Is a well estabilished  property that the curl of a gradient is always zero (i.e. $\nabla\times\nabla\Phi=0$) and it's possible to prove it in many ways. e.g.

If $(\nabla\times\nabla\Phi)_i = \epsilon_{ijk}\partial_j\partial_k\Phi$, where Einstein summation is being used to find the $i$th component...

Using Clairaut's theorem $\partial_{i}\partial_{j}\Phi = \partial_{j}\partial_{i}\Phi$, so $$\epsilon_{ijk}\partial_j\partial_k\Phi = \epsilon_{ijk}\partial_k\partial_j\Phi = -\epsilon_{ikj}\partial_k\partial_j\Phi$$ 

Thus $$\epsilon_{ikj}\partial_k\partial_j\Phi=0 \longrightarrow (\nabla\times\nabla\Phi)=0$$

Now, the definition of [vorticity][1] (in continuum mechanics) is $(\vec{\nabla}\times \vec{u})$ and through the Stokes theorem
$$\oint_{\partial S} d\vec{l}\cdot\vec{u}=\int_Sd\vec{s}\cdot(\vec{\nabla}\times \vec{u})$$
Now, I'm in dealing with vorticity in the context of the [Kosterlitz - Thouless - Berezinsky transition][2], where $\vec {u}=\vec{\nabla}{\theta}$. Thus, the **vorticity $n$**, working with 1 vortex or antivortex is defined through 
$$\oint_{\partial S} d\vec{l}\cdot\vec{\nabla}{\theta}=\int_{ S} d\vec{s}\cdot(\vec{\nabla}\times\vec{\nabla}{\theta})=2\pi n$$
While working with N vortexes or antivortexes is defined through
$$\oint_{\partial S} d\vec{l}\cdot\vec{\nabla}{\theta}=\int_{ S} d\vec{s}\cdot(\vec{\nabla}\times\vec{\nabla}{\theta})=2\pi \sum_j^Nn_j$$

Why in this case is not zero? Because in this context we're dealing with complex scalar fields instead of real scalar fields? How i can prove that is not zero using simple arguments like the first classical used in the beginning?


  [1]: https://en.wikipedia.org/wiki/Vorticity
  [2]: https://en.wikipedia.org/wiki/Berezinskii%E2%80%93Kosterlitz%E2%80%93Thouless_transition


$\theta$ is an angular coordinate and the definition of vorticity in this context comes from the fact that $\sum_{closed loop}(\theta_{i+1}-\theta_i)\neq0$ due to the existence of vortexes or antivortexes.