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Mathematics · Complex Numbers

JEE Main 2026 — 6 April, Evening Shift — Question 23

Let S={z∈C:z2+6iz−3=0}S = \left\{z \in \mathbb{C}: z^2 + \sqrt{6} iz - 3 = 0\right\} . Then ∑z∈Sz8\sum_{z \in S} z^8 is equal to:

  1. Option A:

    162162

    Correct
  2. Option B:

    184184

  3. Option C:

    262262

  4. Option D:

    324324

Answer: A

Step-by-step solution

z=−6i±−6+122z=\frac{-\sqrt{6} i \pm \sqrt{-6+12}}{2} =−6i±62=\frac{-\sqrt{6} \mathrm{i} \pm \sqrt{6}}{2} =62(±1−i)=\frac{\sqrt{6}}{2}( \pm 1-\mathrm{i}) α=−62(1+i),β=62(1−i)\alpha=\frac{-\sqrt{6}}{2}(1+\mathrm{i}), \beta=\frac{\sqrt{6}}{2}(1-\mathrm{i}) α=−62×2eiπ/4,β=+62×2−iπ/4\alpha=\frac{-\sqrt{6}}{2} \times \sqrt{2} \mathrm{e}^{\mathrm{i} \pi / 4}, \beta=+\frac{\sqrt{6}}{2} \times \sqrt{2}-\mathrm{i} \pi / 4 α=−3eiπ/4,β=+3e−iπ/4\alpha=-\sqrt{3} \mathrm{e}^{\mathrm{i} \pi / 4}, \beta=+\sqrt{3} \mathrm{e}^{-\mathrm{i} \pi / 4} α8+β8=81.ei2π+81.ei(−2π)\alpha^{8}+\beta^{8}=81 . \mathrm{e}^{\mathrm{i} 2 \pi}+81 . \mathrm{e}^{\mathrm{i}(-2 \pi)} =81(1+1)=162=81(1+1)=162

Answer key and solution verified before publishing.

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Exam
JEE Main 2026
Paper
6 April, Evening Shift
Subject
Mathematics
Chapter
Complex Numbers
Topic
Introduction to Complex Numbers
Let S = \ z in mathbb C : z 2 + √(6) iz - 3 = 0 \ . Then sum z in S z… | JEE Main 2026 PYQ with Solution · DhiX AI