Given that $z$ is a complex number such that $z+\frac 1z=2\cos 3^\circ,$ find the least integer that is greater than $z^{2000}+\frac 1{z^{2000}}.$
Solution: We have $z=e^{i\theta}$, so $e^{i\theta}+\frac{1}{e^{i\theta}}=\frac{\cos \theta}{\cos ^2\theta +\sin ^2\theta}+\cos \theta +i\sin \theta - \frac{i\sin \theta}{\cos ^2\theta +\sin ^2\theta}=2\cos \theta$. Therefore, $\theta =3^\circ=\frac{\pi}{60}$. From there, $z^{2000}+\frac{1}{z^{2000}}=e^{\frac{100\pi (i)}{3}}+e^{\frac{-100\pi (i)}{3}}=2\cos \frac{4\pi}{3}=-1$, so our answer is $\boxed{0}$.
How to solve this by applying Tchebyshev?
Note that, given $z+\frac1z = 2\cos a$ for any $a$,
$$z^n+\frac1{z^n}=2\cos (na)$$
holds true, which is shown recursively below,
$$z^n+\frac1{z^n}= \left(z+\frac1z\right)\left(z^{n-1}+\frac1{z^{n-1}}\right) -\left(z^{n-2}+\frac1{z^{n-2}}\right)$$ $$=2\cos a \cdot 2\cos [(n-1)a] - 2\cos [(n-2)a] $$ $$=2[\cos (na) + \cos [(n-2)a]]- 2\cos [(n-2)a] =2\cos (na) $$
Thus, for $a=3^\circ=\frac\pi{60}$,
$$z^{2000}+\frac 1{z^{2000}}=2\cos\left(\frac{2000\pi}{60}\right)=-1$$