If z1  and z2  are two complex numbers such that |z1−z2z1+z2|=1,  then

Clearly both z1  and z2  cannot be 0.  Supposez2≠0.  We can write |z1−z2z1+z2|=1,  as |z1/z2−1z1/z2+1|=1  or as |z1z2−1|=|z1z2+1|.This shows that z1/z2  lies on the perpendicular bisector of the segment joining A(−1   +   i0)  and B(1   +   i0)  Thus, z1/z2  lies on the imaginary axis.∴            z1/z2=ia  for some a  ∈  R.⇒          z2/z1=1/ia=−i/a⇒          z2=ikz1   for some k  ∈  RAlternative Solution: O(0),  A(z1),  B(z1+z2)  and c(z2)  are the vertices of a parallelogram.Its diagonals OB(z1+z2)  and CA(z1−z2)  will be equal if and only if OABC  is a rectangle, that is, z2/z1  is purely imaginary.