I am stuck while trying to solve the following system of equations involving complex exponentials:
$$x_1 e^{j\theta_1 m_1} + x_2e^{j\theta_2 m_1} = z_1$$ $$x_1 e^{j\theta_1 m_2} + x_2e^{j\theta_2 m_2} = z_2$$ $$x_1 e^{j\theta_1 m_3} + x_2e^{j\theta_2 m_3} = z_3$$ $$x_1 e^{j\theta_1 m_4} + x_2e^{j\theta_2 m_4} = z_4$$
Here the unknowns are $x_1$, $x_2$, $\theta_1$, and $\theta_2$. Rest of the terms ($z_1, \dots, z_4$, $m_1, \dots, m_4$) are known. Also, $x_i \in \mathbb{C}$, $z_i\in \mathbb{C}$, and $m_i \in \mathbb{Z}_+$.
My attempt:
To solve this system of equations, I am looking at them as being in the following form: $$f_1(x_1, x_2, \theta_1, \theta_2)=0$$ $$f_2(x_1, x_2, \theta_1, \theta_2)=0$$ $$f_3(x_1, x_2, \theta_1, \theta_2)=0$$ $$f_4(x_1, x_2, \theta_1, \theta_2)=0$$
and I am trying to reduce this system of equations to the following form through algebraic manipulations:
$$g_1(x_1)=0$$ $$g_2(x_1, x_2)=0$$ $$g_3(x_1, x_2, \theta_1)=0$$ $$g_4(x_1, x_2, \theta_1, \theta_2)=0$$
Unfortunately, any amount of manipulation did not allow me to express the given system of equations in the above form. Any help/ideas to solve these kinds of equations are appreciated.
(Too long for a comment.) Given that $m_k \in\mathbb N$ the system can be reduced to a polynomial one, but the calculations would be unworkably expensive in the general case.
Let $a=e^{j\theta_1}$ and $b=e^{j\theta_2}$, then the system is $\;x_1a^{m_k} + x_2b^{m_k} = z_k\;$ for $\,k=1,2,3,4\,$.
Eliminating $b$ between the first two equations:
$$ \begin{align} \begin{cases} x_2 b^{m_1} &= z_1 - x_1 a^{m_1} \\ x_2 b^{m_2} &= z_2 - x_1 a^{m_2} \end{cases} \;\;\;\;&\implies\;\;\;\; x_2^{m_2-m_1} = \dfrac{\left(z_1 - x_1 a^{m_1}\right)^{m_2}}{\left(z_2 - x_1 a^{m_2}\right)^{m_1}} \\ &\implies\;\;\;\; x_2^{m_2-m_1} \left(z_2 - x_1 a^{m_2}\right)^{m_1} - \left(z_1 - x_1 a^{m_1}\right)^{m_2} = 0 \end{align} $$
The latter is of the form $P_{12}(a) = 0$ where $P_{12}$ is a polynomial in $a$ with coefficients that depend on $x_1$ and $x_2$. Two more such polynomial equations $P_{23}(a)=0$ and $P_{34}(a)=0$ can be derived in a similar way by pairing the original equations in different combinations.
The polynomial resultant of $P_{12}$ and $P_{23}$ must be $0$ since the two have a common root, which gives a polynomial equation $P_{123}(x_1, x_2) = \text{res}(P_{12}, P_{23}) = 0$, and similarly $P_{234}(x_1, x_2) = 0$. Using resultants again to eliminate $x_2$ between the two gives a polynomial equation in $x_1$ alone.