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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.2011-09-30
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    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.2011-09-30
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    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?2011-09-30
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    The mapping $\dfrac{z-z_0}{-z_0 z+1}$ takes $z_0$ to $0$, $1$ to $1$, and $-1$ to $-1$.2011-09-30
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    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?2011-09-30

2 Answers 2

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${}\qquad\qquad\qquad\dfrac{z-z_0}{-z_0 z+1}$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.

continued....

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.2011-09-30
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    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 $z_0$.2011-09-30
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    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?2011-09-30
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    Is that really true?2011-09-30
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    @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](http://en.wikipedia.org/wiki/Cauchy%27s_integral_formula) vanishes. But then the difference is zero everywhere on the unit disk, so the function must equal the identity function.2011-09-30
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    nice! So now we agree that the function is not the identity on the unit circle.2011-09-30
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    Unfortunately, this map doesn't map the circle to itself. Take for instance $z_0=i/2$ and $z=i$. $$\frac{z-z_0}{1-zz_0}=i/3$$ I think the proper map is $$\frac{z-z_0}{1-z\bar{z}_0}$$2011-09-30
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    I don't see why the circle has to go to itself. I need the unit disc to map to itself. In both cases I can't see this.2011-09-30
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    @shaye: the [Maximum Modulus Principle](http://en.wikipedia.org/wiki/Maximum_modulus_principle) insures that the circle has to be mapped to the circle.2011-09-30
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    @robjohn: "shaye" was right: the answer as originally written would not map the circle into itself. The later correction accomplishes that.2011-10-01
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    @Michael: I agree; that was the point of my first comment.2011-10-01
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    What's the difference between the map given by you and the map given by robjohn and Thomas Andrews?2011-10-01
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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.)

So:

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

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 $$\frac{wz+z_0}{1+wz\bar{z}_0}$$2011-09-30
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    @robjohn: there should be a minus in your last formula right?2011-10-01
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    @shaye: no, if you set $w=1$ in both, it is the same as the previous formula with $z_0\to-z_0$.2011-10-01