 Solution

On the Dahlquist test equation, set $z=h\lambda$. The theta method has stability functionWith the decaying-solution convention, its region of absolute stability is $|R(z)|<1$. Writing $z=x+iy$, squaring numerator and denominator givesFor $0\le \theta <1/2$ this is the open disk with centre $-1/(1-2\theta )$ and radius $1/(1-2\theta )$. At $\theta =1/2$ it is the open left half-plane. For $\theta >1/2$ it is the exterior of the closed disk with centre $1/(2\theta -1)$ and radius $1/(2\theta -1)$. The pole $z=1/\theta$, when present, lies outside the stability region.

Every $z$ with $\mathrm{Re}z<0$ satisfies the inequality exactly for $\theta \ge 1/2$. If $\theta <1/2$, sufficiently large negative real $z$ fails it. ThereforeIf absolute stability is instead defined by bounded amplification, the boundary $|R|=1$ is included; this convention does not change the A-stability conclusion for the open left half-plane.