Finding an inverse of $f(x,y)=(e^x \cos y,e^x \sin y)$ on a neighborhood of a given point

Question

2. 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?

Answer 1

As you've written, we can compute $x(u, v) = \frac{1}{2} \log (u^2 + v^2)$, so it's enough to compute an appropriate function $y(u, v)$. More explicitly, it's enough to

1. express some suitable combination $g(u, v)$ as a function $h(y)$ of $y$, and
2. find an interval $I \ni y_0$ on which $h$ is invertible, where $(x_0, y_0)$ is the point near which we aim to construct a local inverse.

Then, if we declare $y(u, v) := h\vert_I^{-1}(g(u, v))$, $f\vert_{\Bbb R \times I}$ will have inverse $$(u, v) \mapsto \left(\frac{1}{2} \log (u^2 + v^2), y(u, v)\right) .$$

Hint Since $y$ is an angular coordinate, it's typical to compute $\frac{v}{u} = \frac{e^x \sin y}{e^x \cos y} = \tan y$, which holds wherever $u \neq 0$.

First consider the point $(0, \pi)$. Since $\tan$ has period $\pi$, the inverse of $\tan$ on $\left(\frac{\pi}{2}, \frac{3 \pi}{2}\right) \ni \pi$ is $\arctan t + \pi$.

So for $U := \Bbb R \times \left(\frac{\pi}{2}, \frac{3 \pi}{2}\right) \ni (1, \frac{\pi}{2})$, the inverse of $f\vert_U$ is $$(f\vert_U)^{-1} : (-\infty, 0) \times \Bbb R \to \Bbb R \times \left(\frac{\pi}{2}, \frac{3 \pi}{2}\right), \qquad (u, v) \mapsto \left(\frac{1}{2} \log (u^2 + v^2), \arctan \frac{v}{u} + \pi\right) .$$

We can deal with the point $(x, y) = \left(-1, \frac{\pi}{2}\right)$ similarly, but notice that $f\left(-1, \frac{\pi}{2}\right) = (0, \frac{1}{e})$ is on the $u$-axis, where $\arctan \frac{v}{u}$ is undefined. So, instead consider $\frac{u}{v} = \cot y$, which holds wherever $v \neq 0$.

Answer 2

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 $f^{-1}=h^{-1}\circ g^{-1}$