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Use the Wirtinger derivative $f=q_z=(q_x-i{q}_y)/2$. Since $q$ satisfies the Laplace equation, ${\partial }_{\bar{z}}f=\Delta q/4=0$, so $f$ is holomorphic. The differential relation is understood as a one-form identity:The differential $dz$ is needed on its right-hand side. Because this form is closed, its integral is independent of path within the semistrip.

Three spectral primitives are obtained by integrating from the two finite corners and from the infinite end:The last integral is taken horizontally and defines the infinite-end primitive for $\mathrm{Im}k<0$. Their differences are $e^{ikz}$ times the three boundary spectral functions. Using counterclockwise boundary orientation, defineThen ${\mu }_0-{\mu }_{\infty }=e^{ikz}{\rho }_B$, ${\mu }_{\infty }-{\mu }_l=e^{ikz}{\rho }_T$, and ${\mu }_l-{\mu }_0=e^{ikz}{\rho }_L$. Their sum gives the global relation ${\rho }_L+{\rho }_B+{\rho }_T=0$ for $\mathrm{Im}k\le 0$.

To express these spectra in boundary data, write $g_0(x)=q(x,0)$, $g_l(x)=q(x,l)$, $h(y)=q(0,y)$, $n_0(x)=q_y(x,0)$, $n_l(x)=q_y(x,l)$, and $v(y)=q_x(0,y)$. Direct substitution into $f$ givesThese are transforms of tangential derivatives of Dirichlet boundary data and the corresponding Neumann boundary data. The outward derivatives are $-n_0$ at the bottom, $n_l$ at the top, and $-v$ at the left.

The boundary spectral representation of a holomorphic function on a semistrip isAll three $k$-rays are oriented outward from zero. To verify the spectral inversion, usewhenever the exponential decays. The positive real ray works for the bottom, the negative real ray for the top, and the positive imaginary ray for the left. Substitution of the boundary spectra therefore turns the boxed formula into $(2\pi i{)}^{-1}{\int }_{\partial \Omega }f(w)/(w-z)dw=f(z)$ by the Cauchy integral formula. This also fixes all orientation signs.