Given quadratic equation

$x^2-x+1=0$

Let its roots be $\alpha$ and $\beta$.

Find the roots

Using the quadratic formula:

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

So,

$\alpha=\frac{1+i\sqrt{3}}{2},\qquad \beta=\frac{1-i\sqrt{3}}{2}$

Write roots in polar form

Observe that

$\frac{1+i\sqrt{3}}{2}=\cos\frac{\pi}{3}+i\sin\frac{\pi}{3}$ $\frac{1-i\sqrt{3}}{2}=\cos\frac{\pi}{3}-i\sin\frac{\pi}{3}=\cos\frac{\pi}{3}+i\sin\left(-\frac{\pi}{3}\right)$

Hence,

$\alpha=e^{i\pi/3},\qquad \beta=e^{-i\pi/3}$

Raise to the power $2009$

$\alpha^{2009}+\beta^{2009}=e^{i\cdot 2009\pi/3}+e^{-i\cdot 2009\pi/3}$

Using

$e^{i\theta}+e^{-i\theta}=2\cos\theta,$

we get

$\alpha^{2009}+\beta^{2009}=2\cos\left(\frac{2009\pi}{3}\right)$

Reduce the angle modulo $2\pi$

Since $2009\equiv 5\pmod{6}$,

$\frac{2009\pi}{3}\equiv \frac{5\pi}{3} \pmod{2\pi}$

Therefore,

$2\cos\left(\frac{2009\pi}{3}\right)=2\cos\left(\frac{5\pi}{3}\right)$

Now,

$\cos\left(\frac{5\pi}{3}\right)=\frac{1}{2}$

So,

$\alpha^{2009}+\beta^{2009}=2\cdot \frac{1}{2}=1$

Check options

The correct option is:

$\boxed{\text{B: }1}$