Question

求满足｜（z+1）/(z-1)｜=1，且z+2/z∈R的复数z.

RT

Answer 1

1
z=a+bi,z+2/z为实数
a+bi+2/(a+bi)
=a+bi+[2/(a^2+b^2)](a-bi)
b-2b/(a^2+b^2)=0
a^2+b^2=2
|z|=√2
2
z=a+√(2-a^2)i
(z+1)/(z-1)=[(a+1)+√(2-a^2)i]/[(a-1)+√(2-a^2)i]
=[(a+1)+√2-a^2)i][(a-1)-√(2-a^2)i]/[(a-1)^2+(2-a^2)]
=[(a+1)(a-1)+2√(2-a^2)i+(2-a^2)]/(3-2a)
=[1+2√(2-a^2)i]/(3-2a)
1/(3-2a)^2+4√(2-a^2)^2/(3-2a)^2=1
1+4(2-a^2)=(3-2a)^2
9-4a^2=9-6a+4a^2
8a^2-6a=0
2a(4a-3)=0
a=0,b=±√2 或 a=3/4, b=±√23/4
z=±√2i　　　　　　　　　z=(3/4)±(√23/4)i

Answer 2

设z=a+bi (a,b∈R)
则|(a+1+bi)/(a-1+bi)|=1
平方整理得2a+1+2bi=0
故a=-1/2,b=0
此时z=-1/2
且z+2/z∈R，满足
故z=-1/2