Let $f : \mathbb{R}^2 \to \mathbb{R}^2$ be defined by the equation $f(x, y) = (e^x \cos y, e^x \sin y)$

Question

Let $f : \mathbb{R}^2 \to \mathbb{R}^2$ be defined by the equation $f(x, y) = (e^x \cos y, e^x \sin y)$.

(a) Show that $f$ is one-to-one on the set $A$ consisting of all $(x, y)$ with $0 < y < 2\pi$.

(b) What is the set $B = f(A)$?

(c) If $g$ is the inverse function , find $Dg(0, 1)$.

(a) Suppose $f(x_1,y_1)=f(x_2,y_2)$ with which $||f(x_1,y_1)||=||f(x_2,y_2)||$ so $e^{2x_1}\cos^2 y_1+e^{2x_1}\sin^2 y_1=e^{2x_2}\cos^2 y_2+e^{2x_2}\sin^2 y_2$ so $e^{2x_1}=e^{2x_2}$, but $x \mapsto 2x^2 $ is injective, then $x_1=x_2$. But if $x_1=x_2$ then $\cos y_1=\cos y_2$ and $\sin y_1=\sin y_2$ whereupon there must be a $k\in \mathbb{Z}$ such that $y_2=2k\pi +y_1$ but as $0<y_1<2\pi$ and $0<y_2<2\pi$ then necessarily $y_1=y_2$ and thus $(x_1,x_2)=(y_1,y_2)$.

(b) I do not know how to do this, I think you can use polar coordinates but I do not know how, how could this be done? What happens if I identify $\mathbb{C}$ with $\mathbb{R}^2$ and consider the function $f(z)=z^2$?

(c) There is a post very similar to this one and in one of them answer this exercise $f:\mathbb R^2\to\mathbb R^2$ be defined by the equation $f(x,y)=(e^x \cos y,e^x \sin y)$, my question is if that answer also applies here taking into account that the set $A$ is different?

Answer 1

The point of view that helps answer a question like (b) is that, for fixed $x$, $(e^x\cos y,e^x\sin y)$ is circle of radius $e^x$. When you think in this terms, you see that $f(A)$ is $\mathbb R^2\setminus\{0\}$ (any point other than the origin can be described in terms of a radius and an angle).

For part (c), the derivative $Dg$ is, for each point, a linear map $\mathbb R^2\to\mathbb R^2$. You want to use the Inverse Function Theorem, which interestingly has precisely your function as an example. You have $$ Dg(0,1)=J(0,1)^{-1}=\begin{bmatrix} e^0\cos 1&-e^0\sin 1\\ e^0\sin 1& e^0\cos 1\end{bmatrix} ^{-1}=\begin{bmatrix} \cos 1&-\sin 1\\ \sin 1& \cos 1\end{bmatrix} ^{-1}=\begin{bmatrix} \cos1&\sin 1\\-\sin 1&\cos 1\end{bmatrix} $$ (the inverse is really easy because the matrix is orthogonal).