Question

A relation $R$ on the set of complex numbers is defined by, $z_{1}R{z}_2\iff \frac{z_1-z_2}{z_1+z_2}$ is real, then the relation $R$ is

• A. equivalence
• B. only reflexive
• C. only transitive
• D. None of these

Answer 1

Answer: (A)

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
$\Rightarrow \left(a_1+a_2\right)\left(b_1-b_2\right)-\left(a_1-a_2\right)\left(b_1+b_2\right)=0$
$\Rightarrow 2a_2{b}_1-2b_2{a}_1=0$
$\Rightarrow \frac{a_1}{a_2}=\frac{b_1}{b_2}$
or $\frac{a_1}{b_1}=\frac{a_2}{b_2}$
and $\frac{a_2}{b_2}=\frac{a_3}{b_3}$
Similarly,
$z_{2}R{z}_3\Rightarrow \frac{a_2}{b_2}=\frac{a_3}{b_3}$
$z_{1}R{z}_2$ and $z_{2}R{z}_3$
Therefore,
$\Rightarrow \frac{a_1}{b_1}=\frac{a_2}{b_2}$
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.