# 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?

• @AnuragA Euler's Identity Dec 4, 2019 at 21:34
• Do you mean one of these theorems, or something else?
– J.G.
Dec 4, 2019 at 21:59

Note that, given $$z+\frac1z = 2\cos a$$ for any $$a$$,
$$z^n+\frac1{z^n}=2\cos (na)$$
$$z^n+\frac1{z^n}= \left(z+\frac1z\right)\left(z^{n-1}+\frac1{z^{n-1}}\right) -\left(z^{n-2}+\frac1{z^{n-2}}\right)$$ $$=2\cos a \cdot 2\cos [(n-1)a] - 2\cos [(n-2)a]$$ $$=2[\cos (na) + \cos [(n-2)a]]- 2\cos [(n-2)a] =2\cos (na)$$
Thus, for $$a=3^\circ=\frac\pi{60}$$,
$$z^{2000}+\frac 1{z^{2000}}=2\cos\left(\frac{2000\pi}{60}\right)=-1$$