Let n ≥ 2 be an integer and let f(z)=(z-r_1)(z-r_2)… (z-r_n), qquad r_1,… ,r_n∈ C, be a monic polynomial all of whose roots lie in the closed unit disk, i.e. |r_k|≤ 1 for every k ∈ 1,… ,n. Since deg f' = n-1 ≥ 1, the derivative f' has at least one zero in C. Then for every j ∈ 1,… ,n there exists z ∈ C with f'(z)=0 qquad and qquad |z-r_j| ≤ 1; that is, for each root r_j of f the closed disk z ∈ C : |z-r_j| ≤ 1 contains a critical point of f.

Sendov's conjecture

Celebrated

Let $n \ge 2$ be an integer and let

f(z)=(z-r_{1})(z-r_{2})\cdots (z-r_{n}), \qquad r_{1},\dots ,r_{n}\in \mathbb{C},

be a monic polynomial all of whose roots lie in the closed unit disk, i.e. $|r_{k}|\le 1$ for every $k \in \{1,\dots ,n\}$ . Since $\deg f' = n-1 \ge 1$ , the derivative $f'$ has at least one zero in $\mathbb{C}$ .

Then for every $j \in \{1,\dots ,n\}$ there exists $z \in \mathbb{C}$ with

f'(z)=0 \qquad \text{and} \qquad |z-r_{j}| \le 1;

that is, for each root $r_{j}$ of $f$ the closed disk $\{z \in \mathbb{C} : |z-r_{j}| \le 1\}$ contains a critical point of $f$ .

Equivalent formulations 5

Other statements of this same problem, merged from separate entries. Each is equivalent to the statement above — proving any one settles them all.

Sendov's conjecture
In mathematics, Sendov's conjecture, sometimes also called Ilieff's conjecture, concerns the relationship between the locations of roots and critical points of a polynomial function of a complex variable. It is named after Blagovest Sendov.
source: Wikipedia
Sendov's conjecture on critical points of polynomials
Let
$p(z)=\prod_{i=1}^n(z-\lambda_i),\qquad p'(z)=n\prod_{j=1}^{n-1}(z-\mu_j),$
where $|\lambda_i|\leq1$ for all $i$ . Sendov's conjecture. For every root $\lambda_i$ of $p$ , there is a root $\mu_j$ of $p'$ such that
$|\lambda_i-\mu_j|\leq1.$
The source says this conjecture has been open since 1959 and also calls it Ilyeff's conjecture; it gives no resolution.
source: K. M. R. Audenaert and F. Kittaneh, “Problems and Conjectures in Matrix and Operator Inequalities”, arXiv:1201.5232 (2012).
Circulant-matrix formulation of Sendov's conjecture
Let $C\in M_n$ be a circulant matrix with $\lVert C\rVert\leq1$ , let $\lambda$ be an eigenvalue of $C$ , and let $C'$ be the principal submatrix obtained by deleting the first row and first column. Circulant formulation of Sendov's conjecture. The matrix $C'-\lambda I$ has an eigenvalue in the closed unit disk. This is presented as a third formulation of Sendov's conjecture, whose original polynomial statement is described as open in the source.
source: K. M. R. Audenaert and F. Kittaneh, “Problems and Conjectures in Matrix and Operator Inequalities”, arXiv:1201.5232 (2012).
D-companion-matrix formulation of Sendov's conjecture
Let $\lambda_1,\ldots,\lambda_n$ belong to the closed unit disk, let $D=\operatorname{diag}(\lambda_1-\lambda_n,\ldots,\lambda_{n-1}-\lambda_n)$ , and let $I$ and $J$ be respectively the identity matrix and the all-ones matrix of order $n-1$ . D-companion formulation of Sendov's conjecture. The matrix
$D\left(I-\frac{J}{n}\right)$
has an eigenvalue in the closed unit disk. This is presented as a reformulation of Sendov's conjecture; the source gives no resolution beyond stating Sendov's problem as open.
source: K. M. R. Audenaert and F. Kittaneh, “Problems and Conjectures in Matrix and Operator Inequalities”, arXiv:1201.5232 (2012).
Sendov's conjecture in the language of expansivity
Let $P(x)=a_nx^n+a_{n-1}x^{n-1}+\cdots+a_1x+a_0$ be a polynomial of degree $n\geq 2$ . Let $\{b_i\}_{i=1}^{n}$ be the zeros of $P(x)$ satisfying $|b_i|\leq 1$ , and let $\mathcal{S}=(a_nx^n,a_{n-1}x^{n-1},\ldots,a_1x,a_0)$ be a tuple representation of $P(x)$ . For each $b_i$ , let $\mathcal{S}_e$ denote the identity element used in the expansivity construction. Sendov's conjecture. For each $b_i$ , there exists some $\mathcal{S}_a\in\mathcal{Z}[(\gamma^{-1}\circ\beta\circ\gamma\circ\nabla)^1(\mathcal{S})]$ such that
$|\operatorname{Id}_{n+1}(b_i\mathcal{S}_e-\mathcal{S}_a)|\leq 1.$
This is the expansivity formulation of the classical assertion that every zero in the unit disk lies within unit distance of a zero of the derivative. The paper reports partial variants and known degree-specific and asymptotic results, but the conjecture itself is presented as the central problem.
source: Theophilus Agama, “Expansivity theory and Sendov's conjecture”, arXiv:1907.12825 (2026).

Sources & referencesView supporting material

Primary source

Wikipedia

Additional references

B. Sendov, in Research Problems in Function Theory (W. K. Hayman, ed.), Athlone Press (1967), problem 4.5.
T. Tao, "Sendov's conjecture for sufficiently high degree polynomials," Acta Mathematica 229 (2022), 347-392.
G. Schmeisser, "The conjectures of Sendov and Smale," in Approximation Theory, DARBA (2002), 353-369.
Wikipedia, Sendov's conjecture, the article this problem comes from.

analysis unsolved-problem

1 week ago

Progress summary

Refreshed 5 days ago

Solved

A machine-checked proof now appears to settle the conjecture in every degree, although the underlying argument has not yet been independently published.

Sendov’s conjecture asserts that every zero of a polynomial whose zeros lie in the unit disk has a zero of its derivative within distance one. The classical question was known only through partial results before the recent full-generality formalization.

Known results

Degree $n\le 8$ : proved by Brown and Xiang, with Xiang’s result dated 1999.
Arbitrary degree with at most eight distinct zeros: proved by Brown.
Sufficiently large degree: proved by Tao; the bound is effective but not the exact threshold $n=8$ .
Further special cases include all-real zeros and zeros on the unit circle.

August 2026 Lean-checked proof

On August 12, 2026, Tao reported that Lech Mazur’s AI-assisted argument resolves the conjecture for all $n\ge 2$ and has been verified in Lean. The accompanying formalization and proof document state the exact full-generality theorem, so this is best treated as resolved, though not yet as a publication-ready human proof.

Current status (as of August 2026): Sendov’s conjecture is settled for every degree by a Lean-checked formalization, while a publication-ready human proof remains forthcoming.

Sources

No solutions have been posted yet.

Equivalent formulations 5

Sendov's conjecture

Sendov's conjecture on critical points of polynomials

Circulant-matrix formulation of Sendov's conjecture

D-companion-matrix formulation of Sendov's conjecture

Sendov's conjecture in the language of expansivity

Primary source

Additional references

Progress summary