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The Cayley transform gives a conformal bijection between an unbounded half-plane and the unit disc. Two conventions are useful. For the right half-plane, useFor the upper half-plane, useRotating the upper half-plane onto the right half-plane by $z\mapsto -iz$ explains the relationship $C(z)=R(-iz)$. The factors of $i$ distinguish which half-plane is mapped onto the unit disc. ^[ref]

In particular, $C(i)=0$, $C(2i)=1/3$ and $C(1+i)=(1-2i)/5$. The whole positive imaginary axis becomes the real diameter: $C(iy)=(y-1)/(y+1)$ for $y>0$. Horizontal grid lines become circles tangent to the unit circle at $1$ when completed by their endpoint at infinity; they are not hyperbolic geodesics.

On the Riemann sphere, $C$ maps the extended real line onto the unit circle:Every finite real $x$ has $|C(x)|=1$; the remaining point $w=1$ corresponds to infinity. The pole $z=-i$ is outside the upper half-plane, and the pole of $C^{-1}$ at $w=1$ is on the boundary of the unit disc. Neither pole obstructs the interior conformal bijection.

For a short verification, subtract the squared moduli:These identities put each interior image in the required domain; solving $w(z+i)=z-i$ gives the stated inverse function. Both maps are holomorphic, with $C^{\prime }(z)=2i/(z+i{)}^2\ne 0$, so the conformal bijection preserves Euclidean angles between tangent directions.

Euclidean distance is not preserved: $i$ and $2i$ are one unit apart, whereas $0$ and $1/3$ are only one third of a unit apart. The striking feature is that $C$ does preserve length for the hyperbolic metrics of the Poincaré half-plane model and Poincaré disk model:Consequently it is an isometry for these hyperbolic metrics, including their induced hyperbolic distances. For the same example, the corresponding hyperbolic geodesic segments have equal length:Thus the Cayley transform lets us choose the most convenient picture: a straight boundary for calculations, or a bounded unit disc for visualization. It also transports automorphisms of the unit disk into upper half-plane automorphisms by function composition $C^{-1}\circ g\circ C$.

For points $ib$ and $a+ib$ with $a,b>0$, the hyperbolic distance is $2\mathrm{arsinh}(a/(2b))$. It equals their Euclidean distance $a$ exactly when the displayed relation holds. The function of $a$ decreases continuously from one to zero: its derivative has the sign of $\sinh (a/2)-(a/2)\cosh (a/2)<0$. Hence every height $b\in (0,1)$ has exactly one positive matching horizontal separation.

A geodesic segment is the restriction of a geodesic to a compact parameter interval. In the hyperbolic plane, two points determine a unique geodesic segment, which minimizes the hyperbolic distance between them.

A hyperbolic circle of centre $o$ and radius $R>0$ is the level set $\{x:d_{\mathbb{H}}(o,x)=R\}$ for the hyperbolic distance. Equally spaced directions from its centre give vertices of a regular geodesic polygon.

A hyperbolic disc consists of points within a specified hyperbolic distance of its centre. In the Poincare disc model, the disc centred at zero with radius $\rho$ has Euclidean radius $\tanh (\rho /2)$; an isometry transports this description to any other centre.

For $z_1,z_2$ in the Poincare disc model, putTheir hyperbolic distance is

Under the Cayley transform between the half-plane and disk, a radial segment ending at a boundary point becomes a vertical half-ray. If a point is within hyperbolic distance $k$ of that half-ray, it lies in a fixed angular sector. Indeed, in the complex upper half-plane, $\cosh \rho (w,iy)=(|w{|}^2+y^2)/(2y\mathrm{Im}w)\ge |w|/\mathrm{Im}w$. This converts bounded hyperbolic approach to a boundary point into nontangential approach, allowing normal-family proof of angular boundary convergence to transfer radial limits of bounded holomorphic functions.

The curvature-minus-one Poincaré disk model has Riemannian metricThe geodesics through zero are Euclidean diameters. Indeed, for a path written in polar coordinates, its length is at least $\int 2|\dot{r}|/(1-r^2)dt$, and hence at least $2\mathrm{artanh}r$ between zero and a point of modulus $r$. The radial segment attains this lower bound. By uniqueness of a geodesic with given initial tangent, these radial paths describe all geodesics through the origin. The radial hyperbolic distance isThus a hyperbolic circle in the Poincare disc is the indicated Euclidean circle.

An isometry from the Poincaré half-plane model to the Poincaré disk model is the Cayley transform between the half-plane and disk,Writing $z=x+iy$, one has $1-|w{|}^2=4y/|z+i{|}^2$ and $|dw{|}^2=4|dz{|}^2/|z+i{|}^4$. Substitution into the disk Riemannian metric gives $d{s}^2=|dz{|}^2/y^2$, the half-plane metric, and $i$ maps to zero.

Put $r=\tanh (\rho /2)$. The inverse image of $|w|=r$ obeysSince $(1+r^2)/(1-r^2)=\cosh \rho$, completing the square yieldsThis proves that the hyperbolic circle in the upper half-plane has Euclidean centre $i\cosh \rho$ and radius $\sinh \rho$.

In the upper half-plane model, a hyperbolic line is either a vertical line or a Euclidean semicircle perpendicular to the real axis. For the vertical line $x=a$, set $R_L(z)=2a-\bar{z}$. For the circle $|z-a|=r$ with real $a$, setThe vertical formula preserves $y$ and Euclidean speed. For the circular formula, $\mathrm{Im}{R}_L(z)=r^2\mathrm{Im}z/|z-a{|}^2$, while its local Euclidean length multiplier is $r^2/|z-a{|}^2$. Thus both preserve $|dz|/\mathrm{Im}z$, the hyperbolic metric, and consequently all curve lengths and distances. Each is an involution preserving the half-plane; each fixes its specified line pointwise and moves points off that line. This establishes the required hyperbolic reflection explicitly.

To factor an arbitrary isometry $g$, first recall why a bisector reflection exchanges any distinct points $P,Q$. The hyperbolic distance formula isEquating the distances to $P$ and $Q$ yields a vertical line or a circle with centre on the real axis, hence a hyperbolic line. Substitution in the reflection formulas above shows that its reflection exchanges $P$ and $Q$. Choose a base point $P$. If $gP\ne P$, compose $g$ on the left with their bisector reflection, obtaining $h$ that fixes $P$; otherwise take $h=g$.

Choose $Q\ne P$. The points $hQ,Q$ have the same distance from $P$, so their bisector contains $P$. If they differ, its reflection corrects $hQ$ while retaining $P$. We obtain an isometry $k$ fixing both $P,Q$ after at most two reflections.

An isometry fixing two points is either the identity or reflection in their joining line. To see this directly, a point with prescribed distances from $P,Q$ lies at the intersection of two hyperbolic circles; these are Euclidean circles in this model and have at most two intersections. When there are two, reflection in the line $PQ$ interchanges them. Pick a third point off that line. Either $k$ fixes it, or composing $k$ with that line reflection fixes it. An isometry fixing three noncollinear points is the identity: the first two distances leave only the reflected pair, and the distance to the third point distinguishes that pair. Thus $k$ is the identity or one additional reflection. Reversing the compositions proves every isometry is a product of at most three hyperbolic reflections.

There are two necessary qualifications to the printed formula. Critical points satisfy $g^{\prime }=0$, not $g^{\prime }=1$. At a point with $g^{\prime }=1$, the local degree is one and the printed summand is zero. In addition, $\log \Vert g^{\prime }(0)\Vert$ is finite only when $g^{\prime }(0)\ne 0$; constant maps and origin-critical maps need separate treatment.

These are genuine issues even for otherwise ordinary disc maps. The map $g(z)=(z-1/2{)}^2/4$ maps the unit disc into itself and has $g^{\prime }(0)\ne 0$. Its derivative never equals one in the disc, so the printed count is zero, but it has a ramification point at $1/2$ of multiplicity one. For $R>1/2$ the correct identity has the additional term $\log (2R)$, proving that the displayed identity with the literal printed count is false. Also $g(z)=z^2$ shows the undefined basepoint logarithm in an origin-critical example.

Prove the intended identity first for a nonconstant map with $g^{\prime }(0)\ne 0$. Let $H(z)=2|g^{\prime }(z)|/(1-|g(z){|}^2)$ be its hyperbolic derivative density and $u=\log H$. Away from derivative zeros, $\log |g^{\prime }|$ is harmonic, and direct differentiation givesAt a zero $a$ of $g^{\prime }$ of multiplicity $m_a$, factor $g^{\prime }=(z-a{)}^{m_a}h$ with $h(a)\ne 0$. Its logarithm contains $m_a\log |z-a|$, whose distributional Laplacian is $2\pi m_a{\delta }_a$. ThusThe local map degree is $m_a+1$, so this is the required excess-degree weighting.

For a smooth function, differentiating its circular mean and applying the divergence theorem gives ${\overline{u}}^{\prime }(r)=(2\pi r{)}^{-1}{\int }_{|z|<r}\Delta udA$. Integrating in $r$ gives the logarithmic Green-Jensen formula. Factoring the isolated logarithmic singularities extends it to the present $u$:Substitution yields the critical-point Jensen identity for the hyperbolic derivativewhere $N_{\mathrm{crit}}={\sum }_{|a|<R:g^{\prime }(a)=0}{m}_a\log (R/|a|)$. This is the requested formula with the critical-point typo corrected. Initially choose a circle containing no ramification point on its boundary; continuity and integrability of logarithmic boundary singularities extend the result to the remaining radii.

By Schwarz-Pick theorem, $H(z)\le 2/(1-|z{|}^2)$. ThereforeThe normalized hyperbolic distance is $\rho (R)=\log ((1+R)/(1-R))$, andConsequentlyOne can take $C_1=1$ and $C_2=\log (2/H(0))\ge 0$. Since the corrected count is nonnegative in the noncritical-basepoint case, this also proves the growth bound for $T_{\mathbb{D}}$ with the literal printed, identically zero count, though not its false identity.

For completeness, if $g^{\prime }$ vanishes to order $m\ge 1$ at zero, put $c={\lim }_{z\to 0}H(z)/|z{|}^m>0$ andApply the same formula to $\log H-m\log |z|$, which is regular at zero. It givesThe origin contribution is a regularized Jensen term, not the undefined $\log (R/0)$. If one instead omits the origin term, the upper bound gains $-m\log R$; this is bounded for $R\ge 1/2$, while the characteristic and off-origin count are bounded on any smaller compact disc. Thus a bound $C_1\rho +C_2$ still holds for that nonnegative-count convention after enlarging $C_2$. For a constant map $H=0$ and $T_{\mathbb{D}}=0$, so its characteristic growth bound is trivial; the logarithmic identity and a finite critical-divisor count are not defined. These qualifications resolve all maps allowed by the original opening sentence without asserting an undefined equality.

In the Poincare disc model, hyperbolic lines are diameters and circular arcs meeting the boundary circle orthogonally. In the upper half-plane model, they are vertical lines and semicircles with centres on the real axis. The metric is conformal, so hyperbolic perpendicularity agrees with Euclidean perpendicularity.

Use an isometry of the hyperbolic plane to send an arbitrary line to the positive imaginary axis. For $P=x+iy$ with $x\ne 0$, a point on that axis is $Q=iv$ with $v>0$. The hyperbolic distance formula givesPut $R=\sqrt{x^2+y^2}$. The last expression is $(v+R^2/v)/(2y)$, uniquely minimized at $v=R$, since $v+R^2/v\ge 2R$ with equality only there. Hence the unique closest point is $Q_0=iR$. The semicircle $|z|=R$ passes through $P$ and $Q_0$ and meets the imaginary axis at a right angle. Conversely a semicircle centred at a real number $b$ can meet that axis orthogonally only if $b=0$; a vertical line cannot furnish such a crossing. Thus the perpendicular is unique. Isometries transfer both uniqueness and minimization back to the original line.

For $P$ on the specified second semicircle, $a-r\le x\le a+r$ and $R=|P|\le a+r$. Its perpendicular to the axis is the semicircle just found, with radius $R$. MoreoverSince $t\le \mathrm{artanh}t$ for $0\le t<1$, we obtainThis is uniform over the whole semicircle, including points arbitrarily close to the ideal endpoints.

Finally send any first line of an ultraparallel pair to the axis by an isometry of the hyperbolic plane. The second cannot be a vertical line, since that would share the ideal point at infinity. Its two finite endpoints lie on the same side of zero; otherwise its semicircle would meet the axis, or share the endpoint zero. Reflect horizontally if necessary, so those endpoints are $a-r,a+r$ with $a>r>0$. The uniform estimate above then givesThus ultraparallel hyperbolic lines have strictly positive separation; disjointness in the interior alone would not suffice.

Extend the Möbius transformations to isometries of three-dimensional hyperbolic space. For any interior point $o$, the limit set of a Kleinian group is the set of boundary accumulation points of $Go$. Its definition is independent of $o$: two interior orbits stay at a bounded hyperbolic distance, so have the same boundary limits. The Kleinian limit set is closed and $G$-invariant. Iterating a loxodromic Möbius transformation in the positive and negative directions shows that its two fixed points belong to the Kleinian limit set.

The two fixed-point sets in this problem cannot have exactly one common point. Otherwise conjugate that point to infinity and the other fixed point of the first transformation to zero. The transformations then have the forms $f(z)=\alpha z$, with $|\alpha |>1$, and $g(z)=\beta z+b$, with $b\ne 0$. But $f^{-n}g{f}^n(z)=\beta z+b{\alpha }^{-n}$ is a sequence of distinct group elements converging in the Möbius group. Taking quotients of consecutive elements gives nonidentity elements converging to the identity, contrary to discrete subgroup. Consequently the Kleinian limit set contains at least three points.

Let $\xi$ be any point of the Kleinian limit set, and choose distinct $g_n$ with $g_{n}o\to \xi$. The rank-one convergence of divergent Möbius transformations supplies a subsequence converging to $\xi$ uniformly away from one exceptional boundary point $\eta$. To see the lemma directly, normalize determinant-one representing matrices by their norms. The norms tend to infinity, and a subsequential limit has rank one: its kernel is $\eta$ and its image is the attracting point. In the positive-Hermitian-matrix model of hyperbolic space, $g_{n}o$ has this same image limit, so the attracting point is indeed $\xi$.

Choose distinct ${\zeta }_1,{\zeta }_2$ in the Kleinian limit set other than $\eta$. Then $g_n{\zeta }_1$ and $g_n{\zeta }_2$ both tend to $\xi$, belong to the Kleinian limit set, and remain distinct because $g_n$ is injective. If $\xi$ were isolated, both would eventually equal $\xi$, an impossibility. The limit set has no isolated points.

Use the curvature $-1$ hyperbolic metric $ds=2|dz|/(1-|z{|}^2)$. For the disk Möbius transformation ${\varphi }_w(z)=(z-w)/(1-\overline{w}z)$, set $r_w(z)=|{\varphi }_w(z)|$. The hyperbolic distance satisfiesThe triangle inequality givesThus convergence of the exponential sum is independent of the base point. Taking the base point zero, it is equivalent to the Blaschke condition ${\sum }_n(1-|z_n|)<\infty$. This condition implies that only finitely many terms lie in any compact subdisk, including finitely many repetitions of any point.

To construct the required holomorphic function, for $a\ne 0$ define the Blaschke factorFor $|z|\le R<1$, writing $a=|a|e^{i\alpha }$ givesConsequently, after the finitely many factors corresponding to small $|a|$ are removed, the sum of $|1-b_a|$ converges uniformly on each compact subdisk. Those factors are uniformly close to one, so their logarithms form an absolutely convergent series there. Their product tail is holomorphic and nonvanishing. Including the finitely many removed factors gives the Blaschke productwith exactly the requested zero multiplicities. Each finite product has modulus at most one in the disk, so the limit does too. Hence $f=B/2$ maps into the open disk and has exactly the prescribed zeros, proving the implication from the summability condition.

A representative of an element of $G$ is a determinant-one matrix $\left(\begin{array}{cc}a& b\\ \overline{b}& \overline{a}\end{array}\right)$ over the Gaussian integers, considered up to its common sign. It acts by a Möbius transformation of the disk. Since $T(0)=b/\overline{a}$ and $|a{|}^2-|b{|}^2=1$, we haveOnly finitely many Gaussian-integer pairs $(a,b)$ can have $T(0)$ in a fixed compact subdisk. More generally, if $T$ carries some point of a compact set $K$ into $K$, and $K$ lies in the hyperbolic ball of radius $R$ about zero, the triangle inequality gives $\rho (0,T0)\le 2R$. Thus there are only finitely many such $T$. This proves the properly discontinuous group action, and in particular discreteness. The specified $A,B$ belong to $G$, so their subgroup $H$ is discrete as well.

The fixed-point equations reduce to $-i(z-1{)}^2=0$ for $A$ and $i(z+1{)}^2=0$ for $B$. Thus their unique fixed points are respectively $1$ and $-1$ on the ideal boundary. Neither transformation is the identity; their displayed determinant-one representatives have trace two. They are therefore parabolic Möbius transformations.

To find the Dirichlet domains, use the Möbius transformationIt takes the disk to the upper half-plane, and direct substitution givesFor $w=x+iy$ and $v$ in the upper half-plane, the hyperbolic distance obeys $\cosh \rho (w,v)=1+|w-v{|}^2/(2y\mathrm{Im}v)$. Since $C(A^{k}0)=i-2k$, the distance inequalities are equivalent toThese reduce to $kx+k^2\ge 0$ for all integers $k$. The cases $k=1,-1$ imply $-1\le x\le 1$, which also suffices for every other $k$. Therefore the cyclic Dirichlet domain becomes the strip $|\mathrm{Re}w|\le 1$.

The two vertical hyperbolic geodesics pull back to arcs joining $1$ to $i$ and $1$ to $-i$. In disk coordinates they are the circles $|z-(1+i)|=1$ and $|z-(1-i)|=1$, restricted to the disk. ThusConjugating by $z\mapsto -z$ gives $B$, so its two hyperbolic geodesics join $-1$ to $i$ and $-1$ to $-i$, andThe requested drawing and their intersection are shown below.

For any $z_0$, the orbit points satisfying $\rho (0,T{z}_0)\le \rho (0,z_0)$ form a nonempty finite set, by proper discontinuity and compactness of closed hyperbolic balls. Choose a closest one, $q=T{z}_0$. For every integer $k$, its minimality givesand the same inequality with $B$. Hence $q$ lies in the ideal hyperbolic quadrilateralIts ideal vertices are $1,i,-1,-i$.

For the quotient identification, it remains to justify that this covering region is a fundamental polygon. The four open corner caps excluded by $P$ are pairwise disjoint inside the disk. The transformation $A$ carries the exterior of the lower-right cap into the upper-right cap; $A^{-1}$ reverses the pairing. Similarly $B$ carries the exterior of the upper-left cap into the lower-left cap, and $B^{-1}$ reverses this. These assertions follow either from the upper-half-plane formulas or by mapping their boundary circles and testing zero: $A0=(1+i)/2$ and $B0=(-1-i)/2$.

Apply a reduced word in $A^{\pm 1},B^{\pm 1}$ from right to left to a point in the interior of $P$. The first letter puts it into its target cap. Every subsequent letter can do the same, since its inverse cap is distinct from the preceding target cap unless the word cancels. Thus the final point lies in the target cap of the leftmost letter, never in the interior of $P$. This ping-pong lemma argument proves freeness and disjointness of distinct translated interiors. Together with orbit coverage it makes $P$ a fundamental polygon. The corresponding argument on closed caps shows that boundary identifications are generated by the stated side pairings.

The generator $A$ pairs the side from $1$ to $-i$ with the side from $1$ to $i$, fixing the ideal endpoint $1$ and sending $-i$ to $i$. The generator $B$ pairs the side from $-1$ to $i$ with the side from $-1$ to $-i$, fixing $-1$ and sending $i$ to $-i$. There are three ideal-vertex classes: $\{1\}$, $\{-1\}$ and $\{i,-i\}$. The first two give parabolic ends from $A,B$; the third does too, since $AB$ is a nonidentity parabolic transformation fixing $i$. Adding one point at each end produces an oriented closed surface with one face, two edges and three vertices. Its Euler characteristic is $1-2+3=2$, so it is a sphere. Removing the three added points gives a thrice-punctured sphere, proving

The Schwarz-Pick lemma says that every holomorphic self-map $F$ of the unit disc contracts pseudohyperbolic distance and therefore hyperbolic distance:Its infinitesimal form isEquality for distinct points in the distance inequality, or at one point in the derivative inequality, holds exactly when $F$ is a disc Möbius transformation.

For completeness, first prove the Schwarz lemma. If $H:\mathbb{D}\to \mathbb{D}$ is holomorphic and $H(0)=0$, the quotient $H(z)/z$ extends holomorphically at zero. On $|z|=r<1$ its modulus is at most $1/r$. The maximum modulus principle, followed by $r\uparrow 1$, gives $|H(z)|\le |z|$ and $|H^{\prime }(0)|\le 1$. Equality at a nonzero point or in the derivative forces that quotient to be a constant of modulus one, again by the maximum modulus principle; hence $H(z)=e^{i\theta }z$.

Now let ${\varphi }_a(z)=(z-a)/(1-\overline{a}z)$ and apply this result to $H={\varphi }_{F(w)}\circ F\circ {\varphi }_w^{-1}$, which fixes zero. The modulus inequality is exactly the displayed distance inequality, and $\rho =\log ((1+\delta )/(1-\delta ))$ is increasing in $\delta$, giving hyperbolic contraction. Also $|{\varphi }_a^{\prime }(a)|=(1-|a{|}^2{)}^{-1}$ and $|({\varphi }_w^{-1}{)}^{\prime }(0)|=1-|w{|}^2$, so the chain rule proves the derivative inequality. The equality statement follows from the rotation case of the Schwarz lemma after undoing the two automorphisms. Conversely a disc automorphism preserves the distance and attains equality.

There is a maximum of the derivative modulus in the given annulus family, which is nonempty because it contains the constant function one. Indeed the functions are bounded by $e$, and Cauchy estimates bound their derivatives at zero. Take a sequence whose derivative moduli approach the finite supremum. By Montel theorem it has a subsequence converging locally uniformly to a holomorphic function $f$, with $f(0)=1$ and $e^{-1}\le |f|\le e$. The upper equality cannot occur at an interior point, because the maximum modulus principle would force a constant of modulus $e$, contrary to $f(0)=1$. The lower equality is excluded by applying the same argument to $1/f$, which is holomorphic because $|f|\ge e^{-1}$. Thus the limit still maps into the open annulus. Locally uniform convergence gives convergence of derivatives by the Cauchy integral formula, so this limit attains the supremum.

To compute it and all extremizers, the nonvanishing function $f$ on the simply connected disc has a holomorphic logarithm $h$ with $h(0)=0$. One can construct it by integrating $f^{\prime }/f$ from zero; differentiation shows $e^h/f$ is constant, hence one. Its real part obeys $-1<\mathrm{Re}h<1$. The mapis a biholomorphism from this vertical strip onto the disc. To check the inverse explicitly, for $|z|<1$ the ratio $(1+iz)/(1-iz)$ has positive real part, so its logarithm with value zero at zero has imaginary part in $(-\pi /2,\pi /2)$. Thereforehas real part between minus one and one, and direct substitution gives $T(s(z))=z$. The strip restriction removes the period ambiguity of the tangent, so these are inverse maps.

Applying the Schwarz lemma to $T\circ h$ gives $(\pi /4)|h^{\prime }(0)|\le 1$. Since $f^{\prime }(0)=h^{\prime }(0)$, the exact maximum isIt is attained by $f(z)=\exp ((4/\pi )\arctan z)$. The equality case requires $T(h(z))=e^{i\theta }z$, and hence all maximizing functions areTheir derivatives are $(4/\pi )e^{i\theta }$, so the extremizer is not unique: there is exactly this circle of distinct functions, with the parameter considered modulo $2\pi$. This is the extremal derivative of a disc map into a symmetric annulus for $L=1$.

First obtain the hyperbolic circle in the upper half-plane. For centre $w=u+iv$, the fractional-linear map $\zeta =(z-w)/(z-\overline{w})$ takes the Poincaré half-plane model isometrically to the Poincare disc model, sending $w$ to zero. Direct differentiation transforms the metric to $4|d\zeta {|}^2/(1-|\zeta {|}^2{)}^2$. A radial segment from zero to modulus $\rho$ has length $2\mathrm{artanh}\rho$. Any path to that modulus has at least this length, by $|d\zeta |\ge |d\rho |$; thus it is the hyperbolic distance.

A circle of hyperbolic radius $R$ therefore satisfiesSquaring and rearranging givesHence it is a circle in Euclidean geometry, with Euclidean centre $u+iv\cosh R$ and radius $v\sinh R$. Its lowest point has height $v{e}^{-R}>0$, so it stays entirely in the upper half-plane. The centres differ whenever $R>0$: for example, the hyperbolic circle centred at $i$ has Euclidean centre $i\cosh R$, rather than $i$.

For hyperbolic side-side-side congruence, use the standard fact that hyperbolic isometries act transitively on ordered point pairs with the same positive separation, and that reflection in a hyperbolic line is an isometry. Map $A,B$ to $A^{\prime },B^{\prime }$. The image of $C$ has prescribed distances $b$ from $A^{\prime }$ and $a$ from $B^{\prime }$, just as $C^{\prime }$ does. There are at most two such points, interchanged by reflection in the line $A^{\prime }{B}^{\prime }$.

For completeness, normalize $A^{\prime },B^{\prime }$ to points on the positive imaginary axis. Their distance circles are Euclidean circles with centres on that axis by the formula above. Subtracting their equations determines one height $y$, and then the abscissa is determined up to its sign. Reflection $z\mapsto -\overline{z}$ interchanges these choices. Therefore either the chosen isometry or its composition with that reflection maps all three labelled vertices correctly. Equal three side lengths imply a hyperbolic isometry between the triangles.

Two sides and an unspecified equal angle do not suffice. Here is an explicit hyperbolic side-side-angle ambiguity. Fix the angle at $A$ to be $\pi /6$, let $\cosh a=3/2$ and $\cosh b=2$, and chooseBoth are positive and distinct. The hyperbolic cosine rule at $A$ becomesSetting $u=e^c$, the condition is $u^2-6u+7=0$, whose two roots are $3\pm \sqrt{2}$. Construct each triangle from sides $b,c_{\pm }$ meeting at angle $\pi /6$; its opposite side is the same $a$ by the cosine rule. The triangles therefore share $a,b,A$ but have different third sides and cannot be isometric. If the common angle is specifically the included angle between the two given sides, the cosine rule determines the third side and side-side-side congruence does apply.

The Poincare disc model is the open unit disc $D=\{z=x+iy:|z|<1\}$ equipped with the Riemannian metricThis normalization gives Gaussian curvature $-1$. In particular, a Euclidean curve has hyperbolic length $\int 2|dz|/(1-|z{|}^2)$; the boundary circle is infinitely far away in hyperbolic distance.

Use an isometry of the Poincare disc model to move the centre to zero. A hyperbolic disc of radius $\rho$ then has Euclidean radius $R=\tanh (\rho /2)$ by the hyperbolic distance formula. The area element of a surface of the Riemannian metric is $4rdrd\theta /(1-r^2{)}^2$, soTherefore the area is independent of the centre, and

Use curvature $-1$. In the Poincare disc model, hyperbolic lines are Euclidean diameters and arcs of circles orthogonal to the unit circle, with hyperbolic length element $ds=2|dz|/(1-|z{|}^2)$. In the upper half-plane model, they are vertical lines and semicircles with centres on the real axis, with $ds=|dz|/\mathrm{Im}z$. Both displayed hyperbolic metrics are positive scalar multiples of the Euclidean metric, so their angles agree with Euclidean angles. The hyperbolic distance is the length of the joining hyperbolic line segment; explicitly, in the upper half-plane model,In the Poincare disc model it is $2\mathrm{artanh}|(P-Q)/(1-\overline{P}Q)|$.

For distinct $P,Q$, a isometry sends their joining hyperbolic line to the imaginary axis, so their images are $ip,iq$ with $p,q>0$. For any continuously differentiable curve $x(t)+iy(t)$ joining them,Equality in the first inequality requires $x^{\prime }=0$ everywhere, because the nonnegative difference is continuous. Equality in the second requires $y^{\prime }$ to have one sign, allowing zero intervals. Thus equality holds precisely for a monotone reparametrisation of the joining hyperbolic segment. Conversely every such reparametrisation gives equality. For $P=Q$, equality means length zero and the constant curve; monotonicity is understood non-strictly.

Two distinct hyperbolic lines are parallel hyperbolic lines if they are disjoint in the plane and have exactly one common ideal endpoint; they are ultraparallel hyperbolic lines if they are disjoint and have no common ideal endpoint. To prove the common perpendicular of ultraparallel hyperbolic lines theorem, send one line to the imaginary axis. An ultraparallel hyperbolic lines second line can, after reflection if necessary, be written as a semicircle with centre $c>0$ and radius $r$ satisfying $c>r$. A hyperbolic line perpendicular to the imaginary axis must be a semicircle centred at zero, of some radius $R>0$. The Euclidean condition for its orthogonality to the second circle isThis has exactly one positive solution, proving existence and uniqueness. Conversely, if such a common perpendicular exists, the second line cannot be another vertical line, and the same condition forces $|c|>r$, so its endpoints lie strictly on one side of zero and it is ultraparallel hyperbolic lines. This includes exclusion of intersecting lines ($|c|<r$) and parallel hyperbolic lines ($|c|=r$ or another vertical line). The statement concerns distinct lines: a line coincident with itself would have many perpendiculars.

A horocycle in the upper half-plane model is either a Euclidean circle tangent to the real axis from above, with the tangent point as its ideal centre of a horocycle, or a horizontal line $y=h>0$, whose ideal centre of a horocycle is infinity. A isometry sending this centre to infinity sends the horocycle to a horizontal line. The hyperbolic lines meeting that horizontal line orthogonally are exactly the vertical lines; therefore the hyperbolic lines orthogonal to a horocycle are exactly those with its ideal centre as an endpoint.

If two horocycles have distinct horocycle centres, the unique hyperbolic line with those two ideal endpoints meets both orthogonally. If they have the same horocycle centre, send it to infinity: both become horizontal lines and every vertical hyperbolic line meets both orthogonally. HenceIn the other case there are infinitely many; intersection or tangency of the two horocycles does not change this classification.

The hyperbolic lines $AB$ and $CD$ are both perpendicular to $BC$. They are therefore ultraparallel hyperbolic lines, and $BC$ is their unique common perpendicular of ultraparallel hyperbolic lines.

The common perpendicular realizes the minimum hyperbolic distance between two ultraparallel lines. This follows directly in the Poincaré half-plane model after a Möbius transformation sends the two lines to concentric semicircles: the supplied distance formula is minimized along their common radial geodesic. The segment $AD$ also joins the two lines, but it is different from $BC$ in a nondegenerate quadrilateral. Hence