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:
\[\begin{align}
  & \Rightarrow \left( 1+\omega \right)\left( 1+{{\omega }^{2}}\; \right) \\
 & \Rightarrow \left( -{{\omega }^{2}} \right)\left( -\;\omega \right)\,\,\,\,\,\,\,\,\,\, [\because \left( 1+\omega +{{\omega }^{2}}=0 \right) ] \\
 & \Rightarrow {{\omega }^{3}} \\
 & \Rightarrow 1\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, [\because {{\omega }^{3}}=1 ]\\
\end{align}\]
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:
\[\begin{align}
  & \Rightarrow \left( 1+{{\omega }^{4}}\; \right)\left( 1+{{\omega }^{8}}\; \right) \\
 & \Rightarrow \left( 1+{{\omega }^{3+1}}\; \right)\left( 1+{{\omega }^{2\times 3+2}}\; \right) \\
 & \Rightarrow \left( 1+{{\omega }^{3}}\;\omega \right)\left( 1+{{\omega }^{2\times 3}}\;{{\omega }^{2}} \right) \\
 & \Rightarrow \left( 1+\;\omega \right)\left( 1+\;{{\omega }^{2}} \right)\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, [\because {{\omega }^{3}}=1,{{\omega }^{6}}=1 ] \\
\end{align}\]
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:
\[\begin{align}
  & \Rightarrow \left( 1+\omega \right)\left( 1+{{\omega }^{2}}\; \right)\left( 1+{{\omega }^{4}}\; \right)\left( 1+{{\omega }^{8}}\; \right).....2n\,\,\,\,\,terms \\
 & \Rightarrow \left( 1+\omega \right)\left( 1+{{\omega }^{2}}\; \right)\left( 1+\omega \right)\left( 1+{{\omega }^{2}}\; \right).....2n\,\,\,\,\,terms \\
 & \Rightarrow 1\times 1.............n\,\,\,\,\,terms\,\,\,\,\,\,\,\,\,\,\because \left( 1+\omega \right)\left( 1+{{\omega }^{2}}\; \right)=1 \\
 & \Rightarrow 1 \\
\end{align}\]
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:
\[\begin{align}
  & \Rightarrow \left( 1+\omega \right)\left( 1+{{\omega }^{2}}\; \right)\left( 1+{{\omega }^{4}}\; \right)\left( 1+{{\omega }^{8}}\; \right).....\left( 2n+1 \right)\,\,\,terms \\
 & \Rightarrow 1\times \left( 1+\omega \right)\,\,\,\,\,\,\,\, [\because \left( 1+\omega \right)\left( 1+{{\omega }^{2}}\; \right)\left( 1+{{\omega }^{4}}\; \right)\left( 1+{{\omega }^{8}}\; \right).....2n\,\,\,\,\,terms\,\,\,=\,\,\,1 ]\\
 & \Rightarrow \left( 1+\omega \right) \\
 & \Rightarrow -{{\omega }^{2}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, [\because 1+\omega +{{\omega }^{2}}=0 ]\\
\end{align}\]
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.