Finding a Möbius transformation Let $f$ be a holomorphic mapping from {$z:\Re(z)>0$} into itself. Let $1$ be a fixed point of $f$. 
In addition suppose that $\left|\frac{f(2)-1}{f(2)+1}\right|=\frac13$. I want to show that $f(z)=\frac{az+b}{bz+a}$ where $a=1+e^{i\theta}$ and $b=1-e^{i\theta}$ for some $\theta$. 
I tried looking at this problem in different ways, I just don't know what to do. 
Of course if $f(z)=\frac{az+b}{cz+d}$ then I have the obvious $a+b=c+d$. 
And when I look at  $\left|\frac{f(2)-1}{f(2)+1}\right|=\frac13,$ this actually looks like  $\left|\frac{f(2)-f(1)}{f(2)+f(1)}\right|=\frac13,$ 
but I don't know what to do with this last expression or if it is useful at all to write things this way and compute this expression. In general I know that a Möbius transformation has at most 2 fixed points and can be written as a composition of inversions, rotations, dilations and translations. Thanks.
 A: First of all, I can't work in the right half plane, so let me translate the problem to the unit disk using the transformation $\phi(z) = \frac{z-1}{z+1}$ with inverse $\phi^{-1}(w) = \frac{1+w}{1-w}$. Note that $\phi(1) = 0$, $\phi(2) = \frac{1}{3}$ and $\phi(\frac{1}{2}) = - \frac{1}{3}$. In particular, the real line is mapped to itself by $\phi$. Note that $g = \phi \circ f \circ \phi^{-1}$ is a holomorphic map from the unit disk to itself and fixing the origin, so the problem begs for an application of Schwarz's lemma to $g$.
[Edit: if you don't like geometric arguments skip to the next paragraph.] Observe that the condition $\frac{|f(2)-1|}{|f(2)+1|} = \frac{1}{3}$ means that $f(2)$ lies on a circle intersecting the real line orthogonally (it is the  circle of Apollonius with foci $\pm1$ and ratio of radii $1/3$) — in other words, $f(2)$ must lie on the circle $C$ passing through $2$ and $1/2$ as an easy computation shows. Now $\phi(C)$ is the circle around the origin with radius $1/3$ (since $\phi$ sends $C$ to a circle intersecting the real line orthogonally, $\phi(2) = 1/3$ and $\phi(1/2) = -1/3$).
Therefore [or, as Sam pointed out in a comment below, by checking directly that $|g(1/3)| = \frac{|f(2)-1|}{|f(2) + 1|} = \frac{1}{3}$] we have that $g:\mathbb{D} \to \mathbb{D}$ is a  map fixing the origin and sending $1/3$ to some point $g(1/3) = \phi (f(2)) = \frac{e^{i\theta}}{3} \in \phi(C)$. Now Schwarz's lemma tells us that $g(z) = e^{i\theta}z$, and now $f(z) = \phi^{-1}\circ g \circ \phi (z)$ has the desired form as you can compute  immediately.
A: I got this for you to start with : if $|f(2)-1| = \frac 13 |f(2)+1|$, this means that $f(2)-1 = \frac{e^{i \theta}}{3} (f(2) + 1)$, which leads you to 
$$
f(2) = \frac{3+e^{i\theta}}{3-e^{i \theta}}.
$$
I hope you can get something out of this.
A: EDIT - This answer is off-base - I misread upper half plane when right half plane was the hypothesis. 
OLD - 
You should assume that $f$ is injective on the upper half plane.  In that case $f$ will be a linear fractional transformation and you can make progress.  If you don't assume injectivity then the function $f(z) = 1 + 2(z-1) - (z-1)^2$ is a counterexample.
Ooops!  Even if you assume injectivity, consider the linear fractional transformation 
$$ 
f(z) = (z+2)/(-z+4)
$$
having fixed points at $z = 1, 2$.  This satisfies all hypotheses (injective, preserves the upper half plane, fixes $1$, and $|f(2)-1|/|f(2)+1| = 1/3$).  However $f$ cannot be written in the form $(az+b)/(bz+a)$ because $f$ doesn't fix $z = -1$. 
