JEE Main 2026 Apr 5 Shift 2, Mathematics Q2: Introduction to Complex Numbers And Their Conjugates

JEE Main2026Apr 5, Shift 2Mathematics

Let $z_{1}, z_{2} \in C$ be the distinct solutions of the equation $z^{2} + 4 z - (1 + 12 i) = 0$ . Then $∣ z_{1} ∣^{2} + ∣ z_{2} ∣^{2}$ is equal to:

Solution

By the quadratic formula, $z = \frac{- 4 \pm 16 + 4 ( 1 + 12 i )}{2} = - 2 \pm 5 + 12 i$ .

Compute $5 + 12 i$ : if $(a + bi)^{2} = 5 + 12 i$ then $a^{2} - b^{2} = 5$ and $2 ab = 12$ . Solving gives $a = 3$ , $b = 2$ .

So $z_{1} = - 2 + (3 + 2 i) = 1 + 2 i$ and $z_{2} = - 2 - (3 + 2 i) = - 5 - 2 i$ .

$∣ z_{1} ∣^{2} + ∣ z_{2} ∣^{2} = (1 + 4) + (25 + 4) = 5 + 29 = 34$ .

Concept behind this question

Concept-wise practice with instant solutions on Fundamenthol.