Example 1 Suppose there are $n(n \geqslant 2)$ distinct complex numbers forming a set $S$, and the product of any two numbers (which can be the same) in this set still belongs to this set. Try to find the $n$ elements in this set.
Analysis: Since the set is closed under multiplication, we think of the roots of unity and use the knowledge of roots of unity to solve the problem.
Solution: We discuss in two cases:
(1) If the set $S$ does not contain zero, let its $n$ elements be $a_{1}, a_{2}, \cdots, a_{n}$. Take any $z \in S$, then $z \neq 0$, and form the products $z a_{1}, z a_{2}, \cdots, z a_{n}$.

Since $a_{1}, a_{2}, \cdots, a_{n}$ are distinct and $z \neq 0$, the $n$ complex numbers in (2) are also distinct. Therefore, by the problem's condition, the $n$ complex numbers in (2) are the same as those in (1), so $a_{1} a_{2} \cdots a_{n}=z a_{1} \cdot z a_{2} \cdots z a_{n}$, which implies $z^{n}=1$. Since 1 has exactly $n$ distinct $n$-th roots of unity, $S$ consists of the $n$ $n$-th roots of unity, which are $\cos \frac{2 k \pi}{n}+i \sin \frac{2 k \pi}{n}$ $(k=0,1, \cdots, n-1)$.
(2) If the set $S$ contains 0, by (1), the remaining $n-1$ elements of $S$ are exactly the $n-1$ distinct $(n-1)$-th roots of unity. Thus, the $n$ elements of $S$ are 0 and $\cos \frac{2 k \pi}{n-1}+i \sin \frac{2 k \pi}{n-1}(k=0,1, \cdots, n-2)$.