0
$\begingroup$

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.

$\endgroup$
  • 2
    $\begingroup$ Why do you think that this function has an inverse? $\endgroup$ – José Carlos Santos Aug 14 at 17:58
  • 1
    $\begingroup$ 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
1
$\begingroup$

Hint:

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...

$\endgroup$
  • $\begingroup$ 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
  • 1
    $\begingroup$ @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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.