Let `alpha` and `beta` be two fixed non-zero complex numbers and 'z' a variable complex number. If the lines `alphabarz+baraz+1=0` and `betabarz+barbetaz-1=0` are mutually perpendicular, then

Options

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

Answer

To solve the problem, we need to analyze the given equations and determine the condition under which the lines represented by these equations are mutually perpendicular.

### Step-by-Step Solution:

1. **Identify the Given Equations:**
The two lines are given by:
\[ \alpha \bar{z} + \bar{\alpha} z + 1 = 0 \quad \text{(Equation 1)} \]
\[ \beta \bar{z} + \bar{\beta} z - 1 = 0 \quad \text{(Equation 2)} \]

2. **Rearranging the Equations:**
We can rearrange both equations to express \( z \) in terms of \( \bar{z} \):
- From Equation 1:
\[ \bar{\alpha} z = -\alpha \bar{z} - 1 \implies z = -\frac{\alpha}{\bar{\alpha}} \bar{z} - \frac{1}{\bar{\alpha}} \]
- From Equation 2:
\[ \bar{\beta} z = 1 - \beta \bar{z} \implies z = \frac{1}{\bar{\beta}} - \frac{\beta}{\bar{\beta}} \bar{z} \]

3. **Finding the Slopes:**
The slopes of the lines can be derived from the coefficients of \( z \) and \( \bar{z} \):
- For Equation 1, the slope \( m_1 \) can be found as:
\[ m_1 = -\frac{\alpha}{\bar{\alpha}} \]
- For Equation 2, the slope \( m_2 \) is:
\[ m_2 = \frac{\beta}{\bar{\beta}} \]

4. **Condition for Perpendicular Lines:**
Two lines are perpendicular if the product of their slopes is -1:
\[ m_1 \cdot m_2 = -1 \]
Substituting the values of \( m_1 \) and \( m_2 \):
\[ \left(-\frac{\alpha}{\bar{\alpha}}\right) \left(\frac{\beta}{\bar{\beta}}\right) = -1 \]

5. **Simplifying the Condition:**
This leads to:
\[ \frac{\alpha \beta}{\bar{\alpha} \bar{\beta}} = 1 \]
Which implies:
\[ \alpha \beta = \bar{\alpha} \bar{\beta} \]

6. **Final Result:**
The condition for the lines to be mutually perpendicular is:
\[ \alpha \beta + \bar{\alpha} \bar{\beta} = 0 \]