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NDA 2019 Maths · Mathematics · Complex Numbers and Quadratic Equations · medium

If ω is a non-real cube root of unity, then the value of (1+ω)(1+ω^2(1+ω^4(1+ω^8 is

Single correct option

If ω\omega is a non-real cube root of unity, then the value of (1+ω)(1+ω2)(1+ω4)(1+ω8)(1+\omega)(1+\omega^2)(1+\omega^4)(1+\omega^8) is
  1. A. 11
  2. B. 00
  3. C. −1-1
  4. D. 44
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Answer: (A) 11

Reduce exponents using ω3=1\omega^3=1: ω4=ω\omega^4=\omega and ω8=ω2\omega^8=\omega^2. The product becomes [(1+ω)(1+ω2)]2[(1+\omega)(1+\omega^2)]^2. Now (1+ω)(1+ω2)=1+ω+ω2+ω3=0+1=1(1+\omega)(1+\omega^2)=1+\omega+\omega^2+\omega^3=0+1=1. So the whole product is 12=11^2=1. Treating 1+ω1+\omega as zero is the trap, as it is −ω2-\omega^2, not 00.
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