- Show that $f\left(x,y\right)=\left(e^{x}\cos y,e^{x}\sin y\right)$ is one-to-one around any point of $\mathbb{R}^{2}$. For the points $\left(0,\pi\right)$ and $\left(-1,\frac{\pi}{2}\right)$ find such neighborhoods and the suitable inverse function of $f$.
This is actually the exact same question as: Inverse function theorem question - multivariable calculus
But after being able to do all the steps given there I am a bit stuck on the last part. I have done pretty much the same process and calculations there to find that if $u=e^{x}\cos(y)$ and $v=e^{x}\sin(y)$ then
$$x=\frac{1}{2}\ln\left(u^{2}+v^{2}\right)$$ and for any integer $k$ (and $\arccos$ defined from $[-1,1]$ to $[0,\pi]$ $$y=\pm\arccos\left(\frac{u}{\sqrt{u^{2}+v^{2}}}\right)+2\pi k$$
but I'm having troubles explicitly finding an inverse function and a neighborhood for the given points. e.g. for $(0,\pi)$ if I want to define the inverse function as
$$g\left(u,v\right)=\left(\frac{1}{2}\ln\left(u^{2}+v^{2}\right),\arccos\left(\frac{u}{\sqrt{u^{2}+v^{2}}}\right)\right)$$
so that $g(f(0,\pi))=(0,\pi)$ I'm not sure how to find the neighborhood for which this actually holds (and prove it is true in that neighborhood).
Edit: Actually still stuck on this.. Looking at the point given above $(0,\pi)$, any neighborhood of it will include numbers larger than $\pi$ for the second variable, which I won't be able to "translate back" with $g$, even if I add $2\pi$. i.e. if I check $(0,\frac{3\pi}{2})$ , then
$$ g(f(0,\frac{3\pi}{2}))=g(\cos \frac{3\pi}{2},\sin \frac{3\pi}{2})=(0, \arccos (\cos \frac{3\pi}{2})) = (0, \frac{\pi}{2}) $$ Am I thinking about this wrong?
Hint: define $g:\mathbb {R^2}\to \mathbb {R^2}, g(e^x\cos(y), e^x\sin(y))=(r\cos(y),r\sin(y))$. Is clear that $g$ is bijective and $h=g\circ f$ too. Now, $h^{-1}$ is popular and you can obtain $f^{-1}=h^{-1}\circ g^{-1}$