Question

If
\[ \frac{x-1}{3+i}+\frac{y-1}{3-i}=i, \] then the true statement among the following is

• A. \(x\lt 0,\; y\lt 0\)
• B. \(x\lt 0,\; y\gt 0\)
• C. \(x\gt 0,\; y\lt 0\)
• D. \(x\gt 0,\; y\gt 0\)

Hint:

For equations involving complex numbers, equate the real parts and imaginary parts separately after simplifying.

Answer 1

The Correct Option is B

Step 1: Rationalize the denominators.
Given equation is
\[ \frac{x-1}{3+i}+\frac{y-1}{3-i}=i \]
Now, rationalize each denominator:
\[ \frac{1}{3+i}=\frac{3-i}{(3+i)(3-i)} \] \[ =\frac{3-i}{9+1} \] \[ =\frac{3-i}{10} \]
Similarly,
\[ \frac{1}{3-i}=\frac{3+i}{10} \]
Therefore, the equation becomes
\[ \frac{(x-1)(3-i)}{10}+\frac{(y-1)(3+i)}{10}=i \]
Multiplying both sides by \(10\), we get
\[ (x-1)(3-i)+(y-1)(3+i)=10i \]

Step 2: Expand the terms.
Expanding,
\[ 3(x-1)-i(x-1)+3(y-1)+i(y-1)=10i \]
Group real and imaginary parts:
\[ 3x-3+3y-3+i[-(x-1)+(y-1)]=10i \]
\[ 3x+3y-6+i(-x+y)=10i \]

Step 3: Compare real and imaginary parts.
Since the right-hand side is purely imaginary, its real part is zero.
Comparing real parts:
\[ 3x+3y-6=0 \] \[ x+y=2 \]
Comparing imaginary parts:
\[ -y+x=-10 \] or
\[ -y+x=-10 \] \[ y-x=10 \]

Step 4: Solve the equations.
We have
\[ x+y=2 \] and
\[ y-x=10 \]
Adding both equations:
\[ 2y=12 \] \[ y=6 \]
Substituting into \(x+y=2\):
\[ x+6=2 \] \[ x=-4 \]
Thus,
\[ x\lt 0,\qquad y\gt 0 \]

Step 5: Final conclusion.
Hence, the correct statement is
\[ \boxed{x\lt 0,\; y\gt 0} \]