Question

Find the real number $x$ and $y$ if$(x–iy)(3+5i)$is the conjugate of $–6–24i.$

Answer 1

Assume variable

Let’s assume $z=(x-iy)(3+5i)$

$z=3x+5xi-3yi-5y(i{)}^2$

$z=3x+5xi–3yi+5y$

$z=(3x+5y)+i(5x–3y)$

Finding conjugate

$\overline{z}=(3x+5y)-i(5x-3y)$

Compare real & imaginary parts

$\text{Given}\overline{z}=-6-24i$

$\text{then},\overline{z}=(3x+5y)-i(5x-3y)=-6-24i$

On equating real and imaginary parts, we have

$3x+5y=–6\dots ..\left(1\right)$

$5x–3y=24\dots ..\left(2\right)$

Finding the intersection point

On solving equation $\left(1\right)\times 3+\left(2\right)\times 5$

$\Rightarrow (9x+15y)+(25x–15y)=–18+120$

$\Rightarrow 34x=102$

$\Rightarrow x=\frac{102}{34}=3$

Putting the value of x in equation (1), we get

$\Rightarrow 3\left(3\right)+5y=–6$

$\Rightarrow 5y=–6–9=–15\Rightarrow y=–3$