A question on maps having polar co-ordinates

Let $f(r,\theta)=(r\cos\theta ,r\sin\theta)$ for $(r,\theta)$ $\in \mathbb R^2$ with $r\ne0$. Pick out the true statements?

1.$f$ is one-one on {$(r,\theta)$$\in\mathbb R^2:r\ne0} 2.for any (r,\theta)$$\in\mathbb R^2$ with $r\ne0$, $f$ is one-one 0n a neighborhood of $(r.\theta)$

I think Statement 1. is false since $r\ne0$, $f(r,2\pi)$=$f(r,4\pi)$ does not imply $(r,2\pi)$=$(r,4\pi)$

similar reason for 2

Am I right

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Let $\langle r,\theta\rangle$ be a point at which $r\ne 0$. Let $$N=\{\langle r_1,\theta_1\rangle:|r-r_1|<|r|\text{ and }|\theta-\theta_1|<\pi\}\;;$$ $N$ is a neighborhood of $\langle r,\theta\rangle$. Try to show that $f$ is one-to-one on $N$; I’ve complete the solution below, but I’ve left it spoiler-protected to give you a chance to try it your own. (Mouse-over to see it.)
Suppose that $\langle r_1,\theta_1\rangle\in N$ and $f(\langle r,\theta\rangle)=f(\langle r_1,\theta_1\rangle)$. Then $r\cos\theta=r_1\cos\theta_1$, and $r\sin\theta=r_1\sin\theta_1$. Let $a=\frac{r_1}r\ne 0$; then $\cos\theta_1=a\cos\theta$, and $\sin\theta_1=a\sin\theta$. At least one of $\sin\theta$ and $\cos\theta$ is non-zero; suppose that $\cos\theta\ne 0$. Then we can divide the equation $\sin\theta_1=a\sin\theta$ by $\cos\theta_1=a\cos\theta$ to get $\tan\theta_1=\tan\theta$. Since $|\theta-\theta_1|<\pi$, this implies that $\theta=\theta_1$. But then $\cos\theta=\cos\theta_1=a\cos\theta$, so $a=1$, and $r_1=r$. Thus, $\langle r_1,\theta_1\rangle=\langle r,\theta\rangle$, and $f$ is one-to-one on $N$. The argument if $\sin\theta\ne0$ is very similar.
You are right. Intuitively, if you had a finite (and non-zero) rod on a plane and you moved it in some angular displacement $\theta$, there could be multiple ways of achieving same Cartesian position (x,y), by just moving in complete circle(s) which is adding multiple of $2\pi$ as you suggested. However if you had to move very slightly in any angular position, you would always get a unique Cartesian position. Hence 2 is correct and 1 is not.