The product of all `n^(th)` root of unity is always

Options

A. 1
B. `-1`
C. 1 or –1 depending on n is even or odd
D. 1 or –1 depending on n is odd or even

Answer

To solve the problem of finding the product of all \( n^{th} \) roots of unity, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding \( n^{th} \) Roots of Unity**:
The \( n^{th} \) roots of unity are the solutions to the equation \( z^n = 1 \). These roots can be expressed in the form:
\[ z_k = e^{2\pi i k/n} \quad \text{for } k = 0, 1, 2, \ldots, n-1 \]

2. **Writing the Product of Roots**:
The product of all \( n^{th} \) roots of unity can be represented as:
\[ P = z_0 \cdot z_1 \cdot z_2 \cdots z_{n-1} = e^{2\pi i \cdot 0/n} \cdot e^{2\pi i \cdot 1/n} \cdot e^{2\pi i \cdot 2/n} \cdots e^{2\pi i \cdot (n-1)/n} \]

3. **Simplifying the Product**:
This product can be simplified using properties of exponents:
\[ P = e^{2\pi i (0 + 1 + 2 + \ldots + (n-1))/n} \]
The sum \( 0 + 1 + 2 + \ldots + (n-1) \) can be calculated using the formula for the sum of the first \( n-1 \) integers:
\[ \text{Sum} = \frac{(n-1)n}{2} \]
Thus, we have:
\[ P = e^{2\pi i \cdot \frac{(n-1)n/2}{n}} = e^{\pi i (n-1)} \]

4. **Evaluating \( e^{\pi i (n-1)} \)**:
The expression \( e^{\pi i (n-1)} \) can be evaluated based on whether \( n-1 \) is even or odd:
- If \( n-1 \) is even (i.e., \( n \) is odd), then \( e^{\pi i (n-1)} = 1 \).
- If \( n-1 \) is odd (i.e., \( n \) is even), then \( e^{\pi i (n-1)} = -1 \).

5. **Conclusion**:
Therefore, the product of all \( n^{th} \) roots of unity is:
\[ P = (-1)^{n-1} \]
This means:
- If \( n \) is odd, the product is \( 1 \).
- If \( n \) is even, the product is \( -1 \).

### Final Answer:
The product of all \( n^{th} \) roots of unity is \( 1 \) if \( n \) is odd and \( -1 \) if \( n \) is even.

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