设z∈C,且|z-i|＝|z-1|

z可表示为z=x+iy,|z-i|＝|x+i(y-1)|,|z-1|＝|x-1+iy|又|z-i|＝|z-1|即模相等,x^2+(y-1)^2=(x-1)^2+y^2解得y=x|z+i|＝|x+i(y+1)|=x^2+(y+1)^2=x^2+(x+1)^2=2x^2+2x+1=2(x+1/2)^2+1/2根据二次函数性质,x∈R,则|z+i|>=1...