A relation R on the set of complex numbers is defined by `z_1 R z_2` if and oly if `(z_1-z_2)/(z_1+z_2)` is real Show that R is an equivalence relation.

Explanation

`z_(1)Rz_(1)implies(z_(1)-z_(1))/(z_(1)+z_(1)),AAz_(1)inCimplies0` is real
`therefore R` is reflexive.
(ii) `z_(1)Rz_(2)implies(z_(1)-z_(2))/(z_(1)+z_(2))` is real
`implies -((z_(2)-z_(1))/(z_(1)+z_(2)))` is real `implies ((z_(2)-z_(1))/(z_(1)+z_(2)))` is real
`impliesz_(2)Rz_(1),AAz_(1),z_(2)inC`
`therefore` R is symmetric.
(iii) `because z_(1)Rz_(2)implies(z_(1)-z_(2))/(z_(1)+z_(2))` is real
`implies ((bar(z_(1)-z_(2)))/(z_(1)+z_(2)))=-((z_(1)-z_(2))/(z_(1)+z_(2)))`
`implies ((barz_(1)-barz_(2))/(barz_(1)+barz_(2)))+((z_(1)-z_(2))/(z_(1)+z_(2)))=0`
`implies 2(z_(1)barz_(1)-z_(2)barz_(2))=0implies|z_(1)|^(2)=|z_(2)|^(2)" ... (i)"`
Similarly, `z_(2)Rz_(2)implies|z_(2)|^(2)=|z_(3)|^(2)" ... (ii)"`
From Eqs. (i) and (ii), we get
`z_(1)Rz_(2),z_(2)Rz_(3)`
`implies |z_(1)|^(2)=|z_(3)|^(2)`
`implies z_(1)Rz_(3)`
`therefore` R is transitive.
Hence, R is an equivalence relation.