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Question
A relation R defined on the set of non-zero complex numbers as z1Rz2 iff z1−z2z1+z2 is real, then R is
A relation R defined on the set of non-zero complex numbers as z1Rz2 iff z1−z2z1+z2 is real, then R is
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Solution
The correct option is D an equivalence relation
Given (z1,z2)∈R⇒z1−z2z1+z2 is real, z1,z2∈C−{0}
∵z1−z2z1+z2=02z1=0 for all non zero z1
⇒(z1,z1)∈R ∀ z1∈C−{0}
So, R is reflexive.
Let (z1,z2)∈R⇒z1−z2z1+z2 is real,
Let, z1−z2z1+z2=k (k∈R)
⇒z2−z1z2+z1=−k (−k∈R)
⇒(z2,z1)∈R
If (z1,z2)∈R⇒(z2,z1)∈R
so, R is symmetric
Let (z1,z2) and (z2,z3)∈R
⇒z1−z2z1+z2=k1 and z2−z3z2+z3=k2
⇒2z12z2=−(k1+1k1−1) and 2z22z3=−(k2+1k2−1)
⇒z1=−(k1+1k1−1)z2 and
⇒z3=−(k2−1k2+1)z2
Consider, z1−z3z1+z3=−(k1+1k1−1)z2+(k2−1k2+1)z2−(k1+1k1−1)z2−(k2−1k2+1)z2
=k1+k2k1k2+1∈R
⇒(z1,z3)∈R
∴R is transitive.
So,R is an equivalence relation
Given (z1,z2)∈R⇒z1−z2z1+z2 is real, z1,z2∈C−{0}
∵z1−z2z1+z2=02z1=0 for all non zero z1
⇒(z1,z1)∈R ∀ z1∈C−{0}
So, R is reflexive.
Let (z1,z2)∈R⇒z1−z2z1+z2 is real,
Let, z1−z2z1+z2=k (k∈R)
⇒z2−z1z2+z1=−k (−k∈R)
⇒(z2,z1)∈R
If (z1,z2)∈R⇒(z2,z1)∈R
so, R is symmetric
Let (z1,z2) and (z2,z3)∈R
⇒z1−z2z1+z2=k1 and z2−z3z2+z3=k2
⇒2z12z2=−(k1+1k1−1) and 2z22z3=−(k2+1k2−1)
⇒z1=−(k1+1k1−1)z2 and
⇒z3=−(k2−1k2+1)z2
Consider, z1−z3z1+z3=−(k1+1k1−1)z2+(k2−1k2+1)z2−(k1+1k1−1)z2−(k2−1k2+1)z2
=k1+k2k1k2+1∈R
⇒(z1,z3)∈R
∴R is transitive.
So,R is an equivalence relation
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