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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 ∈ C0.
Show that R is an equivalence relation.
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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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