复数 三点共线的充要条件

hbghlyj

复数$z_1,z_2,z_3$共线的充要条件为$$\bar{z}_{1} z_{2}+\bar{z}_{2} z_{3}+\bar{z}_{3} z_{1}\in\Bbb R$$
证明:
三角形$z_1,z_2,z_3$的有向面积为
$A\left(z_{1}, z_{2}, z_{3}\right)=\frac{i}{4}\left|\begin{array}{lll}z_{1} & \bar{z}_{1} & 1 \\ z_{2} & \bar{z}_{2} & 1 \\ z_{3} & \bar{z}_{3} & 1\end{array}\right|$
复数$z_1,z_2,z_3$共线$⇔A\left(z_{1}, z_{2}, z_{3}\right)=0⇔\bar{z}_{1} z_{2}+\bar{z}_{2} z_{3}+\bar{z}_{3} z_{1}=z_1\bar{ z}_2+z_2\bar{z}_3+z_3\bar{z}_1$

hbghlyj

复数$z_1,z_2,z_3$共线$⇔∃λ_1,λ_2∈\Bbb R:λ_1(z_1-z_3)=λ_2(z_2-z_3)$且$λ_1,λ_2$不全为0.
令$λ_3=-λ_1-λ_2,$ 则$λ_1z_1+λ_2z_2+λ_3z_3=0$.
所以存在不全为0的$λ_1,λ_2,λ_3$满足$$\cases{λ_1+λ_2+λ_3=0\\λ_1z_1+λ_2z_2+λ_3z_3=0\\λ_1\bar z_1+λ_2\bar z_2+λ_3\bar z_3=0}$$
所以$$\begin{vmatrix}1&1&1\\z_1&z_2&z_3\\\bar z_1&\bar z_2&\bar z_3\end{vmatrix}=0$$即\[\bar{z}_{1} z_{2}+\bar{z}_{2} z_{3}+\bar{z}_{3} z_{1}=z_1\bar{ z}_2+z_2\bar{z}_3+z_3\bar{z}_1\]
即\[\bar{z}_{1} z_{2}+\bar{z}_{2} z_{3}+\bar{z}_{3} z_{1}\in\Bbb R\]