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

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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?