Find the real values of x and y for which `((x-1)/(3+i)+(y-1)/(3-i))=i`.

Options

A. `-4,6`
B. `4,-6`
C. `-2,3`
D. `-3,3`

Explanation

To solve the equation \(\frac{x-1}{3+i} + \frac{y-1}{3-i} = i\), we will follow these steps:

### Step 1: Multiply by the conjugate
We start by multiplying each term by the conjugate of the denominator to eliminate the complex numbers in the denominators.

\[ \frac{x-1}{3+i} \cdot \frac{3-i}{3-i} + \frac{y-1}{3-i} \cdot \frac{3+i}{3+i} = i \]

This gives us:

\[ \frac{(x-1)(3-i)}{(3+i)(3-i)} + \frac{(y-1)(3+i)}{(3-i)(3+i)} = i \]

### Step 2: Simplify the denominators
Now, we simplify the denominators:

\[ (3+i)(3-i) = 3^2 - i^2 = 9 + 1 = 10 \]

So we can rewrite the equation as:

\[ \frac{(x-1)(3-i)}{10} + \frac{(y-1)(3+i)}{10} = i \]

### Step 3: Combine the fractions
Now we can combine the fractions:

\[ \frac{(x-1)(3-i) + (y-1)(3+i)}{10} = i \]

### Step 4: Multiply both sides by 10
To eliminate the fraction, multiply both sides by 10:

\[ (x-1)(3-i) + (y-1)(3+i) = 10i \]

### Step 5: Expand the left side
Now, we expand both terms on the left side:

\[ (3(x-1) - i(x-1)) + (3(y-1) + i(y-1)) = 10i \]

This simplifies to:

\[ (3x - 3 + 3y - 3) + (-i(x-1) + i(y-1)) = 10i \]

### Step 6: Combine like terms
Combining the real and imaginary parts gives us:

\[ (3x + 3y - 6) + i(-x + y) = 10i \]

### Step 7: Set real and imaginary parts equal
Now, we can equate the real and imaginary parts:

1. Real part: \(3x + 3y - 6 = 0\)
2. Imaginary part: \(-x + y = 10\)

### Step 8: Solve the system of equations
Now we solve the system of equations:

From the first equation:

\[ 3x + 3y = 6 \implies x + y = 2 \quad \text{(1)} \]

From the second equation:

\[ y = x + 10 \quad \text{(2)} \]

Substituting (2) into (1):

\[ x + (x + 10) = 2 \implies 2x + 10 = 2 \implies 2x = -8 \implies x = -4 \]

Now substitute \(x = -4\) back into (2):

\[ y = -4 + 10 = 6 \]

### Final Answer
The real values of \(x\) and \(y\) are:

\[ x = -4, \quad y = 6 \]