If the complex numbers Z1, Z2 and origin form an isosceles triangle with vertical angle (2π/3), then prove that (Z1)^(2) + (Z2)^(2) +Z1*Z2=0?

　
I'll assume you meant vertex angle = 2π/3
I'll also assume vertex is at origin (not at z₁ or z₂)

Since triangle is isosceles with vertex at origin, then |z₁| = |z₂|

Since vertex angle = 2π/3, then either z₁ or z₂ is a rotation of the other by an angle of 2π/3. Without loss of generality, we'll assume z₂ is z₁ rotated 2π/3 radians about the origin:

z₂ = z₁ (cos 2π/3 + i sin 2π/3)
z₂ = z₁ (−1/2 + i√3/2)

z₁² + z₂² + z₁z₂
= z₁² + (z₁ (−1/2 + i√3/2))² + z₁ (z₁ (−1/2 + i√3/2))
= z₁² + z₁² (−1/2 + i√3/2)² + z₁² (−1/2 + i√3/2)
= z₁² (1 + (−1/2 + i√3/2)² + (−1/2 + i√3/2))
= z₁² (1 + 1/4 − i√3/2 − 3/4 −1/2 + i√3/2)
= z₁² (0)
= 0