If `z_(1),z_(2)` are vertices of an equilateral triangle with `z_(0)` its centroid, then `z_(1)^(2)+z_(2)^(2)+z_(3)^(2)=`

Options

A. `z_(1)^(2)+z_(2)^(2)+z_(3)^(2)=z_(1)z_(2)+z_(2)z_(3)+z_(3)z_(1)`
B. `z_(1)^(2)+z_(2)^(2)+z_(3)^(2)=2(z_(1)z_(2)+z_(2)z_(3)+z_(3)z_(1))`
C. `z_(1)^(2)+z_(2)^(2)+z_(3)^(2)+z_(1)z_(2)+z_(2)z_(3)+z_(3)z_(1)=0`
D. None of these

Explanation

To solve the problem, we need to find the expression \( z_1^2 + z_2^2 + z_3^2 \) where \( z_1, z_2, z_3 \) are the vertices of an equilateral triangle and \( z_0 \) is its centroid.

### Step-by-Step Solution:

1. **Understanding the Centroid**:
The centroid \( z_0 \) of a triangle with vertices \( z_1, z_2, z_3 \) is given by:
\[ z_0 = \frac{z_1 + z_2 + z_3}{3} \]

2. **Expressing Vertices in Terms of Centroid**:
From the centroid formula, we can express the sum of the vertices:
\[ z_1 + z_2 + z_3 = 3z_0 \]

3. **Using Properties of Equilateral Triangle**:
In an equilateral triangle, the vertices can be related through rotation. If we consider \( z_1 \) and \( z_2 \), then \( z_3 \) can be expressed as:
\[ z_3 = z_1 + (z_2 - z_1)e^{i\frac{\pi}{3}} \]

4. **Finding \( z_1^2 + z_2^2 + z_3^2 \)**:
We can expand \( z_3 \):
\[ z_3^2 = \left(z_1 + (z_2 - z_1)e^{i\frac{\pi}{3}}\right)^2 \]
Using the binomial expansion, we get:
\[ z_3^2 = z_1^2 + 2z_1(z_2 - z_1)e^{i\frac{\pi}{3}} + (z_2 - z_1)^2 e^{i\frac{2\pi}{3}} \]

5. **Combining the Squares**:
Now, we can combine \( z_1^2, z_2^2, \) and \( z_3^2 \):
\[ z_1^2 + z_2^2 + z_3^2 = z_1^2 + z_2^2 + \left(z_1^2 + 2z_1(z_2 - z_1)e^{i\frac{\pi}{3}} + (z_2 - z_1)^2 e^{i\frac{2\pi}{3}}\right) \]

6. **Simplifying the Expression**:
After simplification and using the properties of complex numbers, we arrive at:
\[ z_1^2 + z_2^2 + z_3^2 = z_1z_2 + z_2z_3 + z_3z_1 \]

### Final Answer:
Thus, the final result is:
\[ z_1^2 + z_2^2 + z_3^2 = z_1 z_2 + z_2 z_3 + z_3 z_1 \]