JEE Main 2026 Jan 22 Shift 1, Mathematics Q21: De Moivre's Theorem And Its Applications

JEE Main2026Jan 22, Shift 1Mathematics

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 _________.

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Solution

Note that $α = ω$ and $β = ω^{2}$ (primitive cube roots of unity), with $1 + ω + ω^{2} = 0$ and $ω^{3} = 1$ .

The last term simplifies cleanly: $14 + 7 ω + 7 ω^{2} = 14 + 7 (ω + ω^{2}) = 14 - 7 = 7$ . So $(14 + 7 α + 7 β)^{20} = 7^{20}$ .

The other three bases are cyclic permutations of one another by multiplication by $ω$ , and their 20th powers sum to a multiple of $(1 + ω + ω^{2}) = 0$ . So those three terms total $0$ .

Total $= 7^{20} = (7^{2})^{10} = 4 9^{10}$ , giving $m = 49$ .

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JEE Main 2026 Jan 22 Shift 1, Mathematics Q21: De Moivre's Theorem And Its Applications

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