If ` - 3 + i x ^(2) y and x^(2) + y + 4i` be conjugate complex, then (x, y) is

Options

A. (1, - 4)
B. (-1, - 4)
C. (2, 1)
D. (-2, 1)

Answer

To solve the problem, we need to find the values of \(x\) and \(y\) given that the complex numbers \(-3 + i x^2 y\) and \(x^2 + y + 4i\) are conjugate complex numbers.

### Step-by-step Solution:

1. **Understanding Conjugate Complex Numbers**:
Two complex numbers \(a + bi\) and \(c + di\) are conjugates if \(a = c\) and \(b = -d\).

2. **Identify the Real and Imaginary Parts**:
Let:
\[ z_1 = -3 + i x^2 y \]
\[ z_2 = x^2 + y + 4i \]
Here, the real part of \(z_1\) is \(-3\) and the imaginary part is \(x^2 y\). For \(z_2\), the real part is \(x^2 + y\) and the imaginary part is \(4\).

3. **Set Up the Equations**:
From the definition of conjugates, we have:
\[ -3 = x^2 + y \quad \text{(1)} \]
\[ x^2 y = -4 \quad \text{(2)} \]

4. **Express \(y\) in Terms of \(x^2\)**:
From equation (1):
\[ y = -3 - x^2 \quad \text{(3)} \]

5. **Substitute \(y\) into Equation (2)**:
Substitute equation (3) into equation (2):
\[ x^2(-3 - x^2) = -4 \]
Simplifying this gives:
\[ -3x^2 - x^4 = -4 \]
Rearranging:
\[ x^4 + 3x^2 - 4 = 0 \]

6. **Let \(u = x^2\)**:
Rewrite the equation as:
\[ u^2 + 3u - 4 = 0 \]

7. **Solve the Quadratic Equation**:
Using the quadratic formula \(u = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\):
\[ u = \frac{-3 \pm \sqrt{3^2 - 4 \cdot 1 \cdot (-4)}}{2 \cdot 1} \]
\[ u = \frac{-3 \pm \sqrt{9 + 16}}{2} \]
\[ u = \frac{-3 \pm 5}{2} \]
This gives:
\[ u = 1 \quad \text{or} \quad u = -4 \]

8. **Find \(x^2\)**:
Since \(u = x^2\):
- If \(u = 1\), then \(x^2 = 1\) which gives \(x = 1\) or \(x = -1\).
- If \(u = -4\), then \(x^2 = -4\) which is not valid for real \(x\).

9. **Find Corresponding \(y\) Values**:
Using equation (3):
- For \(x = 1\):
\[ y = -3 - 1^2 = -4 \]
- For \(x = -1\):
\[ y = -3 - (-1)^2 = -4 \]

10. **Final Pairs**:
The pairs \((x, y)\) are:
\[ (1, -4) \quad \text{and} \quad (-1, -4) \]

### Conclusion:
Thus, the values of \((x, y)\) are \((1, -4)\) and \((-1, -4)\).