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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?

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    The latter two smaller regions are disjoint. There is no way to join a point in one region with one in the other. – 2017-01-07

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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.

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    How did you get the path $\gamma(t)$? – 2017-01-07
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    In $(r,\theta)$-coordinates it is just the formula for a line between 2 points. – 2017-01-07
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    that set isnt convex... for example $z=-1$ and $z=4$ cannot be connected by your path – 2017-01-07
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    @DavidP It's a line in $(r,\theta)$ coordinates. It's not a line in $z$-coordinates as you can see from the formula. – 2017-01-07
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    i misunderstood your use of line. it is actually a circle. +1 – 2017-01-07