Paragraph: Let $S=S_{1} \cap S_{2} \cap S_{3}$, where $S_{1}=\{z \in \mathbb{C}:|z|<4\}, \quad S_{2}=\left\{z \in \mathbb{C}: \operatorname{Im}\left[\frac{z-1+\sqrt{3} i}{1-\sqrt{3} i}\right]>0\right\}$ and $S_{3}=\{z \in \mathbb{C}: \operatorname{Re} z>0\}$.
Question: Area of $S=$

• A

  $\frac{10\pi }{3}$
• Correct answer
  B

  $\frac{20\pi }{3}$
• C

  $\frac{16\pi }{3}$
• D

  ​$\frac{32\pi }{3}$

Solution

S₁ : x² + y² < 16
${\text{S}}_2\text{ : }\frac{\text{z}-1+\sqrt{3\text{i}}}{1-\text{i}\sqrt{3}}=\frac{\left(\text{x}-1\right)+\text{i }\left(\text{y}+\sqrt{3}\right)}{1-\sqrt{3}\text{i}}$
$=\frac{\left\{\left(\text{x}-1\right)+\text{i }\left(\text{y}+\sqrt{3}\right)\right\} \left\{1+\sqrt{3 }\text{i}\right\}}{1+3}$


${\text{S}}_2\text{ : }\frac{\left(\text{x}-1\right)\sqrt{3}+\text{y}+\sqrt{3}}{9}>0$
${\text{S}}_2\text{ : }\sqrt{3}\text{x}+\text{y}>0 \& {\text{S}}_3\text{ : }\text{x}>0$
$\text{A}=\frac{1}{2}{\text{r}}^2\theta =\frac{1}{2}\times 16\times \frac{5\pi }{6}=\frac{40\pi }{6}=\frac{20\pi }{3}$

Hence, the correct answer is option: 'B'.