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Equivalence Relation

A relation R ...

Question

A relation $R$ on the set of non-zero complex numbers defined by $z_{1} R z_{2} ⟺ \frac{z_{1} - z_{2}}{z_{1} + z_{2}}$ is real, then which of the following is not true?

R

is reflexive

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R

is symmetric

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R

is transitive

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R

is not equivalance

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Solution

The correct option is D $R$ is not equivalance
Given, $z_{1} R z_{2} ⟺ \frac{z_{1} - z_{2}}{z_{1} + z_{2}}$ is real

$(1)$ For reflexive
$z_{1} R z_{1} ⟺ \frac{z_{1} - z_{1}}{z_{1} + z_{1}} = 0,$ which is real.
Therefore, $R$ is reflexive.

$(2)$ For symmetric
$z_{1} R z_{2} ⟺ \frac{z_{1} - z_{2}}{z_{1} + z_{2}}$ is real
$\Rightarrow \frac{- (z_{2} - z_{1})}{z_{1} + z_{2}}$ is real
$\Rightarrow z_{2} R z_{1}$ is real
$∴ z_{1} R z_{2} \Rightarrow z_{2} R z_{1}$
Hence, $R$ is symmetric.

$(3)$ For transitive
Let $z_{1} = a_{1} + i b_{1}$ , $z_{2} = a_{2} + i b_{2}$ and $z_{3} = a_{3} + i b_{3}$

$z_{1} R z_{2} ⟺ \frac{z_{1} - z_{2}}{z_{1} + z_{2}}$
$\Rightarrow \frac{z_{1} - z_{2}}{z_{1} + z_{2}} = \frac{a_{1} + i b_{1} - a_{2} - i b_{2}}{a_{1} + i b_{1} + a_{2} + i b_{2}}$
$= \frac{(a_{1} - a_{2}) + i (b_{1} - b_{2})}{(a_{1} + a_{2}) + i (b_{1} + b_{2})}$
On rationalizing, we get
$= \frac{[(a_{1} - a_{2}) + i (b_{1} - b_{2})] [(a_{1} + a_{2}) - i (b_{1} + b_{2})]}{(a_{1} + a_{2})^{2} + (b_{1} + b_{2})^{2}}$
Since, $\frac{z_{1} - z_{2}}{z_{1} + z_{2}}$ is real.
Therefore, the imaginary part of numerator must be zero.
$\Rightarrow (b_{1} - b_{2}) (a_{1} + a_{2}) - (a_{1} - a_{2}) (b_{1} + b_{2}) = 0$
$\Rightarrow 2 a_{2} b_{1} - 2 b_{2} a_{1} = 0$
$\Rightarrow \frac{a_{1}}{b_{1}} = \frac{a_{2}}{b_{2}} . . . (1)$

Similarly,
$z_{2} R z_{3} \Rightarrow \frac{a_{2}}{b_{2}} = \frac{a_{3}}{b_{3}} . . . (2)$
From equation $(1)$ and $(2),$ we have
$\frac{a_{1}}{b_{1}} = \frac{a_{3}}{b_{3}} \Rightarrow z_{1} R z_{3}$

Thus, $z_{1} R z_{2}$ and $z_{2} R z_{3} \Rightarrow z_{1} R z_{3}$
Therefore, $R$ is transitive.

Hence, $R$ is an equivalance relation.

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