How does one move a point in $B(0,1)$ to the origin with a Möbius transformation 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?
 A: ${}\qquad\qquad\qquad\dfrac{z-z_0}{-z_0 z+1}$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.
A: 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 $|a+z_0\bar{z}| = |a| |1+z_0a^{-1}\bar{z}| = |1 + z_0\bar{a}\bar{z}| = |1+a\bar{z_0}z|$)
