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Question 75 of 122

Q.The value of [−1+i32]100+[−1−i32]100\left[\dfrac{-1+i\sqrt3}{2}\right]^{100} + \left[\dfrac{-1-i\sqrt3}{2}\right]^{100} is :

(a) 22
(b) 00
(c) −1-1
(d) 11
Puducherry TnboardTamil Nadu HSC (DGE) Board 2016MCQ· 1mImportance★★★★★
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The expression equals ω+ω2=−1\omega+\omega^2=-1, where ω\omega is a complex cube root of unity.

  1. Let ω=−1+i32\omega=\dfrac{-1+i\sqrt3}{2}; then ω2=−1−i32\omega^2=\dfrac{-1-i\sqrt3}{2}, with ω3=1\omega^3=1 and 1+ω+ω2=01+\omega+\omega^2=0.
  2. ω100=ω99⋅ω=(ω3)33⋅ω=133⋅ω=ω\omega^{100}=\omega^{99}\cdot\omega=(\omega^3)^{33}\cdot\omega=1^{33}\cdot\omega=\omega.
  3. (ω2)100=ω200=ω198⋅ω2=(ω3)66⋅ω2=ω2(\omega^2)^{100}=\omega^{200}=\omega^{198}\cdot\omega^2=(\omega^3)^{66}\cdot\omega^2=\omega^2. …

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