Advanced Math Problem of the Week 12/20/2017

[PF PotW Robot]

Here is this week's advanced math problem of the week. We have several members who will check solutions, but we also welcome the community in general to step in. We also encourage finding different methods to the solution. If one has been found, see if there is another way. Occasionally there will be prizes for extraordinary or clever methods. Spoiler tags are optional.

Consider the open unit disk $\Bbb D\subset \Bbb C$ with Riemannian metric $ds^2 = \dfrac{\lvert dz\rvert^2}{(1 - \lvert z\rvert^2)^2}$. Find a formula for the (Riemannian) distance between two points in $\Bbb D$, and use it to find the distance between $-\frac{1}{2}e^{i\pi/4}$ and $\frac{1}{2}e^{i\pi/4}$

(PotW thanks to our friends at http://www.mathhelpboards.com/)

 

[I like Serena]

Consider the open unit disk $\Bbb D\subset \Bbb C$ with Riemannian metric $ds^2 = \dfrac{\lvert dz\rvert^2}{(1 - \lvert z\rvert^2)^2}$. Find a formula for the (Riemannian) distance between two points in $\Bbb D$, and use it to find the distance between $-\frac{1}{2}e^{i\pi/4}$ and $\frac{1}{2}e^{i\pi/4}$

(PotW thanks to our friends at http://www.mathhelpboards.com/)

Here's my attempt.

Let $c: [a,b] → \Bbb D$ be a parametrized curve in $\Bbb D$, which is differentiable with velocity vector $c'$.
And let $g$ be the given metric.
Then:
$$d(z_1,z_2) = \inf_{c(a)=z_1, c(b)=z_2} \int_a^b \sqrt{g(c'(t),c'(t))}\,dt
= \inf_{c(a)=z_1, c(b)=z_2} \int_a^b \frac{|c'(t)|\,dt}{1-|c(t)|^2}
$$

Due to symmetry and the global minimum of the metric at 0, the shortest path between $-\frac 12e^{i\pi/4}$ and $\frac 12e^{i\pi/4}$ is a straight line.
For that we can pick $c(t)=te^{i\pi/4}$, so that $|c(t)|=|t|$, $|c'(t)| = |e^{i\pi/4}| = 1$, and:
$$\DeclareMathOperator{\artanh}{artanh}
d\left(\!-\frac 12e^{i\pi/4},\frac 12e^{i\pi/4}\!\right) = \int_{-1/2}^{1/2} \frac{dt}{1-|t|^2}
= \artanh t\bigg|_{-1/2}^{1/2} = 2\artanh \frac 12 = 2\cdot \frac 12\log\frac{1+\frac 12}{1 - \frac 12} = \log 3
$$