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If $z_1$ and $z_2$ are complex numbers such that $z_1^2+z_2^2 \in\mathbb R$ and $$z_1(z_1^2-3z_2^2)=2,\qquad z_2(3z_1^2-z_2^2)=11,$$ then find the value of $(z_1^2+z_2^2)^2$. Given answer is $25$.

I have tried many things but I am not getting the answer.

I subtracted two equations to observe that $11/z_2 - 2/z_1$ must be real. Also if $z_1=x_1+iy_1$ and $z_2=x_2+i y_2$, then using the fact that $z_1^2+z_2^2 \in\mathbb R$, we get $x_1 y_1+x_2y_2=0$ but I am not able to compile these results to get the desired value.

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    I think you can also assume them to be conjugates – 2017-01-26
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    This is not an exact duplicate, but a close paraphrase of [if: $x^2-\frac{2}{x}=3y^2,\;y^2-\frac{11}{y}=3x^2$ then : $x^2+y^2=?$](http://math.stackexchange.com/questions/2113062/if-x2-frac2x-3y2-y2-frac11y-3x2-then-x2y2/2113904#2113904). – 2017-01-26

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I suggest that we square both equations. $$z_1^2(z_1^2-3z_2^2)^2=4\\ z_2^2(3z_1^2-z_2^2)^2=121$$

Now add them together and simplify. $$125=z_1^2(z_1^2-3z_2^2)^2+z_2^2(3z_1^2-z_2^2)^2=z_1^2(z_1^4-6z_1^2z_2^2+9z_2^4)+z_2^2(9z_1^4-6z_1^2z_2^2+z_2^4)=\\ =z_1^6+3z_1^4z_2^2+3z_1^2z_2^4+z_2^6=(z_1^2+z_2^2)^3$$ The rest is trivial.

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$$z_{1}(z^2_{1}-3z^2_{2}) = 2\Rightarrow z^3_{1}-3z_{1}z^2_{2} = 2\cdots \cdots (1)$$

Similarly $$z_{2}(3z^2_{1}-z_{2}^2) = 11\Rightarrow 3z^2_{1}z_{2}-z^3_{2} = 11\cdots (2)\times i$$

Now Add and Subtract these two equation, we get $$(z_{1}+iz_{2})^3 = 2+11i$$

$$(z_{1}-iz_{2})^3 = 2-11i$$

So $$(z^2_{1}+z^2_{2})^3 = (2+11i)(2-11i) = 125\Rightarrow z^2_{1}+z^2_{2}=5\Rightarrow (z^2_{1}+z^2_{2}) = 25$$