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JEE Advanced 2020 Paper 1 · Mathematics · Complex Numbers · hard

Let ω=e^2π i/7. The value of prodk=1^6(1+2ω^k is

Numerical answer

Let ω=e2πi/7\omega=e^{2\pi i/7}. The value of ∏k=16(1+2ωk)\displaystyle\prod_{k=1}^{6}\left(1+2\omega^{k}\right) is
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Answer: 43

The seven numbers ωk\omega^k (k=0,…,6k=0,\dots,6) are the roots of x7−1=∏k=06(x−ωk)x^7-1=\prod_{k=0}^{6}(x-\omega^k). Put x=−12x=-\tfrac12: ∏k=06(−12−ωk)=−1128−1=−129128\prod_{k=0}^{6}\left(-\tfrac12-\omega^k\right)=-\tfrac1{128}-1=-\tfrac{129}{128}. Since 1+2ωk=−2(−12−ωk)1+2\omega^k=-2\left(-\tfrac12-\omega^k\right), the full product ∏k=06(1+2ωk)=(−2)7(−129128)=129\prod_{k=0}^{6}(1+2\omega^k)=(-2)^7\left(-\tfrac{129}{128}\right)=129. The factor for k=0k=0 is 1+2=31+2=3, so the required product over k=1,…,6k=1,\dots,6 is 129/3=43129/3=43. This is the cyclotomic-type value 1−(−2)71−(−2)=1293\dfrac{1-(-2)^7}{1-(-2)}=\dfrac{129}{3}, and a direct numerical multiplication of the six complex factors gives 4343.
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