Find Minimum Distance from Complex Point to Set

To find the minimum distance from the point $z_0 = 1 - 3i$ to the set $S = S_1 \cap S_2 \cap S_3$, we first analyze each set in the complex plane $\mathbb{C}$.

Step 1: Characterize the Sets $S_1, S_2,$ and $S_3$

Let $z = x + iy$, where $x, y \in \mathbb{R}$.

1. Set $S_1$: $S_1 = \{z \in \mathbb{C} : |z| < 4\}$ This represents the open region inside a circle centered at the origin $(0, 0)$ with a radius of $4$: $x^2 + y^2 < 16$

2. Set $S_2$: $S_2 = \left\{z \in \mathbb{C} : \text{Im}\left[\frac{z - 1 + \sqrt{3}i}{1 - \sqrt{3}i}\right] > 0\right\}$ Simplifying the expression inside the imaginary part operator: $\frac{z - (1 - \sqrt{3}i)}{1 - \sqrt{3}i} = \frac{z}{1 - \sqrt{3}i} - 1 = \frac{(x + iy)(1 + \sqrt{3}i)}{(1 - \sqrt{3}i)(1 + \sqrt{3}i)} - 1 = \frac{(x - \sqrt{3}y) + i(\sqrt{3}x + y)}{4} - 1$ Taking the imaginary part: $\text{Im}\left[\frac{z - 1 + \sqrt{3}i}{1 - \sqrt{3}i}\right] = \frac{\sqrt{3}x + y}{4} > 0 \implies \sqrt{3}x + y > 0$ This represents the half-plane lying strictly above the line $L: \sqrt{3}x + y = 0$.

3. Set $S_3$: $S_3 = \{z \in \mathbb{C} : \text{Re } z > 0\}$ This represents the open right half-plane ($x > 0$).

Thus, the set $S = S_1 \cap S_2 \cap S_3$ is the region in the right half-plane ($x > 0$) inside the circle of radius 4 ($x^2 + y^2 < 16$) that lies above the line $\sqrt{3}x + y = 0$.

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Step 2: Position of $z_0 = 1 - 3i$ Relative to $S$

For $z_0 = 1 - 3i$, we have $x_0 = 1$ and $y_0 = -3$:

• Check $S_1$: $|z_0| = \sqrt{1^2 + (-3)^2} = \sqrt{10} < 4$ (Inside $S_1$).
• Check $S_3$: $x_0 = 1 > 0$ (Inside $S_3$).
• Check $S_2$: $\sqrt{3}x_0 + y_0 = \sqrt{3}(1) - 3 = \sqrt{3} - 3 < 0$.

Since $\sqrt{3}x_0 + y_0 < 0$, $z_0$ lies outside $S_2$ (below the line $L: \sqrt{3}x + y = 0$).

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Step 3: Calculate the Minimum Distance

The shortest distance from $z_0 = 1 - 3i$ to the set $S$ is given by the perpendicular distance from $z_0(1, -3)$ to the line boundary $L: \sqrt{3}x + y = 0$, provided that the foot of the perpendicular lies within the valid domain defined by $S_1$ and $S_3$.

1. Perpendicular distance formula: $d = \frac{|\sqrt{3}x_0 + y_0|}{\sqrt{(\sqrt{3})^2 + 1^2}} = \frac{|\sqrt{3}(1) + (-3)|}{\sqrt{3 + 1}} = \frac{| \sqrt{3} - 3 |}{2} = \frac{3 - \sqrt{3}}{2}$

2. Verify Foot of Perpendicular: The line perpendicular to $L$ passing through $(1, -3)$ is: $x - \sqrt{3}y = 1 + 3\sqrt{3}$ Solving this with $\sqrt{3}x + y = 0 \implies y = -\sqrt{3}x$: $x - \sqrt{3}(-\sqrt{3}x) = 1 + 3\sqrt{3} \implies 4x = 1 + 3\sqrt{3} \implies x_p = \frac{1 + 3\sqrt{3}}{4} > 0$ $y_p = -\sqrt{3}x_p = -\frac{3 + \sqrt{3}}{4}$ The distance of the foot of perpendicular $P(x_p, y_p)$ from the origin is: $|z_p| = \sqrt{x_p^2 + y_p^2} = \sqrt{4x_p^2} = 2x_p = \frac{1 + 3\sqrt{3}}{2} \approx 3.098 < 4$

Since $x_p > 0$ and $|z_p| < 4$, the foot of the perpendicular lies on the boundary segment of $S$. Therefore, the minimum distance from $z_0$ to $S$ is:

$\min_{z \in S} |1 - 3i - z| = \frac{3 - \sqrt{3}}{2}$

Correct Answer: C