因为|z|=1,故设z=cosx+isinx,所以|(z²-2z+2)/(z-1+i)|=|[(z-1)^2-i^2]/(z-1+i)|=|z-1-i|=|(cosx-1)+i(sinx-1)|.又(cosx-1)^2+(sinx-1)^2=3-2(sinx+cosx)=3-2√2*sin(x+π/4),而-1