264. Find $\sqrt[3]{z}$, if $z=1-i$.
Solution. Let's write the complex number $z$ in trigonometric form. Find $r=\sqrt{a^{2}+b^{2}}=\sqrt{1^{2}+(-1)^{2}}=\sqrt{2}$. Since $a=1$ and $b=-1$, the point corresponding to this number is located in the IV quadrant.

Form the ratios

$$
\cos \varphi=a / r=1 / \sqrt{2}=\sqrt{2} / 2 ; \quad \sin \varphi=b / r=-1 / \sqrt{2}=-\sqrt{2} / 2
$$

Considering that the point $z$ is located in the IV quadrant, we find $\varphi=2 \pi-$ $-\frac{\pi}{4}=\frac{7 \pi}{4}$.

Now we use the formula

$$
\sqrt[3]{z_{1}}=\sqrt[3]{\sqrt{2}}\left(\cos \frac{\frac{7 \pi}{4}+2 \pi k}{3}+i \sin \frac{\frac{7 \pi}{4}+2 \pi k}{3}\right)
$$

where $k$ takes the values 0, 1, and 2. Then we get:

if $k=0$, then

$$
\begin{aligned}
& z^{(1)}=\sqrt[6]{2}\left(\cos \frac{\frac{7 \pi}{4}}{3}+i \sin \frac{\frac{7 \pi}{4}}{3}\right)=\sqrt[6]{2}\left(\cos \frac{7 \pi}{12}+i \sin \frac{7 \pi}{12}\right) \\
& z^{(2)}=\sqrt[6]{2}\left(\cos \frac{\frac{7 \pi}{4}+2 \pi}{3}+i \sin \frac{\frac{7 \pi}{4}+2 \pi}{3}\right)=\sqrt[6]{2}\left(\cos \frac{15 \pi}{4}+\right. \\
& \left.+i \sin \frac{15 \pi}{4}\right)
\end{aligned}
$$

if $k=2$, then

$$
\begin{gathered}
z^{(3)}=\sqrt[6]{2}\left(\cos \frac{\frac{7 \pi}{4}+2 \pi \cdot 2}{3}+i \sin \frac{\frac{7 \pi}{4}+2 \pi \cdot 2}{3}\right)=\sqrt[6]{2}\left(\cos \frac{23 \pi}{12}+\right. \\
\left.+i \sin \frac{23 \pi}{12}\right)
\end{gathered}
$$