Consider the Möbius map $$m(z) = e^{i\theta}\frac{z-\alpha}{\bar{\alpha}z-1},$$ where $\theta$ is a real number. Let $z$ be in the open unit disc $D=\{z\in \mathbb{C}:\left|z\right|<1\}$. What I need to show is that $m(D)=D$. I'm having trouble with inequalities here.
I've tried the following approaches:
$\alpha = a+ib$, $z=x+iy$ and then try to show that $\left|{m(z)}\right|<1$, but failed. Then I tried to use the matrix representation of Möbius maps, multiply by $\begin{bmatrix} x\\ y \end{bmatrix}$ and show that the norm of the resulting vector is less than $1$, but indeed failed anyway.
Can someone please help me solve this problem? Either I don't see how to manipulate the inequality in a certain way, or there's a better way to show this, without inequalities.
Take $|z|<1$; \begin{align*} |m(z)|<1 &\Longleftrightarrow|z-\alpha|^2<|\bar{\alpha}z-1|^2\\ &\Longleftrightarrow(z-\alpha)(\bar z-\bar{\alpha})<(\bar{\alpha}z-1)(\alpha\bar z-1)\\ &\Longleftrightarrow|z|^2+|\alpha|^2-2\Re(\bar z\alpha)<|\alpha|^2|z|^2-2\Re(\bar z\alpha)+1\\ &\Longleftrightarrow(|z|^2-1)(1-|\alpha|^2)<0 \end{align*} which is true iff $|\alpha|<1$, and we assume that. Thus we have proved that $m(D)\subseteq D$.
It can be easily seen that $$ m:\Bbb C\setminus\left\{\frac1{\bar{\alpha}}\right\}\to\Bbb C\setminus\left\{\frac{e^{i\theta}}{\bar{\alpha}}\right\} $$ is invertible, thus, being $\frac1{\bar{\alpha}}\notin D$, $m:D\to m(D)$ is invertible and $m^{-1}:m(D)\to D$ is again a Mobius transformation; and with some computations you will find that your "new" $\alpha$ stays again in $D$, thus, exactly as above, you can see that $m^{-1}(D)\subseteq D$.
Thus applying $m$ to this last one, we get $D\subseteq m(D)$. So we have proved that $m(D)=D$ as wanted.