If p is a multiple of n , then the sum of pth powers of nth roots of unity is

Options

A. p
B. n
C. 0
D. none of these

Answer

To solve the problem, we need to find the sum of the p-th powers of the n-th roots of unity, given that \( p \) is a multiple of \( n \).

### Step-by-step Solution:

1. **Understanding n-th Roots of Unity**:
The n-th roots of unity are given by the formula:
\[ z_k = e^{i \frac{2\pi k}{n}} \quad \text{for } k = 0, 1, 2, \ldots, n-1 \]
These roots satisfy the equation \( z^n = 1 \).

2. **Finding the p-th Powers**:
We need to calculate the p-th power of each n-th root of unity:
\[ z_k^p = \left(e^{i \frac{2\pi k}{n}}\right)^p = e^{i \frac{2\pi k p}{n}} \]

3. **Sum of p-th Powers**:
The sum \( S \) of the p-th powers of the n-th roots of unity can be expressed as:
\[ S = \sum_{k=0}^{n-1} z_k^p = \sum_{k=0}^{n-1} e^{i \frac{2\pi k p}{n}} \]

4. **Using the Property of Roots of Unity**:
Since \( p \) is a multiple of \( n \), we can write \( p = mn \) for some integer \( m \). Thus, we have:
\[ S = \sum_{k=0}^{n-1} e^{i \frac{2\pi k (mn)}{n}} = \sum_{k=0}^{n-1} e^{i 2\pi k m} \]
Because \( e^{i 2\pi k m} = 1 \) for any integer \( m \), we can simplify:
\[ S = \sum_{k=0}^{n-1} 1 = n \]

5. **Conclusion**:
Therefore, the sum of the p-th powers of the n-th roots of unity is:
\[ S = n \]

### Final Answer:
The sum of the p-th powers of the n-th roots of unity, when \( p \) is a multiple of \( n \), is \( n \).

---