Does the map
$$(x,y) \rightarrow (\frac{x^2-y^2}{x^2+y^2},\frac{xy}{x^2+y^2})$$ have a local inverse near $(0,1)$?
Here is my attempt:
treat this as a mapping $f:\mathbb{R^2}\rightarrow\mathbb{R^2}$, with $$Jf = \begin{vmatrix}\frac{4xy^2}{\left(x^2+y^2\right)^2}&-\frac{4x^2y}{\left(x^2+y^2\right)^2}\\\frac{y\left(-x^2+y^2\right)}{\left(x^2+y^2\right)^2}& \frac{x\left(x^2-y^2\right)}{\left(x^2+y^2\right)^2}\end{vmatrix}$$
So $Jf(1,0)=0$, and we cannot use the implicit function theorem, but this does not mean that there is no local inverse, does it? How do I determine if there is a local inverse near a point?
Best Answer
In polar coordinates, $f(r,\theta)=\left(\cos 2\theta,\frac{1}{2}\sin 2\theta\right)$ so $f$ is constant on the ray $(0,\infty)$ which means that $f$ is not injective on any neighborhood of $(1,0)$ in $\mathbb R^2$, and so it is not locally invertible.