每日一题[3160]对对碰

已知复数 $z_1,z_2,z_3,z_4$ 满足 $\left|z_1\right|^2+\left|z_2\right|^2=\left|z_3\right|^2+\left|z_4\right|^2=4$，且 $z_1 \overline{z_3}+z_2 \overline{z_4}=0$，则 $\left|z_1 z_4-z_2 z_3\right|=$ _______．

答案    $4$．

解析    根据题意，有

$$\begin{split} |z_1z_4-z_2z_3|^2&=(z_1z_4-z_2z_3)\cdot \left(\overline{z_1}\overline{z_4}-\overline{z_2}\overline{z_3}\right)\\ &=|z_1z_4|^2+|z_2z_3|^2-z_1\overline{z_2}\overline{z_3}z_4-\overline{z_1}z_2z_3\overline{z_4}\\ &=|z_1|^2\cdot |z_4|^2+|z_2|^2\cdot |z_3|^2-z_1\overline{z_3}\cdot \left(-\overline{z_1}z_3\right)-\overline{z_2}z_4\cdot \left(-z_2\overline{z_4}\right)\\ &=|z_1|^2\cdot |z_4|^2+|z_2|^2\cdot |z_3|^2+|z_2|^2|z_4|^2+|z_1|^2|z_3|^2\\ &=\left(|z_1|^2+|z_2|^2\right)\left(|z_3|^2+|z_4|^2\right)\\ &=4\cdot 4\\ &=16,\end{split}$$ 因此 $|z_1z_4-z_2z_3|=4$．