= Solution
For a complex harmonic field use <complex conjugation> in its <wave-field correlation>: $m=\langle\psi_{\rm sc}(x_1,z)\psi_{\rm sc}(x_2,z)^*\rangle$. Put $A(\nu)=\widehat\psi_{\rm sc}(\nu)$ and let $C(\nu,\nu')=\langle A(\nu)A(\nu')^*\rangle$. Direct substitution of part (a), before imposing any homogeneity, gives
$$
\langle\psi_{\rm sc}(x_1,z)\psi_{\rm sc}(x_2,z)^*\rangle
=\frac1{(2\pi)^2}\iint C(\nu,\nu')
e^{i\nu x_1-i\nu'x_2+i[q(\nu)-q(\nu')^*]z}\,d\nu\,d\nu'.
$$
This is the general answer. It need not depend only on $\xi=x_1-x_2$: even a deterministic sum of two distinct outgoing <plane waves> has cross terms depending on the midpoint. The requested separation-only <wave-field correlation> therefore presumes translation-homogeneous two-point statistics.
Under that assumption define the uncentered spectrum by $C(\nu,\nu')=2\pi S(\nu)\delta(\nu-\nu')$. Then the <height propagation of an angular-spectrum correlation> is
$$
\boxed{m(\xi,z)=\frac1{2\pi}\int_{\mathbb R}
S(\nu)e^{i\nu\xi-2\operatorname{Im}q(\nu)z}\,d\nu}.
$$
Equivalently, for the <spectral measure of a stationary random field> $F(d\nu)=S(\nu)d\nu/(2\pi)$, write $m=\int e^{i\nu\xi-2\operatorname{Im}qz}F(d\nu)$. This version includes coherent spectral atoms and avoids squaring <Dirac delta functions>. The propagating spectrum $|\nu|\le k$ is independent of height, while an <evanescent> contribution is multiplied by $e^{-2\sqrt{\nu^2-k^2}z}$. At $\xi=0$ the formula gives the <mean wave intensity>.
If the brackets instead mean the spatial autocorrelation of a finite-energy deterministic field, define $m_L(\xi,z)=\int\psi_{\rm sc}(x+\xi,z)\psi_{\rm sc}(x,z)^*dx$. <Parseval's identity> gives the same integral with $S(\nu)=|\widehat\psi_{\rm sc}(\nu)|^2$. That finite-energy formula must not be applied unmodified to an infinite stationary field or an infinite <plane wave>, whose transforms require a spectral-measure convention.
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