If `1, omega, omega^2 ,…..,omega^(n-1)` are the nth roots of unity, then `(2- omega)(2-omega^2)….(2-omega^(n-1) )` equals

Options

A. `2^(n)-1`
B. `""^(n) C_(1) +^(n) C_(2) +….+^(n) C_(n)`
C. `[^(2n+1) C_(0) +^(2n+1) C_(1) + ….+^(2n+1) C_(n)]^(1//2)-1`
D. None of these

Answer

To solve the problem, we need to evaluate the expression \((2 - \omega)(2 - \omega^2) \cdots (2 - \omega^{n-1})\), where \(\omega\) is a primitive \(n\)th root of unity.

### Step-by-Step Solution:

1. **Understanding Roots of Unity**:
The \(n\)th roots of unity are the solutions to the equation \(z^n = 1\). These roots are given by \(1, \omega, \omega^2, \ldots, \omega^{n-1}\), where \(\omega = e^{2\pi i/n}\).

2. **Formulating the Polynomial**:
The polynomial whose roots are the \(n\)th roots of unity can be expressed as:
\[ P(z) = z^n - 1 = (z - 1)(z - \omega)(z - \omega^2) \cdots (z - \omega^{n-1}). \]

3. **Rearranging the Polynomial**:
We can express \(P(z)\) in the form:
\[ P(z) = (z - 1)Q(z), \]
where \(Q(z) = (z - \omega)(z - \omega^2) \cdots (z - \omega^{n-1})\).

4. **Evaluating at \(z = 2\)**:
We need to evaluate \(Q(2)\):
\[ Q(2) = (2 - \omega)(2 - \omega^2) \cdots (2 - \omega^{n-1}). \]
From the polynomial \(P(z)\), we know:
\[ P(2) = 2^n - 1. \]
Therefore,
\[ P(2) = (2 - 1)Q(2) = Q(2). \]
Hence,
\[ Q(2) = 2^n - 1. \]

5. **Final Result**:
Thus, we conclude that:
\[ (2 - \omega)(2 - \omega^2) \cdots (2 - \omega^{n-1}) = 2^n - 1. \]

### Summary:
The value of \((2 - \omega)(2 - \omega^2) \cdots (2 - \omega^{n-1})\) is \(2^n - 1\).