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

Options

A. R is reflexive
B. R is symmetric
C. R is transitive
D. R is not equivalence

Answer

D

Explanation

`Z_(1)RZ_(1)implies(Z_(1)-Z_(1))/(Z_(1)+Z_(1))=0`
` :. R ` is Reflexive because 0 is real.
`Z_(1)RZ_(2)implies(Z_(1)-Z_(2))/(Z_(1)+Z_(2))` is real `implies -((Z_(1)-Z_(2))/(Z_(1)+Z_(2)))` is real
` :. R` is symmetric
`Z_(1)=a_(1)+ib_(1), Z_(2)=a_(2)+ib_(2)`
`Z_(1)RZ_(2)implies(Z_(1)-Z_(2))/(Z_(1)+Z_(2))` is real
`implies (a_(1)+a_(2))(b_(1)-b_(2))-(a_(1)-a_(2))(b_(1)+b_(2))=0`
`implies 2a_(1)b_(1)-2b_(1)a_(1)=0`
`implies (a_(1))/(b_(1))=(a_(2))/(b_(2))`
` :. Z_(2)RZ_(3)implies(a_(2))/(b_(2))=(a_(3))/(b_(3))`
` :. Z_(1)RZ_(3)implies R` is transitive.