Solving two complex independent variables from 4 scalar values

Question

I have these equations where I know all the $k_{nm}$. I'm trying to find the complex value of $\Gamma_1$ and $\Gamma_2$ with only four scalar values. Essentially, I'm expending on my previous questions (Solving two complex equation using two real values) and (Solving three complex equation using three real values). $p_n=\left|\frac{1}{\Gamma_1 k_{n5}+k_{n6}} \frac{\Gamma_2 k_{n2}+k_{n1}}{\Gamma_2 k_{n4}+k_{n3}}\right|^2$ for $n=3,4,5,6$

$p_n=\left|\frac{1}{\Gamma_1 k_{n5}+k_{n6}} \frac{\Gamma_2 k_{n2}+k_{n1}}{\Gamma_2 k_{n4}+k_{n3}}\right|^2=\left|\frac{1}{\Gamma_1 k_{n5}+k_{n6}}\right|^2 \left| \frac{\Gamma_2 k_{n2}+k_{n1}}{\Gamma_2 k_{n4}+k_{n3}}\right|^2=\psi_{n1} \psi_{n2}$

$\psi_{n1}= \frac{1}{\left| k_{n5} k_{n6} \Gamma_1\right|^2} \to r_{n1}=\left|\Gamma_1 -c_{n1}\right|^2$ with $r_{n1}^2= \frac{1}{\psi_{n1} \left| k_{n6} \right|^2}$ and $c_{n1}=-\frac{k_{n5}}{k_{n6}}$

$\psi_{n1}= \left|\frac{k_{n1} + k_{n2} \Gamma_2}{ k_{n3}+ k_{4} \Gamma_2}\right|^2 \to r_{n2}=\left|\Gamma_2 -c_{n2}\right|^2$ with $r_{n2}=\frac{(\left| k_{n1} \right|^2-\psi_{n2} \left| k_{n3} \right|^2) (\left| k_{n2} \right|^2-\psi_{n2} \left| k_{n4} \right|^2)-\left|k_{n1} {k_{n2}}^*-k_{n3}{k_{n4}}^* \psi _2\right|^2 }{\left| k_{n2} \right|^2-\left| k_{n4} \right|^2}$and $c_{2}=-\frac{k_{n1} {k_{n2}}^*-k_{n3}{k_{n4}}^* \psi _2 }{\left| k_{n2} \right|^2-\left| k_{n4} \right|^2}$

I feel that this form is somewhat part of the solution since it allows the two equations to be decoupled one from the other and it would then be possible to place it in a matrix and solve it from there. But, I haven't figured how yet.

An other idea I had was to make a 6*6 matrix with $\left( \begin{array}{c} \Gamma _1 \\ {\Gamma _1}^* \\ {\left| \Gamma _1\right|}^2 \\ \Gamma _2 \\ {\Gamma _2}^* \\ {\left| \Gamma _2\right|}^2 \\ \end{array} \right).$I was hoping from there I would be able to solve it with Cramer rule like was suggest in my previous question, but I can't figure how to prevent $\Gamma_1$ and $\Gamma_2$ from multiplying. If they do multiply, I get 9 equations. I don't think I need that many degree of freedom (but might be wrong).

I also tried to use the exact technique from Solving three complex equation using three real values but considering $\Gamma_1$ as $p_{n1}$ but the degree of freedom is too low and I can't figure how to increase it.

Do you have any suggestion to solve this problem? Maybe I started in the wrong direction. I'm not sure what to do.

Thanks!

Answer 1

(Too long for a comment.)

Eliminating denominators and expanding the squared magnitudes, similarly to this answer, the system can be written as

$$ a_{k,1} \Gamma_2 + a_{k,2} \overline{\Gamma_2} + a_{k,3} \left|\Gamma_2\right|^2 + a_{k,4} = 0 \qquad \text{for }\; k = 1,2,3,4 $$

where the coefficients $\,a_{i,j}\,$ are complex polynomials in $\,\Gamma_1, \overline{\Gamma_1}$

Considering the four equations as a linear homogeneous system in $\,\Gamma_2, \overline{\Gamma_2}, \left|\Gamma_2\right|^2, 1\,$, the main determinant must be zero for the non-trivial solution to exist:

$$ \det \big[\,a_{i,j}\,\big] = 0 $$

The latter is an algebraic equation of the form $\,p(\Gamma_1, \overline{\Gamma_1}) = 0\,$ with $\,p\,$ being a bivariate complex polynomial. Taking the conjugate $\,\overline{p}(\overline{\Gamma_1}, \Gamma_1) = 0\,$, and eliminating $\,\overline{\Gamma_1}\,$ between the two (using resultants, for example) gives an algebraic equation in $\,\Gamma_1\,$ alone. The calculations are laborious, though, and there will be no closed-form algebraic solutions in general.