Let z ∈ C be such that (z²+3 i)/(z-2+i)=2+3 i. Then the sum of all possible values of z² is :

Solution

Maths sir · 4 steps

Cross-multiply to get a quadratic in z, then use sum and product of roots to find the sum of their squares.

1. Clear the fraction. Multiply both sides by z-2+i. The constant term is (2+3i)(-2+i) = -4+2i-6i+3i² = -7-4i, so the right side becomes (2+3i)z - 7 - 4i.

   z² - (2+3i)z + (7+7i) = 0

2. Sum and product. Call the roots α and β. Then α+β = 2+3i and αβ = 7+7i. Neither root is 2-i, because putting z = 2-i into the left side gives 3-i ≠ 0. So both roots are allowed.
3. Sum of squares. α²+β² = (α+β)² - 2αβ. Here (2+3i)² = 4+12i-9 = -5+12i, so α²+β² = (-5+12i) - (14+14i) = -19-2i.
4. So. The sum of all possible values of z² is -19-2i, which is option (b).

The idea to keep. You never need the roots themselves to find α²+β². Use (α+β)² - 2αβ, and check that no root makes the original denominator zero.

Answer from the NTA final answer key. Solution by PrepWiser.