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.