设z=a+bi,则z/(z-1)=(a+bi)/(a-1+bi)=(a+bi)(a-1-bi)/(a-1+bi)(a-1-bi)=[a^2-a+b^2+(2a-1)bi]/[(a-1)^2-b^2]是纯虚数则a^2-a+b^2=0即(a-1/2)^2+b^2=1/4则可设a=1/2+1/2cosθ,b=1/2sinθ|z-i|^2=a^2+(b-1)^2=1/4+1/2c...