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[函数] 2020年全国卷2理科第15题 复数的模

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isee Posted at 2020-7-9 19:49:58 |Read mode
2020年全国卷2理科第15题

设复数$z_1$,$z_2$满足$|z_1|=|z_2|=2$,$z_1+z_2=\sqrt{3}+i$,则$|z_1-z_2|$=__________.


答:$\sqrt 3$

偶被其坑了,用代数形式硬算的,后来,在网上见到解答,$|z_1+z_2|=|z_1|=|z_2|=2$,再利用几何意义,或者三角形式,哎,瞬间不好了

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facebooker Posted at 2020-7-9 20:24:41
回复 1# isee

第一时间想到的是平行四边形法则公式 考过太多次了

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 Author| isee Posted at 2020-7-9 20:32:00
Last edited by isee at 2020-7-10 14:44:00回复 2# facebooker


是的是的,刚刚想到这个,这也许才是出题人本意,具体如下:

一般情形,可以利用共轭复数处理,即用$$\abs{z}^2=z\bar z.$$

于是
$$\abs{z_1-z_2}^2=(z_1-z_2)\overline{(z_1-z_2)}=(z_1-z_2)(\bar{z_1}-\bar{z_2})=\abs{z_1}^2+\abs{z_2}^2-\left(z_1\bar{z_2}+\bar{z_1}z_2\right).$$
同样的
$$\abs{z_1+z_2}^2=(z_1+z_2)\overline{(z_1+z_2)}=(z_1+z_2)(\bar{z_1}+\bar{z_2})=\abs{z_1}^2+\abs{z_2}^2+\left(z_1\bar{z_2}+\bar{z_1}z_2\right).$$

两式相加,即

$$\abs{z_1+z_2}^2+\abs{z_1-z_2}^2=2\abs{z_1}^2+2\abs{z_2}^2.$$

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