Roots of Unity Set Count
The sets $ A = \{z : z^{18} = 1\} $ and $ B = \{w : w^{48} = 1\} $ are both sets of complex roots of unity. The set $ C = \{zw : z \in A ~ \mbox{and} ~ w \in B\} $ is also a set of complex roots of unity. How many distinct elements are in $ C^{}_{} $?
- 1
- 2
- 3
- +
- 4
- 5
- 6
- -
- 7
- 8
- 9
- $\frac{a}{b}$
- .
- 0
- =
- %
- $a^n$
- $a^{\circ}$
- $a_n$
- $\sqrt{}$
- $\sqrt[n]{}$
- $\pi$
- $\ln{}$
- $\log$
- $\theta$
- $\sin{}$
- $\cos{}$
- $\tan{}$
- $($
- $)$
- $[$
- $]$
- $\cap$
- $\cup$
- $,$
- $\infty$
Solution
Note that $ \operatorname{lcm}(18,48) = 144 $. So,
\[(zw)^{144} = z^{144} w^{144} = (z^{18})^8 \cdot (w^{48})^3 = 1.\]Hence, every element in $ C $ is a 144th root of unity.
Conversely, consider an arbitrary 144th root of unity, say
\[\operatorname{cis} \frac{2 \pi k}{144}.\]Note that $ \operatorname{cis} \frac{2 \pi (2k)}{18} \in A $ and $ \operatorname{cis} \frac{2 \pi (-5k)}{48} \in B, $ and their product is
\[\operatorname{cis} \frac{2 \pi (2k)}{18} \cdot \operatorname{cis} \frac{2 \pi (-5k)}{48} = \operatorname{cis} \frac{2 \pi (16k)}{144} \cdot \operatorname{cis} \frac{2 \pi (-15k)}{144} = \operatorname{cis} \frac{2 \pi k}{144}.\]Therefore, every 144th root of unity lies in $ C, $ which means $ C $ is precisely the set of 144th roots of unity. It follows that $ C $ contains $ \boxed{144} $ elements.
Freeflows