Question

Real numbers $x$ and $y$ satisfy the equation
$\frac{x}{1+i}+\frac{y}{1-i}=\frac{148}{12+2i}.$
What is the value of $xy$ ?

Answer 1

$\frac{x}{1+i}+\frac{y}{1-i}=\frac{148}{12+2i}=\frac{148}{2(6+i)}$

$\Rightarrow \frac{x}{1+i}+\frac{y}{1-i}=\frac{74}{6+i}$

$\Rightarrow \frac{x}{1+i}\times \frac{1-i}{1-i}+\frac{y}{1-i}\times \frac{1+i}{1+i}=\frac{74}{6+i}\times \frac{6-i}{6-i}$

$\Rightarrow \frac{x(1-i)}{1-i^2}+\frac{y(1+i)}{1-i^2}=\frac{74(6-i)}{36-i^2}$

$i^2=-1$

$\Rightarrow \frac{x(1-i)}{2}+\frac{y(1+i)}{2}=\frac{74(6-i)}{37}$

$\Rightarrow \frac{x(1-i)+y(1+i)}{2}=2(6-i)$

$\Rightarrow x-xi+y+yi=4(6-i)$
$\Rightarrow (x+y)+(y-x)i=24-4i$

Comparing the real and imaginary parts

$\Rightarrow x+y=24;$ and

$y-x=-4$

adding both

$\Rightarrow 2y=20$

$\Rightarrow y=10$

$\Rightarrow x=24-y=24-10$

$\Rightarrow x=14$ and $y=10$

$\Rightarrow xy=14\left(10\right)=140$