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NDA 2024
Exam
NDA
Year
2024
Subject
Mathematics
Chapter
Complex Number

NDA 2024 · Mathematics · Complex Number

If ω ≠ 1 is a cube root of unity, then what is (1+ω-ω^2)^ 100...

If ω≠1\omega \neq 1 is a cube root of unity, then what is \left(1+\omega-\omega^2\right)^{100}+\left(1-\omega+\omega^2\right)^{100} equal to?

Options

  1. A2100ω22^{100} \omega^2
  2. B2100ω2^{100} \omega
  3. C21002^{100}
  4. D−2100-2^{100}

Correct answer

D. −2100-2^{100}

Step-by-step solution

 Let I=(1+ω−ω2)100+(1−ω+ω2)100=(−ω2−ω2)100+(−ω−ω)100{∵1+ω+ω2=0}=2100(ω200+ω100)=2100(ω2+ω){∵ω3=1}=−2100.\begin{aligned} & \text{ Let } I=\left(1+\omega-\omega^2\right)^{100}+\left(1-\omega+\omega^2\right)^{100} \\ & =\left(-\omega^2-\omega^2\right)^{100}+(-\omega-\omega)^{100}\left\{\because 1+\omega+\omega^2=0\right\} \\ & =2^{100}\left(\omega^{200}+\omega^{100}\right) \\ & =2^{100}\left(\omega^2+\omega\right)\left\{\because \omega^3=1\right\} \\ & =-2^{100}.\end{aligned}
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