`zbarz+abarz+baraz+b`=0 where `binR` represents a real circle of nonzero radius if

Options

A. `|bar a|^2gt b`
B. `|a|^2ltb`
C. `|bara|^2geb`
D. `|a|^2leb`

Answer

To solve the equation \( \overline{z}z + \overline{a}z + a\overline{z} + b = 0 \), where \( z \) is a complex number and represents a circle of non-zero radius, we can follow these steps:

### Step 1: Substitute \( z \) and \( a \)
Assume \( z = x + iy \) and \( a = \alpha + i\beta \), where \( x \) and \( y \) are real numbers, and \( \alpha \) and \( \beta \) are real parts of \( a \).

### Step 2: Calculate \( \overline{z} \) and \( \overline{a} \)
The conjugate of \( z \) is \( \overline{z} = x - iy \) and the conjugate of \( a \) is \( \overline{a} = \alpha - i\beta \).

### Step 3: Substitute into the equation
Substituting these values into the equation gives:
\[ (x - iy)(x + iy) + (\alpha - i\beta)(x + iy) + (\alpha + i\beta)(x - iy) + b = 0 \]

### Step 4: Simplify \( \overline{z}z \)
Using the property \( \overline{z}z = |z|^2 = x^2 + y^2 \):
\[ x^2 + y^2 + (\alpha - i\beta)(x + iy) + (\alpha + i\beta)(x - iy) + b = 0 \]

### Step 5: Expand the terms
Expanding the terms:
1. \( (\alpha - i\beta)(x + iy) = \alpha x + i\alpha y - i\beta x - \beta y \)
2. \( (\alpha + i\beta)(x - iy) = \alpha x - i\alpha y + i\beta x - \beta y \)

Combining these gives:
\[ x^2 + y^2 + 2\alpha x + 2\beta y + b = 0 \]

### Step 6: Rearranging the equation
Rearranging the equation, we have:
\[ x^2 + 2\alpha x + y^2 + 2\beta y + b = 0 \]

### Step 7: Completing the square
To write this in standard form, complete the square:
1. For \( x \): \( x^2 + 2\alpha x = (x + \alpha)^2 - \alpha^2 \)
2. For \( y \): \( y^2 + 2\beta y = (y + \beta)^2 - \beta^2 \)

Thus, we have:
\[ (x + \alpha)^2 - \alpha^2 + (y + \beta)^2 - \beta^2 + b = 0 \]
This simplifies to:
\[ (x + \alpha)^2 + (y + \beta)^2 = \alpha^2 + \beta^2 - b \]

### Step 8: Condition for a circle of non-zero radius
For this to represent a circle, the right side must be positive:
\[ \alpha^2 + \beta^2 - b > 0 \]
This implies:
\[ \alpha^2 + \beta^2 > b \]

### Final Result
Thus, the condition for the equation to represent a circle of non-zero radius is:
\[ |\overline{a}|^2 > b \]
where \( |\overline{a}|^2 = \alpha^2 + \beta^2 \).