The correct option is D R is not equivalance
Given, z1 R z2⟺z1−z2z1+z2 is real
(1) For reflexive
z1 R z1⟺z1−z1z1+z1=0, which is real.
Therefore, R is reflexive.
(2) For symmetric
z1 R z2⟺z1−z2z1+z2 is real
⇒−(z2−z1)z1+z2 is real
⇒z2 R z1 is real
∴z1 R z2⇒z2 R z1
Hence, R is symmetric.
(3) For transitive
Let z1=a1+ib1, z2=a2+ib2 and z3=a3+ib3
z1 R z2⟺z1−z2z1+z2
⇒z1−z2z1+z2=a1+ib1−a2−ib2a1+ib1+a2+ib2
=(a1−a2)+i(b1−b2)(a1+a2)+i(b1+b2)
On rationalizing, we get
=[(a1−a2)+i(b1−b2)][(a1+a2)−i(b1+b2)](a1+a2)2+(b1+b2)2
Since, z1−z2z1+z2 is real.
Therefore, the imaginary part of numerator must be zero.
⇒(b1−b2)(a1+a2)−(a1−a2)(b1+b2)=0
⇒2a2b1−2b2a1=0
⇒a1b1=a2b2 ...(1)
Similarly,
z2 R z3⇒a2b2=a3b3 ...(2)
From equation (1) and (2), we have
a1b1=a3b3⇒z1 R z3
Thus, z1 R z2 and z2 R z3⇒z1 R z3
Therefore, R is transitive.
Hence, R is an equivalance relation.