Find x and y if `(x^4+2x i)-(3x^2+y i)=(3-5i)+(1+2y i)`

Options

A. `x=2, y=3`
B. `x=-2, y=(1)/(3)`
C. `x= pm 2 and y=3, (1)/(3)`
D. None of these

Answer

To solve the equation \((x^4 + 2xi) - (3x^2 + yi) = (3 - 5i) + (1 + 2yi)\), we will first simplify and rearrange the equation step by step.

### Step 1: Rearranging the equation
We start with the equation:
\[ (x^4 + 2xi) - (3x^2 + yi) = (3 - 5i) + (1 + 2yi) \]

Combine the terms on the right side:
\[ (3 - 5i) + (1 + 2yi) = 4 - 5i + 2yi \]

Now, rewrite the equation:
\[ x^4 + 2xi - 3x^2 - yi = 4 - 5i + 2yi \]

### Step 2: Grouping real and imaginary parts
Now, we can group the real and imaginary parts:
\[ x^4 - 3x^2 + (2x - y)i = 4 + (2y - 5)i \]

### Step 3: Setting up equations
From the equation, we can equate the real parts and the imaginary parts:
1. Real part:
\[ x^4 - 3x^2 - 4 = 0 \]
2. Imaginary part:
\[ 2x - y = 2y - 5 \]

### Step 4: Solving the imaginary part equation
Rearranging the imaginary part equation:
\[ 2x + 5 = 3y \quad \text{(Equation 1)} \]

### Step 5: Solving the real part equation
Now, we solve the real part equation:
\[ x^4 - 3x^2 - 4 = 0 \]

Let \(u = x^2\), then the equation becomes:
\[ u^2 - 3u - 4 = 0 \]

### Step 6: Factoring the quadratic
Factoring the quadratic:
\[ (u - 4)(u + 1) = 0 \]

This gives us:
\[ u = 4 \quad \text{or} \quad u = -1 \]

Since \(u = x^2\), we have:
\[ x^2 = 4 \quad \Rightarrow \quad x = \pm 2 \]
(Note: \(x^2 = -1\) does not yield real solutions.)

### Step 7: Finding y for each x
Now, we substitute \(x = 2\) and \(x = -2\) into Equation 1 to find \(y\).

1. For \(x = 2\):
\[ 2(2) + 5 = 3y \implies 4 + 5 = 3y \implies 9 = 3y \implies y = 3 \]

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

### Final Solution
Thus, the solutions are:
\[ (x, y) = (2, 3) \quad \text{and} \quad (-2, \frac{1}{3}) \]