Analyticity domains of critical points of polynomials. A proof of Sendov's conjecture

Abstract

Let ${\P }_n^c(\bar{\mu },\bar{\nu })$ be the set of all complex polynomials $p(z)={\prod }_{i=1}^m(z-z_i{)}^{{\mu }_i}$, ${\sum }_{i=1}^m{\mu }_i=n$, with derivatives of the form $$p^{\prime }(z)=n\prod _{i=1}^m(z-z_i{)}^{{\mu }_i-1}\prod _{j=1}^k(z-{\xi }_j{)}^{{\nu }_j},\sum _{j=1}^k{\nu }_j=m-1.$$ In this note we prove the following:\par\medskip {\it \noindent For a fixed ordering $\alpha =(1,2,\dots ,m)$, the distinct zeros $\{z_i{\}}_{i=1}^m$ and the distinct critical points of the second kind $\{{\xi }_j{\}}_{j=1}^k$ of polynomials from ${\P }_n^c(\bar{\mu },\bar{\nu })$ are analytic functions $\{z_i^{\alpha \beta }{\}}_{i=1}^m$ and $\{{\xi }_j^{\alpha \beta }{\}}_{j=1}^k$, resp., $\beta =(i_1,i_2,\dots ,i_{k+1})$, of any of the variables $(z_{i_1},z_{i_2},\dots ,z_{i_{k+1}})$ in the domain $$ \{(z_{i_1},z_{i_2},\ldots,z_{i_{k+1}})\in \C^{k+1}~\vert~p\in ¶_{n}^c(\barμ,\barν) \}, $$ being also continuous on its boundary.}\par\medskip\noindent This statement gives an immediate proof to the well-known conjecture of Bl. Sendov \cite{sen}: \par\medskip {\it \noindent If $n\ge 2$ and $p(z)={\prod }_{i=1}^n(z-z_i)$ is a polynomial of degree $n$ such that $z_i\in \C$, $|z_i|\le 1$, $i=1,2,\dots ,n$, then for every $i=1,2,\dots ,n$, the disk $\{z\in \C\,|\,\vert z_i-z\vert \le 1\}$ contains at least one zero of $p^{\prime }(z)$.}