= Solution
With the <Wirtinger derivatives>, $q_{z\bar z}=(q_{xx}+q_{yy})/4$. Put $W=e^{-ikz-\lambda\bar z/(ik)}$, $k\ne0$. The PDF has the factor $\lambda$ in this exponential; omitting it, as the TeX does, would give the wrong <local relation>. Choose
$$
\boxed{A=q_z,\qquad B=-\frac{\lambda}{ik}q.}
$$
Indeed, $W_{\bar z}=-\lambda W/(ik)$ and $W_z=-ikW$, so direct differentiation gives
$$
(WA)_{\bar z}-(WB)_z=W(q_{z\bar z}-\lambda q).
$$
The unwanted $q_z$ terms cancel exactly. Thus the divergence identity is equivalent to the <modified Helmholtz equation>, including its <Laplace equation> case $\lambda=0$. Equivalently,
$$
\eta=W\left(q_zdz-\frac{\lambda}{ik}q\,d\bar z\right)
$$
is a <closed differential one-form>. Its oriented boundary integral supplies the <polygonal modified Helmholtz global relation>.
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