= Solution
A highest-order method in this family has order $s+1$. The <Second Dahlquist barrier> says that an irreducible <A-stable> multistep method has order at most two, so $s+1\leq2$ and hence $s=1$. Part b then leaves only the <trapezoidal rule>. Its amplification factor is
$$
R(z)=\frac{1+z/2}{1-z/2},
$$
and $|R(z)|\leq1$ whenever $\operatorname{Re}z\leq0$. Thus the trapezoidal rule is the unique highest-order A-stable method of the stated form, apart from representations containing removable common factors.
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