My teacher said without proving that the region $\{z:|2\bar{z}-3|>4\}$ is path-connected but the region $\{z:|z-1|<1\}\cup\{z:|z+1|<1\}$ is not path-connected. Is there a way to prove this?
2026-03-27 10:31:36.1774607496
How to show that a region is path-connected/not path-connected?
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The set $$D = \{z: |2\overline{z} - 3| > 4\} = \{z: |z-3/2| > 2\}$$ is path-connected since any two points $3/2 + r_1 e^{i \theta_1}$ and $3/2 + r_2 e^{i \theta_2}$, $r_1,r_2 > 2$ are connected by the path $$\gamma : [0,1] \rightarrow D, \; \; \gamma(t) := 3/2 + ((1-t)r_1 + t r_2) e^{i (1-t)\theta_1 + it \theta_2}.$$
The set $$U_1 \cup U_2 = \{z: \, |z-1| < 1\} \cup \{z: \, |z+1| < 1\}$$ is not even connected in the topological sense (in particular not path-connected) because it consists of two open subsets $U_1,U_2$ that do not intersect.