= Solution
On the <Dahlquist test equation>, set $z=h\lambda$. The <theta method> has <stability function>
$$
R(z)=\frac{1+(1-\theta)z}{1-\theta z},\qquad y_n=R(z)^n.
$$
With the decaying-solution convention, its region of <absolute stability> is $|R(z)|<1$. Writing $z=x+iy$, squaring numerator and denominator gives
$$
\boxed{\mathcal S_\theta=\{z:2x+(1-2\theta)|z|^2<0\}.}
$$
For $0\leq\theta<1/2$ 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 $\operatorname{Re}z<0$ satisfies the inequality exactly for $\theta\geq1/2$. If $\theta<1/2$, sufficiently large negative real $z$ fails it. Therefore
$$
\boxed{\text{The method is A-stable exactly when }1/2\leq\theta\leq1.}
$$
If <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.
\Image[/past-exam-of-the-numerical-solution-of-differential-equations-course-of-the-university-of-oxford/2011-theta-stability.png]
{title=Absolute stability regions of the theta method at zero, one half and one; green regions damp the scalar test solution}
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