  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 $p'(z)$.}
