满足特殊条件的复数问题

anhcanhsat97

给定复数 $z_{1}, z_{2}, z_{3}$ 满足 $2\left|z_{1}\right|=2\left|z_{2}\right|=\left|z_{ 3}\right|=2$ 和 $\left(z_{1}+z_{2}\right) z_{3}=3 z_{1} z_{2}$. 令$A,B,C$分别为$z_{1},z_{2},z_{3}$在坐标平面上的表示点. 求三角形 $A B C$ 的面积

hbghlyj

$$\begin{rcases}\overline{z_1}=\frac1{z_1}\\\overline{z_2}=\frac1{z_2}\\\overline{z_3}=\frac4{z_3}\\\left(z_{1}+z_{2}\right) z_{3}=3 z_{1} z_{2}\end{rcases}⇒\frac{\overline{z_1}+\overline{z_2}}2=\frac38\overline{z_3}$$ $\begin{rcases}|\overline{z_1}|=|\overline{z_2}|\\\left|\frac{\overline{z_1}+\overline{z_2}}2\right|=\frac34\end{rcases}⇒$ 经过$\overline{z_1},\overline{z_2}$的直线到$0$的距离为$\frac34⇒$ $|\overline{z_1}-\overline{z_2}|=2\sqrt{1-\left(3\over4\right)^2}=\frac{\sqrt7}2$
而$\overline{z_3}$在经过$0,\frac{\overline{z_1}+\overline{z_2}}2$的直线上.
所以$\operatorname{Area}(z_1,z_2,z_3)=\operatorname{Area}(\overline{z_1},\overline{z_2},\overline{z_3})=\frac12·\frac{\sqrt7}2·\left(2-\frac34\right)=\frac{5\sqrt7}{16}$