Let `z in C` and if `A={z:"arg"(z)=pi/4}`and `B={z:"arg"(z-3-3i)=(2pi)/3}`. Then `n(A frown B)=`

Options

A. `1`
B. `2`
C. `3`
D. `0`

Explanation

To solve the problem, we need to find the intersection of two sets defined by the arguments of complex numbers.

### Step 1: Define the sets A and B
- Set A is defined as \( A = \{ z \in \mathbb{C} : \text{arg}(z) = \frac{\pi}{4} \} \).
- This means that all complex numbers \( z \) in set A lie on the ray that makes an angle of \( \frac{\pi}{4} \) (or 45 degrees) with the positive real axis.

- Set B is defined as \( B = \{ z \in \mathbb{C} : \text{arg}(z - 3 - 3i) = \frac{2\pi}{3} \} \).
- This means that all complex numbers \( z \) in set B lie on the ray that starts from the point \( 3 + 3i \) and makes an angle of \( \frac{2\pi}{3} \) (or 120 degrees) with the positive real axis.

### Step 2: Geometric Interpretation
- For set A, the ray at angle \( \frac{\pi}{4} \) can be represented in the complex plane as:
\[ z = r \cdot e^{i\frac{\pi}{4}} \quad \text{for } r > 0 \]
This ray passes through the origin and extends infinitely in the direction of \( \frac{\pi}{4} \).

- For set B, the ray starting from \( 3 + 3i \) at an angle of \( \frac{2\pi}{3} \) can be represented as:
\[ z = (3 + 3i) + r \cdot e^{i\frac{2\pi}{3}} \quad \text{for } r > 0 \]
This ray starts from the point \( (3, 3) \) in the complex plane and extends infinitely in the direction of \( \frac{2\pi}{3} \).

### Step 3: Finding the Intersection of A and B
- To find the intersection \( A \cap B \), we need to determine if there are any points that satisfy both conditions simultaneously.
- The ray from the origin at \( \frac{\pi}{4} \) has the direction vector \( (1, 1) \) (since \( \tan(\frac{\pi}{4}) = 1 \)).
- The ray from \( 3 + 3i \) at \( \frac{2\pi}{3} \) has the direction vector \( (-\frac{1}{2}, \frac{\sqrt{3}}{2}) \) (since \( \tan(\frac{2\pi}{3}) = -\sqrt{3} \)).

### Step 4: Analyze the Rays
- The ray from \( 3 + 3i \) does not intersect with the ray from the origin because:
- The ray from the origin (set A) extends in the first quadrant, while the ray from \( 3 + 3i \) (set B) extends in the second quadrant.
- Therefore, there are no points that satisfy both conditions.

### Conclusion
- The number of points in the intersection \( n(A \frown B) = 0 \).