JEE Main 2026 · 5 Apr · Shift 2 · Mathematics Q2

MathematicsSingle correct+4 / -1Question 2 of 75

Let $z_1,z_2\in\mathbb C$ be the distinct solutions of the equation $z^2+4z-(1+12i)=0$ . Then $|z_1|^2+|z_2|^2$ is equal to :

Answer and worked solution

Official answer: D

Roots.
$z=\frac{-4\pm\sqrt{16+4+48i}}2=-2\pm\sqrt{5+12i}=-2\pm(3+2i),$
so $z=1+2i$ or $z=-5-2i$ .

Sum: $5+29=34$ .

(D)

Topics:Algebra of complex numbers Modulus and argument

JEE Main 2026 · 5 Apr · Shift 2, Mathematics question 2