English

Let C be the set of all complex numbers and C0 be the set of all no-zero complex numbers. Let a relation R on C0 be defined as z1R z2 ⇔z1-z2z1+z2 is real for all z1, z2 ∈ C0. - Mathematics

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Question

Let C be the set of all complex numbers and Cbe the set of all no-zero complex numbers. Let a relation R on Cbe defined as

`z_1 R  z_2  ⇔ (z_1 -z_2)/(z_1 + z_2)` is real for all z1, z2 ∈ C0.

Show that R is an equivalence relation.

Sum
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Solution

(i) Test for reflexivity: 

Since, `(z_1 -z_1)/(z_1 + z_1)`= 0, which is a real number.

So, (z1, z1) ∈ R

Hence, R is relexive relation.

(ii) Test for symmetric:

Let ( z1, z2 ) ∈ R.

Then  `(z_1 -z_2)/(z_1 + z_2) =x`, where x is real

⇒ − `(z_1 -z_2)/(z_1 + z_2) = -x `

⇒ `(z_2 -z_1)/(z_2 + z_1)` = −x, is also a real number

So, (z2, z1) ∈ R

Hence, R is symmetric relation. 

(iii) Test for transivity:

Let (z1, z2) ∈ R and  (z2, z3) ∈ R.

Then, 

`(z_1 -z_2)/(z_1 + z_2) x,`where x is a real number.

⇒ z1 − z2 = xz1 + xz2

⇒ z1 − xz1 = z2 + xz2

⇒ z1 (1 − x) = z2 (1 + x)

⇒ `z_1/z_2 = (1 +x )/(1-x)`    ...(1)

Also, 

`(z_2 -z_3)/(z_2+ z_3)`= y, where y is a real number.

⇒ z2 − z3= yz2 + Yz3

⇒z2 − yz2 = z3 + yz3

⇒ z2 (1 − y) = z3 (1 + y)

⇒ `z_2/z_3 = ((1+y))/((1 -y))`    ...(2)

Dividing (1) and (2), we get

`z_1/z_3= ((1+x)/(1-x)) xx ((1-y)/(1 +y))` = z, where z is a real number.

`(z_1 -z_3)/(z_1+ z_3) = (z-1 ) /(z+1),  which is real `

⇒ (z1, z3) ∈ R

Hence, R is transitive relation

From (i), (ii), and (iii),

R is an equivalenve relation.

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Chapter 1: Relations - Exercise 1.2 [Page 27]

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RD Sharma Mathematics [English] Class 12
Chapter 1 Relations
Exercise 1.2 | Q 17 | Page 27

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