"Local" Gauss-Lucas theorem?

Question

The Gauss-Lucas theorem relates the location of zeros of a polynomial to the location of zeros of its derivative:

Suppose $f(z)\in \mathbb{C}[z]$ is a non-constant polynomial with roots $\alpha_1,\ldots,\alpha_n$, and let $$K = K(\alpha_1,\ldots,\alpha_n) \subset \mathbb{C}$$ denote the convex hull of these points. Then all roots of $f'(z)$ lie inside $K$.

If we instead consider a polynomial with real coefficients, Rolle's theorem gives a stronger condition on the location of zeros of the derivative:

Suppose $f(x)\in\mathbb{R}[x]$ is a non-constant polynomial with real roots $\alpha_1\leq \cdots \leq \alpha_n$, counted with multiplicity. Then for any $i < j$, the closed interval $$ I = [\alpha_i, \alpha_j] \subset \mathbb{R} $$ contains some root of $f'(x)$.

The second statement in "local" in the sense that knowing only two roots of $f$ gives us some information about where the roots of $f'$ lie. In the first statement, knowing a subset of the roots will not in general determine the convex hull.

Question: For $f(z)\in\mathbb{C}[z]$, is there any information we get about the location of roots of $f'(z)$ if we know only the locations of (say) 3 non-collinear* roots of $f(z)$?

Some guesses which are false: Given three roots $\alpha_1, \alpha_2, \alpha_3$ of $f(z)$,

1. $f'(z)$ must contain a root inside the triangle spanned by $\alpha_1, \alpha_2, \alpha_3$.

2. $f'(z)$ must contain a root inside the circle passing through $\alpha_1, \alpha_2, \alpha_3$.

*as suggested by Gerry Myerson, the case when $\alpha_i$ are collinear may require separate analysis. But even this case is unresolved up to my understanding.

Answer 1

Following is information from Marden's Geometry of the Zeros, Sections 25 and 26. Note: Some further information can be found on my paper "Approximate Gauss--Lucas Theorems" (also linked in the comments).

Theorem: (Kakeya) For $1\leq p\leq n$, there is a function $\varphi(n,p)$ such that if a degree $n$ polynomial has $p$ zeros lying in a disk with radius $R$, then that polynomial has at least $p-1$ critical points lying in the concentric disk with radius $R\cdot\varphi(n,p)$.

Thus for your example, let $R$ denote one half the diameter of the set of three zeros, and let $C$ denote the disk containing the three zeros, with radius $R$. Then the concentric disk with radius $R\cdot\varphi(n,p)$ contains at least two critical points of the polynomial. Note that in general, for the known bounds, the degree $n$ of the polynomial must be known.

Various estimates on this function $\varphi$ have been given:

Kakeya: $\varphi(n,2)=\csc(\pi/n)$.

Biernacki: $\varphi(n,n-1)\leq(1+1/n)^{1/2}$.

Marden: $\varphi(n,p)\leq\csc(\pi/(n-p+1))$ and $\varphi(n,p)\leq\displaystyle\prod_{k=1}^{n-p}[(n+k)/(n-k)]$.

Richards: $\varphi(n,p)\leq1+\dfrac{8(n-p)^2}{p-8(n-p)^2}$ (subject to the assumption that $p>8(n-p)^2$). (This result is in the linked paper, definitely not in Marden's awesome book!)

Complete citations may be found in my linked paper. Sorry I did not think of this earlier as the answer, I was working on trying to find the smallest radius which would catch at least one critical point, not all $n-p$. And in any case I have no excuse for my (now deleted) earlier answer!