The form z + 1/z

$z+\frac{1}{z}=2 \cos 6°$ We get the sense of De Moivré. Indeed $z=\cos 6°\pm \iota \sin 6°$ $\Rightarrow z^{1000}+\frac{1}{z^{1000}}= 2 \cos 6000°=2(-\frac{1}{2})=\boxed{-1}$

For the general equation $z + \frac{1}{z} = 2\cos(\theta)$ , (where $\theta$ is in radians instead of degrees), we can first rearrange and then apply the quadratic equation to find that

$z^{2} - 2\cos(\theta)z + 1 = 0$

$\Longrightarrow z = \dfrac{2\cos(\theta) \pm \sqrt{4\cos^{2}(\theta) - 4}}{2}$

$= \cos(\theta) \pm \sqrt{-1*(1 - \cos^{2}(\theta))} = \cos(\theta) \pm i*|\sin(\theta)| = e^{\pm i\theta}$ .

Now regardless of whether we take the positive or negative root, the result for $z^{1000} + \frac{1}{z^{1000}}$ will be the same, namely

$e^{i*1000*\theta} + e^{-i*1000*\theta} = 2\cos(1000*\theta)$ ,

which for $\theta = 6^{\circ} = \frac{\pi}{30}$ radians is

$2\cos(\frac{100\pi}{3}) = 2\cos(\frac{4\pi}{3}) = 2*(-\frac{1}{2}) = \boxed{-1}$ .