If $\dpi{100} \alpha\: and \: \beta$ are the roots of the equation$\dpi{100} x^{2}-x+1= 0,$ then $\dpi{100} \alpha ^{2009}+\beta ^{2009}=$ Option 1) -2 Option 2) -1 Option 3) 1 Option 4) 2

As we have learned

$\alpha = \frac{-b+\sqrt{b^{2}-4ac}}{2a}$

$\beta = \frac{-b-\sqrt{b^{2}-4ac}}{2a}$

- wherein

$ax^{2}+bx+c= 0$

is the equation

$a,b,c\in R,\: \: a\neq 0$

Cube roots of unity -

$z=\left ( 1 \right )^{\frac{1}{3}}\Rightarrow z=\cos \frac{2k\pi }{3}+i\sin \frac{2k\pi }{3}$

k=0,1,2 so z gives three roots

$\Rightarrow 1,\frac{-1}{2}+i\frac{\sqrt{3}}{2}\left ( \omega \right ),\frac{-1}{2}-i\frac{\sqrt{3}}{2}\left ( \omega^{2} \right )$

- wherein

$\omega=\frac{-1}{2} +\frac{i\sqrt{3}}{2},\omega^{2}=\frac{-1}{2} -\frac{i\sqrt{3}}{2},\omega^{3}=1, 1+\omega+\omega^{2}=0$

$1,\omega,\omega^{2}$ are cube roots of unity.

$\alpha ,\beta = \frac{1\pm \sqrt {-3}}{2}= \frac{1 \pm \sqrt{3i}}{2}= -\left ( \frac{-1\pm\sqrt{3i}}{2} \right )= -\omega ,-\omega ^2$

$\alpha ^{2009}+\beta ^{2009}$

$-\omega ^{2009}+(-\omega^2) ^{2009}$

$= -\omega ^2-\omega = -(\omega ^2+\omega )= -(-1)= 1$

Option 1)

-2

Option 2)

-1

Option 3)

1

Option 4)

2

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