Complex Numbers Ques 50
A relation $R$ on the set of complex numbers is defined by $z _1 R z _2$, if and only if $\frac{z _1-z _2}{z _1+z _2}$ is real.
Show that $R$ is an equivalence relation.
(1982, 2M)
Show Answer
Solution:
- Here, $z _1 R z _2 \Leftrightarrow \frac{z _1-z _2}{z _1+z _2}$ is real
(i) Reflexive $z _1 R z _1 \Leftrightarrow \frac{z _1-z _1}{z _1+z _2}=0 \quad$ [purely real]
$ \therefore \quad z _1 R z _1 \text { is reflexive. } $
(ii) Symmetric $z _1 R z _2 \Leftrightarrow \frac{z _1-z _2}{z _1+z _2}$ is real
$ \begin{aligned} & \Rightarrow \quad \frac{-\left(z _2-z _1\right)}{z _1+z _2} \text { is real } \Rightarrow z _2 R z _1 \\ & \therefore \quad z _1 R z _2 \Rightarrow z _2 R z _1 \end{aligned} $
Therefore, it is symmetric.
(iii) Transitive $z _1 R z _2$
$\Rightarrow \frac{z _1-z _2}{z _1+z _2} \text { is real } $
$\text { and } z _2 R z _3 $
$\Rightarrow \frac{z _2-z _3}{z _2+z _3} \text { is real }$
Here, let $z _1=x _1+i y _1, z _2=x _2+i y _2$ and $z _3=x _3+i y _3$
$\therefore \frac{z _1-z _2}{z _1+z _2}$ is real $\Rightarrow \frac{\left(x _1-x _2\right)+i\left(y _1-y _2\right)}{\left(x _1+x _2\right)+i\left(y _1+y _2\right)}$ is real
$\Rightarrow \quad \frac{\{\left(x _1-x _2\right)+i\left(y _1-y _2\right) \}\{\left(x _1+x _2\right)-i\left(y _1+y _2\right) \}}{\left(x _1+x _2\right)^{2}+\left(y _1+y _2\right)^{2}}$
$\Rightarrow \quad\left(y _1-y _2\right)\left(x _1+x _2\right)-\left(x _1-x _2\right)\left(y _1+y _2\right)=0$
$\Rightarrow \quad 2 x _2 y _1-2 y _2 x _1=0$
$\Rightarrow \quad \frac{x _1}{y _1}=\frac{x _2}{y _2}$ $\quad$ …….(i)
Similarly,
$z _2 R z _3$
$\frac{x _2}{y _2}=\frac{x _3}{y _3}$ $\quad$ …….(ii)
From Eqs. (i) and (ii), we have $\frac{x _1}{y _1}=\frac{x _3}{y _3} \Rightarrow z _1 R z _3$
Thus, $z _1 R z _2$ and $z _2 R z _3 \Rightarrow z _1 R z _3$.
Hence, $R$ is an equivalence relation.
BETA
