If  $z_1,z_2\in C$,  $z_1^2+z_2^2\in R$, $z_1(z_1^2-3z_2^2)=2$  and $z_2(3z_1^2-z_2^2)=11$ , the value of $z_1^2+z_2^2$ is

We have,  $z_1(z_1^2-3z_2^2)=2\mathrm{...........}\left(i\right)$
and  $z_2(3z_1^2-z_2^2)=11\mathrm{...........}\left(ii\right)$
multiplying equation (ii) by $i\left(\sqrt{-1}\right)$  and then adding in equation (i), we 

get
 $$\begin{gathered} z_1^3-3z_1{z}_2^2+i(3z_1^2{z}_2-z_2^3)=2+11i \\ \Rightarrow {(z_1+i{z}_2)}^3=2+11i\mathrm{............}\left(iii\right) \end{gathered}$$
 Again, multiplying equation (ii) by and then adding in equation (i), we get$$\begin{gathered} z_1^3-3z_1{z}_2^2+i(3z_1^2{z}_2-z_2^3)=2-11i \\ \Rightarrow {(z_1-i{z}_2)}^3=2-11i\mathrm{............}\left(iv\right) \end{gathered}$$
 Now, on multiplying equation (iii) and (iv), we get
 $$\begin{gathered} {(z_1^2+z_2^2)}^3=4+121=125=5^3 \\ \therefore z_1^2+z_2^2=5 \end{gathered}$$