Question

If we have the complex cube root of unity $\omega $, then the value of $\omega + {\omega ^{\left( {\dfrac{1}{2} + \dfrac{3}{8} + \dfrac{9}{{32}} + \dfrac{{27}}{{128}} + ....} \right)}}$ is
(A). -1
(B). 1
(C). $ - i$
(D). $i$

Answer 1

Hint- Firstly, we will solve the given series using geometric progression formula for sum of infinite terms as ${S_\infty } = \dfrac{a}{{1 - r}}$ ; a being first term and r being the common ratio. After that minimize the given complex term.

Complete step-by-step solution -
It is given that $\omega + {\omega ^{\left( {\dfrac{1}{2} + \dfrac{3}{8} + \dfrac{9}{{32}} + \dfrac{{27}}{{128}} + ....} \right)}}$
Now, solving series as $\dfrac{1}{2} + \dfrac{3}{8} + \dfrac{9}{{32}} + \dfrac{{27}}{{128}} + ....$
Here, first term, ${a_1} = \dfrac{1}{2}$ and second term, ${a_2} = \dfrac{3}{8}$
Now, common ratio, $r = \dfrac{{{a_2}}}{{{a_1}}} = \dfrac{{\left( {\dfrac{3}{8}} \right)}}{{\left( {\dfrac{1}{2}} \right)}} = \dfrac{6}{8} = \dfrac{3}{4}$
Sum up to $\infty $ terms $ = \dfrac{a}{{1 - r }} = \dfrac{{\dfrac{1}{2}}}{{1 - \dfrac{3}{4}}} = \dfrac{{\dfrac{1}{2}}}{{\dfrac{{4 - 3}}{4}}} = \dfrac{{\dfrac{1}{2}}}{{\dfrac{1}{4}}} = \dfrac{4}{2} = 2$
Now substituting the value of $\dfrac{1}{2} + \dfrac{3}{8} + \dfrac{9}{{32}} + \dfrac{{27}}{{128}} + ....$ as 2 we get,
$ \Rightarrow \omega + {\omega ^2}$
We know that $ \Rightarrow \omega + {\omega ^2} + 1 = 0$
$ \Rightarrow \omega + {\omega ^2} = - 1$
Hence, the value of $\omega + {\omega ^{\left( {\dfrac{1}{2} + \dfrac{3}{8} + \dfrac{9}{{32}} + \dfrac{{27}}{{128}} + ....} \right)}}$ is -1
$\therefore $ Option A. -1 is the correct answer.

Note- Always remember the sum of three cube roots of unity is 0 i.e. $\omega + {\omega ^2} + 1 = 0$ . For these types of complex problems, the basic trick is to simplify the given terms for understanding the direction of question.