Past exam of the mathematics course of the University of Cambridge / 2005 / iii / Paper 69 / 2 / c

For the Dahlquist test equation, write $z=h\lambda$. The amplification roots solveA root on $|\zeta |=1$ would require the boundary-locus valueIts denominator cannot vanish on that circle. Setting $\zeta =e^{i\theta }$ and simplifying the real part givesThus no amplification root can cross the unit circle while $\mathrm{Re}z<0$. The leading coefficient is also nonzero throughout that half-plane. At $z=-1$, the roots are $(6\pm \sqrt{3})/11$, both strictly inside the unit disk. Connect any other left-half-plane point to $-1$: continuity of the roots, without a boundary crossing or a root escaping through infinity, keeps both roots inside.

On the imaginary axis a boundary root can occur only at $\theta =0$, which gives $z=0$; there the roots $1$ and $5/7$ satisfy the simple-unit-root condition. Hence $\boxed{\text{the method is A-stable}}$, including the usual boundary root convention at zero.