令z=x+yi
则√(x^2+y^2)+x^2-y^2+2xyi=0
所以2xy=0
若x=0,则|y|=y^2,y>=0则y=1或0,y<0则y=-1
若y=0,则|x|+x^2=0,x=0
综上,(x,y)=(0,0)(0,1)(0,-1)
即z=0,±i