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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.

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Ryker
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1 Answers1

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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 $b=\o c$.

David
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  • Thanks! But how did you get to that inverse image, I don't quite understand how to approach it. – Ryker Feb 04 '15 at 04:10
  • $|(az+b)/(cz+d)|<1$, multiply out, convert modulus to conjugates, do the algebra. – David Feb 04 '15 at 04:16
  • I'm trying to do the algebra now, but what do you mean by multiply out and convert modulus to conjugates? Do you mean starting with $|f(z) \bar{f(z)}| < 1$? That's what I'm trying, but I don't see how you get the form that you do. – Ryker Feb 05 '15 at 00:10
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    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. – David Feb 05 '15 at 00:35
  • I guess you're right, it does hold. I don't quite see how you excluded the possibility that $|a| = |c|$, though. – Ryker Feb 05 '15 at 01:59
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    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. – David Feb 05 '15 at 02:17
  • Ah, yes, basically the set ${z \in \mathbb{C}: Re(z) < -\frac{\delta}{2}}$, right? I guess the last thing that still eluded me was how you obtained $b \bar{b} - d \bar{d} = -1$. I can see $\beta = -1$, but not why the numerator would necessarily be equal to $1$. – Ryker Feb 05 '15 at 02:30
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    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. – David Feb 05 '15 at 03:08