If $\mathrm{\alpha }\text{ and }\mathrm{\beta }$ are the roots of the equation $x^2-x+1=0,$ then ${\mathrm{\alpha }}^{2009}+{\mathrm{\beta }}^{2009}=$

$\begin{array}{l}{\mathrm{x}}^2-\mathrm{x}+1=0\\ \mathrm{x}=\frac{1\pm \sqrt{3}\mathrm{i}}{2}\\ \mathrm{\alpha }=\frac{1}{2}+\mathrm{i}\frac{\sqrt{3}}{2},\mathrm{\beta x}=\frac{1\pm \sqrt{1-4}}{2}\\ \mathrm{\alpha }=\cos \frac{\mathrm{\pi }}{3}+\mathrm{isin}\frac{\mathrm{\pi }}{3},\mathrm{\beta }=\frac{1}{2}-\frac{\mathrm{i}\sqrt{3}}{2}\\ {\mathrm{\alpha }}^{2009}+{\mathrm{\beta }}^{2009}=2\cos 2009\left(\frac{\mathrm{\pi }}{3}\right)\\ =2\cos \left[668\mathrm{\pi }+\mathrm{\pi }+\frac{2\mathrm{\pi }}{3}\right]=2\cos \left(\mathrm{\pi }+\frac{2\mathrm{\pi }}{3}\right)\\ =-2\cos \frac{2\mathrm{\pi }}{3}=-2\left(-\frac{1}{2}\right)=1\end{array}$