If `z_(1),z_(2),z_(3)` are three complex numbers and `A=|{:("arg z"_(1),"arg z"_(2),"arg z"_(3)),("arg z"_(2),"arg z"_(3),"arg z"_(1)),("arg z"_(3),"arg z"_(1),"arg z"_(2)):}|` then A is divisible by

Options

A. `"arg "(z_(1)+z_(2)+z_(3))`
B. `"arg "(z_(1)z_(2)z_(3))`
C. for all number
D. can't say

Answer

To solve the problem, we need to evaluate the determinant of the matrix formed by the arguments of three complex numbers \( z_1, z_2, z_3 \). The matrix \( A \) is given as:

\[ A = \begin{vmatrix} \arg z_1 & \arg z_2 & \arg z_3 \\ \arg z_2 & \arg z_3 & \arg z_1 \\ \arg z_3 & \arg z_1 & \arg z_2 \end{vmatrix} \]

Let’s denote:
- \( \theta_1 = \arg z_1 \)
- \( \theta_2 = \arg z_2 \)
- \( \theta_3 = \arg z_3 \)

Thus, we can rewrite the matrix \( A \) as:

\[ A = \begin{vmatrix} \theta_1 & \theta_2 & \theta_3 \\ \theta_2 & \theta_3 & \theta_1 \\ \theta_3 & \theta_1 & \theta_2 \end{vmatrix} \]

### Step 1: Calculate the Determinant
To calculate the determinant, we can use the property of determinants that allows us to add rows or columns. We will add all three rows together.

\[ R_1 \rightarrow R_1 + R_2 + R_3 \]

This gives us:

\[ A = \begin{vmatrix} \theta_1 + \theta_2 + \theta_3 & \theta_1 + \theta_2 + \theta_3 & \theta_1 + \theta_2 + \theta_3 \\ \theta_2 & \theta_3 & \theta_1 \\ \theta_3 & \theta_1 & \theta_2 \end{vmatrix} \]

### Step 2: Factor Out Common Terms
Now, we can factor out \( \theta_1 + \theta_2 + \theta_3 \) from the first row:

\[ A = (\theta_1 + \theta_2 + \theta_3) \begin{vmatrix} 1 & 1 & 1 \\ \theta_2 & \theta_3 & \theta_1 \\ \theta_3 & \theta_1 & \theta_2 \end{vmatrix} \]

### Step 3: Evaluate the Remaining Determinant
Now we need to evaluate the remaining determinant:

\[ B = \begin{vmatrix} 1 & 1 & 1 \\ \theta_2 & \theta_3 & \theta_1 \\ \theta_3 & \theta_1 & \theta_2 \end{vmatrix} \]

Using the properties of determinants, we can expand this determinant. Notice that if we subtract the first column from the second and third columns, we get:

\[ B = \begin{vmatrix} 1 & 0 & 0 \\ \theta_2 & \theta_3 - \theta_2 & \theta_1 - \theta_2 \\ \theta_3 & \theta_1 - \theta_3 & \theta_2 - \theta_3 \end{vmatrix} \]

### Step 4: Calculate the Final Result
The determinant \( B \) can be calculated, but we can also see that the first column has a common factor of 1, and the remaining terms will yield a result involving differences of the arguments.

Thus, we conclude that:

\[ A = (\theta_1 + \theta_2 + \theta_3) \cdot \text{(some expression involving differences of } \theta_i \text{)} \]

### Conclusion
Since \( \theta_1 + \theta_2 + \theta_3 \) is a common factor, the determinant \( A \) is divisible by \( \theta_1 + \theta_2 + \theta_3 \).