If `z_(1),z_(2),z_(3)` be vertices of an equilateral triangle occurig in the anticlockwise sense, then

Options

A. `z_(1)^(2)+z_(2)^(2)+z_(3)^(2)=2(z_(1)z_(2)+z_(2)z_(3)+z_(3)z_(1))`
B. `1/(z_(1)+z_(2))+1/(z_(2)+z_(3))+1/(z_(3)+z_(1))=0`
C. `z_(1)+omegaz_(2)+omega^(2)z_(3)=0`
D. none of these

Answer

To solve the problem, we will verify the condition for the vertices \( z_1, z_2, z_3 \) of an equilateral triangle occurring in the anti-clockwise sense.

### Step-by-step Solution:

1. **Understanding the Geometry**:
Let \( z_1, z_2, z_3 \) be the vertices of the equilateral triangle. We can represent these points in the complex plane.

2. **Defining the Vectors**:
The vectors representing the sides of the triangle can be defined as:
- \( \overrightarrow{AB} = z_2 - z_1 \)
- \( \overrightarrow{BC} = z_3 - z_2 \)
- \( \overrightarrow{CA} = z_1 - z_3 \)

3. **Rotation of Vectors**:
In an equilateral triangle, the angle between any two sides is \( \frac{\pi}{3} \) radians (or 60 degrees). Therefore, we can express the relationship between the vectors:
\[ \frac{z_3 - z_2}{|z_3 - z_2|} = e^{i \frac{\pi}{3}} \cdot \frac{z_2 - z_1}{|z_2 - z_1|} \]
This means that the vector \( z_3 - z_2 \) is a rotation of the vector \( z_2 - z_1 \) by \( \frac{\pi}{3} \).

4. **Using the Exponential Form**:
Using Euler's formula, we can write:
\[ e^{i \frac{\pi}{3}} = \cos\left(\frac{\pi}{3}\right) + i \sin\left(\frac{\pi}{3}\right) = \frac{1}{2} + i \frac{\sqrt{3}}{2} \]

5. **Setting Up the Equation**:
From the above relationship, we have:
\[ z_3 - z_2 = e^{i \frac{\pi}{3}} (z_2 - z_1) \]
Substituting the value of \( e^{i \frac{\pi}{3}} \):
\[ z_3 - z_2 = \left(\frac{1}{2} + i \frac{\sqrt{3}}{2}\right)(z_2 - z_1) \]

6. **Rearranging the Equation**:
Rearranging gives:
\[ z_3 - z_2 = \frac{1}{2}(z_2 - z_1) + i \frac{\sqrt{3}}{2}(z_2 - z_1) \]

7. **Finding the Relationship**:
We can express this relationship in terms of \( z_1, z_2, z_3 \):
\[ z_1 + \omega z_2 + \omega^2 z_3 = 0 \]
where \( \omega = e^{i \frac{2\pi}{3}} \) is a primitive cube root of unity.

8. **Conclusion**:
Thus, we conclude that:
\[ z_1 + \omega z_2 + \omega^2 z_3 = 0 \]
This confirms that the vertices of the equilateral triangle satisfy the given condition.