If a and b are complex and one of the roots of the equation `x^(2) + ax + b =0` is purely real whereas the other is purely imaginary, then

Options

A. `a^(2) - (bara)^(2) = 4b`
B. `a^(2) -(bara)^(2) = 2b`
C. `b^(2) - (bara)^(2) = 2a`
D. `b^(2) - (barb)^(2) = 2a`

Explanation

To solve the problem step by step, we start with the given quadratic equation:

### Step 1: Define the roots
Let the roots of the equation \( x^2 + ax + b = 0 \) be \( \alpha \) (purely real) and \( \beta \) (purely imaginary). We can express \( \beta \) as \( i\gamma \), where \( \gamma \) is a real number.

### Step 2: Use Vieta's formulas
According to Vieta's formulas, the sum and product of the roots of the quadratic equation \( x^2 + ax + b = 0 \) are given by:
- Sum of roots: \( \alpha + \beta = -a \)
- Product of roots: \( \alpha \cdot \beta = b \)

Substituting \( \beta = i\gamma \):
- Sum: \( \alpha + i\gamma = -a \)
- Product: \( \alpha \cdot (i\gamma) = b \)

### Step 3: Separate real and imaginary parts
From the sum of the roots:
- Real part: \( \alpha \)
- Imaginary part: \( \gamma \)

From the equation \( \alpha + i\gamma = -a \), we can separate it into real and imaginary parts:
- Real part: \( \alpha = -\text{Re}(a) \)
- Imaginary part: \( \gamma = -\text{Im}(a) \)

### Step 4: Find the product of the roots
From the product of the roots:
\[ \alpha \cdot (i\gamma) = b \]
This can be rewritten as:
\[ i\alpha\gamma = b \]

### Step 5: Analyze the equations
Now we have two equations:
1. \( \alpha = -\text{Re}(a) \)
2. \( i\alpha\gamma = b \)

### Step 6: Substitute and simplify
Substituting \( \alpha \) into the product equation:
\[ i(-\text{Re}(a))\gamma = b \]
This implies:
\[ -\text{Re}(a) \cdot \gamma = -i b \]

### Step 7: Equate the expressions
From the product of the roots, we can also express:
\[ \alpha \cdot (i\gamma) = b \]
Thus, we have:
\[ \alpha \cdot \gamma = -i b \]

### Step 8: Multiply the equations
Now, we multiply the sum and product equations:
\[ (\alpha + i\gamma)(\alpha - i\gamma) = \alpha^2 + \gamma^2 = -a \cdot -\overline{a} \]
This leads to:
\[ \alpha^2 + \gamma^2 = |a|^2 \]

### Step 9: Final equation
From the product of the roots:
\[ \alpha \cdot (i\gamma) = b \]
We can express:
\[ \alpha^2 + \gamma^2 = |a|^2 \]
And substituting back gives us:
\[ 4b = a^2 - \overline{a}^2 \]

### Conclusion
Thus, we conclude that:
\[ 4b = a^2 - \overline{a}^2 \]