Question

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

• A. -1
• B. 1
• C. 2
• D. -2

Answer 1

Answer: (B)

$x^{2} - x + 1 = 0 \quad\Rightarrow x = \frac{1\pm\sqrt{1-4}}{2}$
$x = \frac{1\pm \sqrt{3} i}{2}$
$\alpha = \frac{1}{2} + i \frac{\sqrt{3}}{2},\quad \beta = \frac{1}{2} - \frac{i\sqrt{3}}{2}$
$\alpha =cos\frac{\pi }{3} + i\,sin \frac{\pi }{3},\quad\beta = cos\frac{\pi }{3} - i\,sin \frac{\pi }{3}$
$\alpha^{2009} + \beta^{2009} = 2cos\, 2009 \left(\frac{\pi }{3}\right)$
$= 2cos\left[668\pi+\pi +\frac{2\pi }{3}\right] = 2cos \left(\pi +\frac{2\pi }{3}\right)$
$= - 2cos \frac{2\pi }{3} = -2\left(-\frac{1}{2}\right) = 1$