If for `z=alpha+ibeta,|z+2|=z+4(1+i)`, then `alpha +beta ` and `alphabeta` are the roots of the equation.

Options

A. `x^(2)+2x-3=0`
B. `x^(2)+3x-4=0`
C. `x^(2)+7x+12=0`
D. `x^(2)+x-12=0`

Answer

To solve the problem, we start with the equation given in the question:

\[ |z + 2| = z + 4(1 + i) \]

Let \( z = \alpha + i\beta \), where \( \alpha \) and \( \beta \) are real numbers. We can rewrite the equation as follows:

1. **Substitute \( z \)**:
\[ |(\alpha + i\beta) + 2| = (\alpha + i\beta) + 4(1 + i) \]
This simplifies to:
\[ |(\alpha + 2) + i\beta| = (\alpha + 4) + i(\beta + 4) \]

2. **Calculate the modulus**:
The modulus on the left side is:
\[ \sqrt{(\alpha + 2)^2 + \beta^2} \]
The right side can be expressed as:
\[ \sqrt{(\alpha + 4)^2 + (\beta + 4)^2} \]

3. **Set the two sides equal**:
\[ \sqrt{(\alpha + 2)^2 + \beta^2} = \sqrt{(\alpha + 4)^2 + (\beta + 4)^2} \]

4. **Square both sides**:
\[ (\alpha + 2)^2 + \beta^2 = (\alpha + 4)^2 + (\beta + 4)^2 \]

5. **Expand both sides**:
Left side:
\[ \alpha^2 + 4\alpha + 4 + \beta^2 \]
Right side:
\[ \alpha^2 + 8\alpha + 16 + \beta^2 + 8\beta + 16 \]

6. **Combine like terms**:
\[ \alpha^2 + 4\alpha + 4 + \beta^2 = \alpha^2 + 8\alpha + 16 + \beta^2 + 8\beta + 16 \]

7. **Cancel \(\alpha^2\) and \(\beta^2\)**:
\[ 4\alpha + 4 = 8\alpha + 8\beta + 32 \]

8. **Rearrange the equation**:
\[ 0 = 4\alpha + 8\beta + 28 \]
This simplifies to:
\[ 4\alpha + 8\beta = -28 \]
Dividing everything by 4:
\[ \alpha + 2\beta = -7 \quad \text{(Equation 1)} \]

9. **From the right side, equate the imaginary parts**:
We have:
\[ \beta + 4 = 0 \implies \beta = -4 \quad \text{(Equation 2)} \]

10. **Substitute \(\beta\) into Equation 1**:
\[ \alpha + 2(-4) = -7 \implies \alpha - 8 = -7 \implies \alpha = 1 \]

11. **Now we have the values**:
\(\alpha = 1\) and \(\beta = -4\).

12. **Calculate \(\alpha + \beta\) and \(\alpha \beta\)**:
\[ \alpha + \beta = 1 - 4 = -3 \]
\[ \alpha \beta = 1 \cdot (-4) = -4 \]

13. **Form the quadratic equation**:
The roots are \(-3\) and \(-4\). The quadratic equation can be formed as:
\[ x^2 - (\text{sum of roots})x + (\text{product of roots}) = 0 \]
\[ x^2 + 3x - 4 = 0 \]

Thus, the final equation is:
\[ x^2 + 3x - 4 = 0 \]