If `z_1, z_2, z_3` are complex numbers such that `(2//z_1)=(1//z_2)+(1//z_3),` then show that the points represented by `z_1, z_2()_, z_3` lie one a circle passing through e origin.

Explanation

`(2)/(z_(1))=(1)/(z_(2))+(1)/(z_(3))`
or `(1)/(z_(1))-(1)/(z_(2))=(1)/(z_(3))-(1)/(z_(1))`
or `(z_(2)-z_(1))/(z_(1)z_(2))=(z_(1)-z_(3))/(z_(3)z_(1))`
or `(z_(2)-z_(1))/(z_(3)-z_(1))=-(z_(2))/(z_(3))`
`impliesarg((z_(2)-z_(1))/(z_(3)-z_(1)))=arg(-(z_(2))/(z_(3)))`
or `arg((z_(2)-z_(1))/(z_(3)-z_(1)))=pi+arg((z_(2))/(z_(3)))`
or `arg((z_(2)-z_(1))/(z_(3)-z_(1)))=pi-arg((z_(2))/(z_(3)))`
`impliesalpha=pi-beta`
or `alpha+beta=pi`
Hence, points `z_(1),z_(2),z_(3)` and 0 are concyclic.