Complex Numbers 2 Question 43
44. 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)
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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 $z _1 R z _2$
$$ \begin{array}{ll} \Rightarrow & \frac{z _1-z _2}{z _1+z _2} \text { is real } \\ \text { and } & z _2 R z _3 \\ \Rightarrow & \frac{z _2-z _3}{z _2+z _3} \text { is real } \end{array} $$
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
$\Rightarrow \quad \frac{{\left(x _1-x _2\right)+i\left(y _1-y _2\right) }{\left(x _1+x _2\right)-i\left(y _1+y _2\right) }}{\left(x _1+x _2\right)^{2}+\left(y _1+y _2\right)^{2}}$
$\Rightarrow \quad\left(y _1-y _2\right)\left(x _1+x _2\right)-\left(x _1-x _2\right)\left(y _1+y _2\right)=0$
$\Rightarrow \quad 2 x _2 y _1-2 y _2 x _1=0$
$\Rightarrow \quad \frac{x _1}{y _1}=\frac{x _2}{y _2}$
Similarly,
$$ \begin{gathered} z _2 R z _3 \\ \frac{x _2}{y _2}=\frac{x _3}{y _3} \end{gathered} $$
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.
