Question

A relation $R$ in the set of non zero complex numbers is defined by $z_{1}R{z}_2\iff \frac{z_1-z_2}{z_1+z_2}$ is real, then $R$ is not

• A. Reflexive
• B. Symmetric
• C. Transitive
• D. Equivalence

Answer 1

Answer: (D)

Equivalence

Let $Z_1\in C-\left\{0\right\}$

$\therefore \frac{Z_1-Z_1}{Z_1+Z_1}=0=0+i0$, which is real

(i) So $Z_{1}R{Z}_1\Rightarrow R$ is reflexive

Hence (a) is correct

(ii) Let $Z_{1}R{Z}_2\Rightarrow \frac{Z_1-Z_2}{Z_2+Z_1}$ is real $\Rightarrow \frac{Z_1-Z_2}{Z_1+Z_2}$ is real
$\Rightarrow \frac{Z_2-Z_1}{Z_2+Z_1}$ is real

$\therefore Z_{2}R{Z}_1\Rightarrow R$ is symmetric

(iii) Let $Z_{1}R{Z}_2$ and $Z_{2}R{Z}_3$

$\therefore \frac{Z_1-Z_2}{Z_1+Z_2}$ and $\frac{Z_2-Z_3}{Z_2+Z_3}$ are real

Let $Z_1=x_1+i{y}_1,$

$Z_2=x_2+i{y}_2,$

$Z_3=x_3+i{y}_3,$

$\therefore \frac{Z_1-Z_2}{Z_1+Z_2}$ is real $\Rightarrow I_m\frac{Z_1-Z_2}{Z_1+Z_2}=0$

$\Rightarrow \frac{x_1}{y_1}=\frac{x_2}{y_2}...\left(A\right)$

Similarly we have $\frac{x_2}{y_2}=\frac{x_3}{y_3}...\left(B\right)$
$\therefore$ From $A$ and $B$

$\Rightarrow \frac{x_1}{y_1}=\frac{x_3}{y_3}$ $\Rightarrow x_1{y}_3=x_3{y}_1$

$\therefore \frac{Z_1-Z_3}{Z_1+Z_3}$ is real

$\therefore Z_{1}R{Z}_3$

$\therefore R$ is transitive Finally R is an equivalence relation.