Find the real numbers $x$ and $y$ if $(x-i y)(3+5 i)$ is the conjugate of $-6-24 i$

Vedclass pdf generator app on play store
Vedclass iOS app on app store

Let $z=(x-i y)(3+5 i)$

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

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

It is given that, $\bar{z}=-6-24 i$

$\therefore(3 x+5 y)-i(5 x-3 y)=-6-24 i$

Equating real and imaginary parts, we obtain

$3 x+5 y=-6$.....$(i)$

$5 x-3 y=24$....$(ii)$

Multiplying equation $(i)$ by $3$ and equation $(ii)$ by $5$ and then adding them, we obtain

$9 x+15 y=-18$

${25 x-15 y=120}$

$\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_$

${34 x=102}$

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

Putting the value of $x$ in equation $(i),$ we obtain

$3(3)+5 y=-6$

$\Rightarrow 5 y=-6-9=-15$

$\Rightarrow y=-3$

Thus, the values of $x$ and $y$ are $3 $ and $-3$ respectively.

Similar Questions

Given $z$ is a complex number such that  $|z| < 2,$ then the maximum value of $|iz + 6 -8i|$ is equal to-

If  $z_1 = a + ib$ and $z_2 = c + id$ are complex numbers such that   $| z_1 | = | z_2 |=1$ and  $R({z_1}\overline {{z_2}} ) = 0$, then the pair of complex numbers $w_1 = a + ic$ and $w_2 = b + id$ satisfies

If $z_1$ is a point on $z\bar{z} = 1$ and $z_2$ is another point on $(4 -3i)z + (4 + 3i)z -15 = 0$, then $|z_1 -z_2|_{min}$ is (where $ i = \sqrt { - 1}$ )

If the conjugate of $(x + iy)(1 - 2i)$ be $1 + i$, then

If $z$ is a purely real number such that ${\mathop{\rm Re}\nolimits} (z) < 0$, then $arg(z)$ is equal to