Show that Möbius transformations that preserve the unit disk are of the matrix form $$\begin{bmatrix}a & b \\ \bar{b} & \bar{a} \end{bmatrix},$$ where $|a|^2 - |b|^2 = 1$ and $a,b \in \mathbb{C}$.

I tried approaching this by noting that these transformations must first and foremost preserve the unit circle. So I looked at whether I'd get anything useful out of seeing what a random matrix $\begin{bmatrix}a & b \\ c & d \end{bmatrix}$ would need to satisfy in order for, say, $1, -1$ and $i$ to stay on the unit circle. However, I got nothing useful out of that.

I looked at some threads here already (for example, Möbius Transforms that preserve the unit disk), but none of them seem to help in getting me this form.

Help would be greatly appreciated.

  • $\begingroup$ A Möbius transformation is a linear-fractional transformation, $f(z) = (az + b)/(cz + d)$. Perhaps your matrix entries are related to the constants appearing this rational expression, but it cannot be said that the matrix "is" the Möbius transformation. $\endgroup$ – hardmath Feb 4 '15 at 3:07

First note that your statement of the result is not quite accurate. For example, the Mobius transformation $$f(z)=\frac{az+b}{cz+d}=\frac{iz+0}{0z+i}=z$$ clearly preserves the unit disc even though the corresponding matrix $$\pmatrix{i&0\cr0&i\cr}$$ does not have the form you state. However, multiplying $a,b,c,d$ by a common factor does not alter $f$, and so we may assume that $ad-bc=1$. If we suppose that this is the case, then the result is true.

Now note that if $\beta$ is real then the inequation $$\def\o#1{\overline#1} z\o z-\alpha z-\o\alpha\o z+\beta<0$$ can be written as $$|z-\o\alpha|^2<|\alpha|^2-\beta\ ;$$ this is the open disc with centre $\o\alpha$ and radius $\sqrt{|\alpha|^2-\beta}$ if $\beta<|\alpha|^2$, the empty set otherwise.

A bit of algebra shows that the inverse image of $|f(z)|<1$ is $$z\o z-\alpha z-\o\alpha\o z+\beta<0$$ with $$\alpha=\frac{c\o d-a\o b}{a\o a-c\o c}\ ,\quad \beta=\frac{b\o b-d\o d}{a\o a-c\o c}\ .$$ Setting $|\alpha|^2-\beta=1$ and simplifying shows that this is the unit disc if and only if $$c\o d-a\o b=0\ ,\quad b\o b-d\o d=-1\ .$$ We can therefore write $$a=a(d\o d-b\o b)=\o d(ad-bc)+b(c\o d-a\o b)=\o d\ ,$$ and similarly $b=\o c$.

  • $\begingroup$ Thanks! But how did you get to that inverse image, I don't quite understand how to approach it. $\endgroup$ – Ryker Feb 4 '15 at 4:10
  • $\begingroup$ $|(az+b)/(cz+d)|<1$, multiply out, convert modulus to conjugates, do the algebra. $\endgroup$ – David Feb 4 '15 at 4:16
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    $\begingroup$ Yes, you can start there, though modulus signs are superfluous as $f(z)\overline{f(z)}$ is always a non-negative real. Then just continue with basic algebra that you learned in elementary school. $\endgroup$ – David Feb 5 '15 at 0:35
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    $\begingroup$ I think you will find that if $|a|=|c|$ then the inverse image of the unit disc is of the form $\gamma z+\o\gamma\o z+\delta<0$, which is not a disc but a half plane. $\endgroup$ – David Feb 5 '15 at 2:17
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    $\begingroup$ Actually, $Re(\gamma z)<-\delta/2$. If you substitute the expressions for $\alpha$ and $\beta$ into $|\alpha|^2-\beta=1$ and simplify, I think you should find that $b\o b-d\o d=-1$ drops out. $\endgroup$ – David Feb 5 '15 at 3:08

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