## Task B-4.3.

Given the complex number $z=\cos \frac{11 \pi}{6}+i \sin \frac{7 \pi}{6}$. Calculate $\sqrt[3]{z^{2020}}$.
## Solution.

Let's write the number $z$ in trigonometric form.

$$
z=\cos \frac{11 \pi}{6}+i \sin \frac{7 \pi}{6}=z=\cos \frac{11 \pi}{6}+i \sin \frac{11 \pi}{6}
$$

Then,

$$
z^{2020}=\cos \frac{11110 \pi}{3}+i \sin \frac{11110 \pi}{3}=\cos \frac{4 \pi}{3}+i \sin \frac{4 \pi}{3}
$$

Now, let's determine the complex numbers $w$ for which $w^{3}=z^{2020}$.

$$
w_{1}=\cos \frac{4 \pi}{9}+i \sin \frac{4 \pi}{9}, \quad w_{2}=\cos \frac{10 \pi}{9}+i \sin \frac{10 \pi}{9}, \quad w_{3}=\cos \frac{16 \pi}{9}+i \sin \frac{16 \pi}{9} . \quad 2 \text { points }
$$

Note: The solution can be written in the form

$$
w_{k}=\cos \left(\frac{1}{3}\left(\frac{4 \pi}{3}+2 k \pi\right)\right)+i \sin \left(\frac{1}{3}\left(\frac{4 \pi}{3}+2 k \pi\right)\right), \quad k=0,1,2
$$