|Z|=1,表示Z平面上以原点为圆心,半径为1的圆周,即x^2 + y^2 = 1,设z=x+iy,w = u+iv; 则z = z+1/z-i = (x^2+y^2+x-y)/(x^ + (y-1)^2) + i(x-y+1)/ (x^ + (y-1)^2) = (x-y+1)/(x^2+(y-1)^2) + i(x-y+1)/(x^2+(y-1)^2) , 所以u=v,表示一条直线,且不过（0,0）.
