The equation `barbz+b barz=c`, where b is a non-zero complex constant and c is a real number, represents

Options

A. a circle
B. a straight line
C. a pair of straight line
D. none of these

Answer

To solve the equation \( \overline{b} z + b \overline{z} = c \), where \( b \) is a non-zero complex constant and \( c \) is a real number, we can follow these steps:

### Step 1: Define the Complex Numbers
Let \( b = p + iq \) where \( p \) and \( q \) are real numbers, and \( z = x + iy \) where \( x \) and \( y \) are also real numbers.

### Step 2: Compute the Conjugates
The conjugate of \( b \) is:
\[ \overline{b} = p - iq \]
The conjugate of \( z \) is:
\[ \overline{z} = x - iy \]

### Step 3: Substitute into the Equation
Substituting \( b \) and \( z \) into the equation gives:
\[ (p - iq)(x + iy) + (p + iq)(x - iy) = c \]

### Step 4: Expand the Products
Now, we expand both products:
1. For \( \overline{b} z \):
\[ (p - iq)(x + iy) = px + piy - iqx - qy = (px + qy) + i(py - qx) \]

2. For \( b \overline{z} \):
\[ (p + iq)(x - iy) = px - piy + iqx - qy = (px + qy) + i(qx - py) \]

### Step 5: Combine the Results
Now, combine the results from both expansions:
\[ (px + qy + i(py - qx)) + (px + qy + i(qx - py)) = c \]
This simplifies to:
\[ 2(px + qy) + i[(py - qx) + (qx - py)] = c \]

### Step 6: Separate Real and Imaginary Parts
Since \( c \) is a real number, the imaginary part must equal zero:
\[ (py - qx) + (qx - py) = 0 \]
This simplifies to:
\[ 0 = 0 \]
The real part gives:
\[ 2(px + qy) = c \]

### Step 7: Rearranging the Equation
From the equation \( 2(px + qy) = c \), we can express it as:
\[ px + qy = \frac{c}{2} \]

### Step 8: Identify the Representation
This equation \( px + qy = \frac{c}{2} \) represents a straight line in the \( xy \)-plane.

### Conclusion
Thus, the equation \( \overline{b} z + b \overline{z} = c \) represents a straight line.

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