Question

If $\dfrac{x - 1}{3 + i} + \dfrac{y - 1}{3 - i} = i$, then the true statement among the following is:

• A. $x<0,\ y<0$
• B. $x<0,\ y>0$
• C. $x>0,\ y<0$
• D. $x>0,\ y>0$

Hint:

Use complex conjugates to simplify expressions with imaginary denominators.

Answer 1

The Correct Option is B

Let us simplify the expression by multiplying numerator and denominator of each term by the conjugate of the denominator: \[ \dfrac{x - 1}{3 + i} = \dfrac{(x - 1)(3 - i)}{(3 + i)(3 - i)} = \dfrac{(x - 1)(3 - i)}{10} \] \[ \dfrac{y - 1}{3 - i} = \dfrac{(y - 1)(3 + i)}{10} \] Adding the two: \[ \dfrac{(x - 1)(3 - i) + (y - 1)(3 + i)}{10} = i \] Multiply both sides by 10: \[ (x - 1)(3 - i) + (y - 1)(3 + i) = 10i \] Expand: \[ (3x - 3 - ix + i) + (3y - 3 + iy - i) = 10i \Rightarrow (3x - 3 + 3y - 3) + (-ix + i + iy - i) = 10i \Rightarrow 3x + 3y - 6 - i(x - y) = 10i \] Equating real and imaginary parts: Real: $3x + 3y - 6 = 0 \Rightarrow x + y = 2$
Imaginary: $-x + y = 10$
Solve: From $x + y = 2$ and $-x + y = 10$, adding both: $2y = 12 \Rightarrow y = 6 \Rightarrow x = -4$
Hence, $x<0,\ y>0$.