多项式临界点的解析域。Sendov猜想的一个证明
Analyticity domains of critical points of polynomials. A proof of Sendov's conjecture
摘要：
设$¶_{n}^c(\barμ,\barν)$是所有复多项式$p(z)=\prod_{i=1}^{m}(z-z_i)^{μ_i}$，$\sum_{i=1}^mμ_i=n$的集合，其导数的形式为$$ p'(z)=n\prod_{i=1}^{m}(z-z_i)^{μ_i-1}\prod_{j=1}^{k}(z-ξ_j)^{ν_j}, ~\sum_{j=1}^kν_j=m-1. $$在本文中，我们证明了如下：\par \medskip { \it \noindent对于固定的序$α=(1,2,\ldots,m)$，来自$¶_{n}^c(\barμ,\barν)$的多项式的第二类$\{ξ_j\}_{j=1}^k$的不同的零点$\{z_i\}_{i=1}^m$和不同的临界点分别是解析函数$\{z_i^{αβ}\}_{i=1}^m$和$\{ξ_j^{αβ}\}_{j=1}^k$，$β=(i_1,i_2,\ldots,i_{k+1})$，域$$ \{(z_{i_1},z_{i_2},\ldots,z_{i_{k+1}})\in \C^{k+1}~\vert~p\in ¶_{n}^c(\barμ,\barν) \}, $$中的任何变量$(z_{i_1},z_{i_2},\ldots,z_{i_{k+1}})$也在其边界上连续。} \par \medskip \noindent这句话给出了一个立即证明著名的猜想B1。Sendov \cite{sen}：\par \medskip { \it \noindent如果$n\ge 2$和$p(z)=\prod_{i=1}^n (z-z_i)$是次数为$n$的多项式，使得$z_i\in \C$，$\vert z_i\vert\le 1$，$i=1,2,\ldots,n$，则对于每个$i=1,2,\ldots,n$，圆盘$\{z\in \C\,|\,\vert z_i-z\vert \le 1\}$至少包含一个$p'(z)$的零。}
英文摘要：
Let $¶_{n}^c(\barμ,\barν)$ be the set of all complex polynomials $p(z)=\prod_{i=1}^{m}(z-z_i)^{μ_i}$, $\sum_{i=1}^mμ_i=n$, with derivatives of the form $$ p'(z)=n\prod_{i=1}^{m}(z-z_i)^{μ_i-1}\prod_{j=1}^{k}(z-ξ_j)^{ν_j}, ~\sum_{j=1}^kν_j=m-1. $$ In this note we prove the following:\par\medskip {\it \noindent For a fixed ordering $α=(1,2,\ldots,m)$, the distinct zeros $\{z_i\}_{i=1}^m$ and the distinct critical points of the second kind $\{ξ_j\}_{j=1}^k$ of polynomials from $¶_{n}^c(\barμ,\barν)$ are analytic functions $\{z_i^{αβ}\}_{i=1}^m$ and $\{ξ_j^{αβ}\}_{j=1}^k$, resp., $β=(i_1,i_2,\ldots,i_{k+1})$, of any of the variables $(z_{i_1},z_{i_2},\ldots,z_{i_{k+1}})$ in the domain $$ \{(z_{i_1},z_{i_2},\ldots,z_{i_{k+1}})\in \C^{k+1}~\vert~p\in ¶_{n}^c(\barμ,\barν) \}, $$ 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 $z_i\in \C$, $\vert z_i\vert\le 1$, $i=1,2,\ldots,n$, 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 $p'(z)$.}
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