6. Let $z_{1}, z_{2}$ 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 $\left|z_{1}-z_{2}\right|=(\quad)$.
(A) $\sqrt{3}$
(B) $\sqrt{6}$
(C) 3
(D) $2 \sqrt{3}$
6. C.

Solution 1 From the problem, we have
$$
\frac{z_{1}^{2}}{z_{2}}=\overline{\left(\frac{z_{1}^{2}}{z_{2}}\right)} \Rightarrow z_{1}^{2} \overline{z_{2}}=z_{2} \overline{z_{1}^{2}} \text {. }
$$

Multiplying both sides by $z_{1} z_{2}$, we get
$$
z_{1}^{3}\left|z_{2}\right|^{2} \doteq \overline{z_{1}} z_{2}^{2}\left|z_{1}\right|^{2} \text {. }
$$

Given $\overline{z_{1}}=z_{2}$, we know $z_{1}^{3}=z_{2}^{3}$.
$$
\text { Hence }\left(z_{1}-z_{2}\right)\left(z_{1}^{2}+z_{1} z_{2}+z_{2}^{2}\right)=0 \text {. }
$$

Since $z_{1}-z_{2} \neq 0$, then $z_{1}^{2}+z_{1} z_{2}+z_{2}^{2}=0$.
$$
\text { Hence }\left(z_{1}-z_{2}\right)^{2}=-3 z_{1} z_{2}=-3\left|z_{1}\right|^{2}=-9 \text {. }
$$

Therefore, $\left|z_{1}-z_{2}\right|=3$.
Solution 2 From $\frac{z_{1}^{2}}{z_{2}} \in \mathbf{R}$, we know
$$
2 \arg z_{1}-\arg z_{2}=k \pi(k \in \mathbf{Z}) \text {. }
$$

Given $z_{1}=\overline{z_{2}}$, we can take $\arg z_{1}=\frac{\pi}{3}$.
As shown in Figure 2, in $\triangle O Z_{1} Z_{2}$, it is easy to get $\left|Z_{1} Z_{2}\right|=3$.