If p is not a multiple of n, then the sum of pth powers of nth roots of unity is

Options

A. 0
B. 1
C. n
D. p

Answer

To solve the problem, we need to find the sum of the p-th powers of the n-th roots of unity when \( p \) is not a multiple of \( n \).

### Step-by-Step Solution:

1. **Understanding n-th Roots of Unity**:
The n-th roots of unity are given by:
\[ \alpha_k = e^{2\pi i k/n} \quad \text{for } k = 0, 1, 2, \ldots, n-1 \]
This means that the n-th roots of unity are \( 1, \alpha, \alpha^2, \ldots, \alpha^{n-1} \), where \( \alpha = e^{2\pi i/n} \).

2. **Setting Up the Sum**:
We want to compute the sum:
\[ S = \sum_{k=0}^{n-1} \alpha_k^p = \alpha_0^p + \alpha_1^p + \alpha_2^p + \ldots + \alpha_{n-1}^p \]
Substituting the expression for \( \alpha_k \):
\[ S = 1^p + \left(e^{2\pi i/n}\right)^p + \left(e^{2\pi i \cdot 2/n}\right)^p + \ldots + \left(e^{2\pi i \cdot (n-1)/n}\right)^p \]
This simplifies to:
\[ S = 1 + e^{2\pi i p/n} + e^{2\pi i \cdot 2p/n} + \ldots + e^{2\pi i \cdot (n-1)p/n} \]

3. **Recognizing the Geometric Series**:
The sum \( S \) is a geometric series with the first term \( 1 \) and common ratio \( e^{2\pi i p/n} \). The number of terms in the series is \( n \).

4. **Using the Formula for the Sum of a Geometric Series**:
The sum of a geometric series can be calculated using the formula:
\[ S = \frac{a(1 - r^n)}{1 - r} \]
where \( a \) is the first term and \( r \) is the common ratio. Here, \( a = 1 \) and \( r = e^{2\pi i p/n} \):
\[ S = \frac{1 - (e^{2\pi i p/n})^n}{1 - e^{2\pi i p/n}} = \frac{1 - e^{2\pi i p}}{1 - e^{2\pi i p/n}} \]

5. **Simplifying the Expression**:
Since \( e^{2\pi i p} = 1 \) (as \( p \) is an integer), we have:
\[ S = \frac{1 - 1}{1 - e^{2\pi i p/n}} = \frac{0}{1 - e^{2\pi i p/n}} = 0 \]

6. **Conclusion**:
Therefore, the sum of the p-th powers of the n-th roots of unity when \( p \) is not a multiple of \( n \) is:
\[ \boxed{0} \]