If I have $f(z_{1},z_{2}) = \displaystyle\frac{|z_{1} - z_{2}|}{\sqrt{1+ |z_{1}|^2} \cdot \sqrt{1 + |z_{2}|^2}}$, for $z_{1}, z_{2} \in \mathbb{C}$, how would I show that $f(z_{1},z_{2})$ is a metric?
Upon reading and researching, this appears to be the chordal distance (or metric). Despite this, I would still like to verify it. So we need show show $4$ things. First off its clearly $\geq$ to zero since we are squaring everything as well as taking absolute value. Its also clearly zero if both $z_{1}$ and $z_{2}$ are zero. Symmetry is also clear. Just like any other metric problem the triangle inequality is giving me some trouble. I have no idea how to go about it. The help would be appreciated!
Hint: This metric is nothing more than the Euclidean metric between the two points that $z_1$ and $z_2$ are mapped to on the unit sphere by stereographic projection.
Let $P_1,P_2\in\Bbb R^3$ be the points on the unit sphere to which $z_1$ and $z_2$ are mapped to respectively, and let $d(P,Q)$ be the standard Euclidean metric on $\Bbb R^3$. then $2f(z_1,z_2)=d(P_1,P_2)$. If you want the details on this, you can look at any standard text on Complex Analysis.
Let $z_3$ be a third point which is projected to $R$ on the unit sphere, this gives
$$2f(z_1,z_3)=d(P,R)\le d(P,Q)+d(Q,R)=2f(z_1,z_2)+2f(z_2,z_3),$$
which implies
$$f(z_1,z_3)\le f(z_1,z_2+f(z_2,z_3).$$