= Solution
Put $E_\lambda=e^{-i\beta(\lambda z-\bar z/\lambda)}$. The <Wirtinger derivatives> give $(E_\lambda)_z=-i\beta\lambda E_\lambda$ and $(E_\lambda)_{\bar z}=i\beta E_\lambda/\lambda$. Write $W=A\,dz+B\,d\bar z$, where
$$
A=E_\lambda(u_z+i\beta\lambda u),\qquad
B=-E_\lambda(u_{\bar z}-i\beta u/\lambda).
$$
Using the <exterior derivative>, $dW=(B_z-A_{\bar z})dz\wedge d\bar z$. The terms involving $u_z$ and $u_{\bar z}$ cancel, leaving
$$
\boxed{dW=-2E_\lambda(u_{z\bar z}-\beta^2u)\,dz\wedge d\bar z
=iE_\lambda(\Delta u-4\beta^2u)\,dx\wedge dy.}
$$
Because $E_\lambda$ never vanishes for $\lambda\ne0$, $dW=0$ if and only if $u_{z\bar z}-\beta^2u=0$. In real coordinates this is the <modified Helmholtz equation> with mass parameter $2\beta$, not $\beta$. In fact any one nonzero spectral parameter suffices for the equivalence; the full family supplies many independent boundary tests.
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