Example 2 Given $z=(\sqrt{3}-3 \mathrm{i})^{n}$. If $z$ is a real number, then the smallest positive integer $n$ is ( ). ${ }^{[1]}$
(A) 3
(B) 4
(C) 5
(D) 6
Solution 1 Notice that; when $n=1$, $z=\sqrt{3}-3 \mathrm{i}$ is not a real number;

when $n=2$, $z=(\sqrt{3}-3 \mathrm{i})^{2}=-6-6 \sqrt{3} \mathrm{i}$ is not a real number;

when $n=3$, $z=(\sqrt{3}-3 \mathrm{i})^{3}=-24 \sqrt{3}$ is a real number.
Thus, the smallest positive integer $n=3$.
Solution 2 Notice that;
$$
z=(\sqrt{3}-3 \mathrm{i})^{n}=(-2 \sqrt{3})^{n} \mathrm{e}^{-\frac{a x i}{3}} \text {. }
$$

Therefore, the necessary and sufficient condition for $z$ to be a real number is $31 n$.
At this time, the smallest positive integer $n=3$.