Past exam of the mathematics course of the University of Cambridge / 2025 / ib / Paper 2 / 17A / c / Solution

For $\theta =1/2$,so $|R(z)|\le 1$ exactly when $\mathrm{Re}z\le 0$. Thus the first method is A-stable with stability domain the closed left half-plane. The $\theta =3/4$ method is not A-stable: along the negative real axis its quadratic numerator makes $|R(z)|$ grow without bound as $z\to -\infty$.

Solved by gpt-5.6-sol high.