If `z_1a n dz_2` both satisfy `z+ z =2|z-1|` and `a r g(z_1-z_2)=pi/4` , then find `I m(z_1+z_2)` .

Options

A. 0
B. 1
C. 2
D. none of these

Explanation

To solve the problem, we need to find the imaginary part of \( z_1 + z_2 \) given the conditions that both \( z_1 \) and \( z_2 \) satisfy the equation \( z + \bar{z} = 2 |z - 1| \) and that \( \arg(z_1 - z_2) = \frac{\pi}{4} \).

### Step-by-Step Solution:

1. **Understanding the Equation**:
The equation \( z + \bar{z} = 2 |z - 1| \) can be rewritten using \( z = x + iy \) (where \( x \) and \( y \) are the real and imaginary parts, respectively):
\[ x + iy + x - iy = 2 |(x - 1) + iy| \]
This simplifies to:
\[ 2x = 2 \sqrt{(x - 1)^2 + y^2} \]
Dividing both sides by 2 gives:
\[ x = \sqrt{(x - 1)^2 + y^2} \]

2. **Squaring Both Sides**:
Squaring both sides results in:
\[ x^2 = (x - 1)^2 + y^2 \]
Expanding the right side:
\[ x^2 = x^2 - 2x + 1 + y^2 \]
Simplifying gives:
\[ 0 = -2x + 1 + y^2 \quad \Rightarrow \quad 2x = 1 + y^2 \]
Rearranging gives:
\[ y^2 = 2x - 1 \tag{1} \]

3. **Applying the Condition for \( z_2 \)**:
Similarly, for \( z_2 \), we can write:
\[ y_2^2 = 2x_2 - 1 \tag{2} \]

4. **Subtracting Equations (1) and (2)**:
From equations (1) and (2):
\[ y_1^2 - y_2^2 = (2x_1 - 1) - (2x_2 - 1) \quad \Rightarrow \quad y_1^2 - y_2^2 = 2(x_1 - x_2) \]
This can be factored as:
\[ (y_1 + y_2)(y_1 - y_2) = 2(x_1 - x_2) \tag{3} \]

5. **Using the Argument Condition**:
Given \( \arg(z_1 - z_2) = \frac{\pi}{4} \), we have:
\[ \frac{y_1 - y_2}{x_1 - x_2} = 1 \quad \Rightarrow \quad y_1 - y_2 = x_1 - x_2 \tag{4} \]

6. **Substituting (4) into (3)**:
Substitute \( y_1 - y_2 = x_1 - x_2 \) into equation (3):
\[ (y_1 + y_2)(x_1 - x_2) = 2(x_1 - x_2) \]
Assuming \( x_1 \neq x_2 \), we can divide both sides by \( x_1 - x_2 \):
\[ y_1 + y_2 = 2 \]

7. **Finding the Imaginary Part**:
The imaginary part of \( z_1 + z_2 \) is \( y_1 + y_2 \). Therefore:
\[ \text{Im}(z_1 + z_2) = 2 \]

### Final Answer:
\[ \text{Im}(z_1 + z_2) = 2 \]