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Write $\chi =e^{-ikz-\lambda \bar{z}/(ik)}$. In terms of Wirtinger derivatives, the elliptic equation is $q_{z\bar{z}}=\lambda q$. If $W=Adz+Bd\bar{z}$, its coefficients are $A=\chi q_z$ and $B=-\lambda \chi q/(ik)$. Direct differentiation givesTherefore $dW=(B_z-A_{\bar{z}})dz\wedge d\bar{z}=0$, proving that $W$ is a closed differential one-form. The parameter restriction $k\ne 0$ makes its coefficients well defined.

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.

Use the Wirtinger derivatives ${\partial }_z=({\partial }_x-i{\partial }_y)/2$ and ${\partial }_{\bar{z}}=({\partial }_x+i{\partial }_y)/2$. In particular $q_{z\bar{z}}=\Delta q/4$. Let $E=e^{-ikz-\lambda \bar{z}/(ik)}$, $A=E(q_z+ikq)$ and $B=E(q_{\bar{z}}+\lambda q/(ik))$, so the printed differential form is $W=Adz-Bd\bar{z}$. Since $E_z=-ikE$ and $E_{\bar{z}}=-\lambda E/(ik)$,Thus the exterior derivative isBecause $E$ never vanishes and $k\ne 0$, this proves both directions: $W$ is a closed differential form if and only if $q_{z\bar{z}}=\lambda q$. The latter is the modified Helmholtz equation $\Delta q-4\lambda q=0$ for positive $\lambda$. The associated global relations on a bounded region $\Omega$ are, for every nonzero spectral parameter,The boundary orientation is positive, by Stokes theorem. Each admissible spectral parameter gives a separate identity between the traces. Interchanging $z$ and $\bar{z}$ gives the dual family, equivalently the same family at $k\mapsto -\lambda /k$ with an overall minus sign. No assumption that $q$ is real is needed.

For later use with $\lambda >0$, as in the remaining parts, make the global relation explicit in the first quadrant. PutLet $h_2(x)=q(x,0)$, $n_2(x)=q_y(x,0)$ and $h_1(y)=q(0,y)$, $n_1(y)=q_x(0,y)$. These normals are inward-coordinate derivatives. The bottom is traversed from zero to infinity and the left side from infinity to zero. Their oriented spectral integrals areHere ${\rho }_x$ is analytic for $\mathrm{Im}k<0$ and ${\rho }_y$ for $\mathrm{Re}k<0$, with suitable boundary limits. Indeed $\mathrm{Im}a=(r+\lambda /r)\sin \theta$ and $\mathrm{Re}b=(r+\lambda /r)\cos \theta$ for $k=r{e}^{i\theta }$. On their common third quadrant, both exponential weights decay. Apply the finite-domain relation to expanding rectangles; sufficient decay of the solution and traces eliminates the distant edges. This derivesThe limiting relations on the negative real and negative imaginary axes will remove the unknown derivative traces.

Use the Wirtinger derivatives $q_z=(q_x-i{q}_y)/2$ and $q_{\bar{z}}=(q_x+i{q}_y)/2$. The Laplace equation gives $q_{z\bar{z}}=0$, so writewhere $\Phi$ is holomorphic in the first quadrant and $\Psi$ is holomorphic in the fourth quadrant. They are independent functions: the data may be complex, so a complex-conjugation relation between them cannot be assumed.

Let $ϵ=(-1{)}^n$, where ${\beta }_1+{\beta }_2=n\pi /2$. The oblique derivative boundary conditions becomeThe hinted complementary traces would giveHere $u_2$ is a function of $x>0$ on the horizontal side; the printed $y$ range in that hint is a variable typo. We shall eliminate the same unknown traces by reflection of the holomorphic derivatives.

The polygonal ray formula can first be viewed as the Cauchy integral formula. For an oriented edge point $\zeta$, choose its spectral ray so that $e^{ik(z-\zeta )}$ decays for interior $z$. ThenInterchanging the edge and ray integrals in the supplied formula gives $\Phi (z)=(2\pi i{)}^{-1}{\int }_{\partial \Omega }\Phi (\zeta )/(\zeta -z)d\zeta$, with counterclockwise edge orientation. Thus a quadrant limit of that formula can be evaluated by the corresponding holomorphic boundary jumps, provided the additional arcs at infinity and at the corner give no contribution.

Define an holomorphic function separately in the four quadrants byThe angle condition gives $e^{2i({\beta }_1+{\beta }_2)}=e^{-2i({\beta }_1+{\beta }_2)}=ϵ$. It makes the reflected definitions consistent, with parity $H(-z)=-ϵH(z)$. On each axis oriented outward from zero, let $J=H_{\mathrm{left}}-H_{\mathrm{right}}$. The two known boundary equations give all four jumps:For example the vertical equation gives $\Phi (iy)-e^{-2i{\beta }_1}\Psi (-iy)=-i{e}^{-i{\beta }_1}{g}_1(y)$; reversing the jump order on the upward axis accounts for its displayed sign. The horizontal equation similarly supplies the positive real-axis jump, and reflection supplies the other two.

Apply the Cauchy integral formula in the four quadrants and combine their common edges. The edge contributions are exactly the known jumps of this scalar Riemann-Hilbert problem. If the corner and large-arc integrals vanish, the representation in the first quadrant isThis is the rational-angle oblique derivative problem on a quadrant representation, with all unknown traces removed. The Cauchy integrals have precisely the displayed jumps by the Sokhotski–Plemelj formula. Their reflection parity then reconstructs $\Psi$ from the fourth-quadrant value of $H$, so the two original boundary equations hold. Integrating $\Phi (z)$ and $\Psi (\bar{z})$ gives a harmonic function $q$ up to an additive constant.

To put the answer back on the spectral rays of the question, defineUse $1/(s-z)=i{\int }_0^{\infty }{e}^{ik(z-s)}dk$, $1/(s+z)=-i{\int }_0^{\infty }{e}^{ik(z+s)}dk$, and the same identities with $s$ replaced by $is$ and the ray $0\to i\infty$. The spectral representation isAll transforms used on these rays are ordinary boundary values of convergent half-line transforms. For $z=x+iy$ with $x,y>0$, the real-ray integrand decays through $e^{-ky}$ and the imaginary-ray integrand through $e^{-\mathrm{Im}kx}$.

There is an admissibility qualification in passing from a bounded polygon to this unbounded quadrant. Sufficient conditions are $H(z)\to 0$ at infinity uniformly away from the axes, with the corresponding large-arc integral tending to zero, and $|z|H(z)\to 0$ uniformly on small corner arcs. The latter allows logarithmic gradient singularities but excludes a pole with nonzero corner contribution. Smooth decaying data alone, as printed, do not impose these conditions on $q$.

For a concrete homogeneous corner ambiguity in a quadrant Laplace problem, with ${\beta }_1={\beta }_2=0$ and zero data, $q=xy$ satisfies both prescribed tangential derivative conditions but has $q_z=-iz/2$, while the displayed particular representation is zero. With ${\beta }_1={\beta }_2=\pi /2$, the zero-data solution $q=\log |z|$ has $q_z=1/(2z)$: it decays at infinity but has an excluded corner pole. Thus no unique representation of every unrestricted solution follows from the printed data alone. Without the arc conditions, one must add a homogeneous holomorphic term $P(z)$ to the displayed $q_z$, with reflected parity $P(-z)=-ϵP(z)$; permitted corner singularities are included if the admissible class allows them. With the stated decay and corner conditions that term vanishes, giving the boxed formulas. This identifies the missing solution-class conditions without assuming real-valued data.

Use the Wirtinger derivatives ${\partial }_z=({\partial }_x-i{\partial }_y)/2$ and ${\partial }_{\bar{z}}=({\partial }_x+i{\partial }_y)/2$, and area measure $dA=dxdy=(i/2)dz\wedge d\bar{z}$. The supplied boundary-integral identity is the planar Generalized Stokes theorem; the usual Poincaré lemma is a different local exactness result.

Let $f\in C^1(\overline{D})$ and $z\in D$. Remove a disk of radius $ϵ$ around $z$, and apply Generalized Stokes theorem to the one-form $f(\zeta )d\zeta /(\zeta -z)$ on the punctured domain. Away from the puncture,The outer boundary is counterclockwise and the small inner circle clockwise. Its counterclockwise integral tends to $2\pi if(z)$. Passing to the limit gives the Cauchy-Pompeiu formulaThe weak $1/|\zeta -z|$ singularity is locally integrable. If the boundary term vanishes on expanding $D$ to the whole plane, the formula becomes $f(z)={\pi }^{-1}{\int }_{\mathbb{C}}{f}_{\bar{\zeta }}(\zeta )/(z-\zeta )dA(\zeta )$. In particular it yields the distributional normalization ${\partial }_{\bar{z}}[1/(\pi z)]={\delta }_0$. For holomorphic functions the area term vanishes and one recovers the Cauchy integral formula.