If ` alpha _(1), alpha_(2), . . . alpha _(100)` are all the 100 th roots of unity, then ` sum sum (alpha_(i) alpha_(j)) ^(5) 1 le i lt j le 100`

Options

A. 20
B. `(20)^(1//20)`
C. 0
D. none

Answer

To solve the problem, we need to find the value of the double summation:

\[ \sum_{1 \leq i < j \leq 100} (\alpha_i \alpha_j)^5 \]

where \(\alpha_1, \alpha_2, \ldots, \alpha_{100}\) are the 100th roots of unity.

### Step-by-step Solution:

1. **Understanding the Roots of Unity**:
The 100th roots of unity are given by:
\[ \alpha_k = e^{2\pi i k / 100} \quad \text{for } k = 0, 1, 2, \ldots, 99 \]
These roots satisfy the property that \(\alpha_k^{100} = 1\).

2. **Expressing the Summation**:
We can rewrite the double summation:
\[ \sum_{1 \leq i < j \leq 100} (\alpha_i \alpha_j)^5 = \frac{1}{2} \sum_{i=1}^{100} \sum_{j=1}^{100} (\alpha_i \alpha_j)^5 \]
This is because each pair \((i, j)\) is counted twice in the double summation.

3. **Simplifying the Inner Summation**:
The inner summation can be expressed as:
\[ \sum_{i=1}^{100} \sum_{j=1}^{100} (\alpha_i \alpha_j)^5 = \sum_{i=1}^{100} \sum_{j=1}^{100} \alpha_i^5 \alpha_j^5 \]
This can be factored into:
\[ \left( \sum_{i=1}^{100} \alpha_i^5 \right) \left( \sum_{j=1}^{100} \alpha_j^5 \right) \]

4. **Calculating the Sum of the Powers**:
The sum of the 100th roots of unity raised to any power \(k\) that is not a multiple of 100 is zero:
\[ \sum_{k=0}^{99} \alpha_k^5 = 0 \quad \text{(since 5 is not a multiple of 100)} \]
Therefore:
\[ \sum_{i=1}^{100} \alpha_i^5 = 0 \]

5. **Final Calculation**:
Since both sums are zero:
\[ \sum_{i=1}^{100} \sum_{j=1}^{100} (\alpha_i \alpha_j)^5 = 0 \cdot 0 = 0 \]
Thus, we have:
\[ \frac{1}{2} \sum_{i=1}^{100} \sum_{j=1}^{100} (\alpha_i \alpha_j)^5 = \frac{1}{2} \cdot 0 = 0 \]

### Conclusion:
The final result is:
\[ \sum_{1 \leq i < j \leq 100} (\alpha_i \alpha_j)^5 = 0 \]