A Proof of Sendov's Conjecture

TL;DR

This paper provides a proof of Sendov's conjecture, which states that for polynomials with zeros in the unit disk, each zero has a critical point within unit distance, confirming the conjecture's validity.

Contribution

The paper presents a complete proof of Sendov's conjecture, resolving a long-standing open problem in polynomial zero and critical point distribution.

Findings

Proof of Sendov's conjecture for all polynomials with zeros in the unit disk

Confirmation that each zero has a critical point within one unit distance

Advancement in understanding polynomial zero and critical point relationships

Abstract

The Sendovs conjecture asserts that if all the zeros of a polynomial p(z) lie in the closed unit disk, then there must be a critical point of p(z) within unit distance of each zero. The conjecture has been proved to be true for many special cases.Here we give a proof of the conjecture.

Equations

$$z_1=z_1-\frac{{\sum }_{i=1}^n{z}_i}{n}=z_1-\frac{{\sum }_{i=1}^{n-1}{\zeta }_i}{n-1}=\frac{{\sum }_{i=1}^{n-1}(z_1-{\zeta }_i)}{n-1}$$

Full text

Department of Mathematics,

Central University of Kashmir,

Ganderbal-191201(India).

Email: $gmsofi@cukashmir.ac.in$

Keywords: Zeros , crictical points.

1 Introduction

The Sendov’s conjecture is an unproven conjecture in complex analysis. This conjecture was included in the collection $Research~{}Problems~{}in~{}Function~{}Theory$ in 1967 by Professor Hayman. The conjecture is due to the Bulgarian Mathematician B. Sendov. Till present the conjecture has been found to be true for many special cases(see[1]). What follows is a proof of the conjecture .

Theorem 1. Let $p(z)=z^{n}+a_{n-1}z^{n-1}+a_{n-2}z^{n-2}+\cdots+a_{1}z+a_{0}$ be a non-constant polynomial with all its zeros $z_{1},z_{2},\cdots,z_{n}$ lying inside the closed unit disk $|z|\leq 1$. Then each of the disks $|z-z_{i}|\leq 1,~{}i=1,2,\cdots,n$ must contain a zero of $p^{\prime}$.

Proof. [we infact prove that if $p$ has all its zeros $z_{1},z_{2},\cdots,z_{n}$ lying inside any closed disk $|z-a|\leq r$. Then each of the disks $|z-z_{i}|\leq r,~{}i=1,2,\cdots,n$ must contain a zero of $p^{\prime}$ and then taking $r=1$ and $a=0$ we get the required conclusion.]

 

Case1: $\sum_{i=1}^{n}z_{i}=0$

     Let $\zeta_{i},~{}i=1,2,\cdots,n-1$ be the critical points of $p$. Assume to the contrary that the conclusion of the theorem does not hold. Then there exists a zero of $p$ say $z_{1}$ such that |z_{1}-\zeta_{i}|>r~{}~{}\mbox{for all 1\leq i\leq n-1}. Therefore by Gauss-Lucas theorem all $\zeta_{i}s$ must lie in the crescent-shaped region outside $|z-z_{1}|\leq r$ but inside $|z-a|\leq r$(as indicated in Figure 1). We can draw a line through $z_{1}$ parallel to the line through the points of intersection of the circles $|z-a|\leq r$ and $|z-z_{1}|\leq r$ (see Figure 1) and it is clear that all the vectors $z_{1}-\zeta_{i},~{}i=1,2,\cdots,n-1$ lie in the same half plane determined by this line.Hence $z_{1}$ cannot be the centroid of these $z_{1}-\zeta_{i}s$.

But

~

[TABLE]

a contradiction.

 

Case2: $\sum_{i=1}^{n}z_{i}\neq 0$

Now by case 1 the Sendov’s conjecture holds for the polynomial $p(z-\frac{a_{n-1}}{n})$. Since a translation $z\rightarrow z+\frac{a_{n-1}}{n}$ does not alter the relative distances between zeros and critical points of $p(z-\frac{a_{n-1}}{n})$ Sendov’s conjecture must be true for $p(z)$ also. ∎