1. **State the problem:** Find the set of points $z = x + yi$ in the complex plane satisfying $\operatorname{Re}(z) < -1$. 2. **Formula and rules:** The real part of $z$ is $x$, so the inequality becomes $x < -1$. 3. **Interpretation:** This describes all points to the left of the vertical line $x = -1$. 4. **Properties:** - (a) Open: Yes, because the inequality is strict ($<$), so the boundary line $x = -1$ is not included. - (b) Closed: No, since the boundary is excluded. - (c) Domain: Yes, the set is open and connected, so it is a domain. - (d) Bounded: No, it extends infinitely to the left. - (e) Connected: Yes, it is a single continuous half-plane. 5. **Summary:** The set is the open half-plane to the left of $x = -1$. $$\{z \in \mathbb{C} : \operatorname{Re}(z) < -1\} = \{x + yi : x < -1, y \in \mathbb{R}\}$$