Calculus integral on Riemann Surface (Stokes' theorem) I made a question: 
Calculus itegral on Riemann surfaces 
So I thought a bit about the solution and would like some help on what I did: 
Following the hint, 
$\int_{\partial \Omega} i \space \partial\Phi =\int_{\partial\Omega} <grad \space\Phi,v>$. 
Therefore, by Stokes' Theorem, 
$\int_{\partial\Omega} <grad \space\Phi,v>= \int_\Omega div\space(\Phi_x,\Phi_y,\Phi_z)$. 
Now I need to use that $\Phi$ is a real-valued function which is positive on     and vanishes on the boundary of $\Omega$ to justify that $\int_\Omega div\space(\Phi_x,\Phi_y,\Phi_z)\ge 0$. And that would solve the problem. 
So if anyone can help me put the pieces together, I appreciate it.  
 A: There's no Stokes's Theorem involved in this. For one thing, you don't know anything about the Laplacian of $\Phi$ on the interior of $\Omega$. However, if you write everything out carefully in $\Bbb C$, you have
\begin{align*}
i\partial\Phi &= \frac i2\big(\frac{\partial\Phi}{\partial x}-i\frac{\partial\Phi}{\partial y}\big)\big(dx+i\,dy\big) \\
&= \frac12\left(i\big(\frac{\partial\Phi}{\partial x}\,dx + \frac{\partial\Phi}{\partial y}\,dy\big) - \big(\frac{\partial\Phi}{\partial x}dy - \frac{\partial\Phi}{\partial y}\,dx\big)\right) = \tfrac12\big(i\,d\Phi - \star(d\Phi)\big).
\end{align*}
The integral of the first term around the closed curve $\partial\Omega$ is $0$. The integral of the second term is the negative of the flux of $\text{grad}\,\Phi$ across $\partial\Omega$. Since $\Phi$ is positive on $\Omega$ and zero on $\partial\Omega$, $\text{grad}\,\Phi$ points inward everywhere (or is zero). Thus, the flux is negative, and we have our answer. (This can be extended to a Riemann surface by working with local (holomorphic) parametrizations. The geometry of the flux works out just fine.)
