 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 $W=Adz+Bd\bar{z}$, whereUsing the exterior derivative, $dW=(B_z-A_{\bar{z}})dz\wedge d\bar{z}$. The terms involving $u_z$ and $u_{\bar{z}}$ cancel, leavingBecause $E_{\lambda }$ never vanishes for $\lambda \ne 0$, $dW=0$ if and only if $u_{z\bar{z}}-{\beta }^{2}u=0$. 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.