For two non-zero complex numbers `z_(1)` and `z_(2)`,if `Re(z_(1)z_(2))=0` and `Re(z_(1)+z_(2))=0`,then which of the following are possible?
A. `Im(z_(1))>0` and `Im(z_(2))>0`
B. `lm(z_(1))>0` and `Im(z_(2))>0`
C. `lm(z_(1))>0` and `lm(z_(2))<0`
D. `Im(z_(1))<0` and `lm(z_(2))<0`
Choose the correct answer from the options given below :

Options

A. `A` and `B`
B. `B` and `C`
C. ` B` and `D`
D. `A` and `C`

Answer

To solve the problem, we start with two non-zero complex numbers \( z_1 \) and \( z_2 \). We know that:

1. \( \text{Re}(z_1 z_2) = 0 \)
2. \( \text{Re}(z_1 + z_2) = 0 \)

Let's express \( z_1 \) and \( z_2 \) in terms of their real and imaginary parts:

\[ z_1 = x_1 + i y_1 \]
\[ z_2 = x_2 + i y_2 \]

where \( x_1, x_2 \) are the real parts and \( y_1, y_2 \) are the imaginary parts of \( z_1 \) and \( z_2 \), respectively.

### Step 1: Analyze \( \text{Re}(z_1 z_2) = 0 \)

Calculating \( z_1 z_2 \):

\[ z_1 z_2 = (x_1 + i y_1)(x_2 + i y_2) = x_1 x_2 + i (x_1 y_2 + y_1 x_2) - y_1 y_2 \]

The real part of \( z_1 z_2 \) is:

\[ \text{Re}(z_1 z_2) = x_1 x_2 - y_1 y_2 \]

Given that \( \text{Re}(z_1 z_2) = 0 \), we have:

\[ x_1 x_2 - y_1 y_2 = 0 \quad \text{(1)} \]

### Step 2: Analyze \( \text{Re}(z_1 + z_2) = 0 \)

Calculating \( z_1 + z_2 \):

\[ z_1 + z_2 = (x_1 + x_2) + i (y_1 + y_2) \]

The real part of \( z_1 + z_2 \) is:

\[ \text{Re}(z_1 + z_2) = x_1 + x_2 \]

Given that \( \text{Re}(z_1 + z_2) = 0 \), we have:

\[ x_1 + x_2 = 0 \quad \text{(2)} \]

From equation (2), we can express \( x_2 \) in terms of \( x_1 \):

\[ x_2 = -x_1 \]

### Step 3: Substitute \( x_2 \) into equation (1)

Substituting \( x_2 = -x_1 \) into equation (1):

\[ x_1 (-x_1) - y_1 y_2 = 0 \]

This simplifies to:

\[ -x_1^2 - y_1 y_2 = 0 \]

Rearranging gives:

\[ y_1 y_2 = -x_1^2 \quad \text{(3)} \]

### Step 4: Analyze the implications of equation (3)

Since \( x_1^2 \) is always non-negative, \( y_1 y_2 \) must be negative. This means that \( y_1 \) and \( y_2 \) have opposite signs.

### Conclusion

From the analysis, we conclude:

- If \( y_1 > 0 \) (i.e., \( \text{Im}(z_1) > 0 \)), then \( y_2 < 0 \) (i.e., \( \text{Im}(z_2) < 0 \)).
- Conversely, if \( y_1 < 0 \), then \( y_2 > 0 \).

Thus, the only possible combinations of the imaginary parts of \( z_1 \) and \( z_2 \) are:

- \( \text{Im}(z_1) > 0 \) and \( \text{Im}(z_2) < 0 \)
- \( \text{Im}(z_1) < 0 \) and \( \text{Im}(z_2) > 0 \)

### Possible Options

Now, let's evaluate the options given:

A. \( \text{Im}(z_1) > 0 \) and \( \text{Im}(z_2) > 0 \) - **Not possible**
B. \( \text{Im}(z_1) > 0 \) and \( \text{Im}(z_2) > 0 \) - **Not possible**
C. \( \text{Im}(z_1) > 0 \) and \( \text{Im}(z_2) < 0 \) - **Possible**
D. \( \text{Im}(z_1) < 0 \) and \( \text{Im}(z_2) < 0 \) - **Not possible**

Thus, the correct answer is **C**.