# Mobius transformations and showing injections, surjections, and bijections

Let $z_0$ be in the open upper half-plane $\mathbb{C}_{+}$. Show that the map $B_{z_0}(z) = \frac{z-z_0}{z-\overline{z_0}}$ is one-to-one from the closed upper half plane onto the closed unit disk. It is bijective from the real line onto the unit circle. And from $\mathbb{C}_{+}$ onto the open unit disk.

For injectivity, $z_1 = z_2 \implies f(z_1) = f(z_2)$. So we can show $\frac{z_2 - z_0}{z_2 - \overline{z_0}} = \frac{z_1 - z_0}{z_1 - \overline{z_0}}$ means $z_1 = z_2$ by algebra probably.

However, how do we relate the points to the closed upper half plane and the closed unit disk? And to the other spaces?

I am still feeling complex analysis frisky.

As I'm too lazy to check, the injectivity of both arguments (I call it "both" because I'm thinking of this as two questions: show that $B_{z_{0}}$ maps $\mathbb{C}_+$ onto the open unit disk, and show that $B_{z_{0}}$ maps the real line bijectively onto the unit circle) is probably straightforward as you described, so I will not show either.

It should be pretty straightforward to show that $B_{z_0}(z)=\tfrac{z-z_0}{z-\bar{z}_0}$ is a map from the open upper half-plane onto the open unit disk as well. Indeed, let $w\in D$ ($D$ being the unit disk). Setting $w=\tfrac{z-z_0}{z-\bar{z}_0}$ and solving for $z$ (should) show $$z=\frac{w\bar{z}_0-z_0}{w-1}.$$ (Here we need not worry that $z$ is undefined, because $w\in D$ implies that $w\neq 1$.) For $z$ to be in the upper half-plane we should have $\text{Im}(z)>0$. Since $\text{Im}(z)=\tfrac{1}{2i}(z-\bar{z})$ we compute \begin{align*} \text{Im}(z)&=\frac{1}{2i}\left(\frac{w\bar{z}_0-z_0}{w-1}-\frac{\bar{w}z_0-\bar{z}_0}{\bar{w}-1}\right)\\&=\frac{1}{2i}\left(\frac{\left(|w|^2\bar{z}_0-\bar{w}z_0-w\bar{z}_0+z_0\right)-\left(|w|^2z_0-\bar{w}z_0-\bar{z}_0w+\bar{z}_0\right)}{|w-1|^2}\right)\\ &=\frac{1}{2i}\frac{|w|^2(\bar{z}_0-z_0)+(z_0-\bar{z}_0)}{|w-1|^2}\\ &=\frac{\text{Im}(z_0)-|w|^2\text{Im}(z_0)}{|w-1|^2}\\ &=\frac{(1-|w|^2)\text{Im}(z_0)}{|w-1|^2}. \end{align*} Since $w\in D$, $(1-|w|^2)>0$, and since $z_0\in\mathbb{C}_+$, $\text{Im}(z_0)>0$. Thus $\text{Im}(z)>0$ and $z\in\mathbb{C}_+$. This shows that $B_{z_0}(z)$ in a bijective map from the open upper half-plane to the open unit disk. Through similar means you should be able to show that the $B_{z_0}$ maps the real line to the unit circle bijectively. These two facts together show that $B_{z_0}$ maps the closed upper half-plane to the closed unit disk.

• Dear Complex Analysis wizard, thank you. Also, TAMU! I'm close to that place :) Commented Dec 10, 2015 at 12:55
• @complexLost Ha. While I appreciate the flattery, just looking at my post history shows there is a lot of complex analysis I do not know! Where are you located? Commented Dec 10, 2015 at 12:56
• Undergrad at UT Dallas and applying to graduate schools now. (I really hope to get into Iowa). Commented Dec 10, 2015 at 13:03
• @complexLost Good luck on your applications and the GRE and all of those horrible things. Deadlines should be approaching soon so hopefully you've got most of it out of the way. Good luck on getting into Iowa, but definitely make sure you have some fall back schools. I wish you the best. Commented Dec 10, 2015 at 13:10