Prove $\oint_C\vec F \cdot \hat n\;ds=4\pi\,(q_1+...+q_n)$

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Let $F:\Bbb R^2-\{p_1,p_2,\dots,p_n\} \to \Bbb R^2$, where $\{p_1,p_2,\dots,p_n\}\in \Bbb R^2$ be defined as $F(x)=\sum_{i=1}^n q_i \nabla\left(ln||x-p_i||^2\right)$ with $\{q_1,q_2,...,q_n\}\in \Bbb R$.

Prove $$\oint_c \vec F \cdot \hat n\;ds=4\pi\,(q_1+\dots+q_n)$$ Where $C$ is a closed curve containing $\{p_1,p_2,\dots,p_n\}$ with counter-clockwise orientation.

Any hints on how to prove this?

In this case I can't use Green's theorem (or the divergence theorem for planes in this case) because the partial derivatives of $F$ are not of class $C^1$ over the region enclosed by any curve containing $\{p_1,p_2,\dots,p_n\}$ (maybe I'm wrong).

I've tried integrating over a circle with radius $R=max\{||p_i||\}$ but that didn't work (the integral was horrible).

Any help will be appreciated.

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Using complex potential function: $\Omega(z)=\phi(x,y)+i\psi(x,y)$

$\displaystyle f(z)= \overline{\Omega'(z)}=\sum_{i} \frac{q_{i}}{2\pi \overline{(z-p_{i})}}$

$\displaystyle \operatorname{Im} \oint_{C} \overline{f(z)} dz=\sum_{i} q_{i}$