'Let `S= S_(1) cap S_(2) cap S_(3)` where `S_(1) {z in C : |z|lt 4}, S_(2) ={z in C : Im [(z-1 + sqrt(3)i)/(1-sqrt(3)i)]gt 0}` and `S_(3) = {z in C : Re (z) gt 0}`''
`("min")/(z in C)|1-3i-z|=`

Options

A. `(2- sqrt(3))/(2)`
B. `(2+ sqrt(3))/(2)`
C. `(3- sqrt(3))/(2)`
D. `(3+ sqrt(3))/(2)`

Answer

To solve the problem, we need to find the minimum value of the expression \(|1 - 3i - z|\) where \(z\) belongs to the set \(S = S_1 \cap S_2 \cap S_3\).

### Step 1: Understand the sets \(S_1\), \(S_2\), and \(S_3\)

1. **Set \(S_1\)**:
- \(S_1 = \{ z \in \mathbb{C} : |z| < 4 \}\)
- This represents the interior of a circle centered at the origin (0,0) with a radius of 4.

2. **Set \(S_2\)**:
- \(S_2 = \{ z \in \mathbb{C} : \text{Im}\left(\frac{z - 1 + \sqrt{3}i}{1 - \sqrt{3}i}\right) > 0 \}\)
- To analyze this, we rewrite \(z\) as \(z = x + iy\). Then, we have:
\[ \frac{(x - 1) + (y + \sqrt{3})i}{1 - \sqrt{3}i} \]
- Multiplying the numerator and denominator by the conjugate of the denominator:
\[ \frac{(x - 1 + (y + \sqrt{3})i)(1 + \sqrt{3}i)}{1 + 3} = \frac{(x - 1) + (y + \sqrt{3})i + \sqrt{3}(x - 1)i - 3(y + \sqrt{3})}{4} \]
- The imaginary part simplifies to:
\[ \frac{\sqrt{3}(x - 1) + y + \sqrt{3}}{4} \]
- Setting this greater than 0 gives:
\[ \sqrt{3}(x - 1) + y + \sqrt{3} > 0 \implies \sqrt{3}x + y > 0 \]

3. **Set \(S_3\)**:
- \(S_3 = \{ z \in \mathbb{C} : \text{Re}(z) > 0 \}\)
- This represents the right half-plane where the real part of \(z\) (which is \(x\)) is greater than 0.

### Step 2: Determine the intersection \(S = S_1 \cap S_2 \cap S_3\)

- The intersection \(S\) consists of points that lie inside the circle of radius 4, above the line defined by \(\sqrt{3}x + y = 0\) (which has a slope of \(-\sqrt{3}\)), and in the right half-plane.

### Step 3: Find the minimum value of \(|1 - 3i - z|\)

- We rewrite the expression:
\[ |1 - 3i - z| = |(1 - z) + (-3i)| \]
- This represents the distance from the point \(1 - 3i\) to the point \(z\).

### Step 4: Identify the point \(1 - 3i\)

- The point \(1 - 3i\) is located at the coordinates \((1, -3)\).

### Step 5: Find the minimum distance from \(1 - 3i\) to the boundary of the region defined by \(S\)

- The minimum distance from a point to a line can be calculated using the formula:
\[ \text{Distance} = \frac{|Ax_0 + By_0 + C|}{\sqrt{A^2 + B^2}} \]
where \(Ax + By + C = 0\) is the equation of the line.

- The line equation from \(S_2\) is \(\sqrt{3}x + y = 0\) or \(\sqrt{3}x + y + 0 = 0\). Here, \(A = \sqrt{3}\), \(B = 1\), \(C = 0\), and the point \((x_0, y_0) = (1, -3)\).

- Plugging in the values:
\[ \text{Distance} = \frac{|\sqrt{3}(1) + 1(-3) + 0|}{\sqrt{(\sqrt{3})^2 + 1^2}} = \frac{|\sqrt{3} - 3|}{\sqrt{3 + 1}} = \frac{|-\sqrt{3} + 3|}{2} \]
\[ = \frac{3 - \sqrt{3}}{2} \]

### Step 6: Final answer

- The minimum value of \(|1 - 3i - z|\) is:
\[ \frac{3 - \sqrt{3}}{2} \]