265. Find $z^{6}$, if $z=-\sqrt{3}+i$.
Solution. Let's write the number $z$ in trigonometric form, considering that $a=-\sqrt{3}, b=1$. We find $r=\sqrt{a^{2}+b^{2}} ; r=\sqrt{3+1}=2$. The point $z$ is located in the second quadrant. We form the ratios

$$
\cos \varphi=a / r=\sqrt{3} / 2, \sin \varphi=b / r=1 / 2
$$

Considering that the point $z$ is located in the second quadrant, we find $\arg z=$ $=\varphi=\pi-\frac{\pi}{6}=\frac{5 \pi}{6}$. Therefore, $z=2\left(\cos \frac{5 \pi}{6}+i \sin \frac{5 \pi}{6}\right)$.

According to the formula for raising to a power, we have $\left|z^{6}\right|=2^{6}$, $\arg \left(z^{6}\right)=$ $=\frac{5 \pi}{6} \cdot 6$ and, thus,

$$
\begin{gathered}
z^{6}=2^{6}\left(\cos \frac{5 \pi}{6} \cdot 6+i \sin \frac{5 \pi}{6} \cdot 6\right)=64(\cos 5 \pi+i \sin 5 \pi)= \\
=64(\cos \pi+i \sin \pi)=64(-1+i 0)=-64
\end{gathered}
$$