The function $f(r,\theta)=(r\cos\theta,r\sin\theta).$ Consider the function $f:\mathbb{R}^{2}\rightarrow\mathbb{R^2}$ given by $$f(r,\theta)=(r\cos\theta,r\sin\theta)$$ I like to show that $f$ is one-to-one in some neighborhood of any non zero point $(r,\theta).$ I tried as
$(r\cos\theta,r\sin\theta)=(s\cos\phi,s\sin\phi)$
 Which gives $r=s$ and $\theta=\phi+2n\pi,$ 
How to show that $\theta=\phi?$ Please help. Thanks.
 A: Consider two points $(r_1, \theta_1)$ and $(r_2, \theta_2)$, the distance between them is given by the metric $\sqrt{r_1^2+r_2^2 - 2r_1r_2 cos(\theta_1 + \theta_2)}$. Given a point $(r, \theta)$, you have to choose a $\delta-neighborhood$ of it, that is the set of points $(\rho, \phi)$ with $\sqrt{r^2+\rho^2 -2r\rho cos(\theta + \phi)} < \delta$. Choose a suitable value of $\delta$.
A: The definition of injectivity is not useful here; instead, using the inverse map theorem is more effective.
Theorem.
Let $A \subset \Bbb{R}^{n}$ lie in the usual topology of $\Bbb{R}^{n}$; let $f \in \mathscr{C}^{1}(A, \Bbb{R}^{n})$; let $x \in A$; let $\det Df(x) \neq 0$. Then there is some open ball $B^{x}$ of center $x$ such that $B^{x}$ is ($\mathscr{C}^{1}$-)diffeomorphic to $f(B^{x})$ via $f$.
If $f: (r,\theta) \mapsto (r\cos \theta, r\sin \theta)$ on $\Bbb{R}^{2}$,
and if $(r,\theta) \in \Bbb{R}^{2}$, then $\det Df(r,\theta) = r,$ which is $\neq 0$ if in addition $r \neq 0$. Hence for every $(r,\theta) \in \Bbb{R}^{2}$ such that $r \neq 0$ the map $f$ is locally bijective near $(r,\theta)$.
However, not every $(r,\theta) \neq (0,0)$ is such that $f$ is necessarily locally bijective near $(r,\theta)$; take $(r,\theta) := (0,\pi)$, for example.
