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JEE MATHEMATICS MCQ: COMPLEX NUMBERS

1 Easy
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Application +4 | -1
If ω(≠1)\omega(\neq 1) is a cube root of unity and (1+ω)2017=(1+\omega)^{2017}= A+BωA+B \omega. Then AA and BB are respectively the numbers
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Correct option: B
1,1
Option B (1,1) is correct because (1+ω)2017(1+\omega)^{2017} simplifies to −ω2-\omega^2, which can be rewritten as 1+ω1 + \omega, thus A=1 and B=1.

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