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The origin has Jacobian $\left(\begin{array}{cc}r& 0\\ 1-b& 0\end{array}\right)$ and is unstable because $r>1$. At a positive equilibrium, $r{e}^{-p_c}=1$, givingThe linearization isIts trace is $\tau =1+b{n}_c/r$ and determinant $n_c$. Elimination gives $n_{t+2}^{\prime }-\tau n_{t+1}^{\prime }+n_c{n}_t^{\prime }=0$. For distinct roots,At a repeated root replace the first expression by $(A+Bt){\lambda }^t$ and use $p_t^{\prime }=(n_t^{\prime }-n_{t+1}^{\prime })/n_c$.

As $r\downarrow 1$, $n_c\to 0$, so the roots approach one and zero. The characteristic polynomial at one is $n_c(1-b/r)>0$, putting the larger real root just below one and the other positive root below it. No real root can ever equal one. Moreoversince its numerator exceeds $r-1-\log r>0$. At $n_c=1$, $\tau =1+b/r<2$, so the roots are conjugate and nonreal. Just above this point their modulus is $\sqrt{n_c}>1$.

The unit-disc conditions are $1-n_c>0$, $1-\tau +n_c>0$, and $1+\tau +n_c>0$. The last two hold automatically here, so stability is equivalent to $n_c<1$. The unique crossing and its eigenvalue argument satisfyIndeed $\log r_{\ast }=1-b/r_{\ast }<1$. This is the stability threshold of a discrete exponential predator-prey model. The sketch uses $b=1/2$: the real moduli merge and then coincide, increasing through one at the threshold. Linearization decides stability below and instability above; it does not determine nonlinear stability exactly at the crossing.