Question

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

• A. $- \alpha$
• B. $- \alpha^2$
• C. $\alpha$
• D. $1$

Answer 1

Answer: (C)

Given, $\alpha^{2}-\alpha+1 = 0$
$\alpha = \frac{-\left(-1\right)\pm\sqrt{\left(-1\right)^{2}-4\left(1\right)\left(1\right)}}{2\times1}$
$= \frac{1\pm\sqrt{1-4}}{2}$
$= \frac{1\pm\sqrt{-3}}{2}$
$= \frac{1\pm\sqrt{3}i}{2}$
$= -\omega, -\omega^{2}$
$\therefore \alpha^{2011} = \left(-\omega\right)^{2011} = \left(-1\right)^{2011} \omega^{2011}$
$= \left(-1\right)\left(\omega^{3}\right)^{670}\omega$
$= \left(-1\right)\left(1\right)^{670} \omega = -\omega = \alpha\quad\left(\because \omega^{3} = 1\right)$
If we take $\alpha = -\omega^{2}$
$\therefore \quad\alpha^{2011} = \left(-\omega^{2}\right)^{2011} = -\left(\omega^{2011}\right)^{2}$
$= -\left[\left(\omega^{3}\right)^{670} \omega\right]^{2}$
$= -\left(\omega\right)^{2}$
$= - \omega^{2} = \alpha$
Hence, in both cases $\alpha^{2011} = \alpha$