Question

已知复数z满足z+1/z属于R,|z-2|=2,求z

Answer 1

z=a+bi

z+1/z=a+bi+(a-bi)/(a^2+b^2)

所以b-b/(a^2+b^2)=0

b=0,或a^2+b^2=1

|z-2|=2

(a-2)^2+b^2=4

a^2+b^2-4a=0

若b=0

则a=0,或a=4

若a^2+b^2=1

则a=0.25

b=sqrt(15)/4

或b=-sqrt(15)/4

z的解集为

z=0

z=4

z=0.25+0.25*sqrt(15)i

z=0.25-0.25*sqrt(15)i