Q76 / 180 · KCET 2012 · Mathematics
If $\alpha$ is a complex number such that $\alpha^{2}-\alpha+1=0$, then $\alpha^{2011}$ is equal to
- A$-\alpha$
- B$\alpha^{2}$
- Correct answerC$\alpha$
- D1
Solution
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 $\alpha^{2011}=\alpha$
Hence, the correct answer is option: 'C'.