Sendov radius constant

Description of constant

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

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

Known upper bounds

Bound Reference Comments
$2$   Trivial geometric bound since all zeroes of $f$ and $f’$ lie in the closed unit disk.

Known lower bounds

Bound Reference Comments
$1$ [Tao2022] Example $f(z)=z^n-1$, $\lambda_0=1$: the critical points are at $0$ and lie on $\partial D(\lambda_0,1)$. [Tao2022-example-zn1]

References

Contribution notes

Prepared with assistance from ChatGPT 5.2 Pro.