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

Options

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

Answer

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

### Step-by-Step Solution:

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

3. **Finding the Desired Product**:
We want to evaluate:
\[ (1 - \omega)(1 - \omega^2) \cdots (1 - \omega^{n-1}) \]
This can be interpreted as evaluating the polynomial \( z^n - 1 \) at \( z = 1 \), excluding the root \( z = 1 \).

4. **Using L'Hôpital's Rule**:
Since substituting \( z = 1 \) directly gives \( 0 \) (as \( 1^n - 1 = 0 \)), we can use L'Hôpital's Rule to evaluate the limit:
\[ \lim_{z \to 1} \frac{z^n - 1}{z - 1} \]
This limit represents the product we are interested in.

5. **Calculating the Derivative**:
The derivative of \( z^n - 1 \) is:
\[ \frac{d}{dz}(z^n - 1) = n z^{n-1} \]
Therefore, applying L'Hôpital's Rule gives:
\[ \lim_{z \to 1} \frac{n z^{n-1}}{1} = n \cdot 1^{n-1} = n \]

6. **Conclusion**:
Thus, we find that:
\[ (1 - \omega)(1 - \omega^2) \cdots (1 - \omega^{n-1}) = n \]

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