Given that $z$ is a complex number such that $z+\frac 1z=2\cos 3^\circ$, find the least integer that is greater than $z^{2000}+\frac 1{z^{2000}}$.
Using the quadratic equation on $z^2 - (2 \cos 3 )z + 1 = 0$, we have $z = \frac{2\cos 3 \pm \sqrt{4\cos^2 3 - 4}}{2} = \cos 3 \pm i\sin 3 = \text{cis}\,3^{\circ}$.
There are other ways we can come to this conclusion. Note that if $z$ is on the [unit circle](https://artofproblemsolving.com/wiki/index.php/Unit_circle) in the complex plane, then $z = e^{i\theta} = \cos \theta + i\sin \theta$  and $\frac 1z= e^{-i\theta} = \cos \theta - i\sin \theta$. We have $z+\frac 1z = 2\cos \theta = 2\cos 3^\circ$ and $\theta = 3^\circ$. Alternatively, we could let $z = a + bi$ and solve to get $z=\cos 3^\circ + i\sin 3^\circ$.

Using [De Moivre's Theorem](https://artofproblemsolving.com/wiki/index.php/De_Moivre%27s_Theorem) we have $z^{2000} = \cos 6000^\circ + i\sin 6000^\circ$, $6000 = 16(360) + 240$, so 
$z^{2000} = \cos 240^\circ + i\sin 240^\circ$.
We want $z^{2000}+\frac 1{z^{2000}} = 2\cos 240^\circ = -1$.
Finally, the least integer greater than $-1$ is $\boxed{000}$.