TITLE: Finding a Mobius transformation which takes any 2 points in upper half plane to points with same imaginary part
QUESTION [1 upvotes]: I have done most of the below question but I am struggling with the justification of the hint given.The problem is:
Given distinct points $z$ and $w$ in $\mathbb H$, we define the perpendicular bisector of the hyperbolic line segment joining $z$ and $w$ to be the set
$P_{zw} = \{s ∈ \mathbb H | d_{\mathbb H}(z, s) = d_{\mathbb H}(w, s)\}$.
Prove that $P_{zw}$ is a hyperbolic line in $\mathbb H$.
[Hint: you can just look at the situation when $Im(z) = Im(w)$.]
So I have proven this is the case for when $Im(z) = Im(w)$ but I cannot see why this is enough. If there was a transformation in $Mob^+(\mathbb H)$ which took and 2 points to 2 different points with the same imaginary part then I can see that the perpendicular bisector, a vertical line, would be taken to a different hyperbolic line by the inverse transformation but I can't see why this transformation must exist!
Any help would be much appreciated

REPLY [2 votes]: Draw the (euclidean) circle that goes through $z$ and $w$ and touches the real line. Say it touches the real line at $a$. Then 
$$
f(x)=\frac{-1}{x-a}
$$will take that euclidean circle to a horizontal euclidean line, and all points on it will therefore have the same imaginary part.
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