Question
If $$\alpha \,\,{\text{and }}\beta $$ are the roots of the equation $${x^2} - x + 1 = 0,\,{\text{then }}{\alpha ^{2009}} + {\beta ^{2009}} = $$
A.
$$- 1$$
B.
$$1$$
C.
$$2$$
D.
$$- 2$$
Answer :
$$1$$
Solution :
$$\eqalign{
& {x^2} - x + 1 = 0 \cr
& \Rightarrow \,\,x = \frac{{1 \pm \sqrt {1 - 4} }}{2} \cr
& x = \frac{{1 \pm \sqrt 3 \,i}}{2} \cr
& \alpha = \frac{1}{2} + i\frac{{\sqrt 3 }}{2} = - {\omega ^2}\,\,\,\,\,\,\,\,\beta = \frac{1}{2} - \frac{{i\sqrt 3 }}{2} = - \omega \cr
& = {\alpha ^{2009}} + {\beta ^{2009}} \cr
& = {\left( { - {\omega ^2}} \right)^{2009}} + {\left( { - \omega } \right)^{2009}} \cr
& \, = - {\omega ^2} - \omega = 1 \cr} $$