Complex numbers whose real and imaginary parts `x` and `y` are integers and satisfy the equation `3x^(2)-|xy|-2y^(2)+7=0`

Options

A. do not exist
B. exist and have equal modulus
C. form two conjugate pairs
D. do not form conjugate pairs

Explanation

To solve the equation \(3x^2 - |xy| - 2y^2 + 7 = 0\) for integer values of \(x\) and \(y\), we can follow these steps:

### Step 1: Rewrite the equation
Start with the original equation:
\[ 3x^2 - |xy| - 2y^2 + 7 = 0 \]
Rearranging gives:
\[ 3x^2 - 2y^2 + 7 = |xy| \]

### Step 2: Consider the cases for \(|xy|\)
Since \(|xy|\) is always non-negative, we need to consider the cases where \(xy \geq 0\) (i.e., both \(x\) and \(y\) are either both non-negative or both non-positive).

### Step 3: Factor the equation
We can rewrite the equation as:
\[ 3x^2 - 2y^2 + 7 - |xy| = 0 \]
This can be factored or rearranged into a more manageable form. We can assume \(xy \geq 0\) for the moment and rewrite:
\[ 3x^2 - xy - 2y^2 + 7 = 0 \]

### Step 4: Set up a system of equations
From the rearranged equation, we can set up two equations based on the factorization:
1. \(3x + 2y = k\)
2. \(x - y = m\)

Where \(k\) and \(m\) are integers that satisfy the equation. We can set \(k = 7\) and \(m = -1\) (since \(-7 = -1 \times 7\)).

### Step 5: Solve the system of equations
We can solve the following two equations:
1. \(3x + 2y = 7\)
2. \(x - y = -1\)

From the second equation, we can express \(y\) in terms of \(x\):
\[ y = x + 1 \]

Substituting this into the first equation:
\[ 3x + 2(x + 1) = 7 \]
This simplifies to:
\[ 3x + 2x + 2 = 7 \]
\[ 5x + 2 = 7 \]
\[ 5x = 5 \implies x = 1 \]

Now substituting \(x = 1\) back to find \(y\):
\[ y = 1 + 1 = 2 \]

### Step 6: Consider negative values
Since \(x\) and \(y\) can also take negative values, we can also consider:
1. \(x = -1\) and \(y = -2\)

### Step 7: Write the complex numbers
The complex numbers corresponding to these integer pairs are:
1. \(1 + 2i\)
2. \(1 - 2i\)
3. \(-1 + 2i\)
4. \(-1 - 2i\)

### Final Answer
The required complex numbers are:
1. \(1 + 2i\)
2. \(1 - 2i\)
3. \(-1 + 2i\)
4. \(-1 - 2i\)