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

Question

Aicharya · Accepted Answer

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

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

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