10.33 Find $\sqrt[3]{-8 i}$

## Group 5
10.33 For the number $z=-8 i$ we have $r=2, \varphi=-\frac{\pi}{2}$. According to formula (10.14), we get

$$
\sqrt[3]{-8 i}=2\left(\cos \frac{-\frac{\pi}{2}+2 \pi k}{3}+i \sin \frac{-\frac{\pi}{2}+2 \pi k}{3}\right), \text { where } k=0,1,2
$$

For $k=0: 2\left(\cos \left(-\frac{\pi}{6}\right)+i \sin \left(-\frac{\pi}{6}\right)\right)=\sqrt{3}-i$;

for $k=1: 2\left(\cos \frac{\pi}{2}+i \sin \frac{\pi}{2}\right)=2 i$;

for $k=2: 2\left(\cos \frac{7 \pi}{6}+i \sin \frac{7 \pi}{6}\right)=-\sqrt{3}-i$.

Answer: $\sqrt{3}-i ; 2 i ;-\sqrt{3}-i$.