2. $C$ is the set of complex numbers, let the set $A=\left\{z \mid z^{18}=1, z \in C\right\}, B=\left\{\omega \mid \omega^{48}=1\right.$, $\omega \in C\}, D=\{z \omega \mid, z \in A, \omega \in B\}$, find the number of elements in $D$.
2. The elements of sets $A$, $B$, and $D$ are
$$
\begin{array}{l}
z=\cos \frac{2 k \pi}{18}+i \sin \frac{2 k \pi}{18}, \omega=\cos \frac{2 t \pi}{48}+i \sin \frac{2 t \pi}{48}, z \omega=\cos \frac{2(8 k+3 t)}{144} \pi+ \\
i \sin \frac{2(8 k+3 t)}{144} \pi .
\end{array}
$$

Since 8 and 3 are coprime, $8 k+3 t$ can be any integer, so $8 k+3 t$ can be $0,1,2, \cdots, 143$, meaning $z \omega$ has 144 different values, thus the set $D$ contains 144 elements.