设z = a + bi则 |z + 1| = √[(a+1)²+b²] = 1所以(a+1)²+b² = 1又 i/(z-1)= i/[(a-1)+ bi]= i[(a-1)- bi]/[(a+1)²+b²]= b + (a -1)i因为是纯虚数所以 b = 0所以(a+1)² =1所以 a ...