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.

  • $\begingroup$ Prove $m$ maps $|z|=1$ to $|z|=1$ $\endgroup$ – Nosrati Feb 2 '17 at 6:57
  • $\begingroup$ I proved that already, but how does that help with the part I need to prove? $\endgroup$ – sequence Feb 2 '17 at 7:07
  • $\begingroup$ According to open mapping theorem. $\endgroup$ – Nosrati Feb 2 '17 at 7:09
  • $\begingroup$ This kind of fraction is an important "building block" in the so called "Blaschke products". When you have solved your problem have a look at them (en.wikipedia.org/wiki/Blaschke_product); you will find there some supplementary understanding/motivation. $\endgroup$ – Jean Marie Feb 2 '17 at 7:18

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.

  • $\begingroup$ $\alpha$ is not supposed to have any restrictions on its absolute value. Is this because $\frac{1}{\bar{\alpha}}\not \in D$? Also, how do you determine that the image of $m$ is $\mathbb{C}\backslash \left\{\frac{e^{i\theta}}{\bar{\alpha}}\right\}$? And how is $m:D\to m(D)$ defined? $\endgroup$ – sequence Feb 2 '17 at 7:54
  • $\begingroup$ Restrictions on $\alpha$ are necessaries to get $m(D)\subseteq D$. If you take $\alpha$ arbitrarily, the result is false. $\alpha\in D\Leftrightarrow|\alpha|<1\Leftrightarrow\left|\frac1{\bar{\alpha}}\right|>1\Rightarrow\frac1{\bar{\alpha}}\notin D$. $\endgroup$ – Joe Feb 2 '17 at 9:26
  • $\begingroup$ Try to invert $m$ directly: write $m(z)=w$ and find $z$ in function of $w$. In this way you see immediately the image of $m$. Finally, $m:D\to m(D)$ is simply a restriction of $m$, but the map is the same. $\endgroup$ – Joe Feb 2 '17 at 9:31

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