# Find locus of points of constant distance from the imaginary axis in the Poincaré upper-half plane model

I have been asked to find the locus of points of constant distance $$d$$ from the vertical axis $$\{ \textrm{Re}(z)=0\}$$ in the Poincaré half-plane model in hyperbolic geometry.

Here are my thoughts on this problem: I believe the geodesics in the Poincaré half plane model are such that the point on the imaginary axis of the shortest distance from the point $$a+bi$$ is the point $$bi$$ (as in the Euclidean case). We know that the distance between two points $$A$$ and $$B$$ is given by $$d(A,B)=|\textrm{log}[A,B,X,Y]|$$, where $$X$$ and $$Y$$ are the points of intersection between the line between $$A$$ and $$B$$, and the absolute, and $$[A,B,X,Y]$$ denotes the cross ratio.

My idea was to somehow combine these two facts to arrive at an equation for the locus, but I seem to be at a loss as to how to progress.

I was given the hint that I might take one point of a given distance from the imaginary axis and apply isometries of the half-plane model to generate other points of equal distance. This seems to be an approach quite different from what I initially had in mind, and I cannot see how we might arrive at a general expression by this method. We would want an isometry that preserved the imaginary axis, so this would have to be a reflection about the imaginary axis, but this just gives us the point $$-a+bi$$ from the point $$a+bi$$, and I am sure we can all agree that there needs no ghost come from the grave to tell us that these two points are of equal distance from the imaginary axis. Which isometries can we apply that preserve the imaginary axis yet generate new non-trivial points of equal distance from the axis as a given point?

All help or input would be highly appreciated.

• A hint: the answer is two Euclidean lines reflexive through the imaginary axis. The picture looks like a cone at 0 – Dynamic Mar 4 at 14:47
• @Dynamic That is hardly a hint. That is the solution without the reasoning. From the reflection isometry about the imaginary axis mentioned above, it is clear that the locus should be the union of two curves reflexive about the imaginary axis. You state that these two curves are in fact Euclidean lines. This is what I would like to show. The crux lies in finding, for a given point on one of the lines on either side of the axis, an isometry that preserves the axis and generates a new point on the same side of the axis, and then showing that these two points are collinear with the origin. – Heinrich Wagner Mar 4 at 19:58
• Now I understand. We take a hyperbolic line perpendicular to the imaginary axis and reflect about it. This preserves the imaginary axis and sends a point on a given side of the imaginary axis and of a given distance from it to a second point on the same side and with the same distance to the axis. These two points are indeed collinear with the origin. Hence we get the desired result. Thank you for putting me on the right track. – Heinrich Wagner Mar 4 at 20:44