Show $f$ locally invertible with continuously differentiable inverse function, but $\#f^{-1}(\{z\}) = 2$ 
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 $z\in\mathbb{R}^2\setminus0$.

 A: 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
$$z_{1/2} = \pm\sqrt{r}e^{i\varphi/2}.$$
