Let \(p\in\CC[z]\) be a nonconstant polynomial. Then every zero of \(p'\) lies in the convex hull of the set of zeros of \(p\).
Write \(p(z)=a\prod_{j=1}^n(z-a_j)\) with \(a\neq0\), \(n\ge1\), and \(a_1,\ldots,a_n\in\CC\) the zeros of \(p\) listed with multiplicity. Let \(z\in\CC\) with \(p'(z)=0\). If \(p(z)=0\), then \(z\) is one of the \(a_j\) and lies in the convex hull. Otherwise \(z\neq a_j\) for every \(j\), and \[ 0=\frac{p'(z)}{p(z)}=\sum_{j=1}^n\frac{1}{z-a_j}=\sum_{j=1}^n\frac{\overline{z}-\overline{a_j}}{\abs{z-a_j}^2}. \] Rearranging gives \[ \Bigl(\sum_{j=1}^n\frac{1}{\abs{z-a_j}^2}\Bigr)\overline{z}=\sum_{j=1}^n\frac{1}{\abs{z-a_j}^2}\overline{a_j}. \] Taking complex conjugates and dividing by \(\sum_{j}\abs{z-a_j}^{-2}>0\) writes \(z=\sum_j t_ja_j\) with \(t_j=\abs{z-a_j}^{-2}/\sum_k\abs{z-a_k}^{-2}\). The weights \(t_j\) are positive and sum to \(1\), so \(z\) lies in the convex hull of \(\{a_1,\ldots,a_n\}\).