Let $$S=\left\{z:3\leq|2z-3(1+i)|\leq7\right\}$$ be a set of complex nwnbers. Then $$\min_{Z \epsilon S}\left|\left(z+\frac{1}{2}(5+3i)\right)\right|$$ is equal to :

We need to find the minimum value of $$\left|z + \frac{1}{2}(5 + 3i)\right|$$ for $$z \in S = \{z : 3 \leq |2z - 3(1+i)| \leq 7\}$$.

Rewrite the set $$S$$ in standard form. The condition $$|2z - 3(1+i)| \leq 7$$ can be written as:

$$\left|2\left(z - \frac{3(1+i)}{2}\right)\right| \leq 7 \implies \left|z - \frac{3(1+i)}{2}\right| \leq \frac{7}{2}$$

Similarly, $$|2z - 3(1+i)| \geq 3$$ gives $$\left|z - \frac{3(1+i)}{2}\right| \geq \frac{3}{2}$$.

So $$S$$ is the annular region (ring) centered at $$C_1 = \frac{3}{2}(1+i) = \frac{3}{2} + \frac{3}{2}i$$ with inner radius $$r = \frac{3}{2}$$ and outer radius $$R = \frac{7}{2}$$.

Next, we identify the target point. We want to minimize $$\left|z + \frac{5+3i}{2}\right| = \left|z - \left(-\frac{5+3i}{2}\right)\right|$$, which is the distance from $$z$$ to the point $$C_2 = -\frac{5}{2} - \frac{3}{2}i$$.

Now we find the distance between the two centers $$C_1$$ and $$C_2$$:

$$|C_1 - C_2| = \left|\frac{3}{2} + \frac{3}{2}i - \left(-\frac{5}{2} - \frac{3}{2}i\right)\right| = \left|\frac{3}{2} + \frac{5}{2} + \left(\frac{3}{2} + \frac{3}{2}\right)i\right|$$

$$= |4 + 3i| = \sqrt{16 + 9} = \sqrt{25} = 5$$

The minimum distance from any point in the annular region to $$C_2$$ occurs along the line connecting $$C_1$$ and $$C_2$$, at the point on the outer boundary of the annulus closest to $$C_2$$. Since $$C_2$$ is outside the annulus (distance between centers = 5 > outer radius = 7/2 = 3.5), the minimum distance is:

$$d_{\min} = |C_1 - C_2| - R = 5 - \frac{7}{2} = \frac{10 - 7}{2} = \frac{3}{2}$$

We confirm that $$C_2$$ lies outside the annulus because $$|C_1 - C_2| = 5 > \frac{7}{2} = 3.5$$, so the closest point on the annulus to $$C_2$$ is on the outer circle (radius 7/2) along the line from $$C_1$$ to $$C_2$$.

The correct answer is Option A: $$\frac{3}{2}$$.