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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

user2973447
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1 Answers1

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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.

Arthur
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