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

Options

A. `2`
B. `4`
C. `6`
D. `8`

Explanation

To solve the problem step by step, we will analyze the given quadratic equation and use the properties of its roots.

### Step 1: Identify the roots
Let the roots of the equation \( x^2 + ax + b = 0 \) be \( \alpha \) (purely real) and \( i\beta \) (purely imaginary).

### Step 2: Use Vieta's formulas
According to Vieta's formulas, the sum of the roots is given by:
\[ \alpha + i\beta = -a \]
This gives us our first equation:
\[ \alpha + i\beta = -a \tag{1} \]

### Step 3: Conjugate of the roots
Taking the conjugate of the roots, we have:
\[ \alpha - i\beta = -\bar{a} \]
This gives us our second equation:
\[ \alpha - i\beta = -\bar{a} \tag{2} \]

### Step 4: Add equations (1) and (2)
Adding equations (1) and (2):
\[ (\alpha + i\beta) + (\alpha - i\beta) = -a - \bar{a} \]
This simplifies to:
\[ 2\alpha = - (a + \bar{a}) \]
Thus, we can express \( \alpha \):
\[ \alpha = -\frac{(a + \bar{a})}{2} \tag{3} \]

### Step 5: Subtract equations (1) and (2)
Now, subtract equation (2) from equation (1):
\[ (\alpha + i\beta) - (\alpha - i\beta) = -a + \bar{a} \]
This simplifies to:
\[ 2i\beta = - (a - \bar{a}) \]
Thus, we can express \( \beta \):
\[ \beta = -\frac{(a - \bar{a})}{2i} \tag{4} \]

### Step 6: Find the product of the roots
The product of the roots \( \alpha \cdot i\beta \) can be expressed as:
\[ \alpha \cdot i\beta = b \]
Substituting the expressions for \( \alpha \) and \( \beta \):
\[ \left(-\frac{(a + \bar{a})}{2}\right) \cdot \left(-\frac{(a - \bar{a})}{2i}\right) = b \]
This simplifies to:
\[ \frac{(a + \bar{a})(a - \bar{a})}{4i} = b \]

### Step 7: Express \( a^2 - \bar{a}^2 \)
Using the identity \( a^2 - \bar{a}^2 = (a - \bar{a})(a + \bar{a}) \), we can relate this to our previous product:
\[ a^2 - \bar{a}^2 = 4ib \]

### Step 8: Relate to the given equation
According to the problem, we have:
\[ a^2 - \bar{a}^2 = kb \]
Comparing both expressions:
\[ 4ib = kb \]
Thus, we can conclude that:
\[ k = 4i \]

### Final Answer
The value of \( k \) is \( 4 \).