If `1,alpha_(1),alpha_(2),alpha_(3),...,alpha_(n-1)` are n, nth roots of unity, then `(1-alpha_(1))(1-alpha_(2))(1-alpha_(3))...(1-alpha_(n-1))` equals to

Options

A. 0
B. 1
C. n
D. `n^(2)`

Answer

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

### Step-by-Step Solution:

1. **Understanding the Roots of Unity**:
The \(n\)th roots of unity are given by the equation \(x^n = 1\). The roots are \(1, \alpha_1, \alpha_2, \ldots, \alpha_{n-1}\) where \(\alpha_k = e^{2\pi i k/n}\) for \(k = 0, 1, 2, \ldots, n-1\).

2. **Formulating the Polynomial**:
The polynomial whose roots are the \(n\)th roots of unity can be expressed as:
\[ P(x) = x^n - 1 = (x - 1)(x - \alpha_1)(x - \alpha_2) \ldots (x - \alpha_{n-1}) \]

3. **Rearranging the Polynomial**:
We can rearrange the polynomial as:
\[ P(x) = (x - 1)Q(x) \]
where \(Q(x) = (x - \alpha_1)(x - \alpha_2) \ldots (x - \alpha_{n-1})\).

4. **Finding the Value at \(x = 1\)**:
To find the product \((1 - \alpha_1)(1 - \alpha_2)(1 - \alpha_3) \ldots (1 - \alpha_{n-1})\), we substitute \(x = 1\) into the polynomial:
\[ P(1) = 1^n - 1 = 0 \]
This means:
\[ (1 - \alpha_1)(1 - \alpha_2) \ldots (1 - \alpha_{n-1}) = Q(1) \]

5. **Calculating \(Q(1)\)**:
The polynomial \(Q(x)\) can be evaluated at \(x = 1\):
\[ Q(1) = (1 - \alpha_1)(1 - \alpha_2) \ldots (1 - \alpha_{n-1}) \]
Since \(Q(x)\) is a polynomial of degree \(n-1\), we can also express it as:
\[ Q(1) = 1 + 1 + 1 + \ldots + 1 \quad (n-1 \text{ times}) = n - 1 \]

6. **Conclusion**:
Therefore, the product \((1 - \alpha_1)(1 - \alpha_2)(1 - \alpha_3) \ldots (1 - \alpha_{n-1})\) equals \(n\).

### Final Answer:
\[ (1 - \alpha_1)(1 - \alpha_2)(1 - \alpha_3) \ldots (1 - \alpha_{n-1}) = n \]