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))

  • $\begingroup$ This is not really a quadratic equation since it is not a polynomial equation. It amounts to finding the intersections of two conics. $\endgroup$ – Bernard Apr 4 '17 at 17:02
  • $\begingroup$ is there a way to isolated $z$ of the first equation? If not, could you advice me a straightforward way to find the intersection, although I plotted the equations and they seem to be more like circles than conics. $\endgroup$ – Bur Nor Apr 4 '17 at 17:35
  • $\begingroup$ The curves are indeed circles (which are conics anyway). You can eliminate $x^2+y^2$ from the two equation to obtain a linear relation between $x$ and $y$, then a quadratic equation in a single unknown. $\endgroup$ – Bernard Apr 4 '17 at 18:04

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 $z\,$.


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