**Step 1: Set up the Möbius map.** Let $C$ be the unique circle in $\mathbb{C}_\infty$ passing through $z_1, z_2, z_4$. By the [Sharp Triple Transitivity of Möbius Group](/theorems/811) theorem, there exists a unique $g \in \mathcal{M}$ with $g(z_1) = 0$, $g(z_2) = 1$, $g(z_4) = \infty$. **Step 2: Identify the image of $C$.** By the [Möbius Maps Preserve Circles](/theorems/813) theorem, $g(C)$ is a circle in $\mathbb{C}_\infty$ passing through $0$, $1$, and $\infty$. The unique such circle is $\mathbb{R} \cup \{\infty\}$. **Step 3: Compute $g(z_3)$ as the cross-ratio.** By the [Cross-Ratio Characterisation](/theorems/814) theorem: \begin{align*} g(z_3) = [0, 1, g(z_3), \infty] = [g(z_1), g(z_2), g(z_3), g(z_4)] = [z_1, z_2, z_3, z_4]. \end{align*} **Step 4: Conclude.** Since $g$ is a bijection of $\mathbb{C}_\infty$: \begin{align*} z_3 \in C \iff g(z_3) \in g(C) = \mathbb{R} \cup \{\infty\} \iff [z_1, z_2, z_3, z_4] \in \mathbb{R} \cup \{\infty\}. \end{align*} Since the four points are distinct and three of them already lie on $C$, the cross-ratio is finite (it cannot be $\infty$, as that would require $z_3 = z_4$). So the condition simplifies to $[z_1, z_2, z_3, z_4] \in \mathbb{R}$.