Task B-4.2. If $z+z^{-1}=2 \cos \frac{\alpha}{2012}$, determine $\alpha$ for which $z^{2012}+z^{-2012}=1$
Solution. Let's solve the equation

$$
\begin{gathered}
z+z^{-1}=2 \cos \frac{\alpha}{2012} \\
z^{2}-2 z \cdot \cos \frac{\alpha}{2012}+1=0 \\
z_{1,2}=\cos \frac{\alpha}{2012} \pm i \sin \frac{\alpha}{2012}
\end{gathered}
$$

Then we have

$$
\begin{gathered}
\frac{1}{z_{1}}=\cos \frac{\alpha}{2012}-i \sin \frac{\alpha}{2012} \quad \text { and } \quad \frac{1}{z_{2}}=\cos \frac{\alpha}{2012}+i \sin \frac{\alpha}{2012} \\
z^{2012}+z^{-2012}=2 \cos \alpha \\
2 \cos \alpha=1
\end{gathered}
$$

and finally

$$
\alpha= \pm \frac{\pi}{3}+2 k \pi, \quad k \in \mathbb{Z}
$$