Let $\mathbb{S} = \{ (X,Y,Z) \in \mathbb{R}^3 :X^2 + Y^2+Z^2=1\}$ denote the Riemann sphere. The following problem is taken from Gamelin "Complex Analysis" chapter one:
What is the image of the spherical cap $ C := \{(X,Y,Z) \in \mathbb{S} : A \leq X \leq 1\}$, where $A$ is some fixed real number between $-1$ and $1$, under the stereographic projection $ \pi(X,Y,Z) = \frac{X+iY}{1-Z}$?
Here is my work so far:
( The case $A=1$ is trivial, so assume $A <1$)
We want to find $\pi(C) = \{z \in \mathbb{C} \mid \exists(X,Y,Z) \in C: \pi(X,Y,Z) = z \}= \{z \in \mathbb{C} \mid \pi^{-1}(z) \in C\}$, where the last equality holds by $\pi$ being a bijection. Write $z=x+iy$.
Here, $ \pi^{-1}(z) \in C\iff A\leq \frac{2x}{1+x^2+y^2} \leq 1 \iff (\star) :A \leq \frac{2x}{1+x^2+y^2}$, since the inequality $2x \leq 1+x^2+y^2$ holds for all $x,y \in \mathbb{R}$.
$A=0$: $( \star)$ reduces to $ 0 \leq x$, so in this case the image is $\{z \in \mathbb{C} \mid \Re(z) \geq 0 \}$.
$A >0$: Since $A \neq 0$ in $(\star)$ we have,
$ \frac{2x}{1+x^2+y^2} \geq A \iff Ax^2-2x+Ay^2+A \leq 0 \iff (x-\frac{1}{A})^2 +y^2 \leq \frac{1}{A^2}-1$. Thus the image is the interior and boundary of the circle centered at $\frac{1}{A}$ with radius $\sqrt{\frac{1}{A^2}-1}$. Note $\frac{1}{A^2}-1 \geq 0$ as $A \in (0,1)$.
$A <0$ : By symmetry and the above, here the image is $ \mathbb{C} / \{ $ interior and boundary of circle centered at $ \frac{1}{A}$ with radius $\sqrt {\frac{1}{A^2}-1} \}$.
Is my working correct? Many Thanks!
Best Answer
In the case $A<0$, the image is $\mathbb{C}\setminus \{$ interior
and boundaryof circle centred at $\frac{1}{A}$ with radius $\sqrt{\frac{1}{A^2}-1} \ \}$.Remember that when $A<0$, the inequality $Ax^2-2x+Ay^2+A\leq 0$ is equivalent to the inequality $x^2-2\frac{x}{A}+y^2+1\geq 0$, which is satisfied on the boundary of said circle.
One further quibble. Either the sphere $\mathbb{S}$ needs to have the point $(0,0,1)$ removed, or we need to add a 'point at infinity' to $\mathbb{C}$. In the latter case, the image of the spherical cap will contain the 'point at infinity' whenever $A\leq 0$.