9. Let $\alpha, \beta$ be a pair of conjugate complex numbers. If $|\alpha-\beta|=2 \sqrt{3}$, and $\frac{\alpha}{\beta^{2}}$ is a real number, then what is $|\alpha|$?
9. Let $\alpha=a+b i, \beta=a-b i, a, b \in \mathrm{R}$, then from $|\alpha-\beta|=2 \sqrt{3}$ we get $|b|=\sqrt{3}$. Also, $\frac{\alpha}{\beta^{2}}=\frac{\alpha^{3}}{(\alpha \beta)^{2}}$ is a real number, and $\alpha \beta$ is a real number, so $\alpha^{3}$ is a real number, i.e., $(a+b i)^{3}=\left(a^{3}-3 a b^{2}\right)+$ $\left(3 a^{2} b-b^{3}\right) i$ is a real number. Therefore, we have $3 a^{2} b-b^{3}=0$, which gives $|a|=1$. Thus, $|a|=\sqrt{a^{2}+b^{2}}=2$.