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

Given : $|z+2|=z+4(1+i)$ <br/><br/>Also, $z=\alpha+i \beta$ <br/><br/>$$ \begin{aligned} & \therefore|z+2|=|\alpha+i \beta+2|=(\alpha+i \beta)+4+4 i \\\\ & \Rightarrow|(\alpha+2)+i \beta|=(\alpha+4)+i(\beta+4) \\\\ & \Rightarrow \sqrt{(\alpha+2)^2+\beta^2}=(\alpha+4)+i(\beta+4) \\\\ & \Rightarrow \beta+4=0 \Rightarrow \beta=-4 \end{aligned} $$ <br/><br/>$$ \begin{aligned} & \text { Now, }(\alpha+2)^2+\beta^2=(\alpha+4)^2 \\\\ & \Rightarrow \alpha^2+4+4 \alpha+\beta^2=\alpha^2+16+8 \alpha \\\\ & \Rightarrow 4+4 \alpha+16=16+8 \alpha \\\\ & \Rightarrow 4 \alpha=4 \Rightarrow \alpha=1 \\\\ & \text { So, } \alpha+\beta=-3 \text { and } \alpha \beta=-4 \\\\ & \therefore \text { Required equation is } \\\\ & x^2-(-3-4) x+(-3)(-4)=0 \\\\ & \Rightarrow x^2+7 x+12=0 \end{aligned} $$