putting z=x+iy in w gives        w=[(1+y)-ix]/[(x-1)+iy]
on solving gives: w=[(1+y)-ix]*[(x-1)+iy] / [(x-1)2+y2]
                          = {(x-y-1) -i(x(x-1)+y(y+1))} / [(x-1)2+y2]
as |w|=1 , so {(x-y-1)2 + (x(x-1)+y(y+1))2} = [(x-1)2+y2]2
on solving the above equation, we have 
2x3+2y3+2xy(x+y)-2x2-2xy+2(x+y) = 0
which is (x+y).(x2+y2-2x+2) =0  is the locus of Z
--
regards
Ramesh

						

							
1. A lot of problems in complex numbers become easy when you do not put z = x+iy. Most of the problems can be solved by geometric interpretation or by algebraic manipulation.
 
2. In the previous post it has been said that the locus is (x+y)(x2+y2-2x+2) = 0. Does this correspond to any familiar geometrical object (be careful with the answer)?
 
Shorter solution: |1-zi| = |zi-1| = |i(z+i)| = |i| |z+i| = |z+i|.
 
So the given equation is |z+i| = |z-1|.
 
This is geometrically interpreted as: the point z is equidistant from (1,0) and (0,-1).
 
The locus of points equidistant from two given points is the perpendicular bisector of the line joining the two points which is nothing but the line x+y = 0
 
So what about the term (x2+y2-2x+2) seen in the previous thread?