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  4. A relation R on the set of complex numbers is defined by, z1 R z2 Leftrightarrow (z1-z2/z1+z2) is real, then the relation R is

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Q. A relation $R$ on the set of complex numbers is defined by, $z_{1} R z_{2} \Leftrightarrow \frac{z_{1}-z_{2}}{z_{1}+z_{2}}$ is real, then the relation $R$ is

Relations and Functions - Part 2

Solution:

Since $\frac{z_{1}-z_{1}}{z_{1}+z_{1}}=0$, which is real $\forall z_{1} \in C$,
therefore $R$ is reflexive.
For $z_{1}, z_{2} \in C, z_{1} R z_{2}$
$ \Rightarrow \frac{z_{1}-z_{2}}{z_{1}+z_{2}}$ is real
$\Rightarrow -\left(\frac{z_{1}-z_{2}}{z_{1}+z_{2}}\right)$ is real
$\Rightarrow \left(\frac{z_{2}-z_{1}}{z_{2}+z_{1}}\right)$ is real
$\Rightarrow z_{2} R z_{1} .$
For
$z _{1}, z_{2}, z_{3} \in C$
let $z_{1}=a_{1}+i b_{1}, z_{2}=a_{2}+i b_{2}$
and $z_{3}=a_{3}+i b_{3}$.
Now, $z_{1} R z_{2} \Rightarrow \frac{z_{1}-z_{2}}{z_{1}+z_{2}}$ is real
$\Rightarrow \frac{\left(a_{1}-a_{2}\right)+i\left(b_{1}-b_{2}\right)}{\left(a_{1}+a_{2}\right)+i\left(b_{1}+b_{2}\right)}$ is real
$\Rightarrow \frac{\left(a_{1}-a_{2}\right)+i\left(b_{1}-b_{2}\right)}{\left(a_{1}+a_{2}\right)+i\left(b_{1}+b_{2}\right)} \times \frac{\left(a_{1}+a_{2}\right)-i\left(b_{1}+b_{2}\right)}{\left(a_{1}+a_{2}\right)-i\left(b_{1}+b_{2}\right)}$ is real
$\Rightarrow \frac{\left(a_{1}-a_{2}\right)\left(a_{1}+a_{2}\right)+\left(b_{1}-b_{2}\right)\left(b_{1}+b_{2}\right)+i\left[\left(b_{1}-b_{2}\right)\left(a_{1}+a_{2}\right)-\left(b_{1}+b_{2}\right)\left(a_{1}-a_{2}\right)\right]}{\left(a_{1}+a_{2}\right)^{2}+\left(b_{1}+b_{2}\right)^{2}}$ is real
$\Rightarrow \left(a_{1}+a_{2}\right)\left(b_{1}-b_{2}\right)-\left(a_{1}-a_{2}\right)\left(b_{1}+b_{2}\right)=0$
$\Rightarrow 2 a_{2} b_{1}-2 b_{2} a_{1}=0$
$\Rightarrow \frac{a_{1}}{a_{2}}=\frac{b_{1}}{b_{2}} $
or $\frac{a_{1}}{b_{1}}=\frac{a_{2}}{b_{2}}$
and $\frac{a_{2}}{b_{2}}=\frac{a_{3}}{b_{3}}$
Similarly,
$z_{2} \,R \,z_{3} \Rightarrow \frac{a_{2}}{b_{2}}=\frac{a_{3}}{b_{3}}$
Therefore,
$z_{1} \,R\, z_{2} $ and $z_{2} \,R\, z_{3}$
$\Rightarrow \frac{a_{1}}{b_{1}}=\frac{a_{2}}{b_{2}} $
and $ \frac{a_{2}}{b_{2}}=\frac{a_{3}}{b_{3}} $
$\Rightarrow \frac{a_{1}}{b_{1}}=\frac{a_{3}}{b_{3}}$
$\Rightarrow z_{1} \,R \, z_{3}$
$\therefore R$ is transitive.
Hence $R$ is an equivalence relation.