Find the real numbers $x$ and $y$ if $(x-i y)(3+5 i)$ is the conjugate of $-6-24 i$
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.
If complex numbers $z_1$, $z_2$ are such that $\left| {{z_1}} \right| = \sqrt 2 ,\left| {{z_2}} \right| = \sqrt 3$ and $\left| {{z_1} + {z_2}} \right| = \sqrt {5 - 2\sqrt 3 }$, then the value of $|Arg z_1 -Arg z_2|$ is
If $\frac{{z - \alpha }}{{z + \alpha }}\left( {\alpha \in R} \right)$ is a purely imaginary number and $\left| z \right| = 2$, then a value of $\alpha $ is
If for $z=\alpha+i \beta,|z+2|=z+4(1+i)$, then $\alpha+\beta$ and $\alpha \beta$ are the roots of the equation
Let $z$, $w \in C$ satisfy ${z^2} + \bar w = z$ and ${w^2} + \bar z = w$ then number of ordered pairs of complex numbers $(z, w)$ is equal to
If $z$ is a complex number such that ${z^2} = {(\bar z)^2},$ then