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 \((x^4 + 2xi) - (3x^2 + yi) = (3 - 5i) + (1 + 2yi)\), we will follow these steps:

### Step 1: Rearranging the Equation
Start by rewriting the equation:
\[ x^4 + 2xi - 3x^2 - yi = 3 - 5i + 1 + 2yi \]
Combine the terms on the right-hand side:
\[ x^4 + 2xi - 3x^2 - yi = 4 - 5i + 2yi \]

### Step 2: Move All Terms to One Side
Now, move all terms to the left side:
\[ x^4 + 2xi - 3x^2 - yi - 4 + 5i - 2yi = 0 \]
This simplifies to:
\[ x^4 - 3x^2 - 4 + (2x - 3y + 5)i = 0 \]

### Step 3: Separating Real and Imaginary Parts
For the equation to hold true, both the real and imaginary parts must equal zero:
1. Real part:
\[ x^4 - 3x^2 - 4 = 0 \]
2. Imaginary part:
\[ 2x - 3y + 5 = 0 \]

### Step 4: Solving the Real Part
Now, solve the real part:
\[ x^4 - 3x^2 - 4 = 0 \]
Let \(u = x^2\), then the equation becomes:
\[ u^2 - 3u - 4 = 0 \]
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 have:
\[ x^2 = 4 \quad \Rightarrow \quad x = 2 \text{ or } x = -2 \]
\(x^2 = -1\) does not yield real solutions.

### Step 5: Solving the Imaginary Part
Now substitute \(x = 2\) and \(x = -2\) into the imaginary part equation \(2x - 3y + 5 = 0\).

1. For \(x = 2\):
\[ 2(2) - 3y + 5 = 0 \quad \Rightarrow \quad 4 - 3y + 5 = 0 \quad \Rightarrow \quad 9 - 3y = 0 \quad \Rightarrow \quad 3y = 9 \quad \Rightarrow \quad y = 3 \]

2. For \(x = -2\):
\[ 2(-2) - 3y + 5 = 0 \quad \Rightarrow \quad -4 - 3y + 5 = 0 \quad \Rightarrow \quad 1 - 3y = 0 \quad \Rightarrow \quad 3y = 1 \quad \Rightarrow \quad y = \frac{1}{3} \]

### Step 6: Conclusion
The ordered pairs \((x, y)\) are:
1. \((2, 3)\)
2. \((-2, \frac{1}{3})\)

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