 Solution

Only $-1<\alpha <1$ needs consideration. On the Dahlquist test equation, the amplification polynomial of a multistep method is $P_z(w)=\rho (w)-z\sigma (w)$, where $z=h\lambda$. A-stability requires all amplification roots to have modulus at most one for every $\mathrm{Re}z\le 0$, with simple unit-modulus roots.

Choose the negative real value $z=-2/(1-\alpha )$. The leading coefficient of $P_z$ iswhereas $P_z(1)=-z\sigma (1)=4>0$. Hence $P_z(w)\to -\infty$ as real $w\to +\infty$. The intermediate value theorem supplies a real amplification root $w>1$, so this negative test value is unstable. ThereforeThis is a direct absolute stability obstruction, independent of the Second Dahlquist barrier, which by itself would not rule out the convergent second-order members.