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If the complex no.s Z1, Z2 and the origin form an isosceles triangle with vertical angle 120* then proove that z12 +z22+z1z2=0

Naman nema , 6 Years ago
Grade 11
anser 1 Answers
Arun

Last Activity: 6 Years ago

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

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