1. 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 $|\alpha|=$
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1.【Analysis and Solution】Let $\alpha=a+b i, \beta=a-b i, a, b \in \mathbf{R}$,
then from $|\alpha-\beta|=2 \sqrt{3}$,
we get $|b|=\sqrt{3}$,
also from $\frac{\alpha}{\beta^{2}}=\frac{\alpha^{3}}{(\alpha \beta)^{2}}$ being a real number and $\alpha \beta$ being a real number, we know that $\alpha^{3}$ is a real number. That is, $(a+b i)^{3}=\left(a^{3}-3 a b^{2}\right)+\left(3 a^{2} b-b^{3}\right) i$ is a real number, thus $3 a^{2} b-b^{3}=0$, hence we get $|a|=1$, $\therefore|\alpha|=\sqrt{a^{2}+b^{2}}=2$.