JEE: complex roots of unity...

Consider the general case: for sets $A_n = \{z : z^n = 1\}$ and $B_m = \{w : w^m = 1\}$ , how many elements are there in $C_{n,m} = \{zw : z \in A_n, w \in B_m\}$ ?

Since the sets $A_n$ and $B_m$ are sets of roots of unity we can rewrite them like this: $\begin{aligned} A_n &= \left\{ e^{\frac{2\pi i}{n}} : i = 1,2, \ldots,n \right\} \\ B_m &= \left\{ e^{\frac{2\pi j}{m}} : j = 1,2, \ldots, m \right\} \end{aligned}$

Now, this means we can rewrite $C_{n,m}$ as: $C_{n,m} = \left\{ e^{2\pi\left(\frac{i}{n} + \frac{j}{m}\right)} : i = 1,2, \ldots, n; j = 1,2, \ldots, m \right\}$

That is, every element $x \in C_{n,m}$ can be expressed as: $x = e^{2\pi\frac{im + jn}{mn}}$ for some $i,j$ . Clearly, then, every element of C is a mn-th root of unity. Let $g = \gcd(m,n)$ , and $N=n/g, M=g/m$ , then this can be rewritten as: $x = e^{2\pi\frac{iM + jN}{MN}}$ That is, C contains only MN-th roots of unity. The question now is whether it contains all such roots. That is, what values can $iM + jN$ take?

Since M and N are coprime, Bézout's identity tells us that $xM + yN = k$ has integer solutions for x,y for any k. Let $i^\prime_k, j^\prime_k$ be one such solution, for each value of $k \in \{1,2, \ldots, MN\}$ . That is, $i^\prime_k M + j^\prime_k N = k$ Let $i = i^\prime_k \pmod{MN}$ and $j = j^\prime_k \pmod{MN}$ , then $i M + j N = k + rMN$ for some $r \in \mathbb{Z}$ . Returning to our expression for $x$ , this then becomes: $\begin{aligned} x &= e^{2\pi\frac{k + rMN}{MN}} \\ &= e^{2\pi\left(\frac{k}{MN} + r\right)} \\ &= e^{2\pi\frac{k}{MN}} \end{aligned}$

That is, if $\alpha = e^{2\pi/MN}$ , $\beta = e^{2\pi/n}$ , and $\gamma = e^{2\pi/m}$ then for any value $k \in \{1,2, \dots, MN\}$ there is a suitable value of $i \in \{1,2,\ldots,n\}$ and $j \in \{1,2,\dots,m\}$ such that $\alpha^k = \beta^i \gamma^j$ . And since $\beta \in A_n$ and $\gamma \in B_m$ , this shows that every possible MN-th root of unity must be in $C_{n,m}$ .

There are MN such roots, and so $\DeclareMathOperator{\lcm}{lcm} |C_{n,m}| = MN = \lcm(n,m)$ .

In this problem, $n = 18, m = 48$ so $\DeclareMathOperator{\lcm}{lcm} |C_{18,48}| = \lcm(18,48) = \boxed{144}$ .

$C$ is just the set of $144$ th roots of unity. So the answer is $\fbox{144}$ .

Clearly any element of $C$ is a $144$ th root of unity. Now let $a$ = ${\rm exp}(2\pi i / 144)$ , so that any $144$ th root of unity is $a^n$ for some $n$ . Write $n = 8x+3y$ for some integers $x,y$ ; this is always possible because $8$ and $3$ are relatively prime. Then $a^n = (a^8)^x (a^3)^y$ shows that $a^n$ is a product of an $18$ th root of unity and a $48$ th root of unity, so $a^n \in C$ . So all the $144$ th roots of unity are in $C$ .

The same argument shows that, if $18$ and $48$ are replaced by $m$ and $n$ , the answer is lcm $(m,n)$ .