Polinom cu radacinile de modul 1
Posted: Fri Nov 09, 2007 6:43 pm
Fie \( z \in \mathbb{C} \) astfel incat \( 11z^{10}+10iz^9+10iz-11=0 \). Atunci aratati ca \( |z|=1. \)
\( z^9(10i+11z)=(11-10iz) \Rightarrow {|z|}^9|10i+11z|=|11-10iz| \)
Presupunem ca \( |10i+11z|>|11-10iz| \), celalalt caz tratandu-se analog.
\( |10i+11z|>|11-10iz| \Rightarrow (10i+11z)(11 \bar{z}-10i)>(11-10iz)(11+10i \bar{z}) \Rightarrow 21{|z|}^2>21 \Rightarrow |z|>1 \Rightarrow
{|z|}^9|10i+11z|>|11-10iz| \) imposibil.