The user wants me to solve a problem involving complex numbers and geometry. The goal is to find the minimum value of $|1 - 3i - z|$ for a complex number $z$ in a specified region $S$. This is equivalent to finding the minimum distance between the point representing $1-3i$ and the region $S$ in the Argand plane.

Step 1: Define the sets $S_1, S_2, S_3$

Let $z = x + iy$. We analyze each set to understand the region $S = S_1 \cap S_2 \cap S_3$.

1. Set $S_1$: ${S_1} = \left\{ {z \in C:\left| z \right| < 4} \right\}$. This represents the set of all complex numbers inside the circle centered at the origin with radius 4. In Cartesian coordinates, this is the open disk defined by $x^2 + y^2 < 16$.

2. Set $S_2$: ${S_2} = \left\{ {z \in C:{\mathop{\rm Im}} \left[ {{{z - 1 + \sqrt 3 i} \over {1 - \sqrt 3 i}}} \right] > 0} \right\}$. Let's simplify the expression inside the imaginary part. Let $w = 1 - \sqrt 3 i$. The expression is $\frac{z - w}{w} = \frac{z}{w} - 1$. The condition becomes ${\mathop{\rm Im}} (\frac{z}{w} - 1) > 0$, which is equivalent to ${\mathop{\rm Im}} (\frac{z}{w}) > 0$. Let's convert to Cartesian coordinates: $\frac{z}{w} = \frac{x+iy}{1-\sqrt{3}i} = \frac{(x+iy)(1+\sqrt{3}i)}{(1-\sqrt{3}i)(1+\sqrt{3}i)} = \frac{x + i\sqrt{3}x + iy - \sqrt{3}y}{1^2 + (\sqrt{3})^2} = \frac{(x - \sqrt{3}y) + i(\sqrt{3}x+y)}{4}$ The imaginary part is $\frac{\sqrt{3}x+y}{4}$. The condition is $\frac{\sqrt{3}x+y}{4} > 0$, which simplifies to $\sqrt{3}x+y > 0$ or $y > -\sqrt{3}x$. This inequality defines the half-plane above the line $y = -\sqrt{3}x$.

3. Set $S_3$: ${S_3} = \left\{ {z \in C:{\mathop{\rm Re}} z > 0} \right\}$. This is the condition $x > 0$, which represents the right half-plane.

Step 2: Characterize the region S

The region $S$ is the intersection of these three sets:

• $|z| < 4$ (inside the circle of radius 4)
• $x > 0$ (right half-plane)
• $y > -\sqrt{3}x$ (above the line $y = -\sqrt{3}x$)

The line $y = -\sqrt{3}x$ for $x>0$ corresponds to the ray with angle $\theta = \arctan(-\sqrt{3}) = -\pi/3$. The condition $y > -\sqrt{3}x$ for $x>0$ means the angle $\theta$ must be greater than $-\pi/3$. The condition $x>0$ restricts the angle to $(-\pi/2, \pi/2)$. Combining these, we get $-\pi/3 < \theta < \pi/2$. So, the region $S$ is an open sector of the disk, described in polar coordinates as $S = \{z = re^{i\theta} : 0 < r < 4, -\pi/3 < \theta < \pi/2\}$.

Step 3: Formulate the minimization problem

We need to find $\min_{z \in S} |1 - 3i - z|$. Let $c = 1 - 3i$. The problem is to find the minimum value of $|c - z|$, which is the distance between the point $A$ representing $c$ and a point $P$ representing $z$ in the region $S$. The coordinates of point $A$ are $(1, -3)$.

First, let's check if $A(1, -3)$ is in the region $S$.

• $|c| = |1-3i| = \sqrt{1^2 + (-3)^2} = \sqrt{10}$. Since $\sqrt{10} < 4$, the point is in $S_1$.
• $\text{Re}(c) = 1 > 0$, so the point is in $S_3$.
• For $S_2$, we check if $y > -\sqrt{3}x$. For $A(1, -3)$, this is $-3 > -\sqrt{3}(1)$, which is $-3 > -\sqrt{3}$. This is false, since $-3 < -1.732...$. So, the point $A$ is not in the region $S$. The minimum distance will be the distance from $A$ to the boundary of $S$.

Step 4: Calculate the distance to the boundary of S

The boundary of S consists of three parts (considering the closure of S):

1. Line segment $L_1$: The y-axis from $z=0$ to $z=4i$. This is $x=0$ for $0 \le y \le 4$.
2. Line segment $L_2$: The line from $z=0$ to $z=4e^{-i\pi/3}$. This is $y = -\sqrt{3}x$ for $0 \le x \le 2$.
3. Circular arc $C$: The arc of the circle $|z|=4$ from $\theta = -\pi/3$ to $\theta = \pi/2$.

Distance from A(1, -3) to $L_1$: The closest point on the line $x=0$ to $A(1, -3)$ is $(0, -3)$. This point is not on the segment $L_1$ (since $y$ must be in $[0,4]$). Thus, the closest point on the segment is the endpoint $(0,0)$. Distance to $(0,0)$ is $d_1 = \sqrt{(1-0)^2 + (-3-0)^2} = \sqrt{10}$.

Distance from A(1, -3) to $L_2$: The line is $\sqrt{3}x + y = 0$. The perpendicular distance from $A(1, -3)$ to this line is: $d = \frac{|\sqrt{3}(1) + (-3)|}{\sqrt{(\sqrt{3})^2 + 1^2}} = \frac{|\sqrt{3} - 3|}{\sqrt{4}} = \frac{3 - \sqrt{3}}{2}$ We must check if the foot of the perpendicular lies on the segment $L_2$. The segment extends from $(0,0)$ to $(2, -2\sqrt{3})$. The equation of the line perpendicular to $L_2$ and passing through $A$ is $y - (-3) = \frac{1}{\sqrt{3}}(x-1)$. The intersection of the two lines is found by solving the system: $y = -\sqrt{3}x$ $\sqrt{3}(y+3) = x-1$ Substituting $y$: $\sqrt{3}(-\sqrt{3}x+3)=x-1 \implies -3x+3\sqrt{3}=x-1 \implies 4x = 1+3\sqrt{3} \implies x = \frac{1+3\sqrt{3}}{4}$. Since $1.7 < \sqrt{3} < 1.8$, we have $1 < \frac{1+3(1.7)}{4} < x < \frac{1+3(1.8)}{4} < 1.6$. Thus $0 < x < 2$, so the foot of the perpendicular lies on the segment $L_2$. Therefore, the minimum distance to $L_2$ is $d_2 = \frac{3 - \sqrt{3}}{2}$.

Distance from A(1, -3) to $C$: The point A is inside the circle $|z|=4$ since $|c|=\sqrt{10}<4$. The closest point on the full circle would be on the line from the origin through A. The angle of A is $\theta_A = \arctan(-3/1) = \arctan(-3)$. The range of angles for the arc is $[-\pi/3, \pi/2]$. $\tan(-\pi/3) = -\sqrt{3} \approx -1.732$. Since $-3 < -\sqrt{3}$, we have $\arctan(-3) < -\pi/3$. So the closest point on the circle is not on the arc $C$. The minimum distance to the arc will be to one of its endpoints. The endpoints are $E_1(0,4)$ and $E_2(2, -2\sqrt{3})$. Distance to $E_1$: $\sqrt{(1-0)^2+(-3-4)^2} = \sqrt{1+49}=\sqrt{50}$. Distance to $E_2$: $\sqrt{(1-2)^2+(-3-(-2\sqrt{3}))^2} = \sqrt{(-1)^2+(2\sqrt{3}-3)^2} = \sqrt{1+12-12\sqrt{3}+9} = \sqrt{22-12\sqrt{3}}$.

Step 5: Compare the distances

We need to find the minimum of the distances calculated:

• $d_1 = \sqrt{10} \approx 3.16$
• $d_2 = \frac{3-\sqrt{3}}{2} \approx \frac{3-1.732}{2} = 0.634$
• $d_3 = \sqrt{22-12\sqrt{3}} \approx \sqrt{22-12(1.732)} = \sqrt{22-20.784} = \sqrt{1.216} \approx 1.1$

The smallest distance is $d_2 = \frac{3 - \sqrt{3}}{2}$.

Thus, the minimum value of $|1 - 3i - z|$ for $z \in S$ is $\frac{3 - \sqrt{3}}{2}$.