 Solution

The Second Dahlquist barrier states that an A-stable linear multistep method has order at most two. Part a shows that every method in this family has order at least three, so no member can be A-stable.

The exceptional reducible case does not evade the conclusion. At $a=1$,Thus $\zeta =1$ is an amplification root for every $z=h\lambda$ with $\mathrm{Re}z<0$, whereas the multistep A-stability criterion requires all such roots to lie strictly inside the unit disk. Hence