Exercise 1.11 of Eisenbud I'm doing the exercises from Eisenbud's Commutative Algebra with a view toward Algebraic Geometry, and I don't understand part of one of them, ex. 1.11 a):

Exercise 1.11 a: Over $\mathbb{C}$, consider a circle and a parabola represented by $\mathbb{C}[x,y]/(x^2+y^2-1) \cap (y-2-x^2)$ [...]. Show that the "projection from the north pole" gives a bijection (given by rational functions) between the circle minus one point and the line minus two points.

The only line appearing in the exercise is the line $x=0$, but since we can't project the circle from the north pole onto the line, let's assume that the line is $y=0$. We know that the stereographic projection gives a bijection between the circle minus one point and the whole line, so this is not the projection that we are looking for.
Taking a look over the Hints and solutions, we have that the projection is precisely $t \mapsto (4t/(t^2+4), (t^2-4)/(t^2+4))$, which satisfies that it is a bijection between the circle minus $(0,1)$ and $\mathbb{C} \setminus \{2i, -2i\}$.
My questions are the following: where does this map come from? Why is it a stereographic projection?
Thanks.
 A: Let $Z=\{(z,w)\in\mathbb C^2\colon z^2+w^2=1\}\setminus\{(0,1)\}$. The idea is that you take the stereographic projection in $\mathbb R^2$ and simply allow the parameter to be complex. Recall--or compute--the stereographic projection in $\mathbb R^2$ can be stated as a bijection of the line $y=-1$ onto $Z\cap \mathbb R^2$, given by $$f:(t,-1)\mapsto\left(\frac{4t}{t^2+4},\frac{t^2-4}{t^2+4}\right).$$ If we allow $t$ to vary in $\mathbb C\setminus\{\pm 2i\}$ the hope is that it is a bijection onto $Z$.
To construct an inverse, use the other direction of the stereographic projection map in $\mathbb R^2$, which one can calculate as being given by $$g:(z,w)\mapsto 2z/(1-w),$$ and extend to the rest of $Z$.
The maps $f:\mathbb C\setminus\{\pm 2i\}\to \mathbb C^2$ and $g:Z\to\mathbb C$ are inverses where ever it makes sense to say that. We just have to show that the image of $f$ is in the domain of $g$ and vice versa. It is easy to show directly that $f$ maps into $Z$. Also, if $z/(1-w)=i$, where $(z,w)\in Z$, then $z^2=-(1-w)^2$, so $2w=2$, a contradiction. Similarly, $-2i\not\in g(Z)$. Therefore, $g=f^{-1}$ as desired.
