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

The correct option is B 1

Given $\alpha$ and $\beta$ be the root of the equation

$x^2-x+1=0$4

$\Rightarrow x=\frac{1\pm \sqrt{1-4}}{4}=\frac{1\pm i\sqrt{3}}{2}$

$\Rightarrow -{\omega }^2=\frac{1+i\sqrt{3}}{4}=\frac{1-i\sqrt{3}}{2}=-\omega$

We know that $\omega ,{\omega }^2=$ cube root of unity

$\Rightarrow {\omega }^{3n}=1\Rightarrow 1+\omega +{\omega }^2=0\Rightarrow (-{\omega }^2{)}^{2009}+(-\omega {)}^{2009}\Rightarrow (-\omega {)}^{4018}+(-\omega {)}^{2009}$

We know that 4017 is multiple of 3 & 2007 is multiple of 3

$\Rightarrow -\omega .{\omega }^{4017}+(-\omega {)}^2{\omega }^{2007}\Rightarrow -(\omega +{\omega }^2)\Rightarrow -1(-1)\Rightarrow 1$

$\therefore {\alpha }^{2009}+{\beta }^{2009}=1$

$\therefore$ option B is correct.