots to see if my conjecture is any good?