Statement 1: A relation R on the set of complex numbers defined by

$Z_{1}R{Z}_2\iff \frac{Z_1-Z_2}{Z_1+Z_2}$  is real then R is equivalence relation.

Statement 2: The ‘R’ relation is reflexive, and symmetric but not transitive

$Z_{1}R{Z}_1\Rightarrow \frac{Z_1-Z_1}{Z_1+Z_1}=0$ 

$\therefore$  R is Reflexive because 0 is real.

$Z_{1}R{Z}_2=\frac{Z_1-Z_2}{Z_1+Z_2}$  is real $=-\left(\frac{Z_1-Z_2}{Z_1+Z_2}\right)$  real

$\therefore$    R is symmetric.

$\frac{R\text{is}\text{transtive}}{Z_1=a_1+i{b}_1{Z}_2=a_2+i{b}_2}$

$Z_{1}R{Z}_2\Rightarrow \frac{Z_1-Z_2}{Z_1+Z_2}$    is real $\Rightarrow \left(a_1+a_2\right)\left(b_1-b_2\right)-\left(a_1-b_2\right)\left(b_1+b_2\right)=0$

$\therefore 2a_2{b}_1-2b_2{a}_1=0\Rightarrow \frac{a_1}{b_1}=\frac{a_2}{b_2}$

$Z_{2}R{Z}_3\Rightarrow \frac{a_2}{b_2}=\frac{a_3}{b_3}$

$\therefore Z_{1}R{Z}_3\Rightarrow R$  is transitive