$\Phi_{\alpha}$ is a complex quantity and $k_{\alpha}=\alpha\pi/(N+1)$. Inserting this ansatz into Eqs. (\[Az\], \[Aexp\]) we obtain $$\label{res}
 \tilde{k}_{\alpha}= k_{\alpha}-
\dfrac{i}{2N}\,
\text{ln}\left[\Omega(k_{\alpha};\eta)\right]+\mathcal{O}\left(\dfrac{1}{N^2}\right),$$ where $$\label{Omega}
\Omega(k;\eta)=
\dfrac{1-\eta \,e^{2i\,k_{\alpha}}}{1-\eta}.$$ Now, up to $\mathcal{O}(N^{-1})$, $\tilde{z}_{\alpha}$ may be written as $$\label{resz}
 \tilde{z}_{\alpha}=-2\cos(k_{\alpha})-
\dfrac{i\sin(k_{\alpha})}{N}\,
\text{ln}(\Omega).$$ The same result can be obtained for the parametric resonances, after repeating the similar steps, but with different $\Omega$: $$\label{Omgpara}
\Omega(k_{\alpha};\eta)= 
\dfrac{1-i\eta \,e^{i\,k_{\alpha}}}
{1-i\eta \,e^{-i\,k_{\alpha}}}.$$

![Scatter plot for exact resonances where $N=100$, $\eta=0.81$. Dense points in the graph represent exact resonances in the disordered chain for $2500$ realizations where $W=0.015$. These are scattered around dots which represent exact resonances in the clean chain.[]{data-label="scatter_exact"}](scatter-exact.eps){width="75.00000%"}

One should bear in mind that there is no resonance for $\eta=1$, as the system is fully coupled to the lead. However, for parametric resonances, one artificially gets resonances even when $\eta=1$. Note that the result (\[resz\]) is symmetric about the imaginary axis for both cases. In Fig. \[ansatz-exact\] and Fig. \[ansatz-parametric\] we compare the numerical solutions of the polynomial equation (\[fincharpol\]) with our results (\[resz\], \[Omega\], \[Omgpara\]), for $N=100$, $\eta=0.50,\,0.81$ and $0.99$ and $N=100$, respectively for exact and parametric resonances. Eq. (\[fincharpol\]) has been solved by using the Newton’s method with the initial guess $\tilde{k}_{\alpha}=k_{\alpha}$. These figures 