[proofplan]
The proof is a direct translation of the A-stability condition through the reduced rational representation of $R$ as the quotient of $P$ by $Q$. First we identify absence of poles in the closed left half-plane with nonvanishing of the denominator there. Then, on that domain, the bound $|R(z)| \le 1$ is exactly the inequality $|P(z)| \le |Q(z)|$ after multiplying by the positive number $|Q(z)|$. The converse repeats the same algebra in reverse.
[/proofplan]
[step:Identify the rational domain with the nonzero set of the denominator]
Define
\begin{align*}
\mathcal{H} := \overline{\mathbb{C}}_{-} = \{z \in \mathbb{C}: \operatorname{Re}(z) \le 0\}.
\end{align*}
By the definition of the rational function $R=P/Q$ as a quotient of polynomials, its domain is
\begin{align*}
\mathcal{D}_R = \{z \in \mathbb{C}: Q(z) \neq 0\}.
\end{align*}
Thus $R$ has no pole in $\mathcal{H}$ precisely when $\mathcal{H} \subset \mathcal{D}_R$, which is precisely the condition
\begin{align*}
Q(z) \neq 0 \qquad \text{for every } z \in \mathcal{H}.
\end{align*}
More explicitly, if $a \in \mathbb{C}$ and $Q(a)=0$, then the [factor theorem](/theorems/3235) gives that $z-a$ divides $Q$ in $\mathbb{C}[z]$. Since $P$ and $Q$ are relatively prime, $z-a$ does not divide $P$, so $P(a) \neq 0$. Hence every zero of $Q$ is a genuine pole of the reduced rational function and is not a removable zero cancelled by $P$.
[/step]
[step:Translate A-stability into the polynomial modulus inequality]
Assume the method is A-stable. By the preceding step, $Q(z) \neq 0$ for every $z \in \mathcal{H}$. Fix $z \in \mathcal{H}$. Since $Q(z) \neq 0$, the real number $|Q(z)|$ is positive. A-stability gives
\begin{align*}
\left|\frac{P(z)}{Q(z)}\right| \le 1.
\end{align*}
Using multiplicativity of the complex modulus,
\begin{align*}
\left|\frac{P(z)}{Q(z)}\right| = \frac{|P(z)|}{|Q(z)|}.
\end{align*}
Multiplying the inequality by $|Q(z)|>0$ gives
\begin{align*}
|P(z)| \le |Q(z)|.
\end{align*}
Since $z \in \mathcal{H}$ was arbitrary, this proves
\begin{align*}
|P(z)| \le |Q(z)| \qquad \text{for every } z \in \mathcal{H}.
\end{align*}
[guided]
Assume the method is A-stable. Under the convention in the statement, this has two pieces: first, $R$ has no pole in the closed left half-plane $\mathcal{H}$, and second, $|R(z)| \le 1$ for every $z \in \mathcal{H}$. Because $R:\mathcal{D}_R \to \mathbb{C}$ is defined on the domain $\mathcal{D}_R=\{z\in\mathbb{C}:Q(z)\neq 0\}$ by the displayed quotient formula in the theorem statement, a point of $\mathcal{H}$ is a pole exactly when it is a zero of $Q$. Therefore the no-pole condition is exactly
\begin{align*}
Q(z) \neq 0 \qquad \text{for every } z \in \mathcal{H}.
\end{align*}
Now fix an arbitrary point $z \in \mathcal{H}$. Because $Q(z) \neq 0$, the quotient $P(z)/Q(z)$ is defined and $|Q(z)|>0$. The A-stability bound gives
\begin{align*}
|R(z)| \le 1.
\end{align*}
Substituting the definition $R(z)=P(z)/Q(z)$ gives
\begin{align*}
\left|\frac{P(z)}{Q(z)}\right| \le 1.
\end{align*}
The complex modulus is multiplicative and satisfies $|1/Q(z)|=1/|Q(z)|$ when $Q(z)\neq 0$, so
\begin{align*}
\left|\frac{P(z)}{Q(z)}\right| = \frac{|P(z)|}{|Q(z)|}.
\end{align*}
Thus
\begin{align*}
\frac{|P(z)|}{|Q(z)|} \le 1.
\end{align*}
Multiplication by the positive real number $|Q(z)|$ preserves the direction of the inequality, hence
\begin{align*}
|P(z)| \le |Q(z)|.
\end{align*}
Because the point $z \in \mathcal{H}$ was arbitrary, the same inequality holds for every point in the closed left half-plane.
[/guided]
[/step]
[step:Recover A-stability from denominator nonvanishing and the modulus inequality]
Conversely, assume that
\begin{align*}
Q(z) \neq 0 \qquad \text{for every } z \in \mathcal{H}
\end{align*}
and
\begin{align*}
|P(z)| \le |Q(z)| \qquad \text{for every } z \in \mathcal{H}.
\end{align*}
The first condition implies $\mathcal{H} \subset \mathcal{D}_R$, so $R$ has no pole in $\mathcal{H}$. Fix $z \in \mathcal{H}$. Since $|Q(z)|>0$, division by $|Q(z)|$ gives
\begin{align*}
\frac{|P(z)|}{|Q(z)|} \le 1.
\end{align*}
Using $R(z)=P(z)/Q(z)$ and multiplicativity of the complex modulus,
\begin{align*}
|R(z)| = \left|\frac{P(z)}{Q(z)}\right| = \frac{|P(z)|}{|Q(z)|} \le 1.
\end{align*}
Thus $R$ has no pole in $\mathcal{H}$ and satisfies $|R(z)|\le 1$ for every $z \in \mathcal{H}$. By the stated convention for A-stability, the method is A-stable.
[/step]
