= Solution
Work first with $q$ in the <Schwartz space>, so that all Fourier manipulations and spectral <contour integrals> are justified; weaker classes follow by the usual density or distribution arguments. Define
$$
E(z,k)=e^{k\bar z-\bar k z},\qquad
Q(k)=\int_{\mathbb C}e^{\bar k\zeta-k\bar\zeta}q(\zeta)\,dA(\zeta).
$$
The phase is purely imaginary. Since $\partial_{\bar z}E=kE$, setting $F=E\Phi$ reduces the spectral equation to $\partial_{\bar z}\Phi=E^{-1}q$. The whole-plane <Cauchy-Pompeiu formula> therefore constructs the solution decaying spatially at infinity:
$$
\boxed{F(z,k)=\frac{E(z,k)}\pi\int_{\mathbb C}
\frac{e^{\bar k\zeta-k\bar\zeta}q(\zeta)}{z-\zeta}\,dA(\zeta).}
$$
The freedom to add $E$ times an entire function is removed by this decay condition. For every fixed $k$, the integral is a spatial <Cauchy-Green operator> applied to a modulated source.
Now differentiate in the conjugate spectral parameter. The two exponential derivatives produce $\zeta-z$, canceling the Cauchy denominator, so
$$
\boxed{\partial_{\bar k}F(z,k)=-\frac1\pi E(z,k)Q(k).}
$$
This is the spectral <dbar equation>: its right-hand side is the forward transform of $q$ multiplied by a known plane wave. Apply the whole-plane <Cauchy-Pompeiu formula> again, now in $k$:
$$
F(z,k)=-\frac1{\pi^2}\int_{\mathbb C}
\frac{E(z,\ell)Q(\ell)}{k-\ell}\,dA(\ell).
$$
The spatial spectral equation also gives $F(z,k)=-q(z)/k+o(k^{-1})$ as $|k|\to\infty$. One way to justify this is to integrate by parts in the first Cauchy integral: $F=-q/k+T_k(q_{\bar z})/k$, where the modulated Cauchy integral $T_k(q_{\bar z})$ tends to zero by the <Riemann-Lebesgue lemma>. Comparing the $1/k$ coefficient of the spectral <contour integral> therefore gives
$$
\boxed{Q(k)=\int_{\mathbb C}e^{\bar k z-k\bar z}q(z)\,dA(z),\qquad
q(z)=\frac1{\pi^2}\int_{\mathbb C}e^{k\bar z-\bar k z}Q(k)\,dA(k).}
$$
This derives the transform pair from two uses of the <Cauchy-Pompeiu formula>, not from an assumed inversion formula.
To identify the usual normalization, write $z=x+iy$ and $k=k_1+ik_2$. Then $\bar kz-k\bar z=2i(k_1y-k_2x)$. Set $\xi_1=2k_2$, $\xi_2=-2k_1$, so $d\xi_1d\xi_2=4\,dA(k)$. The result is exactly the two-dimensional <Fourier transform> pair
$$
\boxed{\widehat q(\xi)=\int_{\mathbb R^2}e^{-i\xi\cdot x}q(x)\,dx,\qquad
q(x)=\frac1{(2\pi)^2}\int_{\mathbb R^2}e^{i\xi\cdot x}\widehat q(\xi)\,d\xi.}
$$
The factor four in the real-frequency change of variables is essential.
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