If `i=sqrt(-1) and (1)^(1//4) = 1, omega , omega^(z) , omega^3` then
`Delta=|(1,omega,omega^2,omega^3),(omega,omega^2,omega^3,1),(omega^2,omega^3,1,omega),(omega^3,1,omega,omega^2)|` is equal to

Options

A. `i`
B. `-i`
C. 1
D. 0

Answer

To solve the problem, we need to calculate the determinant \( \Delta \) given by:

\[ \Delta = \begin{vmatrix} 1 & \omega & \omega^2 & \omega^3 \\ \omega & \omega^2 & \omega^3 & 1 \\ \omega^2 & \omega^3 & 1 & \omega \\ \omega^3 & 1 & \omega & \omega^2 \end{vmatrix} \]

where \( \omega \) is a primitive 4th root of unity, meaning \( \omega = e^{i\frac{\pi}{2}} = i \) and \( \omega^4 = 1 \).

### Step 1: Understanding the Roots of Unity
The roots of unity are:
- \( 1 \)
- \( \omega = i \)
- \( \omega^2 = -1 \)
- \( \omega^3 = -i \)

### Step 2: Setting Up the Determinant
We can rewrite the determinant using the values of \( \omega \):
\[ \Delta = \begin{vmatrix} 1 & i & -1 & -i \\ i & -1 & -i & 1 \\ -1 & -i & 1 & i \\ -i & 1 & i & -1 \end{vmatrix} \]

### Step 3: Performing Column Operations
To simplify the determinant, we can perform column operations. We can add all columns together to the first column:
\[ C_1 \rightarrow C_1 + C_2 + C_3 + C_4 \]

This gives us:
\[ C_1 = 1 + i - 1 - i = 0 \]
\[ C_2 = i - 1 - i + 1 = 0 \]
\[ C_3 = -1 - i + 1 + i = 0 \]
\[ C_4 = -i + 1 + i - 1 = 0 \]

### Step 4: Resulting Determinant
After performing the column operations, we find that the first column becomes zero. Therefore, the determinant simplifies to:
\[ \Delta = \begin{vmatrix} 0 & 0 & 0 & 0 \\ \text{...} & \text{...} & \text{...} & \text{...} \\ \text{...} & \text{...} & \text{...} & \text{...} \\ \text{...} & \text{...} & \text{...} & \text{...} \end{vmatrix} = 0 \]

### Conclusion
Thus, the value of \( \Delta \) is:
\[ \Delta = 0 \]