设z=x+yiz^2+iz-1/4=0(x+yi)^2+i(x+yi)-1/4=0x^2-y^2+2xyi+xi-y-1/4=0(x^2-y^2-y-1/4)+(2xy+x)i=0故有x^2-y^2-y-1/4=0 2xy+x=0x(2y+1)=0x=0,或y=-1/2y=-1/2,或x=0即有z=-1/2i
