Browse Questions

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

$(a)\;3,-3\qquad(b)\;3,-4\qquad(c)\;2,3\qquad(d)\;3,-2$

Answer : $\;3 ,-3$
Explanation :
Let $\;z= (x-iy)(3+5i)\;= 3x+5xi-3iy-5yi^{2}$
$= 3x + 5y+5xi-3iy$
$= (3x+5y )+i(5x-3y)$
$\overline{z} = (3x+5y)-i(5x-3y)$
it is given that $\overline{z} = -6 - 24 i$
$-6 - 24 i = (3x+5y) - i (5x-3y)$
Equating real and imaginary parts , we obtain
$3x+4y = -6$---(i)
$5x-3y=-24$----(ii)
Multiplying equation (i) by 3 and equation (ii) by 5 and then adding them , we obtain
$9x+12y=-18$
$25x-15y=120$
$---------$
$34x \qquad \qquad= 102$
$x = \large\frac{102}{34} =3$
Putting the value of x in the equation , we obtain
$3(3)+5y=-6$
$y=-3$
Thus the value of x and y are 3 and -3 respectively .