(6) Let $z_{1}$ and $z_{2}$ be a pair of unequal conjugate complex numbers, and $\left|z_{1}\right|=\sqrt{3}, \frac{z_{1}^{2}}{z_{2}}$ is a real number, then the value of $\left|z_{1}-z_{2}\right|$ is ( ).
(A) $\sqrt{3}$
(B) $\sqrt{6}$
(C) 3
(D) $2 \sqrt{3}$
$6 \mathrm{C}$ Method one From the problem, we have $\frac{z_{1}^{2}}{z_{2}}=\left(\frac{\overline{z_{1}^{2}}}{z_{2}}\right)$, which means $z_{1}^{2} \overline{z_{2}}=z_{2} \overline{z_{1}^{2}}$. Multiplying both sides by $z_{1} z_{2}$ gives $z_{1}^{3}\left|z_{2}\right|^{2}=\overline{z_{1}} z_{2}^{2}\left|z_{1}\right|^{2}$. Given $\overline{z_{1}}=z_{2}$, then $z_{1}^{3}=z_{2}^{3}$, therefore
$$
\left(z_{1}-z_{2}\right)\left(z_{1}^{2}+z_{1} z_{2}+z_{2}^{2}\right)=0 .
$$

Since $z_{1}-z_{2} \neq 0$, it must be that $z_{1}^{2}+z_{1} z_{2}+z_{2}^{2}=0$, so
$$
\left(z_{1}-z_{2}\right)^{2}=-3 z_{1} z_{2}=-3\left|z_{1}\right|^{2}=-9 \text {. }
$$

Thus, $\left|z_{1}-z_{2}\right|=3$.
Method two Since $\frac{z_{1}^{2}}{z_{2}} \in \mathbf{R}$, then $2 \operatorname{Arg} z_{1}-\operatorname{Arg} z_{2}=$ $k \pi,(k \in \mathbf{Z})$, and because $z_{1}=\overline{z_{2}}$, we can take $\arg z_{1}=$ $\frac{\pi}{3}$. As shown in the figure, in $\triangle O Z_{1} Z_{2}$, it is easy to get $\left|Z_{1} Z_{2}\right|=3$.