Surjectivity of $f_\beta(z)=\frac{z-\beta}{z-\bar{\beta}}$

Question

Let $\mathbb{H}=\{z=x+yi\mid y>0\}$ and $\mathbb{E}=\{z\in\mathbb{C}\mid\mid{z}\mid<1\}$.

Let $\beta\in\mathbb{H}$ and let

$$f_\beta(z):=\frac{z-\beta}{z-\bar{\beta}}$$

I have already shown that $f_\beta:\mathbb{H}\rightarrow\mathbb{E}$ is well defined and injective. But I'm getting nowhere regarding the surjectivity of $f_\beta$.

We have to show that for any $z\in\mathbb{E}$ there exists a $z'\in\mathbb{H}$ so that $f_\beta(z')=z.$

Any help is appreciated!

($i$ is the imaginary unit and $\bar{\beta}$ is the complex conjugate of $\beta$)

Answer 1

Note that$$\frac{z'-\beta}{z'-\overline\beta}=z\iff z'=\frac{\overline\beta z-\beta }{z-1}.$$So, what is need to be proved is that if $z\in\mathbb E$, then $\frac{\overline\beta z-\beta }{z-1}\in\mathbb H$. Note that\begin{align}\frac{\overline\beta z-\beta}{z-1}&=\frac{\left(\overline\beta z-\beta\right)\left(\overline z-1\right)}{\left(z-1\right)\left(\overline z-1\right)}\\&=\frac{\overline\beta\lvert z\rvert^2-\overline\beta z-\beta\overline z+\beta}{\lvert z-1\rvert^2}.\end{align}Since $-\overline\beta z-\beta\overline z,\lvert z-1\rvert^2\in\mathbb R$,$$\operatorname{Im}\left(\frac{\overline\beta\lvert z\rvert^2-\overline\beta z-\beta\overline z+\beta}{\lvert z-1\rvert^2}\right)=\frac{\operatorname{Im}\left(\overline\beta\lvert z\rvert^2+\beta\right)}{\lvert z-1\rvert^2}.$$But if you write $\beta$ as $a+bi$, with $a\in\mathbb R$ and $b\in(0,\infty)$, then$$\operatorname{Im}\left(\overline\beta\lvert z\rvert^2+\beta\right)=b-\lvert z\rvert^2b=b\left(1-\lvert z\rvert^2\right)>0.$$

Answer 2

Pick any $u=x+iy$ inside the open unit disc and let $w=\dfrac{2iu}{1-u}$. Then $$ \operatorname{Im}(w) =\operatorname{Re}(-iw) =2\left(\operatorname{Re}\frac{1}{1-u}-1\right) =2\left(\operatorname{Re}\frac{1-\bar{u}}{|1-u|^2}-1\right) =2\left(\frac{1-x}{(1-x)^2+y^2}-1\right) >-1. $$ By assumption, $b=\operatorname{Im}(\beta)>0$. Let $z=bw+\beta$. Then $$ \frac{z-\beta}{z-\bar{\beta}} =\frac{bw}{bw+\beta-\bar{\beta}} =\frac{bw}{bw+2ib} =\frac{w}{w+2i}=u $$ and $z$ lies on the open upper half plane, because $\operatorname{Im}(z)=b(\operatorname{Im}(w)+1)>0$.