= Solution
Write $P(z)=1+z/2+z^2/12$ and $Q(z)=1-z/2+z^2/12$. The zeros of $Q$ are $3\pm i\sqrt3$, so the <stability function> has no pole in the closed left half-plane. A direct modulus calculation gives
$$
|Q(z)|^2-|P(z)|^2
=-2\operatorname{Re}z\left(1+\frac{|z|^2}{12}\right).
$$
For $\operatorname{Re}z\leq0$ this is nonnegative, and hence $|R(z)|=|P/Q|\leq1$. Therefore \b[the <Lobatto IIIA method> is <A-stable>]. It is not <L-stable>, because $R(z)\to1$ for large negative real $z$; unconditional scalar stability need not strongly damp the stiffest modes.
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