Trigonometry Has Come Back

Nice problem!

We claim that $C$ consists of the 144th roots of unity, so that $C$ has $\boxed{144}$ elements.

If $z$ is in $A$ and $w$ is in $B$ , then $(zw)^{144}=(z^{18})^8(w^{48})^3=1$ , so that $zw$ is indeed a 144th roots of unity.

Conversely, any 144th root of unity can be written as the product of an 18th and a 48th root of unity, $e^{\frac{2k\pi{i}}{144}}=e^{\frac{4k\pi{i}}{18}}e^{\frac{-10k\pi{i}}{48}}$