The denominator is nonzero because \(|\bar a z|<1\). Expand:

Divide by \(|1-\bar a z|^2\) to obtain \(1-|\phi_a(z)|^2>0\). The inverse is \(\phi_{-a}\) and satisfies the same condition, proving onto as well as into.

Move the first point to the origin and rotate the second to the positive real axis. Any orthogonal cline through the origin must be a diameter: a finite circle through the origin has center-distance equal to its radius, whereas orthogonality to the unit circle would require center-distance squared to equal radius squared plus one. The unique diameter through the second point is the real diameter. Transforming it back proves existence and uniqueness.

No. The boundary is not in the underlying set. If the shared point is an ideal endpoint of both geodesics, they are limiting parallel. Their Euclidean drawings meet at a point which is not a hyperbolic point.

The first circle is wholly inside, so it is a hyperbolic circle. The second is tangent to the boundary at one, so it is a horocycle (remove the tangency point). The horizontal chord crosses the boundary nonorthogonally, so it is a hypercycle.

Conversely, \(z=(w-i)/(w+i)\) has modulus less than one when \(\operatorname{Im}w>0\), because \(|w-i|<|w+i|\). This also proves bijectivity.

Write the real center as c. Equality \((1-c)^2+4=(3-c)^2+4\) gives \(c=2\). The squared radius is \((1-2)^2+4=5\). The geodesic is the upper semicircle \((x-2)^2+y^2=5, y>0\); its ideal endpoints are \(2-\sqrt5,2+\sqrt5\). The horizontal segment between the inputs is not a geodesic. Equal heights do not mean the shortest hyperbolic path is horizontal.

Use a half-plane isometry to send each ideal tangency point to infinity. The horocycles are now horizontal lines \(y=h_1\) and \(y=h_2\), with \(h_1,h_2>0\). If the chosen points have real coordinates \(x_1,x_2\), the map

takes the first line and chosen point to the second. It has real coefficients and positive determinant. Undo the two initial changes of model to obtain the desired isometry.

Approach.

Count real parameters, remembering one complex number has two.

A direct Euclidean motion is \(e^{i\theta}z+b\): one real angle and one unrestricted complex displacement, hence three real parameters. A disk automorphism is \(e^{i\theta}(z-a)/(1-\bar a z)\): one real angle and one complex point \(a\) restricted by \(|a|<1\), again three real parameters. In the latter, \(a\) is the point sent to zero and also occurs conjugated in the denominator; it is not an ordinary Euclidean translation vector.

Approach.

Separate the location of a point from a direction at that point.

In either orientation-preserving group, one may move any point to a chosen point and then prescribe one tangent direction by a rotation. After those choices the motion is unique. In the disk, first use \((z-a)/(1-\bar a z)\) to move \(a\) to zero, then rotate. Both groups therefore have three real degrees of freedom. A second point's direction can be chosen, but its distance from the first cannot be changed arbitrarily.

Approach.

Expand the difference of the two squared moduli.

Direct expansion gives \[1-|T(z)|^2=\frac{(1-|a|^2)(1-|z|^2)}{|1-\bar a z|^2}>0\quad(|z|<1).\] The denominator cannot vanish in the disk because \(|\bar a z|<1\). Hence the image lies in the disk. Conversely solving \(w=T(z)\) gives \(z=(a+e^{-i\theta}w)/(1+\bar a e^{-i\theta}w)\); the same identity shows \(|z|<1\) whenever \(|w|<1\). Thus the map is onto, not merely into.

Approach.

Use set images under composition and inversion.

The identity preserves the disk. If \(S(D)=D\) and \(T(D)=D\), then \((T\circ S)(D)=T(D)=D\), and composition remains Möbius. From \(T(D)=D\) and bijectivity, \(T^{-1}(D)=D\); its inverse is Möbius. Associativity is inherited from function composition. These verify all group conditions.

Approach.

Take \(b=-az_0\) and check the conjugated denominator.

Starting with \(e^{i\theta}(z-z_0)/(1-\bar z_0z)\), take \[a=\frac{e^{i\theta/2}}{\sqrt{1-|z_0|^2}},\qquad b=-az_0.\] Then \(|a|^2-|b|^2=1\), and multiplying numerator and denominator of the alternative form by \(e^{i\theta/2}\sqrt{1-|z_0|^2}\) recovers the original formula. Conversely the condition implies \(a\ne0\) and \(|b/a|<1\). Set \(z_0=-b/a\) and \(e^{i\theta}=a/\bar a\). Dividing numerator and denominator by \(\bar a\) gives the disk formula, so it belongs to the hyperbolic group.

Approach.

Multiply the structured matrices and inspect the lower row.

Represent its maps by \(A=\begin{pmatrix}a&b\\\bar b&\bar a\end{pmatrix}\) with determinant \(1\). The product of two such matrices has the same conjugate-symmetric form: its upper entries are \(ae+b\bar f\) and \(af+b\bar e\). Its determinant remains \(1\). The identity has this form, and \(A^{-1}=\begin{pmatrix}\bar a&-b\\-\bar b&a\end{pmatrix}\) also has it. The corresponding maps therefore contain identity and are closed under composition and inverse.

Approach.

Send the shared ideal endpoint to infinity.

Use a disk-to-upper-half-plane Möbius isometry sending their common ideal point to infinity. The two lines become vertical lines \(x=u<v\). A transversal meeting both must be an upper semicircle of center \(c\) and radius \(R\). At the two intersections, the angles between the upward vertical rays and the transversal segment have cosines \((c-u)/R\) and \((v-c)/R\). Both numbers lie strictly between \(-1\) and \(1\), and their sum is \((v-u)/R>0\). If the angles are \(\alpha,\beta\in(0,\pi)\), this gives \(\cos\alpha> -\cos\beta=\cos(\pi-\beta)\). Since cosine decreases on that interval, \(\alpha<\pi-\beta\), proving the claim. Conformality transfers it back to the disk.

Approach.

Perturb a limiting-parallel configuration; a strict inequality survives a small change.

Begin with the limiting-parallel configuration of Exercise 5 and fix the transversal's two intersection points. Its interior-angle sum is strictly below \(\pi\). Keep the first line fixed and rotate the second by a sufficiently small angle about its intersection with the transversal. Angle sums vary continuously, so sufficiently small rotations preserve the strict inequality. On one side of the limiting position, the second line meets the first inside the disk; on the other, their boundary endpoints separate and the lines are hyperparallel. At the original position they share one ideal endpoint. Draw these three neighboring positions on the same disk: the strict angle inequality alone cannot distinguish their incidence.

Approach.

Prescribe the two ideal endpoints \(e^{\pm i\theta}\).

Put the point at the disk origin. Given \(0<\theta<\pi/2\), take the orthogonal circle with Euclidean center \(c=\sec\theta\) on the real axis and radius \(R=\tan\theta\). Since \(c^2-R^2=1\), it defines a hyperbolic line. Its ideal endpoints satisfy \(x=1/c=\cos\theta\), hence are \(e^{\pm i\theta}\). Its perpendicular from zero is the positive real diameter, while the limiting-parallel rays go to these endpoints. Their angles with the perpendicular are exactly \(\theta\).

Approach.

Normalize the vertex and then one direction.

Move the vertex to zero by a disk automorphism. The two geodesics through zero are diameters and their Euclidean tangent angle remains a right angle because the map is conformal. A rotation then puts the two rays onto the positive real and positive imaginary axes, after choosing the rays in positive orientation. Every unoriented right-angle figure therefore has the same normal form. Ordered oriented pairs must have matching orientation when only orientation-preserving maps are allowed.

Approach.

At the origin, all candidate lines are diameters.

Move the point to zero. The given line is then an arc of a Euclidean circle with center \(c\ne0\) and radius \(R\) satisfying \(|c|^2=R^2+1\). Every geodesic through zero is a diameter. Such a diameter is perpendicular to the circle exactly when it passes through its center; hence there is exactly one candidate, the line through \(0,c\). Its nearer intersection has distance \(|c|-R=1/(|c|+R)<1\), so the perpendicular meets the given hyperbolic line inside the disk. Existence and uniqueness follow and are preserved by the inverse motion.

Approach.

Use the geodesic joining the two ideal endpoints as a barrier.

Move the angle's vertex to zero and make its sides the rays at arguments \(\pm\theta\), where \(0<\theta<\pi/4\). Their ideal endpoints are \(e^{\pm i\theta}\). The geodesic joining those endpoints is the circle of center \(\sec\theta\) and radius \(\tan\theta\). Both open side rays lie in the half-plane on its origin side. A hyperbolic half-plane is geodesically convex: a geodesic segment cannot leave and reenter it, since two distinct geodesics intersect at most once. Thus every segment between side points lies on that origin side. But any real point \(x\) with \(\sec\theta-\tan\theta<x<1\) lies inside the angle on the other side of that geodesic. No such point can lie on one of the joining segments.

Approach.

Send one line to the imaginary axis and solve the circle orthogonality equation.

Use the upper-half-plane model and send the ideal endpoints of the first line to \(0,\infty\), making it the positive imaginary axis. Since the second is hyperparallel, its ideal endpoints \(u<v\) lie on the same side of zero; take \(0<u<v\) after reversing the horizontal direction if necessary. It is the semicircle with center \(c=(u+v)/2\) and radius \(r=(v-u)/2\). A geodesic perpendicular to the imaginary axis must be a semicircle centered at zero. Orthogonality to the second requires its radius \(R\) to satisfy \(c^2=R^2+r^2\), so \(R=\sqrt{uv}>0\) is uniquely determined. Because \(u<R<v\), the circles meet in the upper half-plane. This proves both existence and uniqueness.

Approach.

Compare with the Euclidean chord triangle after putting one vertex at zero.

Move one vertex to zero. Its two sides become radial segments to \(B,C\), with the angle \(\alpha\in(0,\pi)\) at zero. Rotate so the bisector of that angle is the positive real axis. The geodesic joining \(B,C\) is an orthogonal-circle arc bowed toward zero relative to the Euclidean chord \(BC\). At each endpoint its tangent into the arc lies strictly between the radial ray toward zero and the ray along the Euclidean chord. To check the strict position, the orthogonal circle has equation \(|z-c|^2=|c|^2-1\) and therefore \(2\operatorname{Re}(z\bar c)=1+|z|^2\); its interior disk arc is the nearer, convex arc, and its center lies on the opposite side of its chord from zero. Consequently the two hyperbolic angles at \(B,C\) are strictly smaller than the corresponding angles of the Euclidean triangle \(0BC\). Its angle at zero is unchanged. Adding gives \(\alpha+\beta+\gamma<\pi\).

An independent analytic check is the hyperbolic area identity \(A=\pi-\alpha-\beta-\gamma\) for curvature \(-1\). A nondegenerate triangle has positive area, yielding the same strict inequality. This latter proof uses the area theorem developed later.

Approach.

Check each type in a model where its diameter is simple.

A point and a tangent direction determine a unique hyperbolic geodesic: move the point to the disk origin, choose the diameter with that direction, and move back. Thus every smooth cycle has a unique tangent geodesic.

Perpendicularity can be checked in standard forms, since Möbius isometries preserve angles. For a hyperbolic circle centered at zero in the disk, its diameters are radial lines, perpendicular to its Euclidean tangents. For a horocycle \(y=h\) in the upper half-plane, its diameters are vertical geodesics and its tangent direction is horizontal. For a hypercycle represented by a ray of fixed argument in the upper half-plane with axis the imaginary axis, its diameters are the semicircles centered at zero; these meet the ray orthogonally. The normalized forms cover all three cycle types and prove the claim.

Approach.

Use the Euclidean cline through the three vertices.

Three distinct points determine a unique Euclidean cline. If its portion in the disk were a hyperbolic geodesic, the three vertices would be hyperbolically collinear, contrary to nondegeneracy. Hence that portion is a cycle through the vertices. This also covers vertices that are Euclidean-collinear: their cline may be a nongeodesic line, yielding a hypercycle.

Examples in the disk: choose the vertices \(1/2,i/2,-1/2\) on the hyperbolic circle \(|z|=1/2\); choose \(0,(1+i)/2,(1-i)/2\) on the horocycle \(|z-1/2|=1/2\); choose \(-1/2+i/2,i/2,1/2+i/2\) on the hypercycle \(\operatorname{Im}z=1/2\). All vertices lie strictly inside the disk and are not on one hyperbolic geodesic. The last triangle is Euclidean-collinear but hyperbolically nondegenerate.

Approach.

For a hyperbolic example, make the tangent and diameter share one ideal endpoint.

(a) On a Euclidean circle centered at zero, endpoints \(R\) and \(iR\) give a quarter-circle arc: the tangent at \(R\) and diameter through \(iR\) are both vertical.

(b) On \(|z|=r\), \(0<r<1\), in the disk, take \(A=r\). Its tangent geodesic has Euclidean center \(c=(1+r^2)/(2r)\) and radius \(\sqrt{c^2-1}\). Its ideal endpoints are \(e^{\pm i\alpha}\), where \(\cos\alpha=1/c=2r/(1+r^2)\). Choose \(B=re^{i\alpha}\). The diameter through \(B\) shares that ideal endpoint with the tangent at \(A\), so they are limiting-parallel. The minor arc \(AB\) is a fundamental arc.

(c) In the upper half-plane take the horocycle \(y=1\), \(A=1+i\), \(B=i\). The tangent geodesic at \(A\) is \(|z-1|=1\); the diameter at \(B\) is the imaginary axis. They share ideal endpoint \(0\), so the horizontal horocycle arc from \(i\) to \(1+i\) is fundamental.

Approach.

Use uniqueness and invariance of the symmetric point.

Let \(J_C\) denote reflection in the circle \(C\). Möbius invariance of symmetry gives \(T(J_Cz)=J_{T(C)}(Tz)\). If \(T(C)=C\) and \(Tz=z\), the right side is \(J_Cz=z^*\). Hence \(Tz^*=z^*\). This argument includes a point at the center and its symmetric point infinity.

Approach.

Separate an interior fixed point from two boundary fixed points.

Exclude the identity first. If a fixed point lies inside the disk, move it to zero. A disk automorphism fixing zero has form \(e^{i\theta}z\), so is elliptic. A fixed point outside would have a symmetric fixed point inside by Exercise 4, so belongs to this case. Otherwise the fixed points lie on the boundary. A sole fixed point gives a parabolic map. If there are two, send them to \(0,\infty\) in the upper-half-plane model. The conjugated map fixes both, so is \(w\mapsto\lambda w\). Preserving the upper half-plane forces \(\lambda\) to be positive real: preservation of its real boundary makes it real, and a negative value would exchange upper and lower half-planes. The nonidentity map is hyperbolic. None of these cases is loxodromic.

Approach.

Conjugate a rotation, a real translation, and a positive dilation in suitable coordinates.

(a) Let \(\phi(z)=(z-1/2)/(1-z/2)\) and \(\phi^{-1}(w)=(w+1/2)/(1+w/2)\). Then \(T=\phi^{-1}\circ(e^{i\theta}\phi)\), \(\theta\not\equiv0\pmod{2\pi}\), gives every nonidentity rotation about \(1/2\).

(b) The Cayley map \(H(z)=i(1-z)/(1+z)\) sends the disk to the upper half-plane and \(-1\) to infinity. Conjugate all nonzero real translations: \(T=H^{-1}(H(z)+t)\), \(t\in\mathbb R\setminus\{0\}\). Explicitly, \[T(z)=\frac{2iz-t(1+z)}{2i+t(1+z)}.\]

(c) The maps \(T(z)=(z+it)/(1-it z)\), \(0<t<1\), fix \(\pm i\), preserve the disk, and move zero to \(it\). On the imaginary diameter their iteration moves toward \(i\), so they are exactly the forward translations along this oriented axis. Negative \(t\) reverses the direction; \(t=0\) is the identity.

Approach.

Send each horocycle's ideal point to infinity.

For each horocycle choose a disk-to-half-plane isometry taking its ideal point to infinity. It becomes a horizontal line \(y=h_j>0\), and its chosen point has form \(x_j+ih_j\). The half-plane isometry \(w\mapsto(h_2/h_1)(w-x_1)+x_2\) maps the first line and chosen point to the second. Conjugating back proves the assertion.

A single hyperbolic circle is also transitive on its points under rotations about its hyperbolic center. A single hypercycle is transitive under translations along its axis. However, circles of different hyperbolic radii need not be congruent, and hypercycles at different absolute distances from their axes need not be congruent. Horocycles have no corresponding radius or offset obstruction.

Approach.

Normalize one point to zero and use internal tangency.

Exactly two. Move the pair to \(0,r\) in the disk, with \(0<r<1\). A Euclidean circle through zero internally tangent to the unit circle must have center \(c\) and radius \(|c|\), with \(2|c|=1\). To also contain \(r\) requires \(|r-c|^2=|c|^2\), hence \(\operatorname{Re}c=r/2\). Therefore \[c=\frac r2\pm\frac i2\sqrt{1-r^2},\qquad R=\frac12.\] There are exactly two distinct solutions, each giving a horocycle through the two points. Applying the inverse isometry preserves this count. Coincident points, excluded here, lie on infinitely many horocycles.

Approach.

An orthogonal circle to a horizontal line has its center on that line.

The answer depends on the point. Normalize the given horocycle to \(y=1\) in the upper half-plane and write the point as \(P=x+iy\), \(y>0\). A horocycle other than a horizontal line is a circle with center \(a+ir\) and radius \(r>0\), tangent to the real boundary. To cross \(y=1\) orthogonally its center must lie on \(y=1\), so \(r=1\). Requiring it to pass through \(P\) gives \[ (x-a)^2+(y-1)^2=1,\qquad a=x\pm\sqrt{y(2-y)}.\] Thus there are two if \(0<y<2\), one if \(y=2\), and none if \(y>2\). Each circle counted really crosses \(y=1\) at right angles. Horizontal horocycles do not add solutions.

Approach.

At zero, find the two horocycle centers on the normal line.

Yes: exactly two tangent horocycles and one tangent geodesic. Normalize the point to the disk origin and rotate its tangent direction to be horizontal. A horocycle through zero has Euclidean center of modulus \(1/2\) and radius \(1/2\). A horizontal tangent at zero forces its center onto the imaginary axis, giving precisely \(i/2\) and \(-i/2\). Both circles are internally tangent to the unit boundary and tangent to the original circle at zero. The unique geodesic with this tangent direction is the real diameter. Undo the isometry.

Approach.

Compare a horizontal horocycle with circles tangent to the real boundary.

If “parallel” means disjoint inside the hyperbolic plane, there are two natural types: horocycles with the same ideal point, and disjoint horocycles with different ideal points. Normalize one to \(y=1\). Same-ideal horocycles are the distinct horizontal lines \(y=h\). A different-ideal horocycle has center \(a+ir\) and radius \(r\): it is disjoint from \(y=1\) when \(2r<1\), tangent to it inside the plane when \(2r=1\), and intersects it twice when \(2r>1\). Interior tangency is not disjointness.

For a point \(x+iy\) with \(0<y<1\), the horizontal horocycle through it is disjoint from \(y=1\), but so are infinitely many circles through it with \(y/2\le r<1/2\): their centers satisfy \((x-a)^2=2ry-y^2\). Thus Playfair uniqueness fails. For \(y>1\), only the horizontal horocycle through the point is disjoint from \(y=1\), since a finite-boundary tangent circle reaching height \(y>1\) must cross \(y=1\). State the convention explicitly rather than assuming geodesic parallelism transfers unchanged.

Approach.

Organize the comparison by congruence, incidence, perpendicularity and ideal endpoints.

Both are smooth homogeneous curves: an isometry preserving the curve can carry any one of its points to any other. All horocycles are mutually congruent, just as all geodesics are. Both have a well-defined tangent direction at every point and extend without endpoints inside the hyperbolic plane.

The differences are substantial. Two distinct points determine one geodesic but two horocycles. A point admits a unique perpendicular geodesic to a geodesic, whereas perpendicular horocycles to a horocycle may number zero, one or two. A horocycle has one ideal point and nonzero geodesic curvature, whereas a complete geodesic has two ideal endpoints and zero geodesic curvature. Horocycles are not shortest paths between their points. Finally, the disjointness theory of Exercise 11 does not satisfy a uniform Playfair uniqueness rule. These similarities concern symmetry, not a claim that horocycles are geodesics.

Write c=u+iv. From \(|1/2-c|^2=R^2=|c|^2-1\), we obtain \(u=5/4\). Similarly the point i/2 gives \(v=5/4\). Thus \(R^2=25/16+25/16-1=17/8\), and the circle is \((x-5/4)^2+(y-5/4)^2=17/8\). Only its arc with \(x^2+y^2<1\) is the hyperbolic line; its two boundary endpoints are excluded ideal endpoints. The center and radius satisfy orthogonality, which a general circle through the two points would not.

For \(-1<x<1\), \(C(x)=iy\) with \(y=(1+x)/(1-x)>0\). Conversely any y>0 gives \(x=(y-1)/(y+1)\), which lies in that interval. This proves onto, not just into. As x approaches -1 the image approaches zero; as x approaches 1 it tends to infinity. Zero and infinity are ideal endpoints of this half-plane geodesic, not finite-distance points within the model.