Let `z_(1),z_(2) and z_(3)` be non-zero complex numbers satisfying `z^(2)=bar(iz), where i=sqrt(-1).`
Consider the following statements:
1. `z_(1)z_(2)z_(3) ` is purely imaginary.
2. `z_(1)z_(2)+z_(2)z_(3)+z_(3)z_(1)` is purely real.
Which of the above statements is/are correct?

Options

A. 1 only
B. 2 only
C. Both 1 and 2
D. neither 1 nor 2

Explanation

To solve the problem, we need to analyze the given equation and the statements related to the complex numbers \( z_1, z_2, z_3 \).

### Step 1: Understand the given equation
The equation given is:
\[ z^2 = i \bar{z} \]
where \( \bar{z} \) is the conjugate of \( z \).

### Step 2: Express \( z \) in terms of its real and imaginary parts
Let \( z = x + iy \), where \( x \) and \( y \) are real numbers. Then the conjugate \( \bar{z} = x - iy \).

### Step 3: Substitute \( z \) and \( \bar{z} \) into the equation
Substituting into the equation gives:
\[ (x + iy)^2 = i(x - iy) \]

### Step 4: Expand both sides
Expanding the left side:
\[ x^2 + 2xyi - y^2 = i(x - iy) \]
Expanding the right side:
\[ ix + y \]

### Step 5: Equate real and imaginary parts
From the left side, we have:
\[ x^2 - y^2 + 2xyi = ix + y \]
This gives us two equations:
1. Real part: \( x^2 - y^2 = y \)
2. Imaginary part: \( 2xy = x \)

### Step 6: Solve the imaginary part equation
From \( 2xy = x \), we can factor out \( x \):
\[ x(2y - 1) = 0 \]
This gives us two cases:
1. \( x = 0 \)
2. \( 2y - 1 = 0 \) which implies \( y = \frac{1}{2} \)

### Step 7: Analyze the cases
**Case 1:** If \( x = 0 \):
- Substitute into the real part equation:
\[ 0 - y^2 = y \implies y^2 + y = 0 \implies y(y + 1) = 0 \]
Thus, \( y = 0 \) or \( y = -1 \). This gives us two complex numbers:
- \( z_1 = 0 \) (not allowed since \( z \) is non-zero)
- \( z_2 = -i \)

**Case 2:** If \( y = \frac{1}{2} \):
- Substitute into the real part equation:
\[ x^2 - \left(\frac{1}{2}\right)^2 = \frac{1}{2} \implies x^2 - \frac{1}{4} = \frac{1}{2} \implies x^2 = \frac{3}{4} \]
Thus, \( x = \pm \frac{\sqrt{3}}{2} \). This gives us two more complex numbers:
- \( z_2 = \frac{\sqrt{3}}{2} + \frac{i}{2} \)
- \( z_3 = -\frac{\sqrt{3}}{2} + \frac{i}{2} \)

### Step 8: List the complex numbers
The three complex numbers are:
- \( z_1 = -i \)
- \( z_2 = \frac{\sqrt{3}}{2} + \frac{i}{2} \)
- \( z_3 = -\frac{\sqrt{3}}{2} + \frac{i}{2} \)

### Step 9: Verify the statements
1. **Statement 1:** \( z_1 z_2 z_3 \) is purely imaginary.
\[ z_1 z_2 z_3 = (-i) \left(\frac{\sqrt{3}}{2} + \frac{i}{2}\right) \left(-\frac{\sqrt{3}}{2} + \frac{i}{2}\right) \]
The product \( z_2 z_3 = \left(\frac{\sqrt{3}}{2}\right)^2 + \left(\frac{1}{2}\right)^2 = \frac{3}{4} + \frac{1}{4} = 1 \).
Thus, \( z_1 z_2 z_3 = -i \cdot 1 = -i \), which is purely imaginary.

2. **Statement 2:** \( z_1 z_2 + z_2 z_3 + z_3 z_1 \) is purely real.
Calculate each product:
- \( z_1 z_2 = -i \left(\frac{\sqrt{3}}{2} + \frac{i}{2}\right) = -\frac{\sqrt{3}}{2}i - \frac{1}{2} \)
- \( z_2 z_3 = \left(\frac{\sqrt{3}}{2} + \frac{i}{2}\right) \left(-\frac{\sqrt{3}}{2} + \frac{i}{2}\right) = 1 \)
- \( z_3 z_1 = -\frac{\sqrt{3}}{2} + \frac{i}{2} \cdot -i = \frac{\sqrt{3}}{2} + \frac{1}{2} \)

Adding these gives:
\[ z_1 z_2 + z_2 z_3 + z_3 z_1 = -\frac{1}{2} + 1 + \frac{1}{2} = 1 \]
which is purely real.

### Conclusion
Both statements are true.