Let α=frac -1+i √(3) 2 and β=frac -1-i √(3) 2 , i=√(-1) . If (7-7 α+9 β) 20 +(9+7 α-7 β) 20 +(-7+9 α+7 β) 20 +(14+7 α+7 β) 20 = m 10 , then m is \ \ \ \ .

Mathematics · Complex Numbers

JEE Main 2026 — 22 January, Morning Shift — Question 20

Let $\alpha=\frac{-1+i \sqrt{3}}{2}$ and $\beta=\frac{-1-i \sqrt{3}}{2}, i=\sqrt{-1}$ . If $(7-7 \alpha+9 \beta)^{20}+(9+7 \alpha-7 \beta)^{20}+(-7+9 \alpha+7 \beta)^{20} +(14+7 \alpha+7 \beta)^{20}=\mathrm{m}^{10}$ , then m is $\_\_\_\_$ .

Answer: 49

Numerical answer — enter this value.

Step-by-step solution

$\left(9+7 \omega-7 \omega^{2}\right)+\omega^{20}\left(9+7 \omega-7 \omega^{2}\right)^{20}+$ $\omega^{40}\left(9+7 \omega-7 \omega^{2}\right)^{20}+\left(14+7\left(\omega+\omega^{2}\right)\right)^{20}$

$\left(9+7 \omega-7 \omega^{2}\right)^{20}\left(1+\omega+\omega^{2}\right)+(14-7)^{20}=7^{20}$

$=(49)^{10}$

Hence, $\mathrm{M}=49$ .

Answer key and solution verified before publishing.

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Exam: JEE Main 2026
Paper: 22 January, Morning Shift
Subject: Mathematics
Chapter: Complex Numbers
Topic: Demoivre's Theorem and Roots of Unity