Solving three complex equation using three real values

Question

I have this problem for a while now and really can't find a solution. I have the following equations:

$p_n=p_1 | \frac{k_{\text{n1}}+\Gamma k_{\text{n2}}}{k_{\text{n3}}+\Gamma k_{\text{n4}}}| {}^2$, for $n=3, 4, 5$

$p_n\in \mathbb{R}$, $p_1\in \mathbb{R}$, $k_{\text{nm}}\in \mathbb{C}$ and $\Gamma \in \mathbb{C}$

Where $p_n$ and $k_{\text{nm}}$ are known. The unknowns are $Re[\Gamma]$, $Im[\Gamma]$ and $p_1$. Since we have 3 frees equations and 3 unknowns, there should be a solution. I posted previously a similar question (Solving two complex equation using two real values) that was similar but with $p_1$ known. And using the Apollonian circle demonstration, it can be found the following equation: $\frac{p_n | k_{\text{n2}} k_{\text{n3}}-k_{\text{n1}}k_{\text{n4}}| {}^2}{p_1 \left(|k_{\text{n2}}| {}^2-\frac{p_n}{p_1} | k_{\text{n4}}|{}^2\right){}^2}=(Im[\Gamma]+Im[\Delta ])^2+(Re[\Gamma]+Re[\Delta])^2$ with $\Delta=\frac{k_{\text{n1}} \left(k_{\text{n2}}\right){}^* -\frac{p_n}{p_1}k_{\text{n3}}\left(k_{\text{n4}}\right){}^*}{| k_{\text{n2}}|{}^2 -\frac{p_n}{p_1} | k_{\text{n4}}| {}^2}]$

If we expand the previous equation, we can find write each equations over the matrix form. $0=P^T.A_n.\Gamma _r+P^T.B_n.\Gamma _i$ where $P=\begin{pmatrix}1\\\ p_1 \\\ p_1^2\end{pmatrix}$, $\Gamma_r=\begin{pmatrix}1\\\ Re[\Gamma] \\\ Re[\Gamma]^2\end{pmatrix}$ and $\Gamma_i=\begin{pmatrix}1\\\ Im[\Gamma] \\\ Im[\Gamma]^2\end{pmatrix}$

In that case, $A_n$ and $B_n$ are known 3*3 matrices generated by the $k_{nm}$. Sadly, from here, I'm not sure where to go. Here is what I tried:

-Using Wolfram Mathematica, I tried to solve the system of equation in either form with or without values. Every one of those scenarios seems to complex to be solved as is.

-I also tried to solve it visually. Essentially, those each of those equation define an exponential cone that will all meet in one or more location for three given values of $p_n$. Using Wolfram, I'm able to plot it and can find quite easily the right answer. But, it is not able to find the numerical value.

Do you have any idea how I can solve this problem?

Thanks

Answer 1

(Too long for a comment.)

With $p_1 \mapsto q, k_{n1} \mapsto a_n, k_{n2} \mapsto b_n, k_{n3} \mapsto c_n, k_{n4} \mapsto d_n$ the equations are $p_n = q\, \left| \frac{a_n + b_n \Gamma}{c_n + d_n \Gamma}\right|^2$, or:

$$ p_n\left(c_n + d_n \Gamma\right)\left(\overline{c_n} + \overline{d_n} \overline{\Gamma}\right) = q\left(a_n + b_n \Gamma\right)\left(\overline{a_n} + \overline{b_n} \overline{\Gamma}\right) $$

Expanding and regrouping:

$$ (p_n\overline{c_n}d_n - q\overline{a_n}b_n) \,\Gamma + (p_nc_n\overline{d_n} - qa_n\overline{b_n}) \,\overline{\Gamma} + (p_n|d_n|^2 - q|b_n|^2) \,|\Gamma|^2 = q|a_n|^2 - p_n|c_n|^2 $$

This can be solved as a linear system of $3$ equations in $\Gamma, \overline{\Gamma}, |\Gamma|^2$ with the solutions being rational functions with both the numerator and denominator having degree $3$ in $q$. Then the equality $\Gamma \cdot \overline{\Gamma} = |\Gamma|^2$ gives a sextic equation in $q$. Assuming it can be solved, the problem then reduces to the one in OP's linked question.

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[ EDIT ] Solving the linear system by Cramer's rule the equation $\Gamma \cdot \overline\Gamma = |\Gamma|^2$ is:

$$ \small\begin{vmatrix} q|a_1|^2 - p_1|c_1|^2 &p_1c_1\overline{d_1} - qa_1\overline{b_1} &p_1|d_1|^2 - q|b_1|^2 \\ q|a_2|^2 - p_2|c_2|^2 &p_2c_2\overline{d_2} - qa_2\overline{b_2} &p_2|d_2|^2 - q|b_2|^2 \\ q|a_3|^2 - p_3|c_3|^2 &p_3c_3\overline{d_3} - qa_3\overline{b_3} &p_3|d_3|^2 - q|b_3|^2 \end{vmatrix} \cdot \small\begin{vmatrix} p_1\overline{c_1}d_1 - q\overline{a_1}b_1 &q|a_1|^2 - p_1|c_1|^2 &p_1|d_1|^2 - q|b_1|^2 \\ p_2\overline{c_2}d_2 - q\overline{a_2}b_2 &q|a_2|^2 - p_2|c_2|^2 &p_2|d_2|^2 - q|b_2|^2 \\ p_3\overline{c_3}d_3 - q\overline{a_3}b_3 &q|a_3|^2 - p_3|c_3|^2 &p_3|d_3|^2 - q|b_3|^2 \end{vmatrix} \\ = \\ \small\begin{vmatrix} p_1\overline{c_1}d_1 - q\overline{a_1}b_1 &p_1c_1\overline{d_1} - qa_1\overline{b_1} &q|a_1|^2 - p_1|c_1|^2 \\ p_2\overline{c_2}d_2 - q\overline{a_2}b_2 &p_2c_2\overline{d_2} - qa_2\overline{b_2} &q|a_1|^2 - p_1|c_1|^2 \\ p_3\overline{c_3}d_3 - q\overline{a_3}b_3 &p_3c_3\overline{d_3} - qa_3\overline{b_3} &q|a_1|^2 - p_1|c_1|^2 \end{vmatrix} \cdot \small\begin{vmatrix} p_1\overline{c_1}d_1 - q\overline{a_1}b_1 &p_1c_1\overline{d_1} - qa_1\overline{b_1} &p_1|d_1|^2 - q|b_1|^2 \\ p_2\overline{c_2}d_2 - q\overline{a_2}b_2 &p_2c_2\overline{d_2} - qa_2\overline{b_2} &p_2|d_2|^2 - q|b_2|^2 \\ p_3\overline{c_3}d_3 - q\overline{a_3}b_3 &p_3c_3\overline{d_3} - qa_3\overline{b_3} &p_3|d_3|^2 - q|b_3|^2 \end{vmatrix} $$

The above is a sextic in $q$, which does not have closed-form algebraic solutions in general, but can be solved numerically using any of the polynomial root-finding algorithms.