10.3.3 * Let $\triangle Z_{1} Z_{2} Z_{3}$ and $\triangle Z_{1}^{\prime} Z_{2}^{\prime} Z_{3}^{\prime}$ be directly similar. Find the necessary and sufficient condition that the corresponding vertices (such as $Z_{1}$ corresponding to $z_{1}$, etc.) satisfy, and derive the equation of the perpendicular bisector from this.
Given that $z_{1}=z_{2} \Rightarrow\left|z_{1}\right|=\left|z_{2}\right|$ and $\arg z_{1}=\arg z_{2}$, the necessary and sufficient condition for $\triangle Z_{1} Z_{2} Z_{3}$ to be similar to $\triangle Z_{1}^{\prime} Z_{2}^{\prime} Z_{3}^{\prime}$ in the same direction is $\frac{z_{3}-z_{1}}{z_{2}-z_{1}}=\frac{z_{3}^{\prime}-z_{1}^{\prime}}{z_{2}^{\prime}-z_{1}^{\prime}}$.
From this, we can deduce that the equation of the perpendicular bisector of $Z_{1}$ and $Z_{2}$ is $\triangle Z Z_{1} Z_{2} \sim \triangle Z Z_{2} Z_{1}$:
$$
\left|\begin{array}{ccc}
1 & 1 & 1 \\
z & z_{1} & z_{2} \\
\bar{z} & \overline{z_{2}} & \overline{z_{1}}
\end{array}\right|=0,
$$

which simplifies to $\frac{z}{z_{1}-z_{2}}+\frac{\bar{z}}{z_{1}-z_{2}}=\frac{\left|z_{1}\right|^{2}-\left|z_{2}\right|^{2}}{\left|z_{1}-z_{2}\right|^{2}}$. Specifically, when $z_{2}-0$, the equation of the perpendicular bisector is $\frac{z}{z_{1}}+$ $\frac{\bar{z}}{\overline{\bar{z}}}=1$.