 Solution

Write $P(z)=1+z/2+z^2/12$ and $Q(z)=1-z/2+z^2/12$. The zeros of $Q$ are $3\pm i\sqrt{3}$, so the stability function has no pole in the closed left half-plane. A direct modulus calculation givesFor $\mathrm{Re}z\le 0$ this is nonnegative, and hence $|R(z)|=|P/Q|\le 1$. Therefore the Lobatto IIIA method is A-stable. It is not L-stable, because $R(z)\to 1$ for large negative real $z$; unconditional scalar stability need not strongly damp the stiffest modes.