= 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 $k^*$ and exchanges those coefficients, so $S(k)^*S(k^*)=1$.
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 $S(0)=1$, choose
$$
\delta_0(k)=-\arctan(k/\lambda)-\arctan(k/(3\lambda)).
$$
Its small-$k$ behavior is $\delta_0(k)=-4k/(3\lambda)+O(k^3)$. Thus
$$
\boxed{a=\frac4{3\lambda},\qquad4\pi a^2=\frac{64\pi}{9\lambda^2}.}
$$
A 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.
Back to article page
