= Solution
Continue the <transmission amplitude> to complex <wavenumber> $q=i\kappa$ with $\kappa>0$, equivalently complex <momentum> $k=i\hbar\kappa$. Then $E=-\hbar^2\kappa^2/(2m)<0$. In the left exterior region the $A$ term grows at negative infinity and the $B$ term decays; in the right exterior region the $D$ term grows and the $C$ term decays. A <bound state> thus requires $A=D=0$, with nonzero $B,C$. The <transfer matrix> relation gives $M_{22}(i\kappa)=0$. Equivalently, a zero of the incoming coefficient is a <bound-state pole> of $t(q)$, whose numerator is nonzero at $q=i\kappa$.
The <denominator> found in part (c) gives
$$
0=(i\kappa+i\mu)^2+\mu^2e^{-4a\kappa}=-(\kappa+\mu)^2+\mu^2e^{-4a\kappa}.
$$
Factorization produces two branches. To identify their <parity>, use an even interior <wavefunction> $A\cosh(\kappa x)$ or an odd interior <wavefunction> $A\sinh(\kappa x)$ for $|x|<a$, joined to a decaying exterior <wavefunction> at $a$. Continuity and the derivative jump give, respectively,
$$
-\kappa\bigl[1+\tanh(\kappa a)\bigr]=2\mu,\qquad -\kappa\bigl[1+\coth(\kappa a)\bigr]=2\mu.
$$
Rearrangement gives the implicit <energy eigenvalue> equations
$$
\boxed{\kappa_{\rm even}=-\mu(1+e^{-2a\kappa_{\rm even}}),\qquad \kappa_{\rm odd}=-\mu(1-e^{-2a\kappa_{\rm odd}}),\qquad E_{\rm even,odd}=-\frac{\hbar^2\kappa_{\rm even,odd}^2}{2m}.}
$$
Only strictly positive <equation roots> $\kappa$ are retained. With $\nu=-\mu>0$, the even <equation root> lies above $\nu$ and the odd <equation root>, when it exists, lies below $\nu$, so the even <bound state> has lower <energy>. The formal zero <equation root> in the odd equation does not itself establish a <normalizable wavefunction>.
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