 Solution

With the Wirtinger derivatives, $q_{z\bar{z}}=(q_{xx}+q_{yy})/4$. Put $W=e^{-ikz-\lambda \bar{z}/(ik)}$, $k\ne 0$. The PDF has the factor $\lambda$ in this exponential; omitting it, as the TeX does, would give the wrong local relation. ChooseIndeed, $W_{\bar{z}}=-\lambda W/(ik)$ and $W_z=-ikW$, so direct differentiation givesThe 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,is a closed differential one-form. Its oriented boundary integral supplies the polygonal modified Helmholtz global relation.