Question

lf $\alpha$ 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

The correct option is B $1$

$x^2-x+1=0$
$\Rightarrow x=\frac{1\pm \sqrt{1-4}}{2}$
$\Rightarrow x=\frac{1\pm \sqrt{3}i}{2}$
$\Rightarrow \alpha =\frac{1}{2}+i\frac{\sqrt{3}}{2},\beta =\frac{1}{2}-\frac{i\sqrt{3}}{2}$
$\Rightarrow \alpha =\cos \frac{\pi }{3}+i\sin \frac{\pi }{3},\beta =\cos \frac{\pi }{3}-i\sin \frac{\pi }{3}$
${\alpha }^{2009}+{\beta }^{2009}=2\cos 2009\frac{\pi }{3}$
$=2\cos [668\pi +\pi +\frac{2\pi }{3}]=2\cos (\pi +\frac{2\pi }{3})$
$=-2\cos \frac{2\pi }{3}=-2(-\frac{1}{2})=1$
Hence, option 'B' is correct.