If $\omega $ is a complex root of the equation ${z^3} = 1,\;\,{\rm{then}}\,\;{\rm{\omega }} + {{\rm{\omega }}^{\left( {\frac{1}{2} + \frac{3}{8} + \frac{9}{{32}} + \frac{{27}}{{128}} + ...} \right)}}\;$ is equal to

JEE MATHEMATICS MCQ: COMPLEX NUMBERS

1 Easy

1 of 1

Analyzing +4 | -1

\omega

is a complex root of the equation

{z^3} = 1,\;\,\mathrm{then}\,\;\mathrm{\omega } + {\mathrm{\omega }^{\left( {\frac{1}{2} + \frac{3}{8} + \frac{9}{{32}} + \frac{{27}}{{128}} + ...} \right)}}\;

is equal to

View answer and explanation

Correct option: A

- 1

Option A (-1) is correct because the expression simplifies to

\omega + \omega^2

, which equals -1 due to the properties of complex roots of unity.

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