Analyticity domains of critical points of polynomials. A proof of Sendov's conjecture (2104.00348v5)

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 $$ p'(z)=n\prod_{i=1}^{m}(z-z_i)^{\mu_i-1}\prod_{j=1}^{k}(z-\xi_j)^{\nu_j}, ~\sum_{j=1}^k\nu_j=m-1. $$ In this note we prove the following:\par\medskip {\it \noindent For a fixed ordering $\alpha=(1,2,\ldots,m)$, 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., $\beta=(i_1,i_2,\ldots,i_{k+1})$, of any of the variables $(z_{i_1},z_{i_2},\ldots,z_{i_{k+1}})$ in the domain $$ {(z_{i_1},z_{i_2},\ldots,z_{i_{k+1}})\in \C^{k+1}~\vert~p\in \P_{n}^c(\bar{\mu},\bar{\nu}) }, $$ 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 $z_i\in \C$, $\vert z_i\vert\le 1$, $i=1,2,\ldots,n$, then for every $i=1,2,\ldots,n$, the disk ${z\in \C\,|\,\vert z_i-z\vert \le 1}$ contains at least one zero of $p'(z)$.}