Calculate Area of Region Defined by Complex Number Sets

To determine the area of the region $S = S_1 \cap S_2 \cap S_3$, we analyze each of the given sets in the complex plane ($z = x + iy = r e^{i\theta}$):

1. Analysis of Set $S_1$: $S_1 = \{z \in \mathbb{C} : |z| < 4\}$ This represents the interior of a circle centered at the origin with radius $R = 4$.

2. Analysis of Set $S_2$: $S_2 = \left\{z \in \mathbb{C} : \text{Im}\left[\frac{z - 1 + \sqrt{3}i}{1 - \sqrt{3}i}\right] > 0\right\}$ We can simplify the fraction inside the imaginary part operator: $\frac{z - 1 + \sqrt{3}i}{1 - \sqrt{3}i} = \frac{z - (1 - \sqrt{3}i)}{1 - \sqrt{3}i} = \frac{z}{1 - \sqrt{3}i} - 1$ Since $-1$ is a purely real number, its imaginary part is zero. Thus: $\text{Im}\left[\frac{z - 1 + \sqrt{3}i}{1 - \sqrt{3}i}\right] = \text{Im}\left[\frac{z}{1 - \sqrt{3}i}\right]$

   Expressing $1 - \sqrt{3}i$ in polar form: $1 - \sqrt{3}i = 2 e^{-i\pi/3}$

   Substituting $z = r e^{i\theta}$: $\frac{z}{1 - \sqrt{3}i} = \frac{r e^{i\theta}}{2 e^{-i\pi/3}} = \frac{r}{2} e^{i(\theta + \pi/3)}$

   Taking the imaginary part gives: $\text{Im}\left[\frac{z}{1 - \sqrt{3}i}\right] = \frac{r}{2} \sin\left(\theta + \frac{\pi}{3}\right)$

   The condition $\text{Im}\left[\frac{z - 1 + \sqrt{3}i}{1 - \sqrt{3}i}\right] > 0$ implies: $\sin\left(\theta + \frac{\pi}{3}\right) > 0 \implies 0 < \theta + \frac{\pi}{3} < \pi \implies -\frac{\pi}{3} < \theta < \frac{2\pi}{3}$

3. Analysis of Set $S_3$: $S_3 = \{z \in \mathbb{C} : \text{Re } z > 0\}$ This represents the right half-plane ($x > 0$), corresponding to the angular range: $-\frac{\pi}{2} < \theta < \frac{\pi}{2}$

4. Intersection $S = S_1 \cap S_2 \cap S_3$: Combining the conditions on $r$ and $\theta$:

   • Radius $r \in (0, 4)$
   • Argument $\theta \in \left(-\frac{\pi}{3}, \frac{2\pi}{3}\right) \cap \left(-\frac{\pi}{2}, \frac{\pi}{2}\right) = \left(-\frac{\pi}{3}, \frac{\pi}{2}\right)$

   The total angle subtended by the sector is: $\Delta \theta = \frac{\pi}{2} - \left(-\frac{\pi}{3}\right) = \frac{5\pi}{6}$

5. Calculation of the Area: The area of the sector of a circle with radius $R = 4$ and central angle $\Delta \theta = \frac{5\pi}{6}$ is: $\text{Area} = \frac{1}{2} R^2 \Delta \theta = \frac{1}{2} \times 4^2 \times \frac{5\pi}{6} = \frac{1}{2} \times 16 \times \frac{5\pi}{6} = \frac{20\pi}{3}$

Correct Answer: (B) $\frac{20\pi}{3}$