For two non - zero complex numbers A and B, if `A+(1)/(B)=barA and (1)/(A)+B=barB`, then

Options

A. A is purely real
B. B is purely real
C. `|AB|=(1)/(2)`
D. `|A|=|B|`

Answer

To solve the problem, we need to analyze the two given equations involving the complex numbers \( A \) and \( B \):

1. \( A + \frac{1}{B} = \bar{A} \)
2. \( \frac{1}{A} + B = \bar{B} \)

Let's denote \( A = x + iy \) and \( B = u + iv \), where \( x, y, u, v \) are real numbers.

### Step 1: Rewrite the first equation

From the first equation:
\[ A + \frac{1}{B} = \bar{A} \]
Substituting \( A \) and \( B \):
\[ (x + iy) + \frac{1}{u + iv} = x - iy \]

### Step 2: Simplify \( \frac{1}{B} \)

To simplify \( \frac{1}{B} \):
\[ \frac{1}{u + iv} = \frac{u - iv}{u^2 + v^2} \]
Thus, the equation becomes:
\[ x + iy + \frac{u - iv}{u^2 + v^2} = x - iy \]

### Step 3: Separate real and imaginary parts

Separating real and imaginary parts:
- Real part:
\[ x + \frac{u}{u^2 + v^2} = x \]
This simplifies to:
\[ \frac{u}{u^2 + v^2} = 0 \implies u = 0 \]

- Imaginary part:
\[ y - \frac{v}{u^2 + v^2} = -y \]
This simplifies to:
\[ y + y = \frac{v}{u^2 + v^2} \implies 2y = \frac{v}{u^2 + v^2} \]

### Step 4: Substitute \( u = 0 \)

Since \( u = 0 \), we substitute it into the imaginary part equation:
\[ 2y = \frac{v}{0 + v^2} \implies 2y = \frac{v}{v^2} \implies 2y = \frac{1}{v} \]
Thus:
\[ v = \frac{1}{2y} \]

### Step 5: Analyze the second equation

Now, let's analyze the second equation:
\[ \frac{1}{A} + B = \bar{B} \]
Substituting \( A \) and \( B \):
\[ \frac{1}{x + iy} + (0 + iv) = 0 - iv \]

### Step 6: Simplify \( \frac{1}{A} \)

To simplify \( \frac{1}{A} \):
\[ \frac{1}{x + iy} = \frac{x - iy}{x^2 + y^2} \]
Thus, the equation becomes:
\[ \frac{x - iy}{x^2 + y^2} + iv = -iv \]

### Step 7: Separate real and imaginary parts again

Separating real and imaginary parts:
- Real part:
\[ \frac{x}{x^2 + y^2} = 0 \implies x = 0 \]

- Imaginary part:
\[ -\frac{y}{x^2 + y^2} + v = -v \]
This simplifies to:
\[ -\frac{y}{x^2 + y^2} = -2v \]

### Step 8: Substitute \( x = 0 \)

Since \( x = 0 \):
\[ -\frac{y}{0 + y^2} = -2v \implies \frac{1}{y} = 2v \]
Substituting \( v = \frac{1}{2y} \):
\[ \frac{1}{y} = 2 \cdot \frac{1}{2y} \implies \frac{1}{y} = \frac{1}{y} \]
This is always true.

### Step 9: Find \( A \) and \( B \)

From our earlier results:
- \( A = iy \)
- \( B = \frac{i}{2y} \)

### Step 10: Check the options

1. **A is purely real**: Incorrect, \( A \) is purely imaginary.
2. **A is not equal to B**: Correct, since \( A \) and \( B \) are different.
3. **The product \( AB = \frac{1}{2} \)**:
\[ AB = (iy) \left(\frac{i}{2y}\right) = \frac{-1}{2} \implies |AB| = \frac{1}{2} \]
Correct.
4. **Both A and B are real**: Incorrect, both are imaginary.

### Conclusion

The correct options are:
- A is not equal to B.
- The modulus of the product \( AB \) is \( \frac{1}{2} \).