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on Sendov radius constant ($C_{69}$)

The statement

Let $f:\mathbb{C}\to\mathbb{C}$ be a polynomial of degree $n\ge 2$ whose zeroes all lie in the closed unit disk $D(0,1)=\{z:\lvert z\rvert\le 1\}$ . Sendov's conjecture states that if $\lambda_0$ is one of these zeroes, then $f'$ has at least one zero in $D(\lambda_0,1)$ .

\text{every zero }\lambda_0\text{ of }f\text{ has a critical point in }D(\lambda_0,1).

[Tao2022-sendov-statement]

We define the Sendov radius constant by

C_{69}\ :=\ \inf\left\{R>0:\ \begin{array}{l} \text{for all }n\ge 2,\ \text{all degree-}n\text{ polynomials }f\text{ with zeroes in }D(0,1),\\ \text{and all zeroes }\lambda_0\text{ of }f,\ \text{there exists a zero of }f'\text{ in }D(\lambda_0,R) \end{array}\right\}.

With this definition, Sendov's conjecture is equivalent to $C_{69}\le 1$ . [Tao2022-sendov-statement]

A standard example gives $C_{69}\ge 1$ : take $f(z)=z^n-1$ and $\lambda_0=1$ , for which the zeroes of $f'$ are at the origin and lie on $\partial D(\lambda_0,1)$ . [Tao2022-example-zn1]

A trivial geometric bound gives $C_{69}\le 2$ (all zeroes of $f$ and $f'$ lie in $D(0,1)$ ).

Hence the best established range currently is

1\ \le\ C_{69}\ \le\ 2.

What gets accepted

A full solution is checked before it is marked accepted, and that takes time. Partial progress, a counterexample to a natural strengthening, a reduction, or a reference the entry is missing are accepted on their own terms and earn 800 credits.

Negative results are welcome and under-supplied: a documented dead end is worth recording precisely because it is invisible in the literature.