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 \( A = \{ a : a = \text{Im}\left(\frac{1 + z^2}{2iz}\right), |z| = 1, z \neq \pm 1 \} \).

### Step-by-Step Solution:

1. **Express \( z \)**: Since \( |z| = 1 \), we can express \( z \) in terms of an angle \( \theta \):
\[ z = e^{i\theta} = \cos \theta + i \sin \theta \]

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

3. **Substitute \( z \) and \( z^2 \) into the expression**:
\[ \frac{1 + z^2}{2iz} = \frac{1 + (\cos(2\theta) + i \sin(2\theta))}{2i(\cos \theta + i \sin \theta)} \]

4. **Simplify the numerator**:
\[ 1 + z^2 = 1 + \cos(2\theta) + i \sin(2\theta) = (1 + \cos(2\theta)) + i \sin(2\theta) \]

5. **Simplify the denominator**:
\[ 2iz = 2i(\cos \theta + i \sin \theta) = 2i \cos \theta - 2 \sin \theta \]

6. **Combine the fractions**:
\[ \frac{(1 + \cos(2\theta)) + i \sin(2\theta)}{2i \cos \theta - 2 \sin \theta} \]

7. **Multiply by the conjugate of the denominator**:
\[ \text{Conjugate} = 2i \cos \theta + 2 \sin \theta \]
\[ \text{Numerator} = ((1 + \cos(2\theta)) + i \sin(2\theta))(2i \cos \theta + 2 \sin \theta) \]

8. **Expand the numerator**:
\[ = (1 + \cos(2\theta))(2i \cos \theta) + (1 + \cos(2\theta))(2 \sin \theta) + i \sin(2\theta)(2i \cos \theta) + i \sin(2\theta)(2 \sin \theta) \]

9. **Separate real and imaginary parts**:
The imaginary part will be:
\[ \text{Im} = 2\sin \theta(1 + \cos(2\theta)) + 2\sin(2\theta)(-\cos \theta) \]

10. **Use trigonometric identities**:
\[ \cos(2\theta) = 2\cos^2(\theta) - 1 \quad \text{and} \quad \sin(2\theta) = 2\sin(\theta)\cos(\theta) \]

11. **Substitute and simplify**:
After substituting and simplifying, we find:
\[ a = -\frac{1}{2}(x + x^3 + xy^2) \quad \text{where } x = \cos \theta, y = \sin \theta \]

12. **Determine the range of \( a \)**:
Since \( z \neq \pm 1 \), \( \theta \) cannot be \( 0 \) or \( \pi \). Thus, \( a \) varies as \( \theta \) varies from \( 0 \) to \( 2\pi \) excluding \( 0 \) and \( \pi \).

13. **Final Result**:
The set \( A \) is equal to \( (-1, 1) \).