Let $f:\mathbb{R}^2\setminus\{0\}\to\mathbb{R}^2\setminus\{0\}$ be defined as $f(x,y)=(x^2-y^2,2xy)$. Show that $f$ is locally invertible with continuously differentiable inverse function, but that $\#f^{-1}(\{z\}) = 2$ for all $z\in\mathbb{R}^2\setminus0$.

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    $\begingroup$ Hint: writing in complex coordinate, the map $f$ is given by $f(z) = z^2$. Is that easier now? $\endgroup$ – user99914 May 19 '14 at 9:43
  • $\begingroup$ What do you mean with writing in complex coordinate? We are talking about a real function here... $\endgroup$ – sj134 May 19 '14 at 9:50
  • $\begingroup$ I mean you can identify $\mathbb R^2$ as $\mathbb C$. $\endgroup$ – user99914 May 19 '14 at 11:15
  • $\begingroup$ @John Yes, thanks, I wrote an answer to my own question. Can I say my invertible function is continuously differentiable since I have $Df(x,y)$ ? $\endgroup$ – sj134 May 19 '14 at 14:36
  • $\begingroup$ You should probably use the inverse function theorem. $\endgroup$ – Najib Idrissi May 19 '14 at 14:38

For all $(x,y)\in\mathbb{R}^2$ we have $$Df(x,y)=\begin{pmatrix}2x & -2y \\ 2y & 2x\end{pmatrix}.$$ $Df(x,y)$ invertible $\iff \det(Df(x,y))\neq0$ and because $\det(Df(x,y))=4x^2+4y^2$, there exists $Df^{-1}(x,y)$ for all $(x,y)\in\mathbb{R}^2\setminus\{0\}$.

So for all $(x,y)\in\mathbb{R}^2\setminus\{0\}$ $$\begin{align}Df^{-1}(x,y) & =\frac{1}{\det(Df(x,y))}\begin{pmatrix}2x & 2y \\ -2y & 2x\end{pmatrix} \\ & = \begin{pmatrix}\frac{x}{2(x^2+y^2)} & \frac{y}{2(x^2+y^2)} \\ \frac{-y}{2(x^2+y^2)} & \frac{x}{2(x^2+y^2)}\end{pmatrix}\end{align}.$$

We now show that every point $(x,y)\in\mathbb{R}^2\setminus\{0\}$ has exactly two pre-images using complex numbers.

Let $z=x+iy$ and $w=u+iv$, then we have $$z^2 = (x+iy)^2 = (x^2-y^2)+2ixy = u+iv = w.$$

Identifying $\mathbb{R}^2$ with $\mathbb{C}$ we get that the given function is just $$\mathbb{C}\to\mathbb{C}, \quad z\mapsto w=z^2.$$

Since every complex number $w\neq0$ has exactly two roots, using the representation $w=re^{i\varphi},r\gt0,0\le\varphi\lt2\pi$ the two solution of the equation are given by $$z_{1/2} = \pm\sqrt{r}e^{i\varphi/2}.$$


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