A stereographic projection related question This might be an easy question, but I haven't been able to up come up with a solution.
The image of the map $$f : \mathbb{R} \to \mathbb{R}^2, a \mapsto (\frac{2a}{a^2+1}, \frac{a^2-1}{a^2+1})$$
is the unit circle take away the north pole. $f$ extends to a function $$g: \mathbb{C} \backslash \{i, -i \} \to \mathbb{C}^2. $$  Can anything be said about the image of $g$?

Note that although $a$ is complex, is valid :
$$\left(\frac{2a}{a^2+1}\right)^2+\left(\frac{a^2-1}{a^2+1}\right)^2= \frac{4a^2}{(a^2+1)^2}+\frac{(a^2-1)^2}{(a^2+1)}=$$
$$\frac{4a^2+a^4-2a^2+1}{(a^2+1)^2}=\frac{(a^4+2a^2+1)}{(a^2+1)^2}=\frac{(a^2+1)^2}{(a^2+1)^2}=1$$
Thus is also an circle
EDIT
Is say, the points of the set $\{g(a)\in \mathbb{C}^2 :a\in \mathbb{C}/ \{\imath,-\imath\}\}$ meet the above.
