JEE Main 2025 Apr 3 Shift 1, Mathematics Q14: Quadratic Equations

JEE Main2025Apr 3, Shift 1Mathematics

Let $z \in C$ be such that $\frac{z ^{2} + 3 i}{z - 2 + i} = 2 + 3 i$ . Then the sum of all possible values of $z^{2}$ is

Solution

Cross-multiplying: $z^{2} + 3 i = (2 + 3 i) (z - 2 + i) .$

Expand $(2 + 3 i) (- 2 + i) = - 4 + 2 i - 6 i + 3 i^{2} = - 7 - 4 i .$ So the RHS is $(2 + 3 i) z - 7 - 4 i .$

Rearrange to a quadratic in $z$ :

$z^{2} - (2 + 3 i) z + (7 + 7 i) = 0.$

Let the two roots be $z_{1}, z_{2} .$ By Vieta: $z_{1} + z_{2} = 2 + 3 i, z_{1} z_{2} = 7 + 7 i .$

Then

$z_{1}^{2} + z_{2}^{2} = (z_{1} + z_{2})^{2} - 2 z_{1} z_{2} = (2 + 3 i)^{2} - 2 (7 + 7 i) .$

$(2 + 3 i)^{2} = 4 + 12 i + 9 i^{2} = - 5 + 12 i .$

$z_{1}^{2} + z_{2}^{2} = - 5 + 12 i - 14 - 14 i = - 19 - 2 i .$

Concept behind this question

Concept-wise practice with instant solutions on Fundamenthol.