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could you solve the following question from complex number topic?
Asked by ashwinskrishna2006 | 13 Apr, 2024, 11:25: AM

mth root of unity is written as

where n = 0, 1, 2 ..........(m-1)

Cube roots of unity  are

cos(0) , cos((2π)/3) + i sin((2π)/3) and cos((4π)/3) + i sin((4π)/3)

If they are 1, ω and ω2 respectively , then

1 = cos(0)

ω = cos((2π)/3) + i sin((2π)/3)

and ω2 = cos((4π)/3) + i sin((4π)/3)

1 + ω + ω2 = [ 1 + cos((2π)/3)+ cos((4π)/3) ] +i [ sin((2π)/3)+ sin((4π)/3) ]

1 + ω + ω2 = 0 + i 0 = 0

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we have the matric equation as

Let us do elementary operation R1 = ( R1+R2+R3 ) in first matrix and second matrix

Second matrix vanishes because all elements of first row are 1 + ω + ω2 = 0

Hence we get

x × 3 ( ω3 - 1 ) = 3

It is not possible to get value of x from above expression because ( ω3 - 1 ) = 0 .

Hence the answer is (D) None of these

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