Problema usoara cu numere complexe
Posted: Sat Oct 20, 2007 8:03 pm
Fie \( a\in\math{R} \). Daca \( z\in\math{C}-\math{R} \) satisface \( z^n+nz+a=0 \), unde \( n\in\math{N}* \), aratati ca \( |z|\ge1 \).
\( z^n +nz + a = 0, a\in \mathbb{R} \) si \( z \in \mathbb{C-R} \). Daca conjugam relatia obtinem \( \bar{z}^n +n\bar{z} + a = 0 \). Daca scadem cele doua relatii obtinemheman wrote:Fie \( a\in\math{R} \). Daca \( z\in\math{C}-\math{R} \) satisface \( z^n+nz+a=0 \), unde \( n\in\math{N}* \), aratati ca \( |z|\ge1 \).
\( z^n -\bar{z}^n = -n(z-\bar{z}) \). Cum \( z \neq \bar{z} \) avem
\( -n = z^{n-1}+z^{n-2}\bar{z}+...+z\cdot \bar{z}^{n-2} +\bar{z}^{n-1} \). Daca aplicam modulul la ultima relatie avem
\( n = \left| \sum_{k=1}^n z^{n-k} \cdot \bar{z}^{k-1} \right| \leq \sum_{k=1}^n |z|^{n-1} \). Deci rezulta ca \( |z|^{n-1} \geq 1 \Longleftrightarrow |z| \geq 1 \).