已知复数z满足|z+1|=1,且i/(z-1)是纯虚数,则复数z的值为

z=a+bi|z+1|²=1所以(a+1)²+b²=1i/(z-1)=i/(a-1+bi)=i(a-1-bi)/[(a-1)²+b²]=[b+(a-1i)]/[(a-1)²+b²]是纯虚数则b=0且a-1≠0代入(a+1)²+b²=1a+1=±1a=-2,a=0所以z=-2,z=0