= Solution
A plane wave returning down the guide has spatial factor $e^{-ik_0x}$ 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 $\gamma\sinh(\gamma b)\sim2bk_0(k-k_0)$, the interior residue is
$$
\operatorname{Res}_{k=k_0}\Phi_{\rm sc}^{\rm in}=\frac{iC_0L^-(k_0)}{4bk_0^2}.
$$
Hence the reflected plane-wave <amplitude>, relative to the incident <amplitude> one, is
$$
\boxed{\mathcal R=\frac{L^+(-k_0)L^-(k_0)}{4bk_0^2},\qquad
\phi_{\rm refl}=\mathcal R\,e^{-ik_0x+i\omega t}.}
$$
Equivalently, because $L^+$ carries the simple zero at $k_0$, $\mathcal R=L^+(-k_0)/(2k_0(L^+)'(k_0))$. 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 $\mathcal R$.
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