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JEE Main 20265 AprilEvening Shift
Exam
JEE Main
Year
2026
Date
5 April
Shift
Evening Shift
Subject
Mathematics
Chapter
Complex Number

JEE Main 2026 · 5 April · Evening Shift · Mathematics · Complex Number

Let z_1, z_2 ∈ ℂ be the distinct solutions of the equation z^2 + 4z...

Let z1,z2∈Cz_1, z_2 \in \mathbb{C} be the distinct solutions of the equation z2+4z−(1+12i)=0z^2 + 4z - (1 + 12i) = 0. Then ∣z1∣2+∣z2∣2|z_1|^2 + |z_2|^2 is equal to :

Options

  1. A1818
  2. B2222
  3. C2929
  4. D3434

Correct answer

D. 3434

Step-by-step solution

Given equation is z2+4z−(1+12i)=0z^2 + 4z - (1 + 12i) = 0

Let the roots be z1z_1 and z2z_2.

Sum of roots: z1+z2=−4z_1 + z_2 = -4

Product of roots: z1z2=−(1+12i)z_1 z_2 = -(1 + 12i)

We know that (z1−z2)2=(z1+z2)2−4z1z2(z_1 - z_2)^2 = (z_1 + z_2)^2 - 4z_1 z_2

(z1−z2)2=(−4)2−4(−(1+12i))=16+4+48i=20+48i(z_1 - z_2)^2 = (-4)^2 - 4(-(1 + 12i)) = 16 + 4 + 48i = 20 + 48i

Taking modulus on both sides:

∣z1−z2∣2=∣20+48i∣=202+482=400+2304=2704=52|z_1 - z_2|^2 = |20 + 48i| = \sqrt{20^2 + 48^2} = \sqrt{400 + 2304} = \sqrt{2704} = 52

Using the parallelogram law for complex numbers:

∣z1+z2∣2+∣z1−z2∣2=2(∣z1∣2+∣z2∣2)|z_1 + z_2|^2 + |z_1 - z_2|^2 = 2(|z_1|^2 + |z_2|^2)

∣−4∣2+52=2(∣z1∣2+∣z2∣2)|-4|^2 + 52 = 2(|z_1|^2 + |z_2|^2)

16+52=2(∣z1∣2+∣z2∣2)16 + 52 = 2(|z_1|^2 + |z_2|^2)

68=2(∣z1∣2+∣z2∣2)68 = 2(|z_1|^2 + |z_2|^2)

∣z1∣2+∣z2∣2=34|z_1|^2 + |z_2|^2 = 34

Answer: 3434
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