Calculating the area of the hyperbolic triangle with vertices at $e^{i\pi/3}$, $i\sqrt2$, $i$ in the Upper Half Plane model

Question

Sketch the geodesic triangle in H with vertices at $e^{i\pi/3}$, $i\sqrt2$, and $i$, and calculate its area.

I'm not sure how to go about this question, any help would be great, thanks.

I know I need to use the formula $\text{Area}= \pi - (\alpha + \beta + \gamma)$.

Answer 1

So all geodesics on the upper half plane are half circles or vertical lines. (Showing that the geodesics in hyperbolic upper half-plane model are half circles)

Then we could do some elementary geometry of circles to measure the three angles.

The three nodes in euclidean coordinates: A $(1/2,\sqrt 3/2)$, B $(0,\sqrt 2)$, C $(0,1)$. Find the three geodesic edges connecting the nodes

1. easy to see $e^{i\pi/3}$ and $i$ has the same norm 1. They are connectec by the unit circle $\|z\|=1$ (Line CA)
2. $i$ and $\sqrt 2i$ are connected by $x=0$ which is a vertical line, it's also a geodesic line. (Line BC)
3. for $\sqrt 2i$ and $e^{i\pi/3}$, we need to find the center of this half-circle. Assume the center coordinate is $(x,0)$ then we have $$ x^2+\sqrt 2^2 = (1/2-x)^2+(\frac{\sqrt 3}{2})^2\\ x=-1 $$ Then we have the (Line AB) a circle centered at Q $(-1,0)$ with radius $\sqrt3$. Let D $(1/2,0)$

Last step is to calculate the angles

1. $\angle C=\pi/2$ unit circle AC, y axis BC intersect at right angle.
2. $\angle B$ is the same as $\angle BQO=\arctan \sqrt 2$
3. $\angle A$ is a difference between two angles, it's equal to $\angle QAO=\angle QAD -\angle OAD = \pi/3 - \pi/6 = \pi/6$.

Now you get all three angles you can compute the area $Area=\pi - (\pi/2+\pi/6+\arctan \sqrt 2)=\pi/3 - \arctan \sqrt 2$