Question

已知虚数z,|z|=√2,且z^2+2z'(z'为z的共轭复数)为实数.求虚数z的值；

Answer 1

1.设z=x+yi,x,y∈R,y≠0,则

z^2+2z'=x^2-y^2+2xyi+2(x-yi)

=x^2-y^2+2x+(2xy-2y)i∈R,

∴2xy-2y=0,

∴x=1.

由|z|=√2得x^2+y^2=2.y^2=1,y=土1.

∴z=1土i.

2.(x-1-i)(x-1+i)=x^2-2x+2=0,为所求.