+. $C$ is the set of complex numbers, let the set $A=\left\{z \mid z^{18}=1, z \in C\right\}, B$ $=\left\{\omega^{2 x}=1, \omega \in C\right\}, D=\{z \omega \mid z \in A, \omega \in B\}$. Then the number of elements in $D$ is ( ) .
(A) 864
(B) 432
(C) 288
(D) 144
4. (D).

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

Since 8 and 3 are coprime, then $8 k+3 t$ can be any integer, so $8 k+3 t$ can be $0,1,2, \cdots, 143$, i.e., $2 \omega$ has a total of 144 different values.