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Question

If i=1 and ω is non real cube root of unity, then (1+i)2n(1i)2n(1+ω4ω2)(1ω4+ω2) is equal to

A
0, when n is even
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B
2n1in, when n is odd
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C
0 nI
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D
2n1in, when n is even
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Solution

The correct option is B 2n1in, when n is odd
(1+i)2n(1i)2n(1+ω4ω2)(1ω4+ω2)=((1+i)2)n((1i)2)n(1+ωω3ω2)(1ωω3+ω2)

=(1+i2+2i)n(1+i22i)n(1+ωω2)(1ω+ω2)

=(2i)n(2i)n(2ω2)(2ω) =2nin[1(1)n]4

=0,when n is even2n+1in22=2n1in,when n is odd

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