Question

The real values of $x$ and $y$ for which $(\frac{x-1}{3+i}+\frac{y-1}{3-i})=i$ is

• A. $x=-4$ and $y=6$
• B. $x=-6$ and $y=3$
• C. $x=2$ and $y=8$
• D. $x=8$ and $y=-6$

Answer 1

The correct option is A $x=-4$ and $y=6$

We have
$(\frac{x-1}{3+i}+\frac{y-1}{3-i})=i$
$\frac{(x-1)(3-i)+(y-1)(3+i)}{3^2-i^2}=i$
$\Rightarrow \frac{3(x+y-2)-i(x-y)}{10}=i$
$\Rightarrow 3(x+y-2)=0$ and $(y-x)=10$
[equating real parts and the imaginary parts seperately]
$\Rightarrow x+y-2=0$ and $y-x=10$
$\Rightarrow x=-4$ and $y=6$