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