Solution
$\mathrm{z}^{2}+4 \mathrm{z}-(1+12 \mathrm{i})=0$.
\[
\begin{aligned}
& \Rightarrow \mathrm{z}=\frac{-4 \pm \sqrt{16+4(1+12 \mathrm{i})}}{2}=-2 \pm \sqrt{5+12 \mathrm{i}} \\
& \because \sqrt{5+12 \mathrm{i}}=\mathrm{x}+\mathrm{iy} \text { where } \mathrm{x}, \mathrm{y} \in \mathrm{R} \\
& \Rightarrow 5+12 \mathrm{i}=\mathrm{x}^{2}-\mathrm{y}^{2}+2 \mathrm{xyi} \\
& \Rightarrow 2 \mathrm{xy}=12 \Rightarrow \mathrm{y}=\frac{6}{\mathrm{x}} \& \mathrm{x}^{2}-\mathrm{y}^{2}=5 \\
& \Rightarrow \mathrm{x}^{2}-\frac{36}{\mathrm{x}^{2}}=5 \Rightarrow \mathrm{x}^{4}-5 \mathrm{x}^{2}-36=0 \\
& \Rightarrow\left(\mathrm{x}^{2}-9\right)\left(\mathrm{x}^{2}+4\right)=0 \Rightarrow \mathrm{x}^{2}=-4 \Rightarrow \mathrm{x}= \pm 2 \mathrm{i} \text { (rejected) } \\
& \Rightarrow \mathrm{x}^{2}=9 \Rightarrow \mathrm{x}= \pm 3 ; \mathrm{y}=\frac{6}{\mathrm{x}}= \pm 2 \\
& \because \sqrt{5+12 \mathrm{i}}= \pm(3+2 \mathrm{i})
\end{aligned}
\]
\[
\begin{aligned}
& \Rightarrow z=-2 \pm(3+2 i) \Rightarrow z=1+2 i,-5-2 i \\
& \therefore\left|z_{1}\right|^{2}+\left|z_{2}\right|^{2}=(1+4)+(25+4)=5+29=34
\end{aligned}
\]
.
