Let `omega` be the complex number `cos((2pi)/3)+isin((2pi)/3)`. Then the number of distinct complex cos numbers z satisfying `Delta=|(z+1,omega,omega^2),(omega,z+omega^2,1),(omega^2,1,z+omega)|=0` is

Explanation

To solve the problem, we need to analyze the determinant given and find the number of distinct complex numbers \( z \) that satisfy the equation \( \Delta = 0 \).

### Step-by-Step Solution

1. **Identify \( \omega \)**:
\[ \omega = \cos\left(\frac{2\pi}{3}\right) + i\sin\left(\frac{2\pi}{3}\right) = -\frac{1}{2} + i\frac{\sqrt{3}}{2} \]
We know that \( \omega^3 = 1 \) and \( 1 + \omega + \omega^2 = 0 \).

2. **Set up the determinant**:
\[ \Delta = \begin{vmatrix} z + 1 & \omega & \omega^2 \\ \omega & z + \omega^2 & 1 \\ \omega^2 & 1 & z + \omega \end{vmatrix} \]
We want to find when this determinant equals zero.

3. **Apply column operations**:
Replace the first column \( C_1 \) with \( C_1 + C_2 + C_3 \):
\[ C_1 = (z + 1) + \omega + \omega^2 = z + 1 + 0 = z + 1 \]
The determinant simplifies to:
\[ \Delta = \begin{vmatrix} z + 1 & \omega & \omega^2 \\ \omega & z + \omega^2 & 1 \\ \omega^2 & 1 & z + \omega \end{vmatrix} \]

4. **Rearranging the determinant**:
Now we can express the determinant as:
\[ \Delta = \begin{vmatrix} z + 1 & \omega & \omega^2 \\ 0 & z + \omega^2 - (z + 1) & 1 - \omega \\ 0 & 1 - \omega^2 & z + \omega - (z + 1) \end{vmatrix} \]
This simplifies to:
\[ \Delta = \begin{vmatrix} z + 1 & \omega & \omega^2 \\ 0 & \omega^2 - 1 & 1 - \omega \\ 0 & 1 - \omega^2 & \omega - 1 \end{vmatrix} \]

5. **Calculate the determinant**:
The determinant can be simplified further:
\[ \Delta = (z + 1) \begin{vmatrix} \omega^2 - 1 & 1 - \omega \\ 1 - \omega^2 & \omega - 1 \end{vmatrix} \]
We can compute this 2x2 determinant:
\[ = (z + 1) \left( (\omega^2 - 1)(\omega - 1) - (1 - \omega)(1 - \omega^2) \right) \]

6. **Setting the determinant to zero**:
For \( \Delta = 0 \):
\[ (z + 1) \cdot \text{(some expression)} = 0 \]
This gives us two cases:
- \( z + 1 = 0 \) which gives \( z = -1 \).
- The second part leads to a polynomial in \( z \).

7. **Finding roots**:
The polynomial formed from the second part will yield distinct roots. Since we know \( \omega^3 = 1 \), we can find the roots of this polynomial.

8. **Count distinct roots**:
The polynomial will have three roots due to the cubic nature, and since \( z = -1 \) is one distinct solution, we conclude that there are a total of 4 distinct solutions.

### Final Answer
The number of distinct complex numbers \( z \) satisfying \( \Delta = 0 \) is **4**.