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 formula:
\[ \alpha_k = e^{2\pi i k/n} \quad \text{for } k = 0, 1, 2, \ldots, n-1 \]
Here, \(1\) corresponds to \(\alpha_0\), and \(\alpha_1, \alpha_2, \ldots, \alpha_{n-1}\) are the other roots.

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

3. **Dividing by \(x - 1\)**:
We can express the polynomial as:
\[ \frac{x^n - 1}{x - 1} = x^{n-1} + x^{n-2} + \ldots + x + 1 \]
This shows that:
\[ \frac{x^n - 1}{x - 1} = (x - \alpha_1)(x - \alpha_2) \ldots (x - \alpha_{n-1}) \]

4. **Evaluating at \(x = 1\)**:
Now, substituting \(x = 1\) into the polynomial gives:
\[ \frac{1^n - 1}{1 - 1} \text{ is undefined, but we can evaluate the limit.} \]
Instead, we can directly evaluate the polynomial:
\[ 1^{n-1} + 1^{n-2} + \ldots + 1 + 1 = n \]
Therefore, we have:
\[ (1 - \alpha_1)(1 - \alpha_2) \ldots (1 - \alpha_{n-1}) = n \]

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

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

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