Complex Numbers Part 2

Concept:The set $S$ is an annulus in the complex plane. The minimum distance from $S$ to a fixed point equals the distance between centers minus the outer radius.Explanation:Divide the inequality $3 \le |2z-3(1+i)| \le 7$ by $2$.Get $\frac{3}{2} \le |z-(\frac{3}{2}+\frac{3}{2}i)| \le \frac{7}{2}$.This represents an annulus centered at $C_1 = \frac{3}{2}+\frac{3}{2}i$ with inner radius $\frac{3}{2}$ and outer radius $\frac{7}{2}$.We need $\min_{z \in S} |z + \frac{1}{2}(5+3i)| = \min_{z \in S} |z - (-\frac{5}{2}-\frac{3}{2}i)|$.Let $C_2 = -\frac{5}{2}-\frac{3}{2}i$.The distance between centers $|C_1-C_2| = \sqrt{(\frac{3}{2}+\frac{5}{2})^2+(\frac{3}{2}+\frac{3}{2})^2} = \sqrt{4^2+3^2}=5$.The minimum distance from $S$ to $C_2$ is this center distance minus the outer radius: $5 - \frac{7}{2} = \frac{3}{2}$.Answer:$\frac{3}{2}$ (Option D).