# 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.

## 2 Answers

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 $f(0,1/2) = f (0,1/4)$.

• its means its incorrect question? – neelkanth Dec 23 '15 at 4:40
• f(1, 2) = f(1, 1+2$\pi$), how come? – user268307 Dec 23 '15 at 4:42
• there should be $f(1,2+2\pi)$ – neelkanth Dec 23 '15 at 4:43
• That's what I am thinking. – user268307 Dec 23 '15 at 4:44
• oh. yes ..i am sorry for that mistake, – Samiron Dec 23 '15 at 4:44

In options 2 and 4, The determinant of jacobian= $$r$$ is nonzero everywhere. Thus inverse function theorem guarantees that, for every point $$p$$ in $$U$$, there exists a neighborhood about $$p$$ over which function is invertible. This does not mean function is invertible over whole domain $$U$$: in this case $$f$$ is not even injective since it is periodic : $$f(x,y)=f(x,y+2 \pi)$$. (And injection is necessary for inverse.)

• I think all options are incorrect??? – neelkanth Nov 27 '18 at 17:39
• @neelkanth of course all options are incorrect. Actually many got confused that using inverse function theorem, options 2 and 4 should be correct. That's why I wrote this. – ramanujan Nov 27 '18 at 17:53