* According to the geometric meaning of complex numbers, derive the formula for representing the distance from a point to a line using complex numbers.
As shown in the figure, let $Z_{1}$ and $Z_{2}$ be two points on the line $l$, and $Z_{3}$ be a point outside the line $l$. Then the unit vector in the direction of $\overrightarrow{Z_{1} Z_{2}}$ is $\frac{z_{2}-z_{1}}{\left|z_{2}-z_{1}\right|}$, and the unit vector in the direction of $\vec{Z}_{1} Z_{3}$ is $\frac{z_{3}-z_{1}}{\left|z_{3}-z_{1}\right|}$. The angle $\varphi$ is determined by the following formula:
$$
\begin{aligned}
\sin \varphi & =\operatorname{Im}\left(\frac{\left|z_{2}-z_{1}\right|}{\left|z_{3}-z_{1}\right|} \cdot \frac{\left(z_{3}-z_{1}\right)\left(\overline{z_{2}}-\overline{z_{1}}\right)}{\left(z_{2}-z_{1}\right)\left(\overline{z_{2}}-\overline{z_{1}}\right)}\right) \\
& =\frac{1}{\left|z_{3}-z_{1}\right|\left|z_{2}-z_{1}\right|} \operatorname{Im}\left(-z_{1} \overline{z_{2}}+\overline{z_{2}} z_{3}-z_{3} \overline{z_{1}}+z_{1} \overline{z_{1}}\right) .
\end{aligned}
$$

Since $\left|z_{1}\right|^{2} \in \mathbf{R},-\operatorname{Im} z_{1} \overline{z_{2}}=\operatorname{Im} \overline{z_{1}} z_{2},-\operatorname{Im} z_{3} \overline{z_{1}}=\operatorname{Im} \overline{z_{3}} z_{1}$, from the above formula, the distance from $Z_{3}$ to the line $Z_{1} Z_{2}$ is $d=\frac{1}{\left|z_{2}-z_{1}\right|} \cdot \operatorname{Im}\left(\overline{z_{1}} z_{2}+\overline{z_{2}} z_{3}+\overline{z_{3}} z_{1}\right)$. To ensure that the distance $d$ is non-negative, the absolute value of the above formula is taken.