$\text{ If for }\mathrm{z}=\mathrm{\alpha }+\mathrm{i\beta },|\mathrm{z}+2|=\mathrm{z}+4(1+\mathrm{i})\text{, then }\mathrm{\alpha }+\mathrm{\beta }\text{ and }\mathrm{\alpha \beta }$ are roots of the equation

$\begin{array}{r}|\mathrm{z}+2|=\mathrm{x}+4+\mathrm{i}(\mathrm{y}+4)\Rightarrow \mathrm{y}=-4\\ \Rightarrow (\mathrm{x}+2{)}^2+16=(\mathrm{x}+4{)}^2\Rightarrow \mathrm{x}=1\end{array}$
$$\begin{gathered} \alpha +\beta =-3, \alpha \beta =-4 \\ sum of roots =-7 product of roots =12 \\ \therefore \alpha +\beta ,\alpha \beta are roots of equation \\ {x}^2+7x+12=0 \end{gathered}$$