Geodesic Uniqueness in the Hyperbolic Plane I am studying Hyperbolic Geometry. At this part, I have proved that semicircles and straight lines orthogonals to the real axis are geodesics in the hyperbolic plane. But how I proof that this geodesics are uniques? That it does not exist others geometric places between two different points that minimizes the hyperbolic length? 
I have seen a topic related to this, but I do not know nothing about Riemannian Geometry. If someone could indicate to me some references to this proof(or the idea behind) I will be very grateful! 
 A: There is a general fact in Riemannian geometry which follows from "standard" existence and uniqueness theorems in the study of ODEs.  Mainly, given any point $p$ in a Riemannian manifold $M$ and given any $v\in T_pM$, there is a unique geodesic defined for some open neighborhood of time around 0 starting at $p$ in the direction of $v$.
Assuming this, it's not too hard to show that you've found all the geodesics.  You just need to verify that for any point $p$ in the hyperbolic plane and tangent vector,  either the vector points up/down (so a vertical line geodesic has $v$ as tangent vector), or some semicircle through $p$ hitting the $x$-axis perpendicularly does.
Finally, to show there is a unique geodesic between any two points, one argues as follows:
Given $p$ and $q$, if $q$ and $p$ have the same $x$ coordinate, a vertical line passes through them.  Further, no semicircle hitting the $x$-axis perpendicularly can go through both $p$ and $q$ since the semicircle is the graph of a function, so passes the vertical line test.
Next, given $p$ and $q$ with different $x$ coordinates, it's clear that a vertical line can't connect them, so we need only find all semicircles passing through both perpendicular to the $x$-axis.  Algebraically, proceed by noting the equation of such a circle is $(x-a)^2 + y^2 = r^2$ for some $a$ and $r$.  Plugging in $(x,y) = (p_1,p_2)$ and $(x,y) = (q_1,q_2)$ respectively and setting the two equal to each other (since they both equal $r^2$), one gets $$(p_1-a)^2 + p_2^2 = (q_1-a)^2 + q_2^2.$$
By simplifying this, one sees the $a^2$ cancels out and so one gets an equation linear in $a$, so $a$ is uniquely determined.  (The assumption that $p_1\neq q_1$ means that the equation has the linear piece).  Knowing $a$ easily implies that $r$ is uniquely determined (keeping in mind that $r \geq 0$).  This shows that there is a unique semicircle perpendicular to the $x$-axis going through $p$ and $q$.
A: You can argue "geometrically" as follows: For two points $p$, $q$ on the imaginary axis it is pretty obvious that the segment $[p,q]$ is the unique shortest curve connecting $p$ and $q$. Now if $p$, $q\in H$ are arbitrary there is a Moebius transformation $T: H\to H$ sending $p$ and $q$ to two points $p'$, $q'$ on the imaginary axis. Since $T$ is a hyperbolic isometry it follows that $p$ and $q$ have a unique shortest connection, and since $T$ (now considered as a map $\bar {\mathbb C}\to \bar {\mathbb C}$) maps circles to circles and preserves angles between curves it follows that this shortest connection lies on a circle intersecting the real axis orthogonally.
