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If a holomorphic function $F$ on the unit disc has positive real part, the Cayley transform between the half-plane and disk $\varphi =(F-F(0))/(F+\overline{F(0)})$ maps the disc into itself and vanishes at zero. The Schwarz lemma gives the displayed derivative bound. Equality holds exactly when $\varphi (z)=\omega z$ with $|\omega |=1$, equivalently $F(z)=i\mathrm{Im}F(0)+\mathrm{Re}F(0)(1+\omega z)/(1-\omega z)$.

The rational function $g(z)=2z/(1-z^2)$ is a conformal bijection from $\{z:\mathrm{Im}z>0,|z|>1\}$ onto $\{w:\mathrm{Im}w>0\}\setminus \{iy:y\ge 1\}$. For a proof, the Cayley transform between the half-plane and disk $f(z)=(z-i)/(z+i)$ maps the source onto the right half of the unit disc. Squaring maps this half-disc bijectively onto $\mathbb{D}\setminus (-1,0]$. Finally $T(s)=i(1-s)/(1+s)$ maps the disc to the upper half-plane and $(-1,0]$ to the excluded ray. The composition $T(f(z{)}^2)$ simplifies to $g(z)$. Each factor is holomorphic with nonzero derivative and bijective on its indicated domain.

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.

Choose the holomorphic square root equal to one at zero on the unit disc. The resulting elliptic integral maps that disc conformally onto the square with vertices $\pm B,\pm iB$, where $B={\int }_0^{1}dt/\sqrt{1-t^4}$. For the inverse Cayley transform between the half-plane and disk $h(z)=i(1+z)/(1-z)$, the square of the derivative of $F_1\circ h$, with $F_1^{\prime }(w)=1/\sqrt{w(1-w^2)}$, is $2i/(1-z^4)$. Thus the two square maps differ by an affine change of coordinate.

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 $\sinh \rho$.

Let $h(\zeta )=i(1+\zeta )/(1-\zeta )$, the inverse Cayley transform between the half-plane and disk. Direct calculation givessoOn the unit disc, choose the holomorphic square root of $1-{\zeta }^4$ equal to one at zero. Both sides define nonvanishing holomorphic derivatives, and their quotient has constant square. The branch chosen in part ii gives $(F_1\circ h{)}^{\prime }(0)=1+i$. ConsequentlyAn affine change of image coordinate preserves conformality, so the lemniscatic integral maps the disc bijectively onto a square.

The normalization specifies the square exactly. Put $B=F(1)={\int }_0^{1}dt/\sqrt{1-t^4}$. The substitution $x=t^2$ gives $A=2B$. The integral satisfies $F(-z)=-F(z)$ and $F(iz)=iF(z)$, so its four corner values are $B,iB,-B,-iB$. ThusIn particular the image is a square rotated through $\pi /4$ relative to the one in part ii, and its side length is $\sqrt{2}B$. These symmetries also give $F_1(i)=(1+i)B$ from the affine identity.

A normal family of holomorphic functions on a domain is one from which every sequence has a subsequence converging locally uniformly to a holomorphic function, or locally uniformly to infinity in the extended convention. For a bounded family, only finite limits occur.

For functions mapping a plane domain $D$ into the unit disc, the Cauchy integral formula supplies uniform derivative bounds on each compact subset. More explicitly, if a closed disc of radius $r$ around $z$ lies in $D$, the Cauchy estimate gives $|f^{\prime }(z)|\le 1/r$. A slightly larger compact neighborhood of any given compact set therefore gives uniform boundedness and equicontinuity there. Apply the Arzelà-Ascoli theorem on a compact exhaustion and take a diagonal subsequence. Its limit is locally uniform; the Cauchy integral formula shows that the limit is holomorphic. This proves normality rather than merely invoking it.

Now let $|g|\le M$ on the complex upper half-plane. Every dilation $g_t(w)=g(tw)$ has the same bound, so the argument just given, after scaling the bound if necessary, proves that $\{g_t:t\ge 1\}$ is a normal family. For any sequence $t_j\to \infty$, every locally uniform subsequential limit $G$ satisfiesThe identity theorem for holomorphic functions forces $G\equiv \ell$. It follows that $g_t\to \ell$ locally uniformly as $t\to \infty$: otherwise a sequence violating convergence on one compact set would have a convergent subsequence with this same constant limit, a contradiction.

For $0<\epsilon <\pi /2$, the arc $K_{\epsilon }=\{e^{i\theta }:\epsilon \le \theta \le \pi -\epsilon \}$ is a compact subset of the complex upper half-plane. Write $z=|z|w$ with $w\in K_{\epsilon }$. Local uniform convergence of the dilations givesFor $\epsilon \ge \pi /2$ the specified sector is empty. This is the same compactness mechanism as the normal-family proof of angular boundary convergence.

To transfer the result to the unit disc, use the Cayley transform between the half-plane and diskIt sends $r\omega$ to $iy$ with $y=(1+r)/(1-r)$, so $g(iy)\to \ell$ as $y\to \infty$. Choose the curvature-minus-one normalization of the hyperbolic metric, for which the disc line element is $2|dz|/(1-|z{|}^2)$ and the half-plane line element is $|dw|/\mathrm{Im}w$. The Cayley transform between the half-plane and disk preserves these metrics. For $w\in \mathbb{H}$ and $y>0$,Thus ${\rho }_{\mathbb{D}}(z,r\omega )<k$ implies $\mathrm{Im}w/|w|>\mathrm{sech}k$. This is the hyperbolic tube around a radial segment estimate. It places $w$ in a fixed sector bounded away from the real axis. Moreover $z\to \omega$ implies $|C_{\omega }(z)|\to \infty$. Applying the sector result givesA different constant normalization of the hyperbolic metric only rescales $k$ and gives the same conclusion.

Let $p:\mathbb{D}\to D$ be a universal covering map. On a small evenly covered neighborhood in $D$, choose a local inverse $s$ of $p$ and setAny two such lifts differ by a deck transformation of the disc. Every automorphism of the unit disk preserves the disc density: apply the derivative Schwarz-Pick lemma to the automorphism and its inverse to get equality. Hence the definition is independent of the local inverse. Uniqueness of the universal cover up to a covering isomorphism gives independence of the chosen cover as well. This is the Poincare metric on a Riemann surface, normalized to curvature $-1$.

For a piecewise continuously differentiable path $\gamma$, define its length by $L_D(\gamma )=\int {\lambda }_D(\gamma )|{\gamma }^{\prime }|$, and the distance by the infimum of lengths of joining paths. Path reversal and concatenation prove symmetry and the triangle inequality. The distance between distinct points is positive: choose a small closed Euclidean disc around one point not containing the other, on which the positive continuous density is bounded below by $c>0$. Every joining path must leave that disc and has length at least $c$ times its radius before doing so. Joining paths of finite length exist because a plane domain is path connected and the density is locally bounded. Thus this construction gives an actual metric, not just a nonnegative path-length expression.

For the punctured unit disc, use $p(w)=e^{iw}$ from the complex upper half-plane. The half-plane line element is $|dw|/\mathrm{Im}w$, while $|dz|=|z||dw|$ and $\mathrm{Im}w=\log (1/|z|)$. ConsequentlyThe domain called an annulus in the paper has inner radius zero, so it is this punctured-disc metric, not the metric of an annulus with positive inner radius.

We now prove the Great Picard theorem. In its holomorphic form, an essential singularity takes every complex value, with at most one exception, infinitely often in every punctured neighborhood. It suffices to prove that a holomorphic function on a punctured disc omitting $0$ and $1$ extends meromorphically through the puncture: if two values occurred only finitely often, a smaller punctured disc would omit both, and an affine change of target would reduce them to $0,1$.

Use the universal covering by the modular lambda function $\lambda :\mathbb{H}\to Y=\mathbb{C}\setminus \{0,1\}$, whose construction is given in the following solution. Its deck transformation group is the projective level-two principal congruence subgroup $\Gamma (2)$. Rescale the source punctured disc to unit radius. Lift $f(e^{2\pi iw})$ to a holomorphic function $L:\mathbb{H}\to \mathbb{H}$. Uniqueness of lifts givesfor a fixed deck transformation $T\in \Gamma (2)$. The Schwarz-Pick lemma givesas $\mathrm{Im}w\to \infty$. A nonidentity deck transformation has no interior fixed point, so it is either a hyperbolic Möbius transformation or a parabolic Möbius transformation. The hyperbolic case is conjugate to a dilation $w\mapsto aw$ with $a>0$, $aeq1$. Since $|d\log \mathrm{Im}w|\le d{s}_{\mathbb{H}}$, its displacement is at least $|\log a|>0$, contradicting this inequality.

If $T$ is the identity, compose $L$ with the Cayley transform between the half-plane and disk. Its periodicity lets the resulting bounded function descend to the punctured disc and extend by the Riemann removable singularity theorem. A nonconstant extension takes its central value inside the disc by the maximum modulus principle. The covering map then gives an extension of $f$; a constant lift gives a constant $f$.

In the parabolic case, its fixed point is a rational modular cusp. Indeed, a parabolic integral matrix of trace $\pm 2$ has a rational eigenvector for its repeated eigenvalue. Conjugate by an element of the modular group sending this cusp to infinity. The level-two subgroup is normal, so the conjugated transformation is $\zeta \mapsto \zeta +2n$ for a nonzero integer $n$. The conjugated lift $P$ consequently satisfiesThe modular transformations permute the three target cusp values $0,1,\infty$, as seen from the lambda transformation formulas below. It is therefore enough to examine $\lambda (P(w))$.

To fix the sign of $n$, the function $\mathrm{Im}P$ is periodic in the real direction. Its mean $M(y)={\int }_0^1\mathrm{Im}P(x+iy)dx$ satisfies $M^{\prime }(y)=2n$ by the Cauchy-Riemann equations and the increment of $\mathrm{Re}P$. Positivity for all $y>0$ forces $n>0$. The functionis well-defined, nonvanishing and bounded by one on the punctured disc, so it extends holomorphically to zero. Its winding number along a circle around the puncture is $n$, because a unit increment in $w$ increases $\pi iP$ by $2\pi in$. The argument principle therefore shows that $Q$ has a zero of order $n$ at the origin.

At the infinite cusp, the modular lambda function has the convergent expansion $\lambda (\zeta )=16q+O(q^2)$ in $q=e^{\pi i\zeta }$. Thus $\lambda (P(w))=16Q(z)+O(Q(z{)}^2)$ extends and tends to zero. Undoing the target permutation gives a removable singularity or a pole for $f$. In every case the puncture is not essential, which proves the holomorphic Great Picard theorem. For a meromorphic function with an essential singularity, if three spherical values occurred only finitely often, a Möbius transformation would make the omitted values $0,1,\infty$. The same proof applies because omission of infinity makes the transformed function holomorphic. Hence the meromorphic version has at most two exceptional spherical values.

The boundary circles intersect at $z=1$ and $z=i$: subtracting their equations gives $x+y=1$, and $x^2+y^2=1$ then gives these two points. Their intersection is the open lens in the first quadrant, bounded by the unit-circle arc from $1$ to $i$ and the corresponding lower-left arc of the circle centred at $1+i$.

To construct a conformal map of a circular lens, apply the Möbius transformationIt sends the intersection points to zero and infinity, so both boundary circles become straight lines through zero. The boundary arc of the circle centred at zero maps to the ray of argument $-\pi /4$; the other arc maps to the ray of argument $+\pi /4$. For example, their arc midpoints $e^{i\pi /4}$ and $(1+i)-e^{i\pi /4}$ give these signs. The interior point $(1+i)/2$ maps to one, selecting the sector $-\pi /4<\arg \zeta <\pi /4$, not the complementary sector. As a Möbius bijection, $\zeta$ maps the whole lens onto that sector.

Squaring maps this sector bijectively onto the right half-plane: arguments double into $(-\pi /2,\pi /2)$, and the inverse is the unique square root with argument in $(-\pi /4,\pi /4)$. ThusThe derivative never vanishes inside the lens: $\zeta$ and its derivative are nonzero there. Finally use the Cayley transform between the half-plane and disk $w\mapsto (w-1)/(w+1)$. Since $|w-1|<|w+1|$ precisely when $\mathrm{Re}w>0$, it maps the right half-plane bijectively to the unit disc. The desired second map isThe denominator cannot vanish in the lens because $F(z)$ has positive real part. The maps send the centre-line test point $(1+i)/2$ to one and zero respectively.

Put $\alpha =-\log |c|>0$. Since $f$ has no zeros and the unit disc is simply connected, it has a holomorphic logarithm. Thus $F=-\log f$ is holomorphic with $\mathrm{Re}F>0$ and $\mathrm{Re}F(0)=\alpha$. The Cayley transform between the half-plane and disk givesThe Schwarz lemma yields $|{\varphi }^{\prime }(0)|\le 1$, so the Carathéodory derivative bound for the right half-plane gives $|F^{\prime }(0)|\le 2\alpha$. Since $f^{\prime }=-F^{\prime }f$, the sharp answer isEquality in the Schwarz lemma holds exactly when $\varphi (z)=\omega z$, $|\omega |=1$. Solving for $F$ and exponentiating gives exactly the extremizersTheir logarithmic real part is $\alpha \mathrm{Re}[(1+\omega z)/(1-\omega z)]>0$, so each is indeed a zero-free map into the unit disc with the prescribed value at zero.

For a fixed $w\in \mathbb{D}$, put $r=|w|$. Apply the Harnack inequality on the unit disk to the positive harmonic function $\mathrm{Re}F$ and then exponentiate its negative. The zero-free Schwarz lemma givesFor $w\ne 0$ both endpoints are attained: take $\omega w=r$ for the lower endpoint and $\omega w=-r$ for the upper endpoint in the displayed extremizers. At zero both are $|c|$. If the question is interpreted as allowing $w$ to vary as well, the sharp infimum and supremum over all admissible pairs $(f,w)$ are respectively $0$ and $1$, approached as $r\uparrow 1$; neither value is attained inside the unit disc. An individual function need not have these global extremal values: the constant function $f=c$ is a simple example.

The intended open region lies inside the larger circle and outside the smaller one. The two circles are tangent at zero, a boundary point, so $w=1/z$ is holomorphic throughout the region. For a circle $|z-ia|=a$, its equation is $|z{|}^2=2a\mathrm{Im}z$, which becomes $\mathrm{Im}w=-1/(2a)$. The side inside that circle becomes $\mathrm{Im}w<-1/(2a)$. Hence the tangent-circle crescent conformal map begins withThe affine coordinate $\zeta =4w+2i$ sends this strip bijectively to $0<\mathrm{Im}\zeta <1$. Exponentiation $\eta =e^{\pi \zeta }$ is then a conformal bijection to the upper half-plane: its argument ranges strictly from zero to $\pi$, and the inverse logarithm has that unique argument. Finally $(\eta -i)/(\eta +i)$ is a Cayley transform between the half-plane and disk onto the unit disc, since for $\mathrm{Im}\eta >0$ its modulus is less than one and the inverse is $\eta =i(1+u)/(1-u)$. As $e^{\pi \zeta }=e^{4\pi /z}$, a complete answer isEvery stage has nonzero derivative and is one-to-one onto its indicated range. The denominator cannot vanish in the upper half-plane. Unlike a region between disjoint nested circles, this tangent crescent has no enclosed hole, so there is no annulus obstruction to mapping it onto a disc.

Apply the Dahlquist test equation and put $z=h\lambda$. Every recurrence mode must satisfyThe amplification polynomial of a multistep method, rather than just its root near one, determines absolute stability. Put $B=2a-1$ and use the Cayley transform between the half-plane and diskFor $\zeta \ne -1$, the characteristic equation becomesIf $B>0$, the denominator cannot vanish at a root of the characteristic equation: $1+B{w}^2=0$ would force $w=0$, a contradiction. Hence $\mathrm{Re}z<0$ forces $\mathrm{Re}w<0$, so all amplification roots have modulus less than one. On the imaginary axis the roots have modulus one and are simple: the transformed quadratic has discriminant $4(1-B{z}^2)>0$ for imaginary $z$. At $z=0$ they are the two simple roots $\pm 1$. Also $\zeta =-1$ cannot be a root when $B>0$ and $z\ne 0$, and the leading coefficient $1-az$ cannot vanish in the closed left half-plane.

If $a<1/2$, the root near $-1$ isFor small negative real $z$ it lies below $-1$, violating absolute stability. The endpoint $a=1/2$ deserves separate treatment:One root is always $-1$; the other is the trapezoidal rule multiplier $(1+z/2)/(1-z/2)$. They are distinct for every finite $z$ in the left half-plane. Consequently, with absolute stability understood as the bounded root condition for a multistep method,There is a convention at this reducible endpoint: if A-stability is defined to require every unreduced recurrence mode to decay for $\mathrm{Re}z<0$, the answer is $a>1/2$, since the $(-1{)}^n$ mode persists at $a=1/2$. Canceling the common factor $\zeta +1$ gives the A-stable trapezoidal rule, but cancellation removes an actual starting-error mode of the original two-step recurrence. The fourth-order member $a=1/3$ is outside either A-stability range.

Fix $z_0\in \mathcal{U}$. We prove the Riemann mapping theorem in this case by maximizing a normalized derivative. Choose $a\notin \mathcal{U}$. Since $\mathcal{U}$ is a simply connected domain and $z-a$ never vanishes, it has a holomorphic square root $g$. This $g$ is injective, and $g(\mathcal{U})$ is disjoint from $-g(\mathcal{U})$: equality up to sign would first force the original points to coincide. Choose $b\in g(\mathcal{U})$ and $r>0$ with $B(b,r)\subset g(\mathcal{U})$. The ball $B(-b,r)$ is omitted, so $1/(g+b)$ is a bounded univalent function. Scaling and composing with an automorphism of the unit disk produces an injective holomorphic function $\varphi :\mathcal{U}\to \mathbb{D}$ with $\varphi (z_0)=0$ and ${\varphi }^{\prime }(z_0)>0$.

Let $\mathcal{F}$ be the family of all such normalized univalent functions. A Cauchy estimate in a small disk about $z_0$ bounds their derivatives, so $M={\sup }_{\varphi \in \mathcal{F}}{\varphi }^{\prime }(z_0)$ is finite and positive. Choose a maximizing sequence in this normal family. Montel theorem gives a subsequence converging uniformly on compact subsets to $\varphi$. Its derivative at $z_0$ is $M>0$. The maximum modulus principle puts its image in $\mathbb{D}$, and Hurwitz's theorem implies that a nonconstant limit of injective holomorphic functions is injective. Thus the maximum is attained.

If $a_1\in \mathbb{D}$ is omitted, then $a_1\ne 0$. Put $T_a(w)=(w-a)/(1-\overline{a}w)$. A holomorphic square root $v$ of $T_{a_1}\circ \varphi$ exists, is injective, and takes values in $\mathbb{D}$. Compose $v$ with $T_{v(z_0)}$ and a rotation to normalize it. Writing $s=|a_1|\in (0,1)$, the new derivative has magnitudea contradiction. Therefore $\varphi$ maps onto $\mathbb{D}$. The Cayley transform between the half-plane and disk now givesIts inverse is holomorphic by the holomorphic inverse function theorem, so this is a biholomorphism. The proper-subset hypothesis was used to choose $a\notin \mathcal{U}$; the whole complex plane cannot be mapped this way, by Liouville theorem.

Because $f(D)$ avoids the closed negative real axis, its principal square root $g=\sqrt{f}$ is holomorphic, satisfies $g(0)=1$, and has $\mathrm{Re}g>0$. The Cayley transform between the half-plane and diskmaps the right half-plane into the unit disk and has $h(0)=0$. By Schwarz lemma, $|h(z)|\le |z|$. When $|z|\le 1/2$,Thusso one may take the universal constant $C=9$.

Let $D=(1-x{)}^2+y^2$. Direct partial differentiation, away from $(1,0)$, givesThese are the Cauchy-Riemann equations. Equivalently, with $z=x+iy$, the analytic mapping is $u+iv=(1+z)/(1-z)$, the inverse Cayley transform between the half-plane and disk.

The supplied harmonicity result can also be checked by the multivariable chain rule. The Cauchy-Riemann equations imply $\Delta u=\Delta v=0$, $\nabla u\cdot \nabla v=0$, and $|\nabla u{|}^2=|\nabla v{|}^2$. HenceThis is conformal invariance of harmonicity for the given mapping.

Let $V=f(U)$, the right half-disc from part (i). Write $w=r{e}^{it}$ with $0<r<1$ and $-\pi /2<t<\pi /2$. Squaring doubles the angle and squares the radius, giving a holomorphic bijectionIts inverse is the unique square root with positive real part, and its derivative $2w$ is nonzero on $V$. The cut includes zero, because $0\notin V$.

Next, the Möbius transformation $T(s)=i(1-s)/(1+s)$ is a conformal bijection from $\mathbb{D}$ onto $\mathbb{H}$: it is the inverse Cayley transform between the half-plane and disk composed with $s\mapsto -s$. For $s=-t$, $0\le t<1$, it givesConsequently $T((-1,0])=\{iy:y\ge 1\}$. Since $T$ is a bijection on the whole disc, deleting this interval deletes exactly that ray in the image.

The required function composition is thereforeEach component is a conformal bijection between the indicated domains. The only possible denominator zeros of the simplified expression are $z=\pm 1$, outside $U$, andsince $i$ is on the excluded semicircle and $-i$ is outside the upper half-plane. This also checks the local conformality of the conformal map from the exterior half-disc to a slit half-plane.

For $z=x+iy\in \mathbb{H}$, direct calculation givesConversely, solving $w=(z-i)/(z+i)$ givesThe denominator cannot vanish inside the unit disc, so these inverse formulas prove $\boxed{f(\mathbb{H})=\mathbb{D}}$. This is the Cayley transform between the half-plane and disk.

For $z\in U$, we also require $|z|>1$. In inverse coordinates this becomesThe equivalence holds in both directions, soThus the exterior of the unit semicircle maps to the right half of the unit disc. The circular arc maps to its imaginary diameter, which is excluded because the source inequality is strict.

The right half-plane consists of complex numbers with positive real part. Its boundary is the imaginary axis. For example, $1+i$ is inside, $i$ is on the boundary, and $-1+i$ is outside. Rotation $z\mapsto iz$ takes the right half-plane onto the upper half-plane; the Cayley transform between the half-plane and disk uses $R(z)=(z-1)/(z+1)$ for this convention.

The upper half-plane is the open setIt contains precisely the complex numbers with positive imaginary part: $1+i$ is inside, $1$ lies on its boundary, and $1-i$ belongs to the lower half-plane. In the complex plane its boundary is the real line; in the Riemann sphere the boundary also includes infinity. It is a convex set, hence a simply connected domain, and the Cayley transform between the half-plane and disk makes it biholomorphic to the unit disc.

With the hyperbolic metric $ds=|dz|/\mathrm{Im}z$, the same domain becomes the Poincaré half-plane model. Its hyperbolic geodesics are vertical lines and upper semicircles whose centres lie on the real line. Under the Cayley transform, these become arcs of circles orthogonal to the unit circle, with diameters as the straight-line cases. ^[ref]

Matching colours identify each hyperbolic geodesic and its image. The vertical line through zero maps to a diameter; the vertical line through one maps to an arc ending at $-i$ and $1$. A semicircle from $-1$ to $1$ also maps to a diameter, whereas the semicircle from $0$ to $2$ maps to a curved arc. Their endpoints lie on the boundary, outside the domains: the endpoints are infinitely far from interior points in hyperbolic distance.

The upper half-plane is useful because its straight boundary and explicit upper half-plane automorphisms make conformal maps and hyperbolic geometry calculable. The picture explains why shortest hyperbolic geodesic routes can look curved in Euclidean coordinates; their lengths are measured with the position-dependent hyperbolic metric.

For historical context, in his letter to Klein of 22 September 1882, Poincaré constructs a quadrilateral bounded by four circles orthogonal to a fundamental circle, and studies transformations generating repeated copies. In modern language, the sides inside that circle are hyperbolic geodesics in a disk model. The letter directly exhibits the connection between transformation groups, geometry and complex functions. ^[ref]