= Solution
Use a consistent number $n$ of vertices, with $z_{n+1}=z_1$, and set $m_j=(z_j+z_{j+1})/2$, $h_j=(z_{j+1}-z_j)/2$. Pull back the <one-form> to $z_j(s)=m_j+sh_j$. Since $dz=h_jds$ and $d\bar z=\bar h_jds$, the side integrand is
$$
\boxed{W_j(s,\lambda)=e^{-i\beta[\lambda(m_j+sh_j)-(\bar m_j+s\bar h_j)/\lambda]}
\left[h_ju_z-\bar h_ju_{\bar z}
+i\beta(\lambda h_j+\bar h_j/\lambda)u\right]_{z=m_j+sh_j}.}
$$
For counterclockwise traversal, put $\ell_j=|h_j|$ and let $q_j=\partial_nu$ be the outward <normal derivative>, while $g_j=u|_{S_j}$ is the <Dirichlet boundary data>. The outward unit normal is $-ih_j/\ell_j$ in complex notation. Therefore $h_ju_z-\bar h_ju_{\bar z}=i\ell_jq_j$, and
$$
\boxed{W_j=i e^{-i\beta[\lambda(m_j+sh_j)-(\bar m_j+s\bar h_j)/\lambda]}
\left[\ell_jq_j(s)+\beta(\lambda h_j+\bar h_j/\lambda)g_j(s)\right].}
$$
There is no tangential-derivative term: it cancels in this particular <one-form>. The <Generalized Stokes theorem> and $dW=0$ now give the polygonal <global relation>
$$
\boxed{\sum_{j=1}^n\int_{-1}^1W_j(s,\lambda)\,ds=0,\qquad\lambda\ne0.}
$$
The same zero identity holds with every side traversed clockwise, but then $h_ju_z-\bar h_ju_{\bar z}=-i\ell_jq_j$ for outward normals. One must change this sign consistently rather than mix the two orientations.
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