JEE 2023 — Mathematics PYQ

Given: $|z+2|=z+4(1+i)$
Also, $z=\alpha +i\beta$
$\therefore |z+2|=|\alpha +i\beta +2|=(\alpha +i\beta )+4+4i$
$\Rightarrow |(\alpha +2)+i\beta |=(\alpha +4)+i(\beta +4)$
$\Rightarrow \sqrt{(\alpha +2{)}^2+{\beta }^2}=(\alpha +4)+i(\beta +4)$
$\Rightarrow \beta +4=0\Rightarrow \beta =-4$

Now, $(\alpha +2{)}^2+{\beta }^2=(\alpha +4{)}^2$
$\Rightarrow {\alpha }^2+4+4\alpha +{\beta }^2={\alpha }^2+16+8\alpha$
$\Rightarrow 4+4\alpha +16=16+8\alpha$
$\Rightarrow 4\alpha =4\Rightarrow \alpha =1$
So, $\alpha +\beta =-3$ and $\alpha \beta =-4$
$\therefore$ Required equation is $x^2-(-3-4)x+(-3)(-4)=0$
$\Rightarrow x^2+7x+12=0$