 Solution

For $|\alpha |>1$, an exterior polynomial root of $\rho$ persists by continuity for sufficiently small negative real $z$ in the stability polynomial $\rho (w)-z\sigma (w)$. Hence these parameters cannot be A-stable.

For $-1\le \alpha <1$, the principal polynomial root at $w=1,z=0$ is simple. Let $s(z)=\log w(z)$ denote its analytic logarithm near zero. Using the expansion in part (b), and ${\rho }^{\prime }(1)=2(1-\alpha )$, solve the polynomial root equation to obtainIndeed the residual at $s=z$ is $-(\alpha +5)z^4/12$, and the correction in $s$ cancels it by multiplication with $2(1-\alpha )$. Choose $z=-C{\epsilon }^4/2+i\epsilon$. This lies strictly in the left half-plane, butfor small positive $\epsilon$. Thus $|w(z)|=e^{\mathrm{Re}s(z)}>1$, proving linear instability within the A-stability domain. This is a direct principal-root obstruction to A-stability, rather than an appeal to an order-barrier theorem alone.

Finally, at $\alpha =1$, the full stability polynomial isIt retains a double unit polynomial root for every negative $z$, allowing growing parasitic solutions. ThereforeThe canceled Backward Euler method recurrence at $\alpha =1$ is A-stable, but its admissible solutions and starting relations differ from those of the original method.