If |z|=1  and z≠±1 , then all the value of z1−z2  lie on :

Let z=cosθ+isinθ ⇒      z1−z2=cosθ+isinθ1−(cos2θ+isin2θ)=cosθ+isinθ2sin2θ−2isinθcosθ=cosθ+isinθ−2isinθ(cosθ+isinθ)=i2sinθHence, z1−z2  lies on the imaginary axis, i.e., x=0Alternative MethodLet E=z1−z2=zzz¯−z2=1z¯−1  which is imaginary