Question

If $\omega$ is complex cube root of unity then find the value of the expression given below
$\left( 1+\omega \right)\left( 1+{{\omega }^{2}}\; \right)\left( 1+{{\omega }^{4}}\; \right)\left( 1+{{\omega }^{8}}\; \right).....\left( 2n+1 \right)$ terms.
Choose the correct option:
(A). 1
(B). $2\omega$
(C). $-2\omega$
(D). $-{{\omega }^{2}}$

Answer 1

Hint: If we have the expression containing $\omega$ (the complex cube root of unity), then it is first required to simplify it and use two properties: ${{\omega }^{3}}=1$ and $1+\omega \;+{{\omega }^{2}}=0$.

Complete step-by-step solution -
In the problem, we have to find the value of the expression
$\left( 1+\omega \right)\left( 1+{{\omega }^{2}}\; \right)\left( 1+{{\omega }^{4}}\; \right)\left( 1+{{\omega }^{8}}\; \right).....\left( 2n+1 \right)$ terms.
Now, it is known that when $\omega$ is given as the cube root of unity, then we have ${{\omega }^{3}}=1$ and $1+\omega \;+{{\omega }^{2}}=0$.
So, in the given expression there are $\left( 2n+1 \right)$ terms, which is the odd number of terms.
As, $\left( 2n \right)$is even so $\left( 2n+1 \right)$is odd.
Now when we take first two terms we will have:

& \Rightarrow \left( 1+\omega \right)\left( 1+{{\omega }^{2}}\; \right) \\\ & \Rightarrow \left( -{{\omega }^{2}} \right)\left( -\;\omega \right)\,\,\,\,\,\,\,\,\,\, [\because \left( 1+\omega +{{\omega }^{2}}=0 \right) ] \\\ & \Rightarrow {{\omega }^{3}} \\\ & \Rightarrow 1\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, [\because {{\omega }^{3}}=1 ]\\\ \end{aligned}$$ So we can say that $$\left( 1+\omega \right)\left( 1+{{\omega }^{2}}\; \right)=1$$ So the value of the first two terms is 1. Next we will take next two terms, as follows: $$\begin{aligned} & \Rightarrow \left( 1+{{\omega }^{4}}\; \right)\left( 1+{{\omega }^{8}}\; \right) \\\ & \Rightarrow \left( 1+{{\omega }^{3+1}}\; \right)\left( 1+{{\omega }^{2\times 3+2}}\; \right) \\\ & \Rightarrow \left( 1+{{\omega }^{3}}\;\omega \right)\left( 1+{{\omega }^{2\times 3}}\;{{\omega }^{2}} \right) \\\ & \Rightarrow \left( 1+\;\omega \right)\left( 1+\;{{\omega }^{2}} \right)\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, [\because {{\omega }^{3}}=1,{{\omega }^{6}}=1 ] \\\ \end{aligned}$$ So now these two terms are the same as the first two terms and thus the value is 1 again. $$\Rightarrow \left( 1+{{\omega }^{4}}\; \right)\left( 1+{{\omega }^{8}}\; \right)=1$$ Similarly, up to $$\left( 2n \right)$$ terms the value of the expression is 1 or we can say that: $$\begin{aligned} & \Rightarrow \left( 1+\omega \right)\left( 1+{{\omega }^{2}}\; \right)\left( 1+{{\omega }^{4}}\; \right)\left( 1+{{\omega }^{8}}\; \right).....2n\,\,\,\,\,terms \\\ & \Rightarrow \left( 1+\omega \right)\left( 1+{{\omega }^{2}}\; \right)\left( 1+\omega \right)\left( 1+{{\omega }^{2}}\; \right).....2n\,\,\,\,\,terms \\\ & \Rightarrow 1\times 1.............n\,\,\,\,\,terms\,\,\,\,\,\,\,\,\,\,\because \left( 1+\omega \right)\left( 1+{{\omega }^{2}}\; \right)=1 \\\ & \Rightarrow 1 \\\ \end{aligned}$$ Or we can say that $$\Rightarrow \left( 1+\omega \right)\left( 1+{{\omega }^{2}}\; \right)\left( 1+{{\omega }^{4}}\; \right)\left( 1+{{\omega }^{8}}\; \right).....2n\,\,\,\,\,terms\,\,\,=\,\,\,1$$ Now, the $${{\left( 2n+1 \right)}^{th}}$$ term is $$\left( 1+\omega \right)$$ So, the value of the given expression: $$\begin{aligned} & \Rightarrow \left( 1+\omega \right)\left( 1+{{\omega }^{2}}\; \right)\left( 1+{{\omega }^{4}}\; \right)\left( 1+{{\omega }^{8}}\; \right).....\left( 2n+1 \right)\,\,\,terms \\\ & \Rightarrow 1\times \left( 1+\omega \right)\,\,\,\,\,\,\,\, [\because \left( 1+\omega \right)\left( 1+{{\omega }^{2}}\; \right)\left( 1+{{\omega }^{4}}\; \right)\left( 1+{{\omega }^{8}}\; \right).....2n\,\,\,\,\,terms\,\,\,=\,\,\,1 ]\\\ & \Rightarrow \left( 1+\omega \right) \\\ & \Rightarrow -{{\omega }^{2}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, [\because 1+\omega +{{\omega }^{2}}=0 ]\\\ \end{aligned}$$ Finally, we can say that option 4) $$-{{\omega }^{2}}$$ is the correct answer. Note: It is important to know that $$\omega $$ is a complex cube root of unity. So we can’t use $$\omega $$ for any other root of unity other than cube root of unity. To find the cube root of unity, we need to solve the equation $${{x}^{3}}=1$$. Here we need to find both the real and complex roots.