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

$a=\mathrm{Im}\left(\frac{1+z^2}{2iz}\right)$

$\text{ Let, }z=x+iy,\left|z\right|=\sqrt{x^2+y^2}=1\Rightarrow x^2+y^2=1$

$\begin{array}{r}\frac{1+z^2}{2iz}=\frac{1+\left(x^2-y^2+2ixy\right)}{2i(x+iy)}\\ =\frac{2x^2+2ixy}{2i(x+iy)}\\ =\frac{2x(x+iy)}{2i(x+iy)}\\ =\frac{x}{i}=-ix\end{array}$