wiz-icon
MyQuestionIcon
MyQuestionIcon
1
You visited us 1 times! Enjoying our articles? Unlock Full Access!
Question

A relation R on the set of non-zero complex numbers defined by z1 R z2⟺z1−z2z1+z2 is real, then which of the following is not true?

A
R is reflexive
No worries! We‘ve got your back. Try BYJU‘S free classes today!
B
R is symmetric
No worries! We‘ve got your back. Try BYJU‘S free classes today!
C
R is transitive
No worries! We‘ve got your back. Try BYJU‘S free classes today!
D
R is not equivalance
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
Open in App
Solution

The correct option is D R is not equivalance
Given, z1 R z2z1z2z1+z2 is real

(1) For reflexive
z1 R z1z1z1z1+z1=0, which is real.
Therefore, R is reflexive.

(2) For symmetric
z1 R z2z1z2z1+z2 is real
(z2z1)z1+z2 is real
z2 R z1 is real
z1 R z2z2 R z1
Hence, R is symmetric.

(3) For transitive
Let z1=a1+ib1, z2=a2+ib2 and z3=a3+ib3

z1 R z2z1z2z1+z2
z1z2z1+z2=a1+ib1a2ib2a1+ib1+a2+ib2
=(a1a2)+i(b1b2)(a1+a2)+i(b1+b2)
On rationalizing, we get
=[(a1a2)+i(b1b2)][(a1+a2)i(b1+b2)](a1+a2)2+(b1+b2)2
Since, z1z2z1+z2 is real.
Therefore, the imaginary part of numerator must be zero.
(b1b2)(a1+a2)(a1a2)(b1+b2)=0
2a2b12b2a1=0
a1b1=a2b2 ...(1)

Similarly,
z2 R z3a2b2=a3b3 ...(2)
From equation (1) and (2), we have
a1b1=a3b3z1 R z3

Thus, z1 R z2 and z2 R z3z1 R z3
Therefore, R is transitive.

Hence, R is an equivalance relation.



flag
Suggest Corrections
thumbs-up
0
Join BYJU'S Learning Program
similar_icon
Related Videos
thumbnail
lock
MATHEMATICS
Watch in App
Join BYJU'S Learning Program
CrossIcon