Question

If $\frac{x+3i}{2+iy}=1-i$, then the value of ${(5x-7y)}^2$ is

• A. $1$
• B. $0$
• C. $2$
• D. $4$

Answer 1

Answer: (B)

$0$

$\frac{x+3i}{2+iy}=1-i$

Or
$x+3i=(1-i)(2+iy)$
$x+3i=2+iy-2i+y$
$x-y-2=i(y-5)$
Or
$(x-y-2)+i(5-y)=0$

Hence
$5-y=0$ or $y=5$ and

$x-y-2=0$ or $x-y=2$

Since $y=5$

Hence
$x=y+2$ or $x=5+2$

$x=7$

Therefore
$x=7$ and $y=5$.

Therefore
$(5x-7y{)}^2$

$=\left(5\right(7)-7(5){)}^2$

$=(35-35{)}^2$

$=0$