How many distinct elements are there in $C=\{zw\mid z∈A$,$w∈B\}, z^{24}=1$ and $w^{54}=1$.

Question

Let $A$ be a set of all complex numbers $z$ such that $z^{24}=1$ and let $B$ be the set of all complex numbers $w$ such that $w^{54}=1$. That is: $A$={$z$|$z^{24}=1$} and $B$={$w$|$w^{54}=1$}

Finally, let $C$ be the set of all complex numbers that can be formed by multiplying an element of $A$ by an element of $B$:

$C$={$zw$|$z∈A$,$w∈B$}.

How many distinct elements are there in $C$?

by writing $z$ and $w$ in polar form, I showed that $(zw)^{216}$=1, so $zw$ is $216th$ root of unity, hence $C$ can contain at most 216 elements. However, I need some ideas as how can I show that it has EXACTLY that many elements. Please, note that I need to solve this problem using only complex analysis, without any abstract algebra concepts (i.e groups and order of a group). Thanks in advance!

Answer 1

The sets you write can be written in the following way: $$A=\left\{e^{\tfrac{i\pi k}{12}}\Big/k\in \Bbb Z\right\},B=\left\{e^{\tfrac{i\pi k}{27}}\Big/k\in \Bbb Z\right\} $$

And hence $C^becomes:

$$C=\left\{e^{\tfrac{i\pi (9k+4j)}{108}}\Big/k,j \in \Bbb Z\right\}$$

and now the question is how many element can be written in the form $9k+4j$ or if you don't like combinatorics you can prove that: $$C=\left\{e^{\tfrac{i\pi (t)}{108}}\big/ t\in \Bbb Z\right\}$$ by double inclusion using Bézout's identity

Answer 2

$A$ is such that if $x, y\in A$, $xy\in A$. That is also true of $B$ and of $C$. If you want to show that it has $216$ elements, you could show that it contains $\omega = e^{i\pi/108}$. You should be able to work out which elements you need to multiply to get $\omega$ by doing some light number theory.