Why is $\frac{\sinh(t)+i}{\cosh(t)}$ an arc-length parametrization of the circle

Question

This came up on page 40 of Hubbard's Teichmuller Theory (Vol 1), in the context of parametrizing geodesics in the upper half.

Where it is essentially implied that $$\frac{\sinh(t)+i}{\cosh(t)}$$ is an arc-length parametrization of the unit circle.

This is my first time seeing such a parametrization and I was wondering there was a (preferably visual or geometric) way to see this.

Answer 1

At least the first part of the following argument is geometric.

Here are two facts about the Poincaré half-plane model with $d$ denoting the metric:

1. if $x$ and $y$ are positive reals, $d(ix, iy) = |\ln(x) - \ln(y)|$
2. The metric is invariant under linear fractional transformations with real coefficients.

If you don't know these facts, see the Wikipedia link above for 1 (which is easy to prove once you know the integral you have to calculate) and see the Wikipedia page on the Poincaré metric tensor for a proof of 2.

It follows from fact 1 that we can define an arc-length parametrization of the upper half of the imaginary axis by: $$ u(t) = ie^t $$ The linear fractional transformation: $$ f(z) = \frac{z-1}{z+1} $$ maps $0, i, \infty$ to $-1, i, 1$ in that order. Hence (using fact 2), $f$ is an isometry from the upper half of the imaginary axis to the upper half of the unit circle. It follows that: $$ v(t) = f(u(t)) $$ gives an arc-length parametrization of the upper half of the unit circle.

The rest of the proof is algebra. For real $t$, we have: $$ \begin{align} v(t) &= \frac{ie^t - 1}{ie^t + 1}\\ &= \frac{(ie^t - 1)(-ie^t + 1)}{e^{2t}+1} \tag{A}\\ &= \frac{(ie^t - 1)(-i + e^{-t})}{e^t + e^{-t}} \tag{B} \\ &= \frac{e^t -e^{-t} +2i}{e^t + e^{-t}} \tag{C}\\ &= \frac{\sinh(t) + i}{\cosh(t)} \tag{D} \end{align} $$ where (A) follows by calculating the reciprocal of the denominator, (B) follows by multiplying top and bottom by $e^{-t}$, (C) follows by multiplying out and (D) follows from the definitions of $\sinh$ and $\cosh$.

Answer 2

Note that

$$\frac{\sinh t+i}{\cosh t}=\frac{\sinh t}{\cosh t}+i\frac{1}{\cosh t}$$

and

$$\left(\frac{\sinh t}{\cosh t}\right)^2+\left(\frac{1}{\cosh t}\right)^2=\frac{\sinh^2t+1}{\cosh^2 t}=1$$