A relation $R$ on the set of complex numbers is defined by $z_{1} R z_{2}$ if and only if $\frac{z_{

Question

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. Then

Professor PC Thomas Classes · Answer

We shall prove that $R$ is reflexive, symmetric and transitive\ Given $z_{1} R z_{2} \Leftrightarrow \frac{z_{1}-z_{2}}{z_{1}+z_{2}}$ is real.\ $	herefore z_{1} R z_{1} \Leftrightarrow \frac{z_{1}-z_{1}}{z_{1}+z_{1}}=\frac{0}{2 z_{1}}=0$ purely real\ $	herefore \quad$ Relation is reflexive\ Let $z_{1} R z_{2} \Leftrightarrow \frac{z_{1}-z_{2}}{z_{1}+z_{2}}$ is real\ $\Leftrightarrow-\left(\frac{z_{1}-z_{2}}{z_{1}+z_{2}}ight)$ is real\ $\Leftrightarrow \frac{z_{2}-z_{1}}{z_{2}+z_{1}}$ is real $\Leftrightarrow z_{2} R z_{1}$\ $	herefore \quad$ Relation is symmetric\ Let $z_{1} R z_{2}$ and $z_{2} R z_{3}$.\ To show that $z_{1} R z_{3}$.\ Let $z_{1}=x_{1}+i y_{1}, z_{2}=x_{2}+i y_{2}, z_{3}=x_{3}+i y_{3}$.\ $\frac{z_{1}-z_{2}}{z_{1}+z_{2}}$ is real\ $\Rightarrow \frac{\left(x_{1}-x_{2}ight)+i\left(y_{1}-y_{2}ight)}{\left(x_{1}+x_{2}ight)+i\left(y_{1}+y_{2}ight)}$ is real\ $\left[\left(x_{1}-x_{2}ight)+i\left(y_{1}-y_{2}ight)ight]\left[\left(x_{1}+x_{2}ight)ight.$\ $\Rightarrow \frac{\left.-i\left(y_{1}+y_{2}ight)ight]}{\left(x_{1}+x_{2}ight)^{2}+\left(y_{1}+y_{2}ight)^{2}}$ is real\ $\Rightarrow\left(y_{1}-y_{2}ight)\left(x_{1}+x_{2}ight)-\left(x_{1}-x_{2}ight)\left(y_{1}+y_{2}ight)=0$ 
$\Rightarrow 2 x_2 y_1-2 y_2 x_1=0 \Rightarrow \frac{x_1}{y_1}=\frac{x_2}{y_2}$  ...(i)
 Similarly $z_{2} R z_{3} \Rightarrow \frac{x_{2}}{y_{2}}=\frac{x_{3}}{y_{3}}$\ $	herefore \quad \frac{x_{1}}{y_{1}}=\frac{x_{3}}{y_{3}} \Rightarrow z_{1} R z_{3}$.\ $	herefore \quad$ Relation is Transitive\ $	herefore \quad$ It is an equivalence relation.\