Let $z,w\in\mathbb{C}$ be complex variables and define the function $f:w\mapsto z$ given by \begin{align} z=w+ aw^2+bw\bar{w}+c\bar{w}^2, \end{align} where $a,b,c\in\mathbb{C}$ are complex constants. What is the inverse function $f^{-1}:z\mapsto w\:$? I am not very familiar with complex analysis, so I greatly appreciate any comment or response.

Note 1: We can assume that the function is invertible, so we do not need to worry about the invertibility.

Note 2: I am reading a textbook that claims that the inverse function is \begin{align} w=z - az^2 - bz\bar{z}-c\bar{z}^2 + O(|z|^3), \end{align} and no more information is given.

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    $\begingroup$ Why do you think that this function has an inverse? $\endgroup$ – José Carlos Santos Aug 14 at 17:58
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    $\begingroup$ You mean $f:w\mapsto z$? $\endgroup$ – rschwieb Aug 14 at 17:59
  • $\begingroup$ @rschwieb Yes. Sorry for the typo. I revised it. $\endgroup$ – Arthur Aug 14 at 18:01
  • $\begingroup$ @JoséCarlosSantos I am reading an ODE book, where the author has claimed that this function has an inverse. Moreover, If $z,w\in\mathbb{R}$, then this function is invertible (at least in some interval). $\endgroup$ – Arthur Aug 14 at 18:07


The given function can be decomposed in the form of a system of real equations $$u=P(x,y),\\v=Q(x,y)$$

where $P,Q$ are bivariate quadratic polynomials.

When you vary $u$ and $v$, you obtain pencils of conics. Hence the solutions in $x,y$ are formed by the intersections of two conics, and this leads to a quartic equation, having up to four distinct solutions.

There is a (complicated) analytical solution, and you will have to select branches...

  • $\begingroup$ Thanks for the response. Is it possible to say that the inverse function has a general form, for instance $w=z-az^2-bz\bar{z}-c\bar{z}^2+O(|z|^3)$? This is all that is mentioned in the textbook that I am reading, but I cannot justify it. $\endgroup$ – Arthur Aug 14 at 18:16
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    $\begingroup$ @Arthur You should have asked this question directly, it is a very different one. $\endgroup$ – Yves Daoust Aug 14 at 18:17
  • $\begingroup$ Sorry about that. I will edit the question. $\endgroup$ – Arthur Aug 14 at 18:19

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