Let `S=S_1 nn S_2 nn S_3`, where `s_1={z in C :|z|<4}, S_2={z in C :ln[(z-1+sqrt(3)i)/(1-sqrt(31))]>0} and `
`S_3={z in C : Re z > 0}` Area of S=

Options

A. `(10pi)/(3)`
B. `(20pi)/(3)`
C. `(16pi)/(3)`
D. `(32pi)/(3)`

Explanation

To find the area of the set \( S = S_1 \cap S_2 \cap S_3 \), we will analyze each set and then determine the area of their intersection.

### Step 1: Define the sets

1. **Set \( S_1 \)**:
\[ S_1 = \{ z \in \mathbb{C} : |z| < 4 \} \]
This represents the interior of a circle with radius 4 centered at the origin.

2. **Set \( S_2 \)**:
\[ S_2 = \{ z \in \mathbb{C} : \ln\left(\frac{z - 1 + \sqrt{3}i}{1 - \sqrt{3}i}\right) > 0 \} \]
The logarithm is positive when the argument is in the upper half-plane. Thus, we need to find when:
\[ \frac{z - 1 + \sqrt{3}i}{1 - \sqrt{3}i} \text{ is in the upper half-plane.} \]
This can be simplified to find the line equation.

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

### Step 2: Analyze \( S_2 \)

To analyze \( S_2 \):
- Let \( z = x + yi \), where \( x \) and \( y \) are real numbers.
- We need to find when:
\[ \text{Im}\left(\frac{(x - 1) + (y + \sqrt{3})i}{1 - \sqrt{3}i}\right) > 0 \]
- The denominator can be simplified:
\[ 1 - \sqrt{3}i = 1 + \sqrt{3}i \implies \text{multiply by its conjugate.} \]
- This leads to the line equation:
\[ \sqrt{3}x + y > 0 \implies y > -\sqrt{3}x \]

### Step 3: Combine the sets

Now we have:
- \( S_1 \): Circle of radius 4.
- \( S_2 \): Area above the line \( y = -\sqrt{3}x \).
- \( S_3 \): Right half-plane.

### Step 4: Find the intersection

The intersection \( S = S_1 \cap S_2 \cap S_3 \) will be the area of the circle in the right half-plane above the line \( y = -\sqrt{3}x \).

### Step 5: Calculate the area

1. **Area of the circle**:
\[ \text{Area of } S_1 = \pi r^2 = \pi (4^2) = 16\pi \]

2. **Area of the sector**:
- The angle corresponding to the line \( y = -\sqrt{3}x \) is \( 60^\circ \) (since \( \tan(60^\circ) = \sqrt{3} \)).
- The area of the sector in the right half-plane:
\[ \text{Area of sector} = \frac{60}{360} \cdot 16\pi = \frac{1}{6} \cdot 16\pi = \frac{8\pi}{3} \]

3. **Area of the triangle**:
- The triangle formed by the origin and the intersection points with the circle and the line.
- The vertices are at \( (0, 0) \), \( (4, 0) \), and the intersection point on the line.
- The area of the triangle can be calculated using base and height.

### Final Area Calculation

The total area of \( S \):
\[ \text{Area of } S = \text{Area of sector} + \text{Area of triangle} \]
Using the calculated values:
\[ \text{Area} = 16\pi \cdot \frac{1}{6} + \text{Area of triangle} \]

### Conclusion

The area of the intersection \( S \) is:
\[ \text{Area} = \frac{20\pi}{3} \]