If `1,alpha_(1),alpha_(2),………..,alpha_(n-1)` are `nk^(th)` roots of unity, then the value of
`(1-alpha_(1))(1-alpha_(2))(1-alpha_(3))……(1-alpha_(n-1))` is equal to

Options

A. `sqrt(3)`
B. `1//2`
C. n
D. 0

Explanation

To solve the problem, we need to find the value of the expression \((1 - \alpha_1)(1 - \alpha_2)(1 - \alpha_3) \cdots (1 - \alpha_{n-1})\), where \(1, \alpha_1, \alpha_2, \ldots, \alpha_{n-1}\) are the \(nk^{th}\) roots of unity.

### Step-by-Step Solution:

1. **Understanding Roots of Unity**:
The \(nk^{th}\) roots of unity are given by the formula:
\[ \alpha_j = e^{\frac{2\pi i j}{nk}} \quad \text{for } j = 0, 1, 2, \ldots, nk-1 \]
Here, \(1\) corresponds to \(j=0\) and \(\alpha_1, \alpha_2, \ldots, \alpha_{n-1}\) correspond to \(j=1, 2, \ldots, n-1\).

2. **Polynomial Representation**:
The \(nk^{th}\) roots of unity can be expressed as the roots of the polynomial:
\[ x^{nk} - 1 = 0 \]
This can be factored as:
\[ x^{nk} - 1 = (x - 1)(x - \alpha_1)(x - \alpha_2) \cdots (x - \alpha_{nk-1}) \]

3. **Using the Polynomial**:
We can express the polynomial as:
\[ x^{nk} - 1 = (x - 1)Q(x) \]
where \(Q(x) = (x - \alpha_1)(x - \alpha_2) \cdots (x - \alpha_{nk-1})\).

4. **Finding the Value at \(x = 1\)**:
We need to evaluate:
\[ (1 - \alpha_1)(1 - \alpha_2)(1 - \alpha_3) \cdots (1 - \alpha_{n-1}) = Q(1) \]
Substitute \(x = 1\) into \(Q(x)\):
\[ Q(1) = (1 - \alpha_1)(1 - \alpha_2) \cdots (1 - \alpha_{n-1}) \]

5. **Calculating \(Q(1)\)**:
Since \(Q(x)\) is a polynomial of degree \(nk - 1\) and has roots at \(\alpha_1, \alpha_2, \ldots, \alpha_{n-1}\):
\[ Q(1) = \frac{1^{nk} - 1}{1 - 1} = 0 \quad \text{(indeterminate form)} \]
We can apply L'Hôpital's Rule or consider the limit as \(x\) approaches \(1\).

6. **Using the Sum of Roots**:
The sum of the \(nk^{th}\) roots of unity is zero:
\[ 1 + \alpha_1 + \alpha_2 + \cdots + \alpha_{nk-1} = 0 \]
Therefore, the product of the terms \((1 - \alpha_j)\) can be computed as:
\[ (1 - \alpha_1)(1 - \alpha_2) \cdots (1 - \alpha_{n-1}) = n \]

7. **Final Result**:
Thus, the value of \((1 - \alpha_1)(1 - \alpha_2)(1 - \alpha_3) \cdots (1 - \alpha_{n-1})\) is:
\[ n^k \]

### Conclusion:
The value of \((1 - \alpha_1)(1 - \alpha_2)(1 - \alpha_3) \cdots (1 - \alpha_{n-1})\) is \(n^k\).

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