Question

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

`z_1 R  z_2  ⇔ (z_1 -z_2)/(z_1 + z_2) ` is real for all z₁, z₂ ∈ C_{0 .}

_{Show that R is an equivalence relation.}

Answer 1

(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 (z_1, 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.