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Mathematics · Complex Numbers

JEE Main 2026 — 5 April, Evening Shift — Question 26

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

  1. Option A:

    1818

  2. Option B:

    2222

  3. Option C:

    2929

  4. Option D:

    3434

    Correct

Answer: D

Step-by-step solution

z2+4z−(1+12i)=0z^{2}+4z-(1+12 i)=0 ⇒z=−4±16+4(1+12i)2\Rightarrow z=\frac{-4 \pm \sqrt{16+4(1+12 \mathrm{i})}}{2} ⇒z=−2±5+12i\Rightarrow z=-2 \pm \sqrt{5+12 i} ⇒z=−2±(3+2i)\Rightarrow \mathrm{z}=-2 \pm(3+2 \mathrm{i}) ⇒z=1+2i,−5−2i\Rightarrow \mathrm{z}=1+2 \mathrm{i},-5-2 \mathrm{i} ∴∣z1∣2+∣Z2∣2\therefore\left|\mathrm{z}_{1}\right|^{2}+\left|\mathrm{Z}_{2}\right|^{2} =5+29=5+29 =34=34

Answer key and solution verified before publishing.

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Exam
JEE Main 2026
Paper
5 April, Evening Shift
Subject
Mathematics
Chapter
Complex Numbers
Topic
Introduction to Complex Numbers
Let z 1, z 2 ∈ C be the distinct solutions of the equation z² + 4z … | JEE Main 2026 PYQ with Solution · DhiX AI