 Solution

A plane wave returning down the guide has spatial factor $e^{-i{k}_{0}x}$ and arises from the pole at $k=k_0$, lying below the inversion contour. For $x>0$ closing below is clockwise, so the inverse transform is minus $i$ times the residue. Since $\gamma \sinh (\gamma b)\sim 2b{k}_0(k-k_0)$, the interior residue isHence the reflected plane-wave amplitude, relative to the incident amplitude one, isEquivalently, because $L^{+}$ carries the simple zero at $k_0$, $\mathcal{R}=L^{+}(-k_0)/(2k_0(L^{+}{)}^{\prime }(k_0))$. Other guided modes, if propagating at the chosen frequency, come from the additional kernel zeros; they do not alter which residue gives the plane-mode amplitude. Evanescent terms near the aperture are also separate. A multiplicative renormalization of the factors cancels in the product defining $\mathcal{R}$.