$If\alpha and\beta aretherootsoftheequation{x}^2-x+1=0,then{\alpha }^{2009}+{\beta }^{2009}=$

$\begin{array}{l}GivenEquation{x}^2-x+1=0\\ x=\frac{1\pm \sqrt{1-4}}{2}=\frac{1\pm i\sqrt{3}}{2}\\ \therefore \alpha =\frac{1+i\sqrt{3}}{2},\beta =\frac{1-i\sqrt{3}}{2}\\ =-\omega ,=-{\omega }^2\\ \therefore {\alpha }^{2009}+{\beta }^{2009}={(-\omega )}^{2009}+{(-{\omega }^2)}^{2009}\\ =-[{\left({\omega }^3\right)}^{669}.{\omega }^2+{\left({\omega }^3\right)}^{1337}.\omega ]\\ =-[{\omega }^2+\omega ][\because {\omega }^3=1]\\ =-(-1)=1\end{array}$