JEE MATHEMATICS MCQ: COMPLEX NUMBERS

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Analyzing +4 | -1

{z^2} + z + 1 = 0

, where

z

is a complex number, then the value of

{\left( {z + \frac{1}{z}} \right)^2} + {\left( {{z^2} + \frac{1}{{{z^2}}}} \right)^2} + \left( {{z^3} + \frac{1}{{{z^3}}}} \right) + ... + {\left( {{z^6} + \frac{1}{{{z^6}}}} \right)^2}

View answer and explanation

Correct option: B

Option B is correct because by substituting

z = \omega

z = \omega^2

, the expression simplifies to 12, where

\omega

is a complex cube root of unity.

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