JEE-Advanced 2015 (19.C/40)

Algebra Level 5

Let ω 1 \omega \neq 1 be a complex root of unity, find the possible value(s) of n n such that ( 3 3 ω + 2 ω 2 ) 4 n + 3 + ( 2 + 3 ω 3 ω 2 ) 4 n + 3 + ( 3 + 2 ω + 3 ω 2 ) 4 n + 3 = 0 (3-3\omega+2\omega^2)^{4n+3}+(2+3\omega-3\omega^2)^{4n+3}+(-3+2\omega+3\omega^2)^{4n+3}=0 ( 1 ) 1 ( 2 ) 2 ( 3 ) 4 ( 4 ) 5 \begin{array}{ll} (1) \, 1 \quad \quad \quad \quad \quad \quad \quad \quad & (2) \, 2 \\ (3) \, 4 & (4) \, 5 \end{array} Note:

  • Submit your answer as the increasing order of the serial numbers of all the correct options.

  • For eg, if your answer is ( 1 ) , ( 2 ) (1),(2) , then submit 12 as the correct answer, if your answer is ( 2 ) , ( 3 ) , ( 4 ) (2),(3),(4) , then submit 234 as the correct answer.


The answer is 1234.

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1 solution

Kushal Dey
Mar 25, 2020

Let 3-3w+2w^2=a 2+3w-3w^2=b=aw -3+2w+3w^2=c=aw^2 Thus a^k+b^k+c^k= (a^k)(1+w^(k)+^w(2k)), which can only be 0 if k is not a multiple of 3.

Small note for the creator of this question: please mention "complex cube root" of unity instead of only "complex root" of unity, otherwise due to which you also know that this question is not solvable.

Kushal Dey - 1 year, 2 months ago

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