## Task B-4.3.

If $z_{1}, z_{2} \in \mathbb{C}$ are solutions to the equation $z+\frac{1}{z}=2 \sin \frac{\pi}{7}$, what is $z_{1}^{770}+z_{2}^{770}$?
## Solution.

If we multiply $z+\frac{1}{z}=2 \sin \frac{\pi}{7}$ by $z$, we get the equation $z^{2}-2 \sin \frac{\pi}{7} \cdot z+1=0$.

The solutions to the equation are:

$$
\begin{aligned}
z_{1,2} & =\frac{2 \sin \frac{\pi}{7} \pm \sqrt{4 \sin ^{2} \frac{\pi}{7}-4}}{2}=\frac{2 \sin \frac{\pi}{7} \pm 2 \sqrt{-\left(1-\sin ^{2} \frac{\pi}{7}\right)}}{2} \\
& =\sin \frac{\pi}{7} \pm i \sqrt{\cos ^{2} \frac{\pi}{7}}=\sin \frac{\pi}{7} \pm i \cos \frac{\pi}{7}
\end{aligned}
$$

Let's write the trigonometric form of the obtained solutions:

$$
\begin{array}{ll}
z_{1}=\cos \left(\frac{\pi}{2}-\frac{\pi}{7}\right)+i \sin \left(\frac{\pi}{2}-\frac{\pi}{7}\right)=\cos \frac{5 \pi}{14}+i \sin \frac{5 \pi}{14} . & 1 \text { point } \\
z_{2}=\cos \frac{5 \pi}{14}-i \sin \frac{5 \pi}{14}=\cos \left(-\frac{5 \pi}{14}\right)+i \sin \left(-\frac{5 \pi}{14}\right)=\cos \frac{23 \pi}{14}+i \sin \frac{23 \pi}{14} . & 1 \text { point } \\
z_{1}^{770}=\cos \left(770 \cdot \frac{5 \pi}{14}\right)+i \sin \left(770 \cdot \frac{5 \pi}{14}\right)=\cos (275 \pi)+i \sin (275 \pi)=-1 . & \\
z_{2}^{770}=\cos \left(770 \cdot \frac{23 \pi}{14}\right)+i \sin \left(770 \cdot \frac{23 \pi}{14}\right)=\cos (1265 \pi)+i \sin (1265 \pi)=-1 . & \\
z_{1}^{770}+z_{2}^{770}=-2 & 2 \text { points }
\end{array}
$$