已知复数Z满足|Z|=1,且Z≠±i,求证:(z+i)/(z-i)是纯虚数

方法有好多呦,给一种好懂的吧：设z=x+yi(x,y∈R),因为|Z|=1,则x^2+y^2=1所以(z+i)/(z-i)=[x+(y+1)i]/[x+(y-1)i]=[x+(y+1)i]*[x-(y-1)i]/[x^2+(y-1)^2]=[(x^2+y^2-1)+2xi]/[x^2+(y-1)^2]=xi/(1-y) 因为Z≠±i,所以x≠...