Question

If \( \frac{x-1}{3+i} + \frac{y-1}{3-i} = i \) then \( (y,x) = \)

• A. \( (-6,-4) \)
• B. \( (-4,-6) \)
• C. \( (6,-4) \)
• D. \( (-4,6) \)

Hint:

In complex equations, always rationalize denominators and equate real and imaginary parts separately.

Answer 1

The Correct Option is C

Step 1: Rationalize denominators.
\[ \frac{1}{3+i}=\frac{3-i}{(3+i)(3-i)}=\frac{3-i}{10},\quad \frac{1}{3-i}=\frac{3+i}{10}. \]

Step 2: Substitute into the equation.
\[ \frac{x-1}{3+i}+\frac{y-1}{3-i} = \frac{(x-1)(3-i)}{10}+\frac{(y-1)(3+i)}{10}. \]
\[ =\frac{1}{10}\left[(x-1)(3-i)+(y-1)(3+i)\right]. \]

Step 3: Expand the expression.
\[ (x-1)(3-i)=3(x-1)-i(x-1), \] \[ (y-1)(3+i)=3(y-1)+i(y-1). \]
Adding:
\[ =3(x-1)+3(y-1)+i[(y-1)-(x-1)]. \]
\[ =3(x+y-2)+i(y-x). \]

Step 4: Substitute into original equation.
\[ \frac{1}{10}\left[3(x+y-2)+i(y-x)\right]=i. \]

Step 5: Multiply both sides by 10.
\[ 3(x+y-2)+i(y-x)=10i. \]

Step 6: Equate real and imaginary parts.
Real part:
\[ 3(x+y-2)=0 \Rightarrow x+y=2. \]
Imaginary part:
\[ y-x=10. \]

Step 7: Solve the system.
From \(y-x=10\),
\[ y=x+10. \]
Substitute into \(x+y=2\):
\[ x+(x+10)=2. \]
\[ 2x+10=2 \Rightarrow x=-4. \]
\[ y=x+10=6. \]
Thus:
\[ (y,x)=(6,-4). \]
Final Answer:
\[ \boxed{(6,-4)} \]