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Apply the Dahlquist test equation and put $z=h\lambda$. Every recurrence mode must satisfyThe amplification polynomial of a multistep method, rather than just its root near one, determines absolute stability. Put $B=2a-1$ and use the Cayley transform between the half-plane and diskFor $\zeta \ne -1$, the characteristic equation becomesIf $B>0$, the denominator cannot vanish at a root of the characteristic equation: $1+B{w}^2=0$ would force $w=0$, a contradiction. Hence $\mathrm{Re}z<0$ forces $\mathrm{Re}w<0$, so all amplification roots have modulus less than one. On the imaginary axis the roots have modulus one and are simple: the transformed quadratic has discriminant $4(1-B{z}^2)>0$ for imaginary $z$. At $z=0$ they are the two simple roots $\pm 1$. Also $\zeta =-1$ cannot be a root when $B>0$ and $z\ne 0$, and the leading coefficient $1-az$ cannot vanish in the closed left half-plane.

If $a<1/2$, the root near $-1$ isFor small negative real $z$ it lies below $-1$, violating absolute stability. The endpoint $a=1/2$ deserves separate treatment:One root is always $-1$; the other is the trapezoidal rule multiplier $(1+z/2)/(1-z/2)$. They are distinct for every finite $z$ in the left half-plane. Consequently, with absolute stability understood as the bounded root condition for a multistep method,There is a convention at this reducible endpoint: if A-stability is defined to require every unreduced recurrence mode to decay for $\mathrm{Re}z<0$, the answer is $a>1/2$, since the $(-1{)}^n$ mode persists at $a=1/2$. Canceling the common factor $\zeta +1$ gives the A-stable trapezoidal rule, but cancellation removes an actual starting-error mode of the original two-step recurrence. The fourth-order member $a=1/3$ is outside either A-stability range.

Fix $z_0\in \mathcal{U}$. We prove the Riemann mapping theorem in this case by maximizing a normalized derivative. Choose $a\notin \mathcal{U}$. Since $\mathcal{U}$ is a simply connected domain and $z-a$ never vanishes, it has a holomorphic square root $g$. This $g$ is injective, and $g(\mathcal{U})$ is disjoint from $-g(\mathcal{U})$: equality up to sign would first force the original points to coincide. Choose $b\in g(\mathcal{U})$ and $r>0$ with $B(b,r)\subset g(\mathcal{U})$. The ball $B(-b,r)$ is omitted, so $1/(g+b)$ is a bounded univalent function. Scaling and composing with an automorphism of the unit disk produces an injective holomorphic function $\varphi :\mathcal{U}\to \mathbb{D}$ with $\varphi (z_0)=0$ and ${\varphi }^{\prime }(z_0)>0$.

Let $\mathcal{F}$ be the family of all such normalized univalent functions. A Cauchy estimate in a small disk about $z_0$ bounds their derivatives, so $M={\sup }_{\varphi \in \mathcal{F}}{\varphi }^{\prime }(z_0)$ is finite and positive. Choose a maximizing sequence in this normal family. Montel theorem gives a subsequence converging uniformly on compact subsets to $\varphi$. Its derivative at $z_0$ is $M>0$. The maximum modulus principle puts its image in $\mathbb{D}$, and Hurwitz's theorem implies that a nonconstant limit of injective holomorphic functions is injective. Thus the maximum is attained.

If $a_1\in \mathbb{D}$ is omitted, then $a_1\ne 0$. Put $T_a(w)=(w-a)/(1-\overline{a}w)$. A holomorphic square root $v$ of $T_{a_1}\circ \varphi$ exists, is injective, and takes values in $\mathbb{D}$. Compose $v$ with $T_{v(z_0)}$ and a rotation to normalize it. Writing $s=|a_1|\in (0,1)$, the new derivative has magnitudea contradiction. Therefore $\varphi$ maps onto $\mathbb{D}$. The Cayley transform between the half-plane and disk now givesIts inverse is holomorphic by the holomorphic inverse function theorem, so this is a biholomorphism. The proper-subset hypothesis was used to choose $a\notin \mathcal{U}$; the whole complex plane cannot be mapped this way, by Liouville theorem.

Because $f(D)$ avoids the closed negative real axis, its principal square root $g=\sqrt{f}$ is holomorphic, satisfies $g(0)=1$, and has $\mathrm{Re}g>0$. The Cayley transform between the half-plane and diskmaps the right half-plane into the unit disk and has $h(0)=0$. By Schwarz lemma, $|h(z)|\le |z|$. When $|z|\le 1/2$,Thusso one may take the universal constant $C=9$.