If `z_(1),z_(2)andz_(3)` represent the vertices of an equilateral triangle such that `|z_(1)|=|z_(2)|=|z_(3)|`, then

Options

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

Answer

To solve the problem, we need to show that if \( z_1, z_2, z_3 \) are the vertices of an equilateral triangle and \( |z_1| = |z_2| = |z_3| \), then the sum of these complex numbers is zero, i.e., \( z_1 + z_2 + z_3 = 0 \).

### Step-by-Step Solution:

1. **Understanding the Geometry**:
- Let \( z_1, z_2, z_3 \) be the vertices of an equilateral triangle centered at the origin (the circumcenter of the triangle).
- Since the triangle is equilateral and centered at the origin, the distances from the origin to each vertex are equal.

2. **Expressing the Vertices**:
- We can express the vertices in terms of a common radius \( r \) (which is the distance from the origin to each vertex):
\[ z_1 = r e^{i\theta}, \quad z_2 = r e^{i(\theta + \frac{2\pi}{3})}, \quad z_3 = r e^{i(\theta + \frac{4\pi}{3})} \]
- Here, \( \theta \) is the angle corresponding to \( z_1 \), and the other vertices are obtained by rotating \( z_1 \) by \( 120^\circ \) and \( 240^\circ \).

3. **Calculating the Sum**:
- Now, we can calculate the sum of these complex numbers:
\[ z_1 + z_2 + z_3 = r e^{i\theta} + r e^{i(\theta + \frac{2\pi}{3})} + r e^{i(\theta + \frac{4\pi}{3})} \]
- Factoring out \( r \):
\[ = r \left( e^{i\theta} + e^{i(\theta + \frac{2\pi}{3})} + e^{i(\theta + \frac{4\pi}{3})} \right) \]

4. **Using the Property of Roots of Unity**:
- The expression inside the parentheses is a sum of the cube roots of unity:
\[ e^{i\theta} + e^{i(\theta + \frac{2\pi}{3})} + e^{i(\theta + \frac{4\pi}{3})} = 0 \]
- This is because the sum of all the cube roots of unity is zero.

5. **Conclusion**:
- Therefore, we have:
\[ z_1 + z_2 + z_3 = r \cdot 0 = 0 \]
- Hence, we conclude that:
\[ z_1 + z_2 + z_3 = 0 \]

### Final Result:
Thus, if \( z_1, z_2, z_3 \) are the vertices of an equilateral triangle with equal magnitudes, then:
\[ z_1 + z_2 + z_3 = 0 \]