JEE MATHEMATICS MCQ: COMPLEX NUMBERS

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Application +4 | -1

Let

{\omega _n} = \cos \left( {\frac{{2\pi }}{n}} \right) + i\sin \left( {\frac{{2\pi }}{n}} \right),\;{i^2} = - 1,

then

\left( {x + y{\omega _3} + z\omega _3^2} \right)\left( {x + y\omega _3^2 + z{\omega _3}} \right)

is equal to

View answer and explanation

Correct option: C

{x^2} + {y^2} + {z^2} - yz - zx - xy

Option C (

{x^2} + {y^2} + {z^2} - yz - zx - xy

) is correct because substituting

\omega_3 = \omega

and

\omega_3^2 = \omega^2

(where

\omega

is a cube root of unity) and expanding the product leads to the given expression.

More Questions from COMPLEX NUMBERS

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