Recall that the convex hull of a set of points \(S\) is the smallest convex set that contains all points of \(S\) (i.e., the intersection of all convex sets which contain each point in \(S\)). Denote the convex hull of all roots of a polynomial \(f\) by \(H(f)\). Then \(H(f') \subset H(f)\). This (kind of unexpected!) result is known as the Gauss-Lucas Theorem.

The canonical proof of this theorem is ~~really really boring~~ perhaps not particularly insightful. Instead, I'd like
to give a physical intuition (a *demiproof*?) for why this is true:

Consider some polynomial \(f \in \mathbb{C}[x]\) with distinct roots \(\alpha_i\), and fix an equally charged particle, say an electron, at each root. This creates an electric field everywhere in the plane which vanishes exactly at the roots of \(f'\). (!)

We'll prove this for a special case of \(f\), in particular, we want \(f(x) \in \mathbb{R}[x]\) to be monic, separable, and such that \(f(0) \neq 0\).

*Proof.* Let \(\alpha_i\) be the roots of \(f\). The electric potential at any point in the plane is given
by
$$ E(z) := \sum_i \frac{\alpha_i - z}{|\alpha_i - z|^2} = \sum_i \frac{1}{\overline{\alpha_i} - \overline{z}}.
$$
Because we can shift the axes without affecting the electric field, we can choose \(z = 0\). So \({E(z = 0) =
\sum_i \frac{1}{\bar{\alpha_i}}}\). Since \(f\) takes coefficients from \(\mathbb{R}\), we have
$$ E(0) = \sum_i \frac{1}{\alpha_i}. $$
Write \(f(z) = \prod_i (z - \alpha_i)\). Then
\begin{align*}
f'(z) &= (z-\alpha_2)(z-\alpha_3)\dots(z-\alpha_n) + \dots + (z-\alpha_1)(z-\alpha_2)\dots(z-\alpha_{n-1})\\ &=
\sum_i\prod_{j\neq i} (z-\alpha_j).
\end{align*}
So,
\begin{align*}
f'(z=0) = \sum_i\prod_{j\neq i} -\alpha_j &= \pm f(0)\sum_i\frac{1}{\alpha_i}\\ &= \pm f(0)E(0).
\end{align*}
So we have \(f'(0) = \pm f(0)E(0)\), i.e., since by hypothesis \(f(0) \neq 0\), we get \(f'(0) = 0 \iff E(0) =
0\). \(\Box\)