"Technical" doubt about Ahlfors proof related to stereographic projection I have a tiny little doubt related to one proof given in Ahlfors' textbook. I'll copy the statement and the first part of the proof, which is the part where my doubt lies on.
Statement
The stereographic projection transforms every straight line in the $z$-plane into a circle on $S$ which passes through the pole $(0,0,1)$ and the converse is also true. More generally, any circle on the sphere corresponds to a circle or straight line in the $z$-plane.
Proof
To prove this we observe that a circle on the sphere lies in a plane $\alpha_1x_1+\alpha_2x_2+\alpha_3x_3=\alpha_0$, where we can assume ${\alpha_1}^2+{\alpha_2}^2+{\alpha_3}^2=1$ and $0\leq \alpha_0 <1$
I don't understand why it is always the case that the condition $0\leq \alpha_0 <1$ can be satisfied. I mean, a plane can be described as:
$$\Pi: \space n.(v-v_0)=0 \tag{1}$$ where $v$ and $v_0$ are two vectors with endpoints lying on $\Pi$. I know that $n$ is a perpendicular vector to the plane, and I understand that if $n=(\alpha_1,\alpha_2,\alpha_3)$ doesn't satisfy $||n||=1$, then the vector $n'=\dfrac{n}{||n||}$ is a unit vector which also satisfies equation (1).
Equation (1) is the same as $$\space n.v=n.v_0 \tag{2}$$
In this problem, $\alpha_0=n.v_0$, I don't understand why we can always choose $n$ and $v_0$ such that all the conditions said in my previous lines are satisfied.
I put the title "complex-analysis" but I am not sure if it is the proper tag, if anyone can think of a better tag for this post, tell me and I'll change it.
 A: i  was searching  right this question  for my complex analysis course because i was stuck on the same page ( guess we are studying the same thing) but my doubt was on ${\alpha_1}^2+{\alpha_2}^2+{\alpha_3}^2=1$. Anyway, now  that i have understood this equation i think i can give  you another  answer for  why  $ 0 <{\alpha_0} <1 $ . my explanation involve the formula that give you the distance between a point and a plane
$$\frac{|{\alpha_1}x_1 + {\alpha_2}y_1 + {\alpha_3}z_1 +{\alpha_0}|}{\sqrt{{\alpha_1}^2+{\alpha_2}^2+{\alpha_3}^2}}$$
remembering that the sphere has centre in (0,0,0) this formula give you 
$ 1 > ||{\alpha_0}|| / \sqrt { ({\alpha_1}^2+{\alpha_2}^2+{\alpha_3}^2)}  $
but the denominator we know to be  1 .
A: $0 \le a_0< 1$ is satisfied because the plane must intersect the Riemann sphere, whereon the maximum component if any point is bounded by 1.
A: Here's another way to say it. Suppose $n \cdot v = n \cdot v_0$ is the equation of a plane (so $v = (x, y, z)$ are the variables, $n$ is a non-zero vector, and $v_0$ is an arbitrary point on the plane). It turns out $n \cdot v_0$ is $|n|$ times the minimum distance from the origin to the plane (times $\pm 1$).
For proof, let $v_1$ be the point on the plane closest to the origin, or equivalently it's the point you hit on the plane if you travel from the origin in a direction parallel to $n$. My previous comment says $v_1$ (as a vector starting at the origin) is parallel to $n$. From the usual formula for the magnitude of a dot product, $n \cdot v_1 = \pm |n||v_1|$, which I've claimed is $n \cdot v_0$. But this is true since $n \cdot v_1 = n \cdot v_0$.
In particular, if $n$ has length $1$, then $|n \cdot v_0|$ is the distance from the plane to the origin. For Ahlfors, $n \cdot v_0 = \alpha_0$. Since the plane in question intersects the unit sphere, $|n \cdot v_0| < 1$. The rest follows.
