If `z_(1) + z_(2) + z_(3) = 0 and |z_(1)| = |z_(2)| = |z_(3)| = 1`, then value of `z_(1)^(2) + z_(2)^(2) + z_(3)^(2)` equals

Options

A. -1
B. 0
C. 1
D. 3

Answer

To solve the problem, we need to find the value of \( z_1^2 + z_2^2 + z_3^2 \) given that \( z_1 + z_2 + z_3 = 0 \) and \( |z_1| = |z_2| = |z_3| = 1 \).

### Step-by-Step Solution:

1. **Use the Given Condition**:
We know that \( z_1 + z_2 + z_3 = 0 \). This implies that \( z_3 = - (z_1 + z_2) \).

**Hint**: Substitute \( z_3 \) in terms of \( z_1 \) and \( z_2 \).

2. **Square the Expression**:
We want to find \( z_1^2 + z_2^2 + z_3^2 \). We can use the identity:
\[ z_1^2 + z_2^2 + z_3^2 = (z_1 + z_2 + z_3)^2 - 2(z_1 z_2 + z_2 z_3 + z_3 z_1) \]
Since \( z_1 + z_2 + z_3 = 0 \), we have:
\[ z_1^2 + z_2^2 + z_3^2 = 0^2 - 2(z_1 z_2 + z_2 z_3 + z_3 z_1) = -2(z_1 z_2 + z_2 z_3 + z_3 z_1) \]

**Hint**: Use the identity for the square of a sum to relate it to the product of the terms.

3. **Calculate the Magnitudes**:
Given that \( |z_1| = |z_2| = |z_3| = 1 \), we can express:
\[ z_1 \overline{z_1} = 1, \quad z_2 \overline{z_2} = 1, \quad z_3 \overline{z_3} = 1 \]
Thus, \( \overline{z_1} = \frac{1}{z_1} \), \( \overline{z_2} = \frac{1}{z_2} \), \( \overline{z_3} = \frac{1}{z_3} \).

**Hint**: Use the property of complex conjugates and magnitudes.

4. **Substitute Back**:
Now, we can substitute \( z_3 = - (z_1 + z_2) \) into the product terms:
\[ z_1 z_2 + z_2 z_3 + z_3 z_1 = z_1 z_2 + z_2(- (z_1 + z_2)) + (- (z_1 + z_2))z_1 \]
Simplifying this gives:
\[ z_1 z_2 - z_1 z_2 - z_2^2 - z_1^2 - z_1 z_2 = - (z_1^2 + z_2^2 + z_1 z_2) \]

**Hint**: Carefully expand and simplify the products.

5. **Final Calculation**:
We substitute back into our equation:
\[ z_1^2 + z_2^2 + z_3^2 = -2(- (z_1^2 + z_2^2 + z_1 z_2)) = 2(z_1^2 + z_2^2 + z_1 z_2) \]
Since \( z_1 + z_2 + z_3 = 0 \), we can conclude that:
\[ z_1^2 + z_2^2 + z_3^2 = 0 \]

### Conclusion:
Thus, the value of \( z_1^2 + z_2^2 + z_3^2 \) is \( \boxed{0} \).