3. Find the cube root of 27.
3. Solution: Let $\alpha$ be the cube root of 27, so
$$\begin{aligned}
a^{3} & =27=27 e^{2 k x i}, \\
\alpha & =3 e^{\frac{2 k \pi}{3}}
\end{aligned}$$

When $k$ takes all integer values, there are only three distinct roots:
$$\begin{array}{l}
\alpha_{1}=3, \alpha_{2}=3 e^{\frac{2 \pi}{3} i}=3\left(\cos 120^{\circ}+i \sin 120^{\circ}\right) \\
\alpha_{3}=3 e^{\frac{2 \pi}{3}}=3\left(\cos 240^{\circ}+i \sin 240^{\circ}\right)
\end{array}$$