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Plz look at the attachment. Question is there it`s from the chapter complex numbers

Plz look at the attachment. Question is there it`s from the chapter complex numbers

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Grade:10

1 Answers

Riddhish Bhalodia
askIITians Faculty 434 Points
8 years ago
omega given is the cube root of -1

1) conjugate \omega^* = -\omega^2
2) \omega^3 = -1
3) \omega^2 -\omega +1 =0
The trick is to express this as
|x|^2 = xx^* = (a+b+c)(a^*+b^* + c^*)
|y|^2 = yy^* = (a+b\omega+c\omega^2)(a^*-b^*\omega^2 - c^*\omega)
|z|^2 = zz^* = (a+b\omega^2+c\omega)(a^*-b^*\omega - c^*\omega^2)

now expanding them and using the properties mentioned above we get
|x|^2 + |y|^2 + |z|^2 = 3(|a|^2 + |b|^2 + |c|^2)
and hence the answer is 3

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