I am studying hyperbolic geometry. But I got stuck a problem. My problem is as follows:
Let $\mathbb{H}$ be the upper half plane model in hyperbolic geometry. Let $a$ and $b$ be two points on the imaginary axis (one point say, $a$ below $i$ and the other point $b$ a bit above) such that each points have same hyperbolic distance from $i$. Now assume that hyperbolic distance from $a$ to $i$ is $ l_{m}$. Now my question is:
How to find an explicit Moebius transformation that preserves the imaginary axis and takes $a$ to $b$. (Hint: you have its hyperbolic translation distance $(2l_{m})$).
Please help me. I am looking for a solution using the hint.
Thanking in advanced.
From your hint I am not sure if the distance from $a$ to $i$ is supposed to be $l_m$ or $2l_m$. Let's assume it is $l_m$ (different from the statement of your problem).
Let's assume $a=(0, a)$ (by abusing notation) is below $i=(0, 1)$, then we compute the distance from $a$ to $i$: $$ \int_a^1\frac{dy}{y}=l_m. $$ That implies $-\log a=l_m$, that is $a=e^{-l_m}$. Similarly we see $b=(0, e^{l_m})$.
Recall that on ${\mathbb H}$, the "dilation" $(x, y)\to (rx, ry)$ is an isometry. So $$ \Phi(x, y)=(e^{2l_m}x, e^{2l_m}y) $$ is the transformation you need.