If `S_1={z:abs(z-3-2i)^2=8} , S_2={z:abs(z-barz)=8} and S_3={z:re(z) ge 5} then S_1capS_2capS_3` has

Options

A. infinite many element
B. only one element
C. No element
D. two element

Answer

To solve the problem, we need to find the intersection of the three sets \( S_1 \), \( S_2 \), and \( S_3 \).

### Step 1: Analyze \( S_1 \)

The set \( S_1 \) is defined by the equation:
\[ |z - (3 + 2i)|^2 = 8 \]
This can be rewritten as:
\[ |z - (3 + 2i)| = \sqrt{8} = 2\sqrt{2} \]
This represents a circle in the complex plane with center at the point \( (3, 2) \) and radius \( 2\sqrt{2} \).

### Step 2: Equation of the Circle

The equation of the circle can be expressed in Cartesian coordinates as:
\[ (x - 3)^2 + (y - 2)^2 = (2\sqrt{2})^2 = 8 \]
This is the equation of a circle centered at \( (3, 2) \) with a radius of \( 2\sqrt{2} \).

### Step 3: Analyze \( S_2 \)

The set \( S_2 \) is defined by the equation:
\[ |z - \bar{z}| = 8 \]
Since \( z = x + iy \) and \( \bar{z} = x - iy \), we have:
\[ |z - \bar{z}| = |(x + iy) - (x - iy)| = |2iy| = 2|y| = 8 \]
This simplifies to:
\[ |y| = 4 \]
Thus, \( y = 4 \) or \( y = -4 \). These represent two horizontal lines in the complex plane.

### Step 4: Analyze \( S_3 \)

The set \( S_3 \) is defined by:
\[ \text{Re}(z) \geq 5 \]
This means that the real part \( x \) of \( z \) must be greater than or equal to 5. This corresponds to the vertical line \( x = 5 \) and the region to the right of this line.

### Step 5: Finding the Intersection \( S_1 \cap S_2 \cap S_3 \)

Now, we need to find the intersection of the three sets. We will check the points where the circle \( S_1 \) intersects the lines \( S_2 \) and the vertical line \( S_3 \).

1. **Intersection of \( S_1 \) and \( S_2 \)**:
- For \( y = 4 \):
\[ (x - 3)^2 + (4 - 2)^2 = 8 \implies (x - 3)^2 + 4 = 8 \implies (x - 3)^2 = 4 \implies x - 3 = \pm 2 \]
This gives \( x = 5 \) or \( x = 1 \).
- For \( y = -4 \):
\[ (x - 3)^2 + (-4 - 2)^2 = 8 \implies (x - 3)^2 + 36 = 8 \implies (x - 3)^2 = -28 \]
This has no real solutions.

Thus, the only intersection point from \( S_1 \) and \( S_2 \) is \( (5, 4) \).

2. **Check against \( S_3 \)**:
- The point \( (5, 4) \) satisfies \( x \geq 5 \).

### Conclusion

The intersection \( S_1 \cap S_2 \cap S_3 \) consists of the single point:
\[ (5, 4) \]

### Final Answer

The intersection \( S_1 \cap S_2 \cap S_3 \) has exactly **one point**: \( (5, 4) \).

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