 Solution

Assume the physical potential is real, as required for the stated conjugation identity, and use the regular radial solution and its analytic continuation. The differential equation depends on $k^2$, so exchanging $k$ and $-k$ exchanges the outgoing and incoming coefficients of the same regular solution. Their ratio consequently gives $S(-k)=S(k{)}^{-1}$. Complex conjugation gives a solution at $k^{\ast }$ and exchanges those coefficients, so $S(k{)}^{\ast }S(k^{\ast })=1$.

For real $k$, $|S(k)|=1$, hence $S=e^{2i{\delta }_0(k)}$ with real phase. The first identity makes the phase odd modulo $\pi$; one can choose odd branches on paired nonzero intervals. A continuous odd branch through zero additionally requires $S(0)=1$, excluding a threshold-resonance exception. For the given rational expression, which has $S(0)=1$, chooseIts small-$k$ behavior is ${\delta }_0(k)=-4k/(3\lambda )+O(k^3)$. ThusA zero of the continued $S$ means that the outgoing coefficient vanishes, leaving an incoming-only asymptotic solution. It is paired by $S(k)S(-k)=1$ with a pole at the opposite wave number. The displayed zeros are $-i\lambda$ and $-3i\lambda$. Genuine bound-state poles on the positive imaginary axis correspond to negative energies and decaying regular solutions; their opposite zeros encode the same spectral information. A bare continued pole outside the physical analyticity region need not be a bound state, so no additional classification of the two poles follows merely from the rational expression without further potential assumptions.