How to solve complex equation with same variable on two sides What is the analytic solution of X for the equation below?
$$conjugate(X)= \frac{-2\times A}{B\times X} $$
A, B and X are complex numbers. 
Would the magnitude of X be given by this? $$ |X| =\sqrt \frac {-2\times A}{B}$$
Then what would be the solution for the angle? Can I get a polynomial solution for X_real and X_imag without having to use atan(.) function?
Thanks in advance for your help.
 A: Because the two equations you wrote are equivalent if $X \neq 0$, every angle gives you a valid solution. You have
$$
\bar X= \frac{-2\times A}{B\times X} 
\Leftrightarrow 
|X|^2 = X \bar X = \frac{-2A}B
$$
There are only solutions, if $\frac{-2 a}B > 0$
A: You want to solve
$$
\bar z=-\frac{2a}{bz}
$$
for $z$, with $z,a,b\in\Bbb C$.
First of all $b,z\neq0$.
Multipliyng both sides by $z$, you get $|z|^2=z\bar z=-\frac{2a}{b}$ thus your thoughts about the magnitude were correct.
Then to solve, just write these complex numbers in their polar form:
$$
a=re^{i\theta}\\
b=se^{i\psi}\\
z=te^{i\vartheta} \\
\mbox{(in particular being $z,b\ne0$ you have $t,s>0$)}
$$
thus you get
$$
t^2=-2\frac rse^{i(\theta-\psi)}
$$
from which you deduce that $e^{i(\theta-\psi)}$ must be real and negative: thus $\theta-\psi\in\pi+2\pi\Bbb Z$, thus the previous equation turns into
$$
t^2=2r/s
$$
so our solutions are all and only $te^{i\vartheta}$ with $t=\sqrt{2r/s}$ and $\vartheta\in\Bbb R$ (i.e. every angle solves the equation).
Then if we want to write the solutions in an "analytic" form, just pass from polar to algebraic form: 
$$
z=\sqrt{2\frac rs}e^{i\vartheta}=\sqrt{2\frac rs}\cos\vartheta+i\sqrt{2\frac rs}\sin\vartheta\;\;.
$$
