$f(G)$ of $f(x,y):= \begin{pmatrix} e^x \cos y \\ e^x \sin y \end{pmatrix} $ with $G = \mathbb{R} \times (-\pi, \pi)$

Question

Let $G \subset \mathbb{R^2}$ given by

$$f(x,y):= \begin{pmatrix} e^x \cos y \\ e^x \sin y \end{pmatrix} $$

Let $G = \mathbb{R} \times (-\pi, \pi)$.

To get $f(G)$ can we just put the values in the function? Meaning:

$$f(G)= \begin{pmatrix} e^{-\pi} \cos (-\pi) \\ e^\pi \sin \pi \end{pmatrix} = \begin{pmatrix} -0.0432 \\ 0 \end{pmatrix} $$

And how can one prove that $f$ has a locally differentiable inverse function in every point $(x,y)^T$? Is there a formula for the derivative of that inverse function?

Answer 1

Let $f(x,y)=(e^x\cos y,e^x\sin y)$ and $G=\mathbb{R}\times(-\pi,\pi)$. I claim that $f(G)=\mathbb{R}^2\setminus\{(0,0)\}$. Indeed, let $(a,b)\in\mathbb{R}^2$. If $a\neq 0$ then consider $y=\tan^{-1}(b/a)$ and $x=\ln\sqrt{a^2+b^2}$ so that $f(x,y)=(a,b)$. If instead $a=0$ then $\cos(y)=0$ and hence $y\in\{\pm\pi/2\}$ so that $\sin(y)\in\{\pm 1\}$. It follows that $b\neq 0$, and that if $x=\ln(|b|)$ with $y$ chosen to give $\sin(y)=\text{sgn}(b)$ then $f(x,y)=(a,b)$.

As Jan already pointed out, the other statement follows from the inverse function theorem.