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；

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.