Let O be the origin, z₁ = a + 3i (where a > 0 ), and z₂ be a complex number such that |z₂| = 2|z₁| and (z₂) - (z₁) = 3 . If the area of the triangle formed by O , z₁ , and z₂ is 6 3 , then the value of a^2 + AB^2 , where AB is the distance between z₁ and z₂ , is

Step-by-step solution

The area of the triangle formed by O , z₁ , and z₂ is given by: Area = 1 2 |z₁||z₂| ( z₁ O z₂) Given |z₂| = 2|z₁| and the angle between them is 3 , we have: 1 2 |z₁|(2|z₁|) ( 3 ) = 6 3 |z₁|^2 ( 3 2 ) = 6 3 |z₁|^2 = 12 Since z₁ = a + 3i , its modulus squared is a^2 + 3^2 = a^2 + 9 . Thus, a^2 + 9 = 12 a^2 = 3 . The distance AB between z₁ and z₂ can be found using the cosine rule in O z₁ z₂ : AB^2 = |z₁|^2 + |z₂|^2 - 2|z₁||z₂| ( 3 ) AB^2 = |z₁|^2 + 4|z₁|^2 - 2|z₁|(2|z₁|) ( 1 2 ) AB^2 = 5|z₁|^2 - 2|z₁|^2 = 3|z₁|^2 Sub