题目

复数问题,要解题过程,急求
适合条件/z/=1,/(z+1)/(z-1)/=1的复数z的集合

提问时间：2021-03-04

答案

设 z = x + yi
|z|^2 = 1
x^2 + y^2 = 1
(z+1)/(z-1)
=(x+1 + yi)/(x-1 + yi)
= (x+1+yi)(x-1 -yi)/[(x-1 + yi)(x-1 -yi)]
= [x^2 - (1+yi)^2]/[(x-1)^2 +y^2]
= [x^2 +y^2 - 2yi -1]/[x^2 + y^2 - 2x + 1]
将 x^2 + y^2 = 1 代入上式
= (1 - 2yi -1)/(1 - 2x + 1)
= -2yi/(2 -2x)
= -yi/(1-x)
|(z+1)/(z-1)|^2 = 1
y^2/(1-x)^2 = 1
y^2 = (1-x)^2
联立
y^2 = (1-x)^2
x^2 + y^2 = 1
1-x^2 = (1-x)^2
1-x^2 = 1 -2x + x^2
2x^2 - 2x = 0
x^2 -x = 0
x(x-1) = 0
x=0 和 x=1
代入到 x^2 + y^2 = 1 中
分别对应 y = ±1 和 y = 0
x = 1 ,y = 0 时
(z+1)/(z-1) 的分母为0,所以舍去
故z的解集合为 { z | z = ±i }