Let $C$ be a curve defined over $\mathbb{Q}$. The regulator is a map
$$
reg: K_2(C)_{\mathbb{Q}} \longrightarrow H^1(C(\mathbb{C}), \mathbb{R}). 
$$ Here $K_2(C)_{\mathbb{Q}}$ is the K-group tensor $\mathbb{Q}$ which admits the description:  
$$
\ker(K_2(\mathbb{Q}(C)) \longrightarrow \bigoplus_{x \in C(\overline{\mathbb{Q}})} \overline{\mathbb{Q}}^\times) \otimes_{\mathbb{Z}} \mathbb{Q}
$$ with $K_2(\mathbb{Q}(C))$ generated by symbols $\{f, g\}$ and the maps given by the tame symbols
$$
T_x(\{f, g\})=(-1)^{ord_x(f)ord_x(g)} [f^{ord_x g}/g^{ord_x f}](x).
$$

Since $H^1(C(\mathbb{C}), \mathbb{Q})$ is the dual of $H_1(C(\mathbb{C}), \mathbb{Q})$, to define the regulator of a symbol $\{f, g\}$ one needs to attach a real number 
$$
reg(\{f, g\})(\gamma) 
$$ to each $\gamma \in H_1(C(\mathbb{C}), \mathbb{Q})$. 

To do so, one first starts defining 
$$
\omega(f, g)=\log |f|d arg(g)-\log |g| d arg(f) 
$$ which is a closed real analytic 1-form on the complement of $S=div(f) \cup div(g)$ on the Riemann surface $C(\mathbb{C})$. We first define the real number
$$
\frac{1}{2\pi i} \int_{\gamma} \omega(f, g) 
$$ for $\gamma$ a cycle on $U=X-S$ (It actually only depends on the cohomology class of $\gamma$). Then we would like to say that this definition remains valid also for $\gamma$ in the whole of $X$. Since the cohomologies of $U$ and $X$ are related by residues maps
$$
0 \to H^1(X) \to H^1(U) \to \bigoplus_{s \in S} \mathbb{R} \to \mathbb{R} \to 0
$$ one needs to show that the residue vanishes at all points of $S$. What I have read is that this residue is given by 
$$
log | T_s(\{f, g\})|
$$ (so since the tame symbol is torsion equal to zero) but I don't understand how to prove this. I guess I should take a small circle $C_s$ around a point in $S$ and show that
$$
\frac{1}{2\pi i} \int_{C_s} \omega(f, g)=\log |T_s(\{f, g\})|. 
$$ Can anybody help me proving this? This should be some kind of Cauchy theorem but since the integrand is not holomorphic I am a bit lost...