If ${\mathrm{z}}_1{\mathrm{z}}_2\in \mathrm{C},{\mathrm{z}}_1^2+{\mathrm{z}}_2^2\in \mathrm{R},{\mathrm{z}}_1\left({\mathrm{z}}_1^2-3{\mathrm{z}}_2^2\right)=2$ and ${\mathrm{z}}_2\left(3{\mathrm{z}}_1^2\right.\left.-{\mathrm{z}}_2^2\right)=11$, then the value of ${\mathrm{z}}_1^2+{\mathrm{z}}_2^2$ is

We have, ${\mathrm{z}}_1\left({\mathrm{z}}_1^2-3{\mathrm{z}}_2^2\right)=2$  ….(1)

and, ${\mathrm{z}}_2\left(3{\mathrm{z}}_1^2-{\mathrm{z}}_2^2\right)=11$    …….(2)

Multiplying (2) by i and adding it to (1), we get

$\begin{array}{l}{\mathrm{z}}_1^3-3{\mathrm{z}}_2^2{\mathrm{z}}_1+\mathrm{i}\left(3{\mathrm{z}}_1^2{\mathrm{z}}_2-{\mathrm{z}}_2^3\right)=2+11\mathrm{i} \\ \Rightarrow {\left({\mathrm{z}}_1+{\mathrm{iz}}_2\right)}^3=2+11\mathrm{i} .....\left(3\right)\end{array}$

Again, multiplying (2) by i and subtracting it from (1), we get

$\begin{array}{l}{{\mathrm{z}}^3}_1-3{\mathrm{z}}_2^2{\mathrm{z}}_1-\mathrm{i}\left(3{\mathrm{z}}_1^2{\mathrm{z}}_2-{\mathrm{z}}_2^3\right)=2-11\mathrm{i}\\ \Rightarrow {\left({\mathrm{z}}_1-{\mathrm{iz}}_2\right)}^3=2-11\mathrm{i} ....\left(4\right)\end{array}$

Multiplying (3) and (4), we get

${\left({\mathrm{z}}_1^2+{\mathrm{z}}_2^2\right)}^3=4+121\Rightarrow {\mathrm{z}}_1^2+{\mathrm{z}}_2^2=5$.