If   $\mathrm{\alpha }$   and   $\mathrm{\beta }$   are the roots of equation $x^2-x+1=0,\text{ then }{\alpha }^{2000}+{\beta }^{2009}$  is equal to 

 

  Given equation is  $x^2-x+1=0$ 

  $\Rightarrow x=\frac{1\pm \sqrt{1-4}}{2}$    $\Rightarrow x=\frac{1\pm \sqrt{-3}}{4}$

  $x=\frac{1\pm \sqrt{3}i}{2}$  

We know that,   $\omega =\frac{-1+\sqrt{3}i}{2}$  and  ${\omega }^2=\frac{-1-\sqrt{3}i}{2}$

$\therefore \alpha =\frac{1+\sqrt{3i}}{2}=-{\omega }^2$   and    $\beta =\frac{1-\sqrt{3}i}{2}=-\omega$

 Now,  ${\alpha }^{2009}+{\beta }^{2009}={\left(-{\omega }^2\right)}^{2009}+(-\mathrm{\omega }{)}^{2009}$

   $\begin{array}{r}={\left(-{\omega }^2\right)}^{3\times 669+2}-(\omega {)}^{3\times 669+2}\\ =-{\left({\omega }^3\right)}^{2\times 669}\cdot {\omega }^4-{\left({\omega }^3\right)}^{699}\cdot {\omega }^2\\ =-{\omega }^4-{\omega }^2\\ =-{\omega }^3\cdot \omega -{\omega }^2=-\omega -{\omega }^2=1\\ \left[\because {\omega }^3=1\right]\end{array}$

    $\begin{array}{r}=-{\omega }^3\cdot \omega -{\omega }^2=-\omega -{\omega }^2=1\\ \left[\because 1+\omega +{\omega }^2=0\right]\end{array}$