Let $$a = \text{Im}\left(\frac{1+z^2}{2iz}\right)$$, where z is any non-zero complex number. The set $$A = \{a : |z| = 1$$ and $$z \neq \pm 1\}$$ is equal to:

Given that $$ a = \text{Im}\left(\frac{1+z^2}{2iz}\right) $$ where $$ z $$ is a non-zero complex number, and the set $$ A = \{a : |z| = 1 \text{ and } z \neq \pm 1\} $$, we need to find the set $$ A $$.

Since $$ |z| = 1 $$, we can write $$ z = \cos \theta + i \sin \theta $$ for some real $$ \theta $$, and because $$ z \neq \pm 1 $$, $$ \theta \neq 0, \pi $$ modulo $$ 2\pi $$. The expression for $$ a $$ simplifies as follows:

First, rewrite the expression inside the imaginary part: $$$ \frac{1 + z^2}{2iz} = \frac{1}{2i} \cdot \frac{1 + z^2}{z} = \frac{1}{2i} \left( z + \frac{1}{z} \right). $$$ Since $$ |z| = 1 $$, $$ \frac{1}{z} = \bar{z} $$, so $$ z + \frac{1}{z} = z + \bar{z} = 2 \operatorname{Re}(z) $$, which is real. Let $$ w = 2 \operatorname{Re}(z) $$, so: $$$ a = \text{Im}\left( \frac{1}{2i} \cdot w \right). $$$ Now, $$ \frac{1}{i} = -i $$, so: $$$ \frac{w}{2i} = w \cdot \frac{1}{2i} = w \cdot \left( -\frac{i}{2} \right) = -\frac{i w}{2}. $$$ Since $$ w $$ is real, $$ -\frac{i w}{2} = 0 - \frac{w}{2} i $$, which is purely imaginary. The imaginary part is $$ -\frac{w}{2} $$, so: $$$ a = -\frac{w}{2}. $$$ Substituting $$ w = 2 \operatorname{Re}(z) $$: $$$ a = -\frac{2 \operatorname{Re}(z)}{2} = -\operatorname{Re}(z). $$$ Thus, $$ a = -\operatorname{Re}(z) $$.

With $$ z = \cos \theta + i \sin \theta $$, $$ \operatorname{Re}(z) = \cos \theta $$, so: $$$ a = -\cos \theta. $$$ As $$ \theta $$ varies over all real numbers except $$ 0 $$ and $$ \pi $$ modulo $$ 2\pi $$, $$ \cos \theta $$ varies over $$ [-1, 1] $$ excluding the endpoints $$ \cos 0 = 1 $$ and $$ \cos \pi = -1 $$. Therefore, $$ \cos \theta \in (-1, 1) $$, and: $$$ a = -\cos \theta \in (-1, 1). $$$ Specifically:

• As $$ \theta $$ approaches $$ 0 $$ (but $$ \theta \neq 0 $$), $$ \cos \theta $$ approaches $$ 1 $$ from below, so $$ a $$ approaches $$ -1 $$ from above.
• As $$ \theta $$ approaches $$ \pi $$ (but $$ \theta \neq \pi $$), $$ \cos \theta $$ approaches $$ -1 $$ from above, so $$ a $$ approaches $$ 1 $$ from below.
• At $$ \theta = \frac{\pi}{2} $$, $$ \cos \theta = 0 $$, so $$ a = 0 $$.

Since $$ \cos \theta $$ takes all values in $$ (-1, 1) $$ as $$ \theta $$ varies in $$ (0, \pi) \cup (\pi, 2\pi) $$, $$ a $$ takes all values in $$ (-1, 1) $$. The endpoints $$ a = -1 $$ and $$ a = 1 $$ are excluded because they correspond to $$ z = 1 $$ and $$ z = -1 $$, which are excluded by the condition $$ z \neq \pm 1 $$.

Therefore, the set $$ A = (-1, 1) $$. Comparing with the options:

• A: $$ (-1, 1) $$
• B: $$ [-1, 1] $$
• C: $$ [0, 1) $$
• D: $$ (-1, 0] $$

The set $$ A $$ matches option A.

Hence, the correct answer is Option A.