$2 \cdot 8$ set $A=\left\{z \mid z^{18}=1\right\}, B=\left\{w \mid w^{48}=1\right\}$ are both sets of complex roots of unity, $C=\{z w \mid z \in A, w \in B\}$ is also a set of complex roots of unity, how many elements does set $C$ contain?
[Solution] The elements of set $A$ are
$$
z=\cos \frac{2 k \pi}{18}+i \sin \frac{2 k \pi}{18} \text {. }
$$

The elements of set $B$ are
$$
\begin{aligned}
w & =\cos \frac{2 t \pi}{48}+i \sin \frac{2 t \pi}{48} . \\
z w & =\cos \left(\frac{2 k \pi}{18}+\frac{2 t \pi}{48}\right)+i \sin \left(\frac{2 k \pi}{18}+\frac{2 t \pi}{48}\right) \\
& =\cos \frac{2(8 k+3 t)}{144} \pi+i \sin \frac{2(8 k+3 t)}{144} \pi
\end{aligned}
$$

Since 8 and 3 are coprime, there exist integers $k$ and $t$ such that
$$
8 k+3 t=1 \text {. }
$$

Thus, there exist integers $k$ and $t$ such that
$$
8 k+3 t=\dot{m}
$$

Taking $m=0,1,2, \cdots, 143$, we get 144 different values for $z w$.
Therefore, the set
$$
C=\{z w \mid z \in A, w \in B\}
$$

contains 144 elements.