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Let $z_0$ be in the open unit disc $B(0,1)\subset \mathbf{C}$.

Is there a general formula for an automorphism of $B(0,1)$ which sends $z_0$ to the origin?

I find it easier to think about the complex upper half plane $\mathcal{H}$ so I guess one could do the following.

Map $B(0,1)$ bijectively to $\mathcal{H}$ via $$\varphi:z\mapsto \frac{z+1}{iz+1}.$$ Let $\tau_0$ be the image of $z_0$ under this isomorphism. Find a Möbius transformation $\mu$ sending $\tau_0$ to $i$ and define

$$f = \varphi^{-1} \circ \mu \circ\varphi.$$ This is the automorphism we're looking for. The only problem is finding $\mu$. How can I do this explicitly?

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If it's going to send $z_0$ to the origin, it should have $z-z_0$ in the numerator. – Michael Hardy Sep 30 '11 at 19:13
There's more than one such mapping, because if you send $z_0$ to $0$ and then rotate, it leaves $0$ at $0$ and clearly maps the ball to itself and the circle that is the ball's boundary to itself. – Michael Hardy Sep 30 '11 at 19:15
Thank you for your comments. Very helpful. So $f$ should have $z-z_0$ in the numerator, but $f$ can't be taken to be $z-z_0$ because this doesn't map $B(0,1)$ onto itself. So, how does one fix this? You should divide by something in such a way that this maps $B(0,1)$ to itself, right? – shaye Sep 30 '11 at 19:19
The mapping $\dfrac{z-z_0}{-z_0 z+1}$ takes $z_0$ to $0$, $1$ to $1$, and $-1$ to $-1$. – Michael Hardy Sep 30 '11 at 19:20
Cool! So all I need to do to show that this is an automorphism of $B(0,1)$ is show that if $\vert z\vert <1$, we have that $\vert z-z_0\vert < \vert z z_0 -1$, right? – shaye Sep 30 '11 at 19:25
up vote 2 down vote accepted

${}\qquad\qquad\qquad\dfrac{z-z_0}{-z_0 z+1}$WRONG ANSWER; SEE BELOW.

Or else
${}\qquad\qquad\qquad\dfrac{e^{i\theta}(z-z_0)}{-z_0 z+1}$WRONG ANSWER; SEE BELOW.

where $\theta$ is real. The first solution above is the case where $\theta=0$. If I'm not mistaken, the first solution above leaves the circle pointwise fixed. The other ones rotate it.

ERRATUM: I realized I mangled this five minutes after posting it, then couldn't get back here till now. I gave the unique l.f.t. that maps $z_0$ to $0$ and fixes $\pm1$. But I should have given the one that fixes $\pm z_0/|z_0|$. That's the one that would leave the circle invariant. But not pointwise fixed---that's silly. If you fix three points with this kind of mapping, then you fix all points.


So if we want to fix $\pm z_0/|z_0|$ we can do this: $$ \pm\frac{z_0}{|z_0|} \mapsto \pm1 \mapsto \pm1 \mapsto \pm\frac{z_0}{|z_0|} $$ where the first arrow is $z \mapsto z/\left(z_0/|z_0|\right)$ (so this takes $\pm z_0/|z_0|$ to $\pm1$ and takes $z_0$ to $|z_0|$), and the second arrow is the unique linear fractional transformation that fixes $\pm1$ and takes $|z_0|$ to $0$, and the third arrow takes $\pm1$ back to $z_0/|z_0|$. So the second arrow is $$ \frac{z-|z_0|}{-|z_0|z+1}. $$ The third arrow can be dispensed with, since it's just a rotation that leaves the circle invariant and leaves $0$ fixed. So we have $$ \frac{\frac{z}{z_0/|z_0|} - |z_0|}{-|z_0|\left(\frac{z}{z_0/|z_0|}\right) +1}. $$

How do we know this leaves the circle invariant? Because the mapping above, with the three arrows, leaves $\pm z_0/|z_0|$ fixed and maps $z_0$, which is on the line between those two points, to $0$, which is also on the line between those three points. So it's just like a mapping that fixes $\pm1$ and takes a real number to another real number. Mappings that do that are of the form $(az+b)/(bz+a)$ where $a$ and $b$ are real. You can check that this leaves the circle invariant by looking at $u+iv$ where $u,v$ are real and $u^2+v^2=1$ and putting it through this mapping and summing the squares of the real and imaginary parts and getting $1$. There's probably also a slick way to do it.

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The point $i$ doesn't go to $i$ unless $z_0 =0$. So the circle isn't pointwise fixed. I'm also not convinced yet (because of my own lack of skill) that this morphism maps $B(0,1)$ onto itself. – shaye Sep 30 '11 at 19:36
If a function is the identity on the unit circle and is holomorphic on the unit disk, then by Cauchy's integral formula it must also map $z_0$ to $z_0$. – Henning Makholm Sep 30 '11 at 19:41
Do you mean that a function which is the identity on the unit circle and is holomorphic on the unit disc is the identity function on the unit disc? – shaye Sep 30 '11 at 19:44
Is that really true? – shaye Sep 30 '11 at 19:47
@shaye, yes. Namely, if given any such function, subtract the identity function. The difference is holomorphic on the disk, and is identically zero on the unit circle, so the integral in Cauchy's integral formula vanishes. But then the difference is zero everywhere on the unit disk, so the function must equal the identity function. – Henning Makholm Sep 30 '11 at 19:47

It's a little easier to find one that sends $0$ to $z_0$. We can easily see that if: $$f(z) = \frac{az+b}{cz+d}$$ and $$f(0)=z_0$$ then $b/d = z_0$. We can therefore take $b=z_0$ and $d=1$. So now we have:

$$f(z) = \frac{az+z_0}{cz+1}$$

Now if $|z|=1$, you want $|cz+1| = |az+z_0| = |z| |a + z_0 z^{-1}| = |a + z_0 \bar{z}|$ We see that if we take $a=1$ and $c = \bar{z_0}$, we satisfy this condition. (This is because the two expressions, $1+z_0\bar{z}$ an $1+\bar{z_0}z$ are compliments.)


$$f(z) = \frac{z+z_0}{\bar{z_0}z+1}$$

This is the inverse of the Moebius function that you want. Writing $w = f(z)$, we can solve for $z$ and get: $$z = f^{-1}(w) = \frac{w-z_0}{-\bar{z_0}w+1}$$

(You could, in fact, have chosen any $a$ such that $|a|=1$, then $c=a\bar{z_0}$. That's because $|a+z_0\bar{z}| = |a| |1+z_0a^{-1}\bar{z}| = |1 + z_0\bar{a}\bar{z}| = |1+a\bar{z_0}z|$)

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As I commented to Michael Hardy's post, the basic function that maps $z_0\mapsto0$ is $$\frac{z-z_0}{1-z\bar{z}_0}$$ It turns out that this also maps $0\mapsto-z_0$. Of course, the full family that maps to $z_0\mapsto0$ is $$w\frac{z-z_0}{1-z\bar{z}_0}$$ where $|w|=1$, whereas the full family that maps $0\mapsto z_0$ is $$\frac{wz+z_0}{1+wz\bar{z}_0}$$ – robjohn Sep 30 '11 at 23:32
@robjohn: there should be a minus in your last formula right? – shaye Oct 1 '11 at 11:59
@shaye: no, if you set $w=1$ in both, it is the same as the previous formula with $z_0\to-z_0$. – robjohn Oct 1 '11 at 12:06

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