If `alpha` is a root of `x^4 = 1` with negative principal argument then the principal argument of `Delta(alpha) = |(1,1,1), (alpha^n, alpha^(n+1), alpha^(n+3)), (1/alpha^(n+1), 1/alpha^n, 0)|` is

Options

A. `(5pi)/(14)`
B. `-(3pi)/(4)`
C. `(pi)/(4)`
D. `-(pi)/(4)`

Explanation

To find the principal argument of the determinant \( \Delta(\alpha) = \begin{vmatrix} 1 & 1 & 1 \\ \alpha^n & \alpha^{n+1} & \alpha^{n+3} \\ \frac{1}{\alpha^{n+1}} & \frac{1}{\alpha^n} & 0 \end{vmatrix} \), where \( \alpha \) is a root of \( x^4 = 1 \) with a negative principal argument, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Roots of \( x^4 = 1 \)**:
The roots of the equation \( x^4 = 1 \) are \( 1, i, -1, -i \). Since we need the root with a negative principal argument, we choose \( \alpha = -i \).

**Hint**: Remember that the principal argument of a complex number is the angle it makes with the positive x-axis, measured in radians.

2. **Substitute \( \alpha \) into the Determinant**:
We substitute \( \alpha = -i \) into the determinant:
\[ \Delta(-i) = \begin{vmatrix} 1 & 1 & 1 \\ (-i)^n & (-i)^{n+1} & (-i)^{n+3} \\ \frac{1}{(-i)^{n+1}} & \frac{1}{(-i)^n} & 0 \end{vmatrix} \]

**Hint**: Recall that \( (-i)^k = e^{-i\frac{\pi}{2}k} \) for integer \( k \).

3. **Calculate the Entries of the Determinant**:
The entries of the determinant become:
- First row: \( 1, 1, 1 \)
- Second row: \( (-i)^n, (-i)^{n+1}, (-i)^{n+3} \)
- Third row: \( \frac{1}{(-i)^{n+1}}, \frac{1}{(-i)^n}, 0 \)

**Hint**: Use properties of exponents to simplify \( (-i)^k \).

4. **Expand the Determinant**:
We can use the determinant formula:
\[ \Delta(-i) = 1 \cdot \begin{vmatrix} (-i)^{n+1} & (-i)^{n+3} \\ \frac{1}{(-i)^{n+1}} & 0 \end{vmatrix} - 1 \cdot \begin{vmatrix} (-i)^n & (-i)^{n+3} \\ \frac{1}{(-i)^{n}} & 0 \end{vmatrix} + 1 \cdot \begin{vmatrix} (-i)^n & (-i)^{n+1} \\ \frac{1}{(-i)^{n}} & \frac{1}{(-i)^{n+1}} \end{vmatrix} \]

**Hint**: Remember that the determinant of a 2x2 matrix \( \begin{vmatrix} a & b \\ c & d \end{vmatrix} \) is calculated as \( ad - bc \).

5. **Calculate Each 2x2 Determinant**:
- For the first determinant:
\[ \begin{vmatrix} (-i)^{n+1} & (-i)^{n+3} \\ \frac{1}{(-i)^{n+1}} & 0 \end{vmatrix} = 0 - (-i)^{n+1} \cdot \frac{1}{(-i)^{n+3}} = \frac{(-i)^{n+1}}{(-i)^{n+3}} = (-i)^{-2} = -1 \]
- For the second determinant:
\[ \begin{vmatrix} (-i)^n & (-i)^{n+3} \\ \frac{1}{(-i)^{n}} & 0 \end{vmatrix} = 0 - (-i)^n \cdot \frac{1}{(-i)^{n+3}} = \frac{(-i)^{n}}{(-i)^{n+3}} = (-i)^{-3} = i \]
- For the third determinant:
\[ \begin{vmatrix} (-i)^n & (-i)^{n+1} \\ \frac{1}{(-i)^{n}} & \frac{1}{(-i)^{n+1}} \end{vmatrix} = \frac{(-i)^{n}}{(-i)^{n+1}} - \frac{(-i)^{n+1}}{(-i)^{n}} = -1 + 1 = 0 \]

**Hint**: Pay attention to the signs and the properties of complex numbers.

6. **Combine the Results**:
Putting it all together:
\[ \Delta(-i) = 1(-1) - 1(i) + 1(0) = -1 - i \]

**Hint**: Ensure you keep track of the signs when combining terms.

7. **Find the Principal Argument**:
The principal argument of \( -1 - i \) can be found using the formula:
\[ \text{arg}(-1 - i) = \tan^{-1}\left(\frac{-1}{-1}\right) + \pi = \tan^{-1}(1) + \pi = \frac{\pi}{4} + \pi = \frac{5\pi}{4} \]

**Hint**: The angle is in the third quadrant where both sine and cosine are negative.

### Final Answer:
The principal argument of \( \Delta(\alpha) \) is \( \frac{5\pi}{4} \).