Polinom cu coeficienti complecsi

[Adriana Nistor]

Fie \( f=X^n+a_{n-1}X^{n-1}+...+a_{1}X+a_{0} \) un polinom cu coeficienti complecsi. Sa se arate ca exista \( z\in\mathbb{C} \) astfel incat \( |z|=1 \) si \( |f(z)|\ge1 \).

[Marius Mainea]

Daca \( 1,\epsilon,\epsilon^2,\epsilon^3,....,\epsilon^n \) sunt radacinile de ordinul n+1 ale unitatii, atunci, presupunand contrariul

\( n+1=|f(1)+\epsilon f(\epsilon)+\epsilon^2f(\epsilon^2)+.....+\epsilon^nf(\epsilon^n)|\le|f(1)|+|\epsilon f(\epsilon)|+...+|\epsilon^nf(\epsilon^n)|< n+1 \), contradictie.

[Theodor Munteanu]

Observatie:Fie polinomul \( f = a_0 + a_1 x + ... + a_n x^n \) cu \( |f(z)| \le 1,\forall z \in C,|z| = 1. \).Demonstreaza ca \( |a_0 |^2 + |a_1 |^2 + .... + |a_n |^2 \le 1 \).
Demonstratie(schita):\( |f(z)| \le 1 \Leftrightarrow f(z)\overline {f(z)} \le 1 \Leftrightarrow \left( {\sum\limits_{k = }^n {a_k z^k } } \right)\left( {\sum\limits_{j = 0}^n {\overline {a_j } } \frac{1}{{z^j }}} \right) \le 1 \Rightarrow \sum\limits_{k,j = 0}^n {a_k \overline {a_j } } z^{k - j} \le 1; \)
Cunoastem relatia (*) deci:\( \left. \begin{array}{l}
\sum\limits_{z \in U_{n + 1} } {\left( {\sum\limits_{k,j = 0}^n {a_k \overline {a_j } } z^{k - j} } \right)} \le n + 1 \\
\sum\limits_{z \in U_n } {z^k } = \left\{ {\begin{array}{*{20}c}
{0,k(!|)n} \\
{n,k|n} \\
\end{array}\left( * \right)} \right. \\
\end{array} \right| \Rightarrow \sum {a_k \overline {a_k } \le 1} \Rightarrow \sum {|a_k |^2 \le 1} \)
Observam acum ca pentru \( a_n = 1 \Rightarrow f = X^k \)

[Theodor Munteanu]

La relatia * se stie ca pe prima ramura e 0 dar nu stiu sa editez semnele alea.