If `z = x + iy and 0 le sin^(-1) ((z-4)/(2i)) le (pi)/(2)` then

Options

A. `x = 4, 0 le y le 2`
B. `0 le x le 4, 0 le y le 2`
C. `x = 0, 0 le y le 2`
D. None of these

Answer

To solve the problem, we need to analyze the given inequality involving the complex number \( z = x + iy \) and the expression \( \sin^{-1} \left( \frac{z - 4}{2i} \right) \).

### Step-by-Step Solution:

1. **Understanding the given inequality**:
We have:
\[ 0 \leq \sin^{-1} \left( \frac{z - 4}{2i} \right) \leq \frac{\pi}{2} \]
This means that the expression \( \frac{z - 4}{2i} \) must lie within the range of the sine function, which is between 0 and 1.

2. **Expressing \( z \)**:
Substitute \( z = x + iy \):
\[ \frac{z - 4}{2i} = \frac{(x + iy) - 4}{2i} = \frac{(x - 4) + iy}{2i} \]
This can be rewritten as:
\[ \frac{(x - 4)}{2i} + \frac{y}{2i} = \frac{y}{2} - \frac{x - 4}{2} i \]

3. **Identifying the real and imaginary parts**:
The real part is \( \frac{y}{2} \) and the imaginary part is \( -\frac{x - 4}{2} \).

4. **Setting the conditions for the sine function**:
Since \( \sin^{-1}(u) \) is defined for \( 0 \leq u \leq 1 \), we need:
\[ 0 \leq \frac{y}{2} \leq 1 \]
This implies:
\[ 0 \leq y \leq 2 \]

5. **Finding the condition on \( x \)**:
The imaginary part must also satisfy:
\[ -\frac{x - 4}{2} \geq 0 \implies x - 4 \leq 0 \implies x \leq 4 \]

6. **Combining the results**:
From the above conditions, we have:
- \( 0 \leq y \leq 2 \)
- \( x \leq 4 \)

7. **Final values**:
The values of \( x \) and \( y \) can be summarized as:
- \( x = 4 \) (fixed value)
- \( y \) can take any value in the range \( [0, 2] \).

### Conclusion:
Thus, the solution to the problem is:
- \( x = 4 \)
- \( 0 \leq y \leq 2 \)