Let a and b be two non- zero complex numbers. If the lines ` a bar(z) + bar(a) z + 1 = 0 and b bar(z) + bar(b) z - 1 = 0` are mutually perpendicular, then a, b are connected by the relation

Options

A. `ab + bar(a) bar(b) = 0 `
B. ` ab - bar(a) bar(b) = 0 `
C. ` bar(a) b - a bar(b) = 0`
D. ` a bar(b) + bar(a) b = 0`

Answer

To solve the problem, we need to analyze the two given equations of lines in the complex plane and use the condition for perpendicularity.

### Step-by-Step Solution:

1. **Write down the equations of the lines**:
- The first line is given by:
\[ a \bar{z} + \bar{a} z + 1 = 0 \]
- The second line is given by:
\[ b \bar{z} + \bar{b} z - 1 = 0 \]

2. **Rearranging the equations**:
- For the first line, we can express it as:
\[ a \bar{z} + \bar{a} z = -1 \]
- For the second line, we can express it as:
\[ b \bar{z} + \bar{b} z = 1 \]

3. **Identify coefficients**:
- From the first line, the coefficients of \(\bar{z}\) and \(z\) are \(a\) and \(\bar{a}\), respectively.
- From the second line, the coefficients of \(\bar{z}\) and \(z\) are \(b\) and \(\bar{b}\), respectively.

4. **Condition for perpendicularity**:
- Two lines are perpendicular if the product of their slopes is \(-1\). In terms of coefficients, this can be expressed as:
\[ \frac{a}{\bar{a}} \cdot \frac{\bar{b}}{b} = -1 \]
- This can be rearranged to:
\[ a \bar{b} + b \bar{a} = 0 \]

5. **Final relation**:
- Thus, we have the relation:
\[ ab + \bar{a} \bar{b} = 0 \]
- This implies:
\[ ab = -\bar{a} \bar{b} \]

### Conclusion:
The relation connecting \(a\) and \(b\) when the lines are mutually perpendicular is:
\[ ab + \bar{a} \bar{b} = 0 \]