Question

If \(\omega\) is an imaginary cube root of unity, then the value of \((1+\omega)(1+\omega^2)(1+\omega^3)(1+\omega^4)(1+\omega^5)..........(1+\omega^{3n})\) is

• A. \(2^{3n}\)
• B. \(2^{2n}\)
• C. \(2^n\)
• D. None of these

Hint:

Properties of cube roots of unity: \(1+\omega+\omega^2=0\), \(\omega^3=1\).

Answer 1

The Correct Option is C

To solve the problem, we need to understand the properties of the imaginary cube roots of unity. The cube roots of unity are given by \(1, \omega, \omega^2\), where:

• \(\omega\) and \(\omega^2\) are the imaginary cube roots of unity.
• \(\omega^3 = 1\) and \(\omega + \omega^2 = -1\).

The given expression is \((1+\omega)(1+\omega^2)(1+\omega^3)(1+\omega^4)(1+\omega^5)\dots(1+\omega^{3n})\).

First, note that:

• \(\omega^3 = 1\)
• \(\omega^4 = \omega\)
• \(\omega^5 = \omega^2\)
• \(\omega^6 = 1\) (because \(\omega^6 = (\omega^3)^2 = 1^2 = 1\))

The pattern repeats every three terms: \((1+\omega), (1+\omega^2), (1+\omega^3 = 2)\).

We can pair the sequence in groups of three successive terms:

• The pair \((1+\omega)(1+\omega^2)(1+1)\) simplifies to:
• \((1+\omega)(1+\omega^2)\)
• \(= 1 + \omega + \omega + \omega^2\)
• \(= 1 + (\omega + \omega^2)\)
• \(= 1 - 1\)
• \(= 0\)

This can be repeated for each set of the three terms until \(3n\). So, the original expression simplifies to \(2 \times 2 \times \dots\) (multiplied \(n\) times). Each group contributes a factor of 2:

• \((1+1)\) from \((1+\omega^3) = (1+1)^n = 2^n\).

Thus, the value of the expression \((1+\omega)(1+\omega^2)\dots(1+\omega^{3n}) = 2^n\).

Therefore, the correct option is:

• \(2^n\)