Let $f:D(f)\subseteq\mathbb C\to\mathbb C$ be defined by $$f(z)=a_0+a_1\frac{1}{z},$$ where $a_0,a_1\in\mathbb R$.
Problem: Is $f$ bounded on the closed, right half-plane $\{z\in\mathbb C:\text{Re}(z)\ge0\}\subseteq\mathbb C$, or, on closed planes of the form $\{z\in\mathbb C:\text{Re}(z)\ge\alpha>0\}\subseteq\mathbb C$? In the affirmative case, what is the infimum choice of $\alpha\in\mathbb R$ for which $f$ is bounded?
Attempt: For $z\in\{z\in\mathbb C: \text{Re}(z)\ge0\}$ we have that $|f(z)|=\left|a_0+a_1\frac1z\right|$, and note that $\lim_{z\to\infty}\left|a_0+a_1\frac1z\right|=\infty$, so that $f$ is not bounded on the half plane.
Let $\text{Re}(z)\ge\alpha>0$ and consider $z\in\{z\in\mathbb C: \text{Re}(z)\ge\alpha\}$. Then, $\sup_{\{z\in\mathbb C: \text{Re}(z)\ge\alpha\}}|f(z)|=M<\infty$ is attained for $w\in\{z\in\mathbb C: \text{Re}(z)\ge\alpha\}$ with $\text{Im}(w)=0$, so that $w$ is closer to zero in absolute value than any other $r\in\{z\in\mathbb C: \text{Re}(z)\ge\alpha\}$.
Hence, $f$ is bounded on closed planes of the form $\{z\in\mathbb C: \text{Re}(z)\ge\alpha\}\subset\mathbb C$, when $\alpha>0$. The infimum choice of $\alpha\in\mathbb R$ for which this holds is then $\alpha=0$.
Is this correct? Is my reasoning well founded and water-tight? Or is there anything on which to be improved?
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Rephrased question in accord with the more succinct and elegant suggestion of @BrevanEllefsen - my thanks for this.