<s>[INST] Classic Complex Numbers - Given $z+\frac 1z=2\cos 3^\circ$, find least integer greater than $z^{2000}+\frac 1{z^{2000}}$ 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?
 [/INST] 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 $a=3^\circ=\frac\pi{60}$,
$$z^{2000}+\frac 1{z^{2000}}=2\cos\left(\frac{2000\pi}{60}\right)=-1$$
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