JEE MATHEMATICS MCQ: COMPLEX NUMBERS

1 Easy

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Analyzing +4 | -1

z^2+z+1=0

, where

z

is a complex number, then values of

S=\left(z+\frac{1}{z}\right)^2+\left(z^2+\frac{1}{z^2}\right)^2+\left(z^3+\frac{1}{z^3}\right)^2+\cdots+\left(z^6+\frac{1}{z^6}\right)^2 \text { is }

View answer and explanation

Correct option: A

Option A (12) is correct because by substituting

z = \omega

(a complex cube root of unity) into the expression and using the properties of

\omega

, the sum simplifies to 12.

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