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

Since, a and 13 are roots of the equation $x^2-x+1=0$

Then $\alpha +\beta =1,\alpha \beta =1$

$\begin{array}{l}\Rightarrow \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}x=\frac{1\pm \sqrt{3}i}{2}\\ \Rightarrow \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}x=\frac{1+\sqrt{3}i}{2}\text{ or }\frac{1-\sqrt{3}i}{2}\\ \Rightarrow \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}x=-\omega \text{ or }-{\omega }^2\end{array}$

Thus,  $\alpha =-{\omega }^2$

Then, $\beta =-\omega$

or $\left.\alpha =-\omega ,\text{ then }\beta =-{\omega }^2\text{ [where, }{\omega }^3=1\right]$

$\begin{array}{l}\therefore {\alpha }^{2009}+{\beta }^{2009}=(-\omega {)}^{2000}+{\left(-{\omega }^2\right)}^{2009}\\ =-\left[{\left({\omega }^3\right)}^{669}\cdot {\omega }^2+{\left({\omega }^3\right)}^{1337}\cdot \omega \right]\\ =-\left[{\omega }^2+\omega \right]=-(-1)=1\end{array}$