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 evaluate the expression \((1+\omega)(1+\omega^2)(1+\omega^3)(1+\omega^4)(1+\omega^5)\cdots(1+\omega^{3n})\), where \(\omega\) is an imaginary cube root of unity.

The known properties of cube roots of unity are:

• \(\omega^3 = 1\)
• \(\omega \neq 1\) and \(\omega^2 \neq 1\)
• \(1 + \omega + \omega^2 = 0\)

Given these properties, observe that \(\omega^3 = 1\), meaning that the cycle repeats every three powers. This implies:

• \(\omega^3 = 1\), \(\omega^4 = \omega\), \(\omega^5 = \omega^2\), etc.

The expression can be simplified by examining the product terms:

• \((1+\omega^3) = (1+1) = 2\)
• \((1+\omega^6) = (1+1) = 2\) and so on.

For any multiple of 3, \( \omega^{3k} = 1 \), hence \(1 + \omega^{3k} = 2\).

Therefore, for every third power of \(\omega\), the term simplifies to 2. Now consider:

• The terms \(1+\omega^1, 1+\omega^2\) repeat every three positions after simplifying, so \(1+\omega^4 = 1+\omega\) and so on.

The full expression contains \(n\) terms where \(1 + \omega^{3k} = 2\). Each such pair contributes a factor of 2 to the final result.

Therefore, the entire product becomes:

• \((2)^{n} = 2^n\)

Thus, the value of the expression is 2^n, making the correct answer: \(2^n\).