9. Let $z_{1}, z_{2}$ be a pair of conjugate complex numbers. If $\left|z_{1}-z_{2}\right|=\sqrt{6}$, and $\frac{z_{1}}{z_{2}^{2}}$ is a real number, then $\left|z_{1}\right|=$ $\qquad$ .
$$
\begin{array}{l}
\frac{z_{1}}{z_{2}^{2}}=\frac{\overline{z_{1}}}{\overline{z_{2}}}=\frac{z_{2}}{z_{1}^{2}} \Rightarrow z_{1}^{3}=z_{2}^{3} \Rightarrow z_{1}^{2}+z_{1} z_{2}+z_{2}^{2}=0 \text {, and }\left|z_{1}-z_{2}\right|^{2}=\left(z_{1}-z_{2}\right)\left(\overline{z_{1}}-\overline{z_{2}}\right) \\
=\left|z_{1}\right|^{2}+\left|z_{2}\right|^{2}-\left(z_{1}^{2}+z_{2}^{2}\right)=\left|z_{1}\right|^{2}+\left|z_{2}\right|^{2}+z_{1} z_{2}=3\left|z_{1}\right|^{2}=6 \Rightarrow\left|z_{1}\right|=\sqrt{2} .
\end{array}
$$