设z=x+yi (x,y为实数)1=|z+1|^2-|z-i|^2=|(x+1)+yi|^2-|x+(y-1)i|^2=(x+1)^2+y^2-[x^2+(y-1)^2]=x^2+2x+1+y^2-(x^2+y^2-2y+1)=2x+2y即：y=-x+1/2|z|=√(x^2+y^2)=√[x^2+(-x+1/2)^2]=√(x^2+x^2-x+1/4)=√[2(x-1/4)^...