What do I know when the curl of a vector field equals 0? I'm currently doing a multiple-choice exercise which includes the following question:

Let $F: \mathbb{R}^2 \setminus \{0\} \to \mathbb{R}^2$ be defined by $F(x,y) = \left( - \frac{y}{x^2+y^2}, \frac{x}{x^2+y^2} \right)$. Which of the following statements are true?
  
  
*
  
*There exists a closed path $\gamma$ in $\mathbb{R}^2 \setminus \{0\}$ such that
  $$\int_\gamma F \cdot \mathrm ds < 0.$$
  
*There exists a $\epsilon \in (0,1)$ such that for all closed $C^1$-paths $\gamma$ with Im $\gamma \subseteq B_\epsilon((1,0))$
  $$\int_\gamma F \cdot \mathrm ds = 0. $$
  
*F is a gradient field.
  

Now up to now I thought that whenever the curl of a vector field equals 0, firstly the vector field is a gradient field and secondly the integral around every closed paths equals 0. So this would make the second and the third statement to be correct whilst the first statement obviously would be wrong.
However, I don't feel like my reasoning is correct. It would be somewhat too easy, so there has to be some mistake in my train of thought, some point I've been missing.
Where did I make a mistake or is it really that easy?
Thanks for any answer in advance.
 A: I'm making my comments into an answer. As you said in a comment, there is the theorem:

If $\Omega \subset \mathbb{R^{2}}$ is an open and star-shaped domain then a $C^1$-vector field $K: \Omega \to \mathbb{R}^2$ is conservative if and only if $\operatorname{rot} K = 0$.

But as $\Omega = \mathbb{R}^2 \smallsetminus \{0\}$ is not star-shaped with respect to any point [proof: if $x$ is any point of $\Omega$ the line connecting $x$ and $-x$ crosses the origin, thus $\Omega$ is not star-shaped], this theorem is not applicable and its conclusion can turn out to be false, as we will see in a moment.
Now let's look at the vector field $F(x,y) = \left( \frac{-y}{x^2 + y^2}, \frac{x}{x^2 + y^2}\right)$. If $(x_{0},y_{0})$ lies on the unit circle, i.e., $x_{0}^2 + y_{0}^2 = 1$, then $F(x_{0},y_{0}) = (-y_{0},x_{0})$. Note that $F(x_{0},y_{0})$ has length one and is perpendicular to the vector $(x_{0},y_{0})$ and points in the counterclockwise direction (seen from the origin). So geometrically, if we move around the origin clockwise, the line integral should evaluate negatively (since the vector field points in the opposite direction of our movement). Formally, if we parameterize the unit circle by $\gamma(t) = (\cos{t}, -\sin{t})$ with $t \in [0,2\pi]$ (signs are chosen so that we walk clockwise) the line integral of $F$ along $\gamma$ evaluates to
\begin{align*}
\int_{\gamma} F\cdot ds & = \int_{0}^{2\pi} F(\gamma(t)) \cdot \dot{\gamma}(t)\,dt = \int_{0}^{2\pi} \begin{pmatrix} \sin{t} \\ - \cos{t} \end{pmatrix} \cdot \begin{pmatrix} -\sin{t} \\ \cos{t} \end{pmatrix}\,dt \\ & = \int_{0}^{2\pi} \left(-\sin^2{t} - \cos^{2}\right)\,dt = -2\pi
\end{align*}
since $\sin^2 + \cos^2 = 1$.
This computation solves the question almost completely: We have found a path such that the integral evaluates negatively and thus the first bullet point is correct. The third bullet point is wrong since our path is closed and a gradient vector field is conservative, hence evaluates to zero on all closed paths.
Finally, note that for every $\varepsilon \in (0,1)$ the ball $B_{\varepsilon}(0,1)$ is entirely contained in $\Omega$ and it is star-shaped (with respect to $(0,1)$). So the theorem is applicable to the restriction $K$ of $F$ to $B_{\varepsilon}(0,1)$. We conclude that the vector field $K$ is conservative on $B_{\varepsilon}(0,1)$ and hence for every closed $C^1$-path $\gamma$ entirely contained in $B_{\varepsilon}(0,1)$ the integral $\int_{\gamma} F \cdot ds = 0$. So the second bullet-point is correct.
To summarize:

  
*
  
*true
  
*true
  
*false
  

