Let S = {z : 3 le |2z - 3(1 + i)| le 7} be a | vedclass

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(B) Given the set $S = \{z : 3 \le |2z - 3(1 + i)| \le 7\}$.Dividing by $2$, we get: $\frac{3}{2} \le |z - \frac{3}{2}(1 + i)| \le \frac{7}{2}$.This r

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$ \frac{3}{2} $

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(B) Given the set $S = \{z : 3 \le |2z - 3(1 + i)| \le 7\}$.Dividing by $2$, we get: $\frac{3}{2} \le |z - \frac{3}{2}(1 + i)| \le \frac{7}{2}$.This represents an annulus centered at $C(\frac{3}{2}, \frac{3}{2})$ with inner radius $r_1 = \frac{3}{2}$ and outer radius $r_2 = \frac{7}{2}$.We want to find the minimum distance from a point $P(-\frac{5}{2}, -\frac{3}{2})$ to the set $S$.The distance $PC = \sqrt{(\frac{3}{2} - (-\frac{5}{2}))^2 + (\frac{3}{2} - (-\frac{3}{2}))^2} = \sqrt{4^2 + 3^2} = \sqrt{16 + 9} = 5$.The minimum distance from $P$ to the annulus is $PC - r_2 = 5 - \frac{7}{2} = \frac{3}{2}$.

Let $S = \{z : 3 \le |2z - 3(1 + i)| \le 7\}$ be a set of complex numbers. Then $\min_{z \in S} |z + \frac{1}{2}(5 + 3i)|$ is equal to:

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