Solution
$\mathrm{z}=\alpha+\mathrm{i} \beta \Rightarrow|\mathrm{z}+2|=\mathrm{z}+4(\mathrm{l}+\mathrm{i}) \Rightarrow \sqrt{(\alpha+2)^{2}+\beta^{2}}=(\alpha+4)+\mathrm{i}(\beta+4)$
\[
\begin{aligned}
& \sqrt{(\alpha+2)^{2}+\beta^{2}}=\alpha+4 \quad \ldots(\mathrm{i}), \beta+4=0 \Rightarrow \beta=-4 \\
& \text { (i) } \Rightarrow \sqrt{(\alpha+2)^{2}+16}=\alpha+4 \Rightarrow \alpha^{2}+4 \alpha+20=\alpha^{2}+8 \alpha+16 \Rightarrow \alpha=1
\end{aligned}
\]
$\alpha+\beta=-3, \alpha \beta=-4$ Equation with roots -3 and -4 is $\mathrm{x}^{2}+7 \mathrm{x}+12=0$
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