6. Let $O$ be the origin in the complex plane, and $Z_{1}$ and $Z_{2}$ be two moving points in the complex plane, satisfying:
(1) The arguments of the complex numbers corresponding to points $Z_{1}$ and $Z_{2}$ are fixed values $\theta$ and $-\theta\left(0<\theta<\frac{\pi}{2}\right)$, respectively;
(2) The area of $\triangle O Z_{1} Z_{2}$ is a fixed value $S$.
Find the equation of the locus of the centroid $Z$ of $\triangle O Z_{1} Z_{2}$ and the minimum value of the modulus of the complex number corresponding to point $Z$.
6. As shown in Figure 18-6, let the complex numbers corresponding to $Z_{1}, Z_{2}$, and $Z$ be $z_{1}, z_{2}$, and $z$, respectively, where
$$
z_{1}=r_{1}(\cos \theta+\mathrm{i} \sin \theta), z_{2}=r_{2}(\cos \theta-\mathrm{i} \sin \theta).
$$

Since $Z$ is the centroid of $\triangle O Z_{1} Z_{2}$, we have $z=\frac{z_{1}+z_{2}}{3}$.
Thus, $3 z=z_{1}+z_{2}$
$$
=\left(r_{1}+r_{2}\right) \cos \theta+\left(r_{1}-r_{2}\right) \mathrm{i} \sin \theta.
$$

Let $z=x+y \mathrm{i} (x, y \in \mathbf{R})$. From (1), we get
$$
\left\{\begin{array}{l}
3 x=\left(r_{1}+r_{2}\right) \cos \theta, \\
3 y=\left(r_{1}-r_{2}\right) \sin \theta.
\end{array}\right.
$$

Thus, $\left\{\begin{array}{l}\frac{3 x}{\cos \theta}=r_{1}+r_{2}, \\ \frac{3 y}{\sin \theta}=r_{1}-r_{2}.\end{array}\right.$
Subtracting (3) from (2) squared, we get $\frac{9 x^{2}}{\cos ^{2} \theta}-\frac{9 y^{2}}{\sin ^{2} \theta}=4 r_{1} r_{2}$.
The area of $\triangle O Z_{1} Z_{2}$ is $S$, i.e., $\frac{1}{2} r_{1} r_{2} \sin 2 \theta=S$. Thus, $r_{1} r_{2}=\frac{2 S}{\sin 2 \theta}$.
From (4) and (5), the trajectory equation of point $Z$ is $\frac{9 x^{2}}{\cos ^{2} \theta}-\frac{9 y^{2}}{\sin ^{2} \theta}=\frac{8 S}{\sin 2 \theta}$.

From (1), we get
$$
\begin{aligned}
|3 z|^{2} & =\left(r_{1}+r_{2}\right)^{2} \cos ^{2} \theta+\left(r_{1}-r_{2}\right)^{2} \sin ^{2} \theta \\
& =\left(r_{1}-r_{2}\right)^{2} \cos ^{2} \theta+4 r_{1} r_{2} \cos ^{2} \theta+\left(r_{1}-r_{2}\right)^{2} \sin ^{2} \theta \\
& =\left(r_{1}-r_{2}\right)^{2}+\frac{8 S \cos ^{2} \theta}{\sin 2 \theta}=\left(r_{1}-r_{2}\right)^{2}+4 S \cot \theta.
\end{aligned}
$$

When $r_{1}=r_{2}=\sqrt{\frac{2 S}{\sin 2 \theta}}$, $|z|_{\text {min }}=\frac{2}{3} \sqrt{S \cot \theta}$.