If `(x^(4)+2x i)-(3x^(2)+yi)=(3-5i)+(1+2yi)`
then the number of ordered pairs (x, y) is/are equal to
`{AA x,y in R and i^(2)=-1}`

Options

A. 0
B. 1
C. 2
D. 3

Answer

To solve the equation given in the problem, we will follow these steps:

1. **Set Up the Equation**: We start with the equation:
\[ (x^4 + 2xi) - (3x^2 + yi) = (3 - 5i) + (1 + 2yi) \]
Simplifying the right-hand side gives:
\[ 3 - 5i + 1 + 2yi = 4 + (2y - 5)i \]
Thus, we rewrite the equation as:
\[ x^4 - 3x^2 + (2x - y)i = 4 + (2y - 5)i \]

2. **Equate Real and Imaginary Parts**: We separate the real and imaginary parts:
- Real part: \(x^4 - 3x^2 = 4\)
- Imaginary part: \(2x - y = 2y - 5\)

3. **Solve the Real Part Equation**: We rearrange the real part equation:
\[ x^4 - 3x^2 - 4 = 0 \]
Let \(u = x^2\), then we have:
\[ u^2 - 3u - 4 = 0 \]
Using the quadratic formula \(u = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\):
\[ u = \frac{3 \pm \sqrt{(-3)^2 - 4 \cdot 1 \cdot (-4)}}{2 \cdot 1} = \frac{3 \pm \sqrt{9 + 16}}{2} = \frac{3 \pm 5}{2} \]
This gives us:
\[ u = 4 \quad \text{or} \quad u = -1 \]
Since \(u = x^2\), we find:
- From \(u = 4\): \(x^2 = 4 \Rightarrow x = \pm 2\)
- From \(u = -1\): No real solutions since \(x^2\) cannot be negative.

4. **Solve the Imaginary Part Equation**: We now solve for \(y\) using the imaginary part equation:
\[ 2x - y = 2y - 5 \implies 2x + 5 = 3y \implies y = \frac{2x + 5}{3} \]

5. **Find Corresponding Values of \(y\)**:
- For \(x = 2\):
\[ y = \frac{2(2) + 5}{3} = \frac{4 + 5}{3} = \frac{9}{3} = 3 \]
- For \(x = -2\):
\[ y = \frac{2(-2) + 5}{3} = \frac{-4 + 5}{3} = \frac{1}{3} \]

6. **Ordered Pairs**: The ordered pairs \((x, y)\) are:
- For \(x = 2\), \(y = 3\) → \((2, 3)\)
- For \(x = -2\), \(y = \frac{1}{3}\) → \((-2, \frac{1}{3})\)

Thus, the number of ordered pairs \((x, y)\) is **2**.