If one root of `z^2 + (a + i)z+ b +ic =0` is real, where `a, b, c in R` , then `c^2 + b-ac=`

Options

A. 0
B. `-1`
C. 1
D. none of these

Explanation

To solve the problem, we need to analyze the quadratic equation given:

\[ z^2 + (a + i)z + (b + ic) = 0 \]

where \( a, b, c \in \mathbb{R} \) and one of the roots is real.

### Step 1: Let the real root be denoted as \( \alpha \).
Since \( \alpha \) is a root of the equation, we can substitute \( z = \alpha \) into the equation:

\[ \alpha^2 + (a + i)\alpha + (b + ic) = 0 \]

### Step 2: Separate the equation into real and imaginary parts.
This gives us:

1. Real part: \( \alpha^2 + a\alpha + b = 0 \)
2. Imaginary part: \( \alpha + c = 0 \)

### Step 3: Solve the imaginary part for \( c \).
From the imaginary part equation \( \alpha + c = 0 \), we can express \( c \) in terms of \( \alpha \):

\[ c = -\alpha \]

### Step 4: Substitute \( c \) back into the real part equation.
Now substitute \( c = -\alpha \) into the real part equation:

\[ \alpha^2 + a\alpha + b = 0 \]

### Step 5: Rearranging the real part equation.
We can rearrange this equation to find \( b \):

\[ b = -\alpha^2 - a\alpha \]

### Step 6: Substitute \( b \) and \( c \) into the expression \( c^2 + b - ac \).
We need to evaluate:

\[ c^2 + b - ac \]

Substituting \( c = -\alpha \) and \( b = -\alpha^2 - a\alpha \):

\[ c^2 = (-\alpha)^2 = \alpha^2 \]

Now substituting into the expression:

\[ c^2 + b - ac = \alpha^2 + (-\alpha^2 - a\alpha) - a(-\alpha) \]

### Step 7: Simplify the expression.
This simplifies to:

\[ \alpha^2 - \alpha^2 - a\alpha + a\alpha = 0 \]

### Conclusion:
Thus, we have shown that:

\[ c^2 + b - ac = 0 \]

### Final Result:
The final result is:

\[ c^2 + b - ac = 0 \]

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