## Task B-4.5.

If $z=-\sin \frac{2022 \pi}{5}-i \cos \frac{2023 \pi}{5}$, what is $z^{10}$?
## Solution.

By applying the periodicity properties of sine and cosine functions and reduction formulas, we have the following sequence of equalities:

$$
\begin{aligned}
& z=-\sin \frac{2022 \pi}{5}-i \cos \frac{2023 \pi}{5} \\
& z=-\sin \left(404 \pi+\frac{2 \pi}{5}\right)-i \cos \left(404 \pi+\frac{3 \pi}{5}\right) \\
& z=-\sin \frac{2 \pi}{5}-i \cos \frac{3 \pi}{5} \\
& z=-\cos \left(\frac{\pi}{2}-\frac{2 \pi}{5}\right)-i \sin \left(\frac{\pi}{2}-\frac{3 \pi}{5}\right) \\
& z=-\cos \frac{\pi}{10}-i \sin \frac{-\pi}{10}
\end{aligned}
$$

Furthermore, using the odd property of the sine function, we conclude that

$$
z=-\cos \frac{\pi}{10}+i \sin \frac{\pi}{10}
$$

and since $\sin \frac{\pi}{10}=\sin \frac{9 \pi}{10} \text{ and } \cos \frac{\pi}{10}=-\cos \frac{9 \pi}{10}$, we finally get

$$
z=\cos \frac{9 \pi}{10}+i \sin \frac{9 \pi}{10}
$$

Therefore,

$$
\begin{aligned}
z^{10} & =\cos 9 \pi+i \sin 9 \pi & & 1 \text { point } \\
& =\cos \pi+i \sin \pi=-1 & & 1 \text { point }
\end{aligned}
$$

Note: The student can leave the solution in trigonometric or standard form.