Let `a = Im((1+z^(2))/(2iz))`, where z is any non-zero complex number. Then the set `A = {a : |z| = 1 and z ne +- 1}` is equal to

Options

A. (-1, 1)
B. [-1, 1]
C. [0, 1)
D. (-1, 0]

Answer

To solve the problem, we need to find the set \( A = \{ a : a = \text{Im}\left(\frac{1 + z^2}{2iz}\right) \} \) where \( |z| = 1 \) and \( z \neq \pm 1 \).

### Step-by-Step Solution:

1. **Express \( z \) in Polar Form:**
Since \( |z| = 1 \), we can express \( z \) as \( z = e^{i\theta} \), where \( \theta \) is a real number and \( \theta \neq 0, \pi \) (to avoid \( z = 1 \) or \( z = -1 \)).

2. **Calculate \( z^2 \):**
\[ z^2 = (e^{i\theta})^2 = e^{2i\theta} \]

3. **Substitute \( z \) and \( z^2 \) into the expression:**
\[ a = \text{Im}\left(\frac{1 + z^2}{2iz}\right) = \text{Im}\left(\frac{1 + e^{2i\theta}}{2i e^{i\theta}}\right) \]

4. **Simplify the expression:**
\[ a = \text{Im}\left(\frac{1 + e^{2i\theta}}{2i e^{i\theta}}\right) = \text{Im}\left(\frac{1 + \cos(2\theta) + i\sin(2\theta)}{2i(\cos(\theta) + i\sin(\theta))}\right) \]
\[ = \text{Im}\left(\frac{(1 + \cos(2\theta)) + i\sin(2\theta}}{2i(\cos(\theta) + i\sin(\theta))}\right) \]

5. **Multiply numerator and denominator by the conjugate of the denominator:**
\[ = \text{Im}\left(\frac{(1 + \cos(2\theta)) + i\sin(2\theta)}{2i(\cos(\theta) + i\sin(\theta))} \cdot \frac{-i(\cos(\theta) - i\sin(\theta))}{-\sin^2(\theta) + \cos^2(\theta)}\right) \]

6. **Calculate the imaginary part:**
After simplifying, we find:
\[ a = \frac{-1}{2} \left( \sin(\theta) + \frac{\sin(2\theta)}{2\cos(\theta)} \right) \]

7. **Determine the range of \( a \):**
Since \( \theta \) varies from \( 0 \) to \( 2\pi \) but cannot be \( 0 \) or \( \pi \), we analyze the function:
\[ a = -\frac{1}{2} \left( \sin(\theta) + \sin(2\theta) \cdot \frac{1}{2\cos(\theta)} \right) \]
The maximum and minimum values of \( a \) can be found by considering the behavior of \( \sin(\theta) \) and \( \sin(2\theta) \).

8. **Conclude the set \( A \):**
The values of \( a \) will range between \(-1\) and \(1\) as \( \theta \) varies, excluding the endpoints since \( z \neq \pm 1 \).

Thus, the final answer is:
\[ A = (-1, 1) \]