SOLUTION 1:

Present complex numbers z₁, z₂, z₃ by three points A, B, C.

We advance the coordinator by vector OA to have the new coordinator AXY where A is the origin.

Let A(0,0) ; B(a,b) ; C(x,y) in the new coordinator AXY.

(z₂ - z₁) / ( z₃ - z₁) = cos(pi/3) + i sin(pi/3)

--> (a + bi) / (x +yi) = 1/2 + √3/2 i

--> a + bi = (1/2 + √3/2 i)(x + yi)

= 1/2x + √3/2 ix + 1/2yi - √3/2y

= (1/2x - √3/2y ) + (√3/2x + 1/2y) i

--> a = 1/2x - √3/2y

b = √3/2x + 1/2y

--> B(1/2x - √3/2y , √3/2x + 1/2y)

Now we will calculate AB, AC, BC

AB² = a² + b² = (1/2x - √3/2y )² + (√3/2x + 1/2y)² = x² + y²

BC² = (1/2x + √3/2y)² + (-√3/2x + 1/2y)² = x² + y²

AC² = x² + y²

--> AB = BC =AC --> Triangle ABC is equilateral.

SOLUTION 2:

--> AB = AC

Similarly, we can prove AB = BC

--> AB = AC = BC

--> Triangle ABC is equilateral