Question

Fie multimea A = { z ∈ C | z^2 + z + 1 = 0 }
Stiind ca z ∈ A sa se calculeze

(z^11+z^10+z^9+z^8+z^7+z^6+z^3-4)/z^2013 + 2

Answer 1

$A = \Big\{ z \in\mathbb{C}\Big| \,z^2 + z + 1 = 0 \Big\}\\ \\ z^2+z+1 = 0\Big|\cdot (z-1),\quad z\neq 1 \\ \\ (z-1)(z^2+z+1) = 0 \Rightarrow z^3-1 = 0 \Rightarrow z^3 = 1,\quad z\neq 1\\ \\ \\\dfrac{z^{11}+z^{10}+z^{9}+z^{8}+z^{7}+z^{6}+z^3-4}{z^{2013}+2} = \\ \\ = \dfrac{z^9 (z^2+z+1)+z^6(z^2+z+1)+z^3-4}{(z^{3})^{671}+2} = \\ \\ = \dfrac{z^9\cdot 0+z^6\cdot 0+1-4}{1^{671}+2} = \dfrac{-3}{3} = \boxed{-1}$