Quadratic complex equation with conjugate

Question

I am trying to solve this equation:

$az = b + c z \bar{z}$

where, $a, b$ and $c$ are complex known values, and $\bar{z}$ is the conjugate of the variable $z$. So far I have done this:

$a=a_1+a_2j $
$b=b_1+b_2j $
$c=c_1+c_2j $
$z=x+yj $

After replacing and doing the operation, I have got two quadratic equations that I cant solve using Matlab's solver (it says: "Warning: Explicit solution could not be found.")

$b_1 - a_1 x + a_2 y + c_1(x^2+y^2) = 0 $
$b_2 - a_1 y + a_2 x + c_2(x^2+y^2) = 0 $

Is there any other way to solve that quadratic complex equation (maybe just from the first equation without using real and imaginary decomposition)? am I missing something? I really would appreciate very much any help you can give me. Thanks a lot in advance))

Answer 1

Hint: if $\,c=0\,$ then the equation reduces to a trivial linear one. Otherwise take the conjugate:

$$ \begin{cases} az = b + c z \bar{z} \\ \bar a \bar z = \bar b + \bar c \bar z z \end{cases} $$

Multiply the first equation by $\bar c$, the second one by $c$ and subtract so that the $z \bar z$ terms cancel out:

$$\require{cancel} a \bar c z - \bar a c \bar z = b \bar c - \bar b c + \cancel{c \bar c z \bar z} - \cancel{\bar c c \bar z z} $$

Other than the trivial case $ac=0\,$, the above gives $\bar z = \cfrac{a \bar c z - b \bar c + \bar b c}{\bar a c}\,$, then substituting $\bar z$ back into the original equation gives a plain quadratic in $z\,$.