If the triangle fromed by complex numbers `z_(1), z_(2)` and `z_(3)` is equilateral then prove that `(z_(2) + z_(3) -2z_(1))/(z_(3) - z_(2))` is purely imaginary number

Explanation

We have to prove that `(z_(2)+z_(3)-2z_(1))/(z_(3)-z_(2))` is purely imaginary number.
Then `arg((z_(2)+z_(3)-2z_(1))/(z_(3)-z_(2)))=+-(pi)/(2)`
or `arg2(((z_(2)+z_(3))/(2)-z_(1))/(z_(3)-z_(2)))=+-(pi)/(2)`
or `arg2+arg(((z_(2)+z_(3))/(2)-z_(1))/(z_(3)-z_(2)))=+-(pi)/(2)`
or `arg(((z_(2)+z_(3))/(2)-z_(1))/(z_(3)-z_(2)))=+-(pi)/(2)" "["as arg"(2)=0]`

Now, complex number `(z_(2)+z_(3))/(2)` represents the mid-point of BC, i.e., point D.
Since triangle ABC is equilateral, `AD_|_BC.`
`:." "arg(((z_(2)+z_(3))/(2)-z_(1))/(z_(3)-z_(2)))=+-(pi)/(2)`