Solving a system of complex equations with conjugates involved I am having difficulty with a system of equations whose unknowns are two complex values. Here it is:
$$A = z_1^2 + z_2^2$$
$$B = \frac{z_1^2 - z_2^2}{z_1 z_1^* - z_2 z_2^*}$$
$$C =i\frac{z_1^* z_2 -z_1 z_2^*}{z_1 z_1^* - z_2 z_2^*}$$
$$D =z_1^* z_2 + z_1 z_2^*$$
$\{A,B,C,D\} \in \mathbb{R}$ and $\{z_1, z_2\} \in \mathbb{C}$. The first four are known, and I am looking for the two others.
I tried to develop the complex numbers by explicitly stating their real and imaginary parts, but the calculations quickly became huge and unsolvable. The same thing happened when developing them according to their module and argument: the results were more interesting however.
$$A = \rho_1^2 e^{2i\theta_1} + \rho_2^2 e^{2i\theta_2}$$
$$B = \frac{\rho_1^2 e^{2i\theta_1} + \rho_2^2 e^{2i\theta_2}}{\rho_1^2 - \rho_2^2}$$
$$C = i\frac{\rho_1 \rho_2 (e^{i(\theta_2 - \theta_1)} - e^{i(\theta_1 - \theta_2)})}{\rho_1^2 - \rho_2^2}$$
$$D = \rho_1 \rho_2 (e^{i(\theta_2 - \theta_1)} + e^{i(\theta_1 - \theta_2)}) $$
I am looking for an advice to solve this system, at best analytically, at worst via a numerical method.
I also tried to write the system of equations in Mathematica, but the software runs for hours without being able to give me a solution (at least not within a few hours).
 A: I couldn't think of any generalized way of solving this (I am curious to know how this set equations was arrived at). Certain observations can be made firstly which can help us solve this set of equations for $z_1$ and $z_2$.
First, observe that both $z_1^2$ and $z_2^2$ are real numbers (Hint: Both $z_1^2+z_2^2$ and $z_1^2-z_2^2$ are real numbers). Hence, the pair $(z_, z_2)$ can have the following possibilities only:

*

*Both $z_1$ and $z_2$ are  real.

*$z_1$ is  real and $z_2$ is purely imaginary.

*$z_1$ is purely imaginary and $z_2$ is real.

*Both $z_1$ and $z_2$ are purely imaginary.

(Note that by purely imaginary I mean a complex number of the form $ik$, where $k$ is a non zero real number.)
Let us now see the conditions in which the given system of equations has got solutions of the first kind. Now, suppose the given system of equations has got solutions $z_1$ and $z_2$ such that both are real. Then it is quite easy to see that $B=1$ and $C=0$. Clearly $D = 2z_1z_2$. And hence
$$ A+D = (z_1+z_2)^2 \quad \text{and} \quad  A-D = (z_1-z_2)^2.$$
The above two equations yield:
$$ z_1+z_2 = \pm\sqrt{A+D} \quad \text{and} \quad  z_1-z_2 = \pm\sqrt{A-D}. \tag{1}$$
Since we require both $z_1$ and $z_2$ to be real, we must have
$$A+D \geq 0 \quad \text{and} \quad A-D\geq 0.$$
If any one of the above conditions that $B=1, C=0, A+D\geq0$ and $A-D\geq 0$, is not satisfied, then the given system of equations cannot have a solution of the first kind. Clearly, the four possible solutions for $z-1$ and $z_2$ we obtain on solving (1), satisfy given system of equations (with the condition that $B=1$, $C=0$, $A+D \geq 0$ and $A-D \geq 0$.
Conversely, I just leave it to you to think that under other three
possibilities for $z_1$ and $z_2$, we will not have $B=1$. Thus the conclusion is:

Under the given conditions that $B=1$, $C=0$, $A+D \geq0, A-D\geq 0$, the given system of equations is going to have exactly four solutions $(z_1, z_2)$ such that both are real and these solutions are the ones satisfying
$$ z_1+z_2 = \pm \sqrt{A+D} \quad \text{and} \quad z_1-z_2 = \pm \sqrt{A-D}.$$

Now, can we move in this direction, when will the given set of equations is going to have second, third and fourth kind of possibilities for $z_1$ and $z_2$?
