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

Options

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

Answer

To solve the problem, we will follow these steps:

### Step 1: Find the roots of the equation \(x^4 - 1 = 0\)
The roots of the equation \(x^4 - 1 = 0\) can be found by factoring it as follows:
\[ x^4 - 1 = (x^2 - 1)(x^2 + 1) = (x - 1)(x + 1)(x - i)(x + i) = 0 \]
The roots are:
\[ x = 1, -1, i, -i \]

### Step 2: Identify the root with a negative principal argument
Among the roots, the complex numbers \(i\) and \(-i\) have arguments. The principal argument of \(i\) is \(\frac{\pi}{2}\) and for \(-i\) it is \(-\frac{\pi}{2}\). Since we need the root with a negative principal argument, we choose:
\[ \alpha = -i = e^{-i\frac{\pi}{2}} \]

### Step 3: Define the matrix \(D(\alpha)\)
The matrix \(D(\alpha)\) is given as:
\[ D(\alpha) = \begin{pmatrix} 1 & 1 & 1 \\ \alpha^n & \alpha^{n+1} & \alpha^{n+3} \\ \frac{1}{\alpha^{n+1}} & \frac{1}{\alpha^n} & 0 \end{pmatrix} \]

### Step 4: Compute the determinant of \(D(\alpha)\)
We will compute the determinant of the matrix \(D(\alpha)\) using the first row for expansion:
\[ \text{det}(D(\alpha)) = 1 \cdot \text{det}\begin{pmatrix} \alpha^{n+1} & \alpha^{n+3} \\ \frac{1}{\alpha^{n+1}} & 0 \end{pmatrix} - 1 \cdot \text{det}\begin{pmatrix} \alpha^n & \alpha^{n+3} \\ \frac{1}{\alpha^{n+1}} & 0 \end{pmatrix} + 1 \cdot \text{det}\begin{pmatrix} \alpha^n & \alpha^{n+1} \\ \frac{1}{\alpha^{n}} & \frac{1}{\alpha^{n+1}} \end{pmatrix} \]

Calculating the determinants:
1. The first determinant:
\[ \text{det}\begin{pmatrix} \alpha^{n+1} & \alpha^{n+3} \\ \frac{1}{\alpha^{n+1}} & 0 \end{pmatrix} = 0 - \alpha^{n+1} \cdot \frac{1}{\alpha^{n+1}} \cdot \alpha^{n+3} = -\alpha^{n+3} \]

2. The second determinant:
\[ \text{det}\begin{pmatrix} \alpha^n & \alpha^{n+3} \\ \frac{1}{\alpha^{n+1}} & 0 \end{pmatrix} = 0 - \alpha^n \cdot \frac{1}{\alpha^{n+1}} \cdot \alpha^{n+3} = -\alpha^{n+2} \]

3. The third determinant:
\[ \text{det}\begin{pmatrix} \alpha^n & \alpha^{n+1} \\ \frac{1}{\alpha^{n}} & \frac{1}{\alpha^{n+1}} \end{pmatrix} = \frac{1}{\alpha^{n}} \cdot \alpha^{n+1} - \alpha^{n} \cdot \frac{1}{\alpha^{n+1}} = \alpha - \frac{1}{\alpha} \]

Combining these results, we have:
\[ \text{det}(D(\alpha)) = -\alpha^{n+3} + \alpha^{n+2} + \left(\alpha - \frac{1}{\alpha}\right) \]

### Step 5: Substitute \(\alpha = e^{-i\frac{\pi}{2}}\)
Now substituting \(\alpha\):
\[ \text{det}(D(\alpha)) = -\left(e^{-i\frac{\pi}{2}}\right)^{n+3} + \left(e^{-i\frac{\pi}{2}}\right)^{n+2} + \left(e^{-i\frac{\pi}{2}} - e^{i\frac{\pi}{2}}\right) \]

### Step 6: Simplify and find the principal argument
The principal argument of the determinant can be found from the final expression. The principal argument will be:
\[ \text{arg}(\text{det}(D(\alpha))) = \text{arg}(-1) + \text{arg}(-i) = -\frac{3\pi}{4} \]

### Final Answer
Thus, the principal argument of \(D(\alpha)\) is:
\[ \text{arg}(D(\alpha)) = -\frac{3\pi}{4} \]

---