If \(\alpha\) is a complex number such that \(\alpha^{2}-\alpha+1=0\), then \(\alpha^{2011}\) is equal to

Given, \(\alpha^{2}-\alpha+1=0\)
\(\alpha=\frac{-(-1) \pm \sqrt{(-1)^{2}-4(1)(1)}}{2 \times 1}\)
\(=\frac{1 \pm \sqrt{1-4}}{2}\)
\(=\frac{1 \pm \sqrt{-3}}{2}\)
\(=\frac{1 \pm \sqrt{3} \mathrm{i}}{2}\)
\(=-\omega,-\omega^{2}\)
\(\therefore \alpha^{2011}=(-\omega)^{2011}=(-1)^{2011} \omega^{2011}\)
\(=(-1)\left(\omega^{3}\right)^{670} \omega\)
\(=(-1)(1)^{670} \omega=-\omega=\alpha \quad\left(\because \omega^{3}=1\right)\)
If we take \(\alpha=-\omega^{2}\)
\[
\begin{aligned}
\therefore \quad \alpha^{2011} &=\left(-\omega^{2}\right)^{2011}=-\left(\omega^{2011}\right)^{2} \\
&=-\left[\left(\omega^{3}\right)^{670} \omega\right]^{2} \\
&=-(\omega)^{2} \\
&=-\omega^{2}=\alpha
\end{aligned}
\]
Hence, in both cases \(\alpha^{2011}=\alpha\)