The Function $f:\mathbb{R}^{2}\rightarrow \mathbb{R}^{2}$ defined by $f(r,\theta)=(r\cos\theta,r\sin\theta)$

Let $f:\mathbb{R}^2\rightarrow \mathbb{R}^2$ be the function defined by $f(r,\theta)=(r\cos\theta,r\sin\theta).$ Then for which of the open subset $U$ of $\mathbb{R}^2$ given below, $f$ restricted to $U$ admit an inverse?

1. $U=\mathbb{R}^2$

2. $U=\{x,y \in\mathbb{R}^2:x>0,y>0\}$

3. $U=\{x,y \in\mathbb{R}^2:x^2+y^2<1\}$

4. $U=\{x,y \in\mathbb{R}^2:x<-1,y<-1\}$

It is clear 1 is not true as $\sin ,\cos$ are periodical function. Similarly 3 is not true. What about 2nd and 4th ? please help.Thanks in advance.

To admit an inverse $f$ restricted to $U$ must be bijective, in particular one-one. $1$ and $2$ are not True since $f(1,2) = f(1,2+2\pi)$. $4$ is not true since $f(-2,-2\pi) = f (-2,-4\pi)$. $3$ is not True since $f(0,1/2) = f (0,1/4)$.