Find the real values of x and y for which:

Solution:

Given:

$\frac{x+3 i}{2+i y}=(1-i)$

$\Rightarrow x+3 i=(1-i)(2+i y)$

$\Rightarrow x+3 i=1(2+i y)-i(2+i y)$

$\Rightarrow x+3 i=2+i y-2 i-i^{2} y$

$\Rightarrow x+3 i=2+i(y-2)-(-1) y\left[i^{2}=-1\right]$

$\Rightarrow x+3 i=2+i(y-2)+y$

$\Rightarrow x+3 i=(2+y)+i(y-2)$

Comparing the real parts, we get

$x=2+y$

$\Rightarrow x-y=2 \ldots$ (i)

Comparing the imaginary parts, we get

$3=y-2$

$\Rightarrow y=3+2$

$\Rightarrow y=5$

Putting the value of y = 5 in eq. (i), we get

$x-5=2$

$\Rightarrow x=2+5$

$\Rightarrow x=7$

Hence, the value of x = 7 and y = 5