ABCD is a rhombus in the Argand plane. If the affixes of the vertices are `z_(1),z_(2),z_(3)` and `z_(4)` respectively, and `angleCBA=pi//3`, then

Options

A. `z_(1)+omegaz_(2)=omega^(2)z_(3)=0`
B. `z_(1)-omegaz_(2)-omega^(2)z_(3)=0`
C. `omegaz_(1)+z_(2)+omega^(2)z_(3)=0`
D. `omega^(2)z_(1)+omegaz_(2)+z_(3)=0`

Explanation

To solve the problem, we need to analyze the properties of the rhombus in the Argand plane and utilize the given angle information. Here’s a step-by-step solution:

### Step-by-Step Solution

1. **Understanding the Rhombus**:
A rhombus has all sides equal and its diagonals bisect each other at right angles. Let the vertices of the rhombus be represented by the complex numbers (affixes) \( z_1, z_2, z_3, z_4 \).

2. **Identifying Angles**:
We are given that \( \angle CBA = \frac{\pi}{3} \). This means that the angle formed at vertex B between the lines BA and BC is \( \frac{\pi}{3} \).

3. **Using Vector Representation**:
We can express the vectors:
- \( \overrightarrow{BA} = z_1 - z_2 \)
- \( \overrightarrow{BC} = z_3 - z_2 \)

4. **Using Rotation**:
Since \( \angle CBA = \frac{\pi}{3} \), we can represent the rotation of vector \( BC \) to get vector \( BA \). This can be expressed in terms of complex exponentials:
\[ \overrightarrow{BA} = e^{i \frac{\pi}{3}} \cdot \overrightarrow{BC} \]

5. **Substituting the Vectors**:
Substitute the expressions for \( \overrightarrow{BA} \) and \( \overrightarrow{BC} \):
\[ z_1 - z_2 = e^{i \frac{\pi}{3}} (z_3 - z_2) \]

6. **Expressing \( e^{i \frac{\pi}{3}} \)**:
We know that:
\[ e^{i \frac{\pi}{3}} = \cos\left(\frac{\pi}{3}\right) + i \sin\left(\frac{\pi}{3}\right) = \frac{1}{2} + i \frac{\sqrt{3}}{2} \]

7. **Setting Up the Equation**:
Thus, we can rewrite the equation:
\[ z_1 - z_2 = \left(\frac{1}{2} + i \frac{\sqrt{3}}{2}\right)(z_3 - z_2) \]

8. **Rearranging the Equation**:
Rearranging gives us:
\[ z_1 - z_2 = \frac{1}{2}(z_3 - z_2) + i \frac{\sqrt{3}}{2}(z_3 - z_2) \]
This can be further simplified to:
\[ z_1 - z_2 = \frac{1}{2}z_3 - \frac{1}{2}z_2 + i \frac{\sqrt{3}}{2}z_3 - i \frac{\sqrt{3}}{2}z_2 \]

9. **Combining Terms**:
Collecting like terms, we can express this as:
\[ z_1 = z_2 + \frac{1}{2}z_3 - \frac{1}{2}z_2 + i \frac{\sqrt{3}}{2}z_3 - i \frac{\sqrt{3}}{2}z_2 \]

10. **Final Expression**:
This leads us to the final expression which relates \( z_1, z_2, \) and \( z_3 \) in the context of the rhombus and the angle given.