Let $S=S_1 \cap S_2 \cap S_3$,where $S_1=\{z \in \mathbb{C}:|z|<4\}$,$S_2=\{z \in \mathbb{C}: \operatorname{Im}[\frac{z-1+\sqrt{3} i}{1-\sqrt{3} i}]>0\}$,and $S_3=\{z \in \mathbb{C}: \operatorname{Re} z>0\}$.
$1.$ Area of $S=$
$(A) \frac{10 \pi}{3} \quad (B) \frac{20 \pi}{3} \quad (C) \frac{16 \pi}{3} \quad (D) \frac{32 \pi}{3}$
$2.$ $\min _{z \in S}|1-3 i-z|=$
$(A) \frac{2-\sqrt{3}}{2} \quad (B) \frac{2+\sqrt{3}}{2} \quad (C) \frac{3-\sqrt{3}}{2} \quad (D) \frac{3+\sqrt{3}}{2}$

(B,C) $1.$ $S_1$ represents the interior of a circle with radius $r=4$ centered at the origin.
$S_2: \operatorname{Im}[\frac{(x-1)+i(y+\sqrt{3})}{1-\sqrt{3} i} \cdot \frac{1+\sqrt{3} i}{1+\sqrt{3} i}] > 0 \implies \operatorname{Im}[\frac{(x-1+i(y+\sqrt{3}))(1+\sqrt{3} i)}{4}] > 0$
$\implies (x-1)\sqrt{3} + (y+\sqrt{3}) > 0 \implies \sqrt{3}x + y > 0$.
$S_3: x > 0$.
The region $S$ is the intersection of the disk $x^2+y^2 < 16$,the half-plane $y > -\sqrt{3}x$,and the half-plane $x > 0$. This forms a circular sector with angle $\theta = 150^\circ = \frac{5\pi}{6}$ radians.
Area $= \frac{1}{2} r^2 \theta = \frac{1}{2} \times 16 \times \frac{5\pi}{6} = \frac{20\pi}{3}$.
$2.$ We need the minimum distance from the point $P(1, -3)$ to the region $S$. The boundary of $S$ includes the line $y = -\sqrt{3}x$ for $x > 0$.
The distance from $(1, -3)$ to the line $\sqrt{3}x + y = 0$ is $d = \frac{|\sqrt{3}(1) + (-3)|}{\sqrt{(\sqrt{3})^2 + 1^2}} = \frac{|\sqrt{3}-3|}{2} = \frac{3-\sqrt{3}}{2}$.