Let the triangle be △ABC. A,B,C are vertices represented by z1,z2 andz3 respectively. Since the triangle is equilateral,
AB=BC=CA and
∠A=∠B=∠C=π3
(z1−z2)(z2−z1)=(z3−z1)(z3−z2)
z1z2−z21−z22+z2z1=z23−z2z3−z1z3+z1z2
z21+z22+z23=z1z2+z2z3+z3z1 ----------(1)
z21+z22+z23−z1z2−z2z3−z3z1=0
2(z21+z22+z23−z1z2−z2z3−z3z1)=0 [ Multiplying equation by 2 ]
(z1−z2)2+(z2−z3)2+(z3−z1)2=0 ---------------(2)
Also
z1−z2z3−z2=z3−z1z2−z1=z2−z3z1−z3
From the above equation we get the following equations
(z3−z1)(z3−z2)=(z2−z1)(z1−z2)
(z1−z2)(z1−z3)=(z2−z3)(z3−z2)
(z2−z1)(z2−z3)=(z3−z1)(z1−z3)
Adding these three equations,
(z3−z1)(z3−z2)+(z1−z2)(z1−z3)+(z2−z1)(z2−z3)=−(z1−z2)2−(z2−z3)2−(z3−z1)2=0 (From (2))
Divide by (z1−z2)(z2−z3)(z3−z1)
−1(z1−z2)+−1(z2−z3)+−1(z3−z1)=0
1(z1−z2)+1(z2−z3)+1(z3−z1)=0 -----(3)