JEE Main 2023 — Complex Number Question with Solution

Given,

$z=\alpha +i\beta \& \left|z+2\right|=z+4\left(1+i\right)$

Now putting the value of $z=\alpha +i\beta$ in $\left|z+2\right|=z+4\left(1+i\right)$ we get,

$\left|z+2\right|=z+4\left(1+i\right)$

$\Rightarrow \sqrt{{\left(\alpha +2\right)}^2+{\beta }^2}=\left(\alpha +4\right)+i\left(\beta +4\right)$

Now on comparing real and imaginary part, we get

$\sqrt{(\alpha +2{)}^2+{\beta }^2}=\alpha +4...\left(i\right)$ and  $\beta +4=0\Rightarrow \beta =-4 ...\left(ii\right)$

Now solving,

$\sqrt{(\alpha +2{)}^2+16}=\alpha +4$

$\Rightarrow {\alpha }^2+4\alpha +20={\alpha }^2+8\alpha +16$

$\Rightarrow \alpha =1$

So, $\alpha +\beta =-3, \alpha \beta =-4$

We know that quadratic equation is given by,

$x^2-\left(\text{sum of roots}\right)x+\left(\text{product of roots}\right)=0$

So, equation with roots $-3$ and $-4$ will be,

$x^2+7x+12=0$