= Solution
Apply the method to the <Dahlquist test equation> $y'=\lambda y$ and set $z=h\lambda$. Its amplification roots satisfy
$$
\left(1-\frac8{11}z+\frac2{11}z^2\right)\xi^2
-\frac{16}{11}\xi+\frac5{11}+\frac2{11}z=0.
$$
At $z=0$ the roots are $1$ and $5/11$. A root can leave the unit disk only through $\xi=e^{i\varphi}$. Substitution gives the boundary equation
$$
2\xi^2z^2+(2-8\xi^2)z+11\xi^2-16\xi+5=0.
$$
Direct separation into real and imaginary parts shows that both branches satisfy $\operatorname{Re}z\geq0$; equality occurs on the branch through $z=0$. Hence no root crosses the unit circle in $\operatorname{Re}z<0$. Moreover both roots tend to zero as $z\to\infty$ in the left half-plane. Consequently the method is <A-stable>, and it also strongly damps the infinitely stiff limit.
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