Describe the image of the set $A=\{z \in \mathbb{C}: \Im(z) >0 \}$ on the Riemann sphere.

Question

I'm trying to solve the following question:

Describe the image of set $A=\{z \in \mathbb{C}: \Im(z) >0 \}$ under stereographic projection onto the Riemann sphere.

I know that if I have a complex number $z$, then the stereographic projection $f: \mathbb{C} \to S^2$ is given by $$ f(z) = \left(\frac{z + \overline{z}}{1 + |z|^2},\frac{z - \overline{z}}{i\left(1 + |z|^2\right)},\frac{|z|^2-1}{|z|^2+1}\right) $$ and similarly, given a point $(a,b,c) \in S^2$ I know that the inverse stereographic projection $f^{-1}:S^2 \to \mathbb{C}$ is given by $$ f^{-1}(a,b,c) = \frac{a+ib}{1-c} $$

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My solution

Since I didn't really have a good idea of where to start with this, I decided to make a GeoGebra plot to get a feel for where the numbers with positive imaginary part landed on the sphere. After playing around with it for a while, It seemed to me that all the points landed on the set $$ B=\{(a,b,c) \in S^2: \ |a|\le 1, \ \color{blue}{0< b< 1} \} \subset S^2 $$ So from here, I tried to prove that $f(A) = B$. To accomplish this, I attempted to use double inclusion.

• To show that $f(A) \subseteq B$ I noticed the following. Since $\Im(z) = \frac{z -\overline{z}}{2i}$, this means that the second entry of $f(z)$ can be seen as $b=\frac{2 \Im(z)}{1 + |z|^2}$, and if $\Im(z) >0$ then clearly under the projection I would have $b >0$, which I think is enough to justify that $f(A) \subseteq B$ since $f(z) \in S^2 \ \forall z \in \mathbb{C}$.
• For the second inclusion, I needed to show that given $(a,b,c) \in S^2$ with $b>0$ then $\frac{b}{1-c} >0$. This basically reduces to show that $1-c>0$, since the numerator is already positive by hypothesis. To do this I argue by contradiction, so I assume $1-c \le 0$ which implies $$ 1\le c \implies 1^2 \le 1-a^2 -b^2 \implies a^2 + b^2 \le 0 $$ but since $a^2 + b^2 >0$ this creates a contradiction, and hence our assumption was wrong. This then shows that $\Im\left(f^{-1}(a,b,c)\right)>0$ if $b >0$.

Notes:

1. I use $\color{blue}{b<1}$ instead of $b \le 1$ since the equiality in the latter case only happens if for $z = x + iy$ we would have $2y = x^2 + y^2 +1$, which (I believe) has no solutions for $x,y \in \mathbb{R}$.
2. I say that $a^2 + b^2 >0$ since I think the equality only happens for $(0,0,1)\in S^2$, which corresponds to the point at infinity, which is not on the complex plane.

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I found this problem to be a bit tricky and with a lot of "special cases" in need of consideration, so I'm not sure if my solution might have missed some of these cases or if I argued correctly in the parts I did write.

Could anyone tell me if my attempt is correct? Thank you!

Answer 1

When I say $S_2$, I'll be excluding the point at infinity for simplicity.

I think that $f(A) = \{(a,b,c) \in S_2, y > 0\} = B$. For any complex number $z$, we have that $z - \overline{z} = 2i\Im(z)$ as you noted so the $b$ value of $f(z)$ would be $$b = \frac{2\Im(z)}{1 + |z^2|}.$$ This shows that $f(A) \subseteq B$.

For the reverse inclusion, suppose we had some $(a,b,c) \in S_2$ so that $y > 0$ (and by assumption, $c \neq 1$). By the surjectivity of stereographic projection, there is some complex number $\zeta$ so that $f(\zeta) = (a,b,c)$. Given the formula for $b$ above, we can then conclude that $\zeta \in A$. Thus, $B \subseteq f(A)$.

In conclusion, your reasoning was right but your definition of $B$ is a little redundant. Since stereographic projection is onto the unit sphere, $|a| \leq 1$ for all $(a,b,c) \in S_2$. Also, $b = 1$ is in the image of $A$. Consider $f(i)$ for example.