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Up to invertible affine changes in source and target, every complex cubic polynomial is exactly one ofThe extension to the Riemann sphere has two finite units of ramification. A single finite critical point has ramification index three and gives the first form; two distinct simple critical points can be moved to $\pm 1$, after which integration gives the second.

Every nonconstant meromorphic function on the complex plane has a simple island over at least one of any five Jordan domains on the Riemann sphere with disjoint closures. The Ahlfors second fundamental theorem, area comparison over target domains, and a length-area exhaustion of the complex plane prove this: without a simple island every counted island has degree at least two, forcing $3S\le (5/2)S+O(L)$.

For a finite simply connected covering surface over the Riemann sphere and $q\ge 3$ fixed separated Jordan domains, let $n_j$ count its simply connected islands over the $j$th domain once each, regardless of mapping degree. Let $S$ be its average sheet number and $L$ its spherical boundary length. The displayed inequality holds with $h$ depending only on the target domains. For regular bordered surfaces $f:X\to Y$, a general formulation is $\max \{-\chi (X),0\}\ge -S\chi (Y)-hL$, where $L$ counts only the boundary mapped into the interior of $Y$ and $\chi$ is the modern Euler characteristic. The boundary error disappears for an actual branched covering. The covering-surface formulation is explained in www.math.purdue.edu/~eremenko/dvi/ahl.pdf ; the simply connected island estimate is also recorded in Lemma 2 of pdfs.semanticscholar.org/d247/562607e237af7c8fc6e683b77be6b94829fa.pdf .

For a nonconstant rational map between round unit Riemann spheres, $R^{\ast }d\Omega =J_{R}d\Omega$. Counting inverse images with multiplicity gives $\int J_{R}d\Omega =4\pi \mathrm{deg}R$. The Jacobian determinant vanishes at ramification points of a holomorphic map, but the integral still records the global degree of a holomorphic map.

For north-pole stereographic projection $p=(X+iY)/(1-Z)$ on the unit sphere, the antipodal pair point has coordinate $q=-(X+iY)/(1+Z)$. Since $X^2+Y^2=1-Z^2$, this gives $p\overline{q}=-1$ when both coordinates are finite. On the whole Riemann sphere, write $q=-1/\overline{p}$, pairing zero with infinity. The two axis endpoints fixed by a sphere rotation obey this relation.

A branch point is a point around which a sufficiently small loop can change a germ of a holomorphic function under analytic continuation. This is a local condition: a large loop around a regular point might also enclose a distant branch point, so that large loop alone cannot classify the regular point. For the complex logarithm and square root, every nonzero point has a local holomorphic branch, but a loop once around $0$ changes the chosen branch. ^[ref]

For $z=r{e}^{it}$ with fixed $r>0$, start at $t=0$ with values $\sqrt{r}$ and $\log r$. Their continuations areOne counterclockwise turn gives $w(2\pi )=-\sqrt{r}$ and $L(2\pi )=\log r+2\pi i$. At $r=4$, the square root therefore changes from $2$ to $-2$, while the logarithm changes from $\log 4$ to $\log 4+2\pi i$. Two turns restore the square root to $2$; no nonzero number of turns restores the logarithm. This distinguishes a two-sheeted algebraic branch point from a logarithmic singularity.

A pole need not be a branch point: $1/z$ is unbounded near $0$, yet $1/(r{e}^{it})=r^{-1}{e}^{-it}$ returns to its starting value after one turn. Conversely, the square-root values tend to $0$ at their branch point. The issue is whether a single holomorphic choice survives going around the point, rather than whether the values become large. On the Riemann sphere, infinity is also a branch point of both $\sqrt{z}$ and $\log z$: in the local coordinate $\zeta =1/z$, these are ${\zeta }^{-1/2}$ and $-\log \zeta$, whose continuations around $\zeta =0$ change sign and subtract $2\pi i$, respectively. ^[ref]

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 $C^{-1}\circ g\circ C$.

The chordal metric on the Riemann sphere is $\chi (z,w)=|z-w|/\sqrt{(1+|z{|}^2)(1+|w{|}^2)}$ for finite points, with $\chi (z,\infty )=(1+|z{|}^2{)}^{-1/2}$. It is half the Euclidean chord distance between stereographic images on the unit sphere, so the triangle inequality follows from the Euclidean one. If $\rho$ is round spherical geodesic distance, then $\rho =2\arcsin \chi$ and $2\chi \le \rho \le \pi \chi$. These inequalities transfer uniform-continuity estimates between spherical and chordal distances, including at poles of meromorphic functions.

If ${\mathbb{C}}^{\ast }$ embeds holomorphically in a compact Riemann surface $R$, the embedding extends across zero and infinity to a degree-one holomorphic map from the Riemann sphere to $R$. Hence $R$ is conformally the sphere.

On $\mathbb{C}\setminus [-1,1]$, take the principal square root with positive real part. The Möbius transformation $(z-1)/(z+1)$ has image $(\mathbb{C}\setminus (-\infty ,0])\setminus \{1\}$. The principal square root therefore maps to the right half-plane with $1$ omitted, and the final Möbius transformation gives exactly $0<|w|<1$. The inverse is $z=-(w+w^{-1})/2$. Adding $\infty$ to the source on the Riemann sphere fills the missing value $w=0$ and gives a conformal map onto the full disk. Without $\infty$, a loop surrounding the segment has nonzero winding number, so the source is not a simply connected space and cannot be conformally equivalent to a disk.

For $f(z)=(az+b)/(cz+d)$ with distinct finite fixed points of a Möbius transformation $\alpha ,\beta$, the fixed-point equation gives $f(z)-\alpha =(a-c\alpha )(z-\alpha )/(cz+d)$. Dividing the two identities gives $k=(a-c\alpha )/(a-c\beta )$, which is nonzero because $(a-c\alpha )(a-c\beta )=ad-bc\ne 0$. The identities extend over the Riemann sphere. This conjugates the transformation to a dilation.

For four distinct points of the Riemann sphere, their cross-ratio iswith the evident limiting conventions at infinity. Every Möbius transformation preserves it.

Normalize three distinct points on a generalized circle by a Möbius transformation $f$ taking them to $1,0,\infty$. The inverse image of the extended real line is their generalized circle, and $J=f^{-1}\circ c\circ f$, with $c(z)=\overline{z}$, fixes that circle pointwise and squares to the identity. Changing the three normalizing points replaces $f$ by $M\circ f$ with a real-coefficient Möbius map $M$. Since $M$ commutes with conjugation, $J$ is unchanged. Thus this construction depends only on the circle. It is inversion in a circle for an ordinary circle and Euclidean reflection for a straight line, each extended to the Riemann sphere.

An even elliptic function descends through the degree-two quotient map $\wp$ to a rational function $R$ on the Riemann sphere. If the original function also has degree two, composition of map degrees gives $2=2\mathrm{deg}R$, so $R$ has degree one. Consequently it is a Möbius transformation with $ad-bc\ne 0$. The shift $\mathrm{sn}(z+K,k)$ is even and has degree two, providing an example for the common period lattice of the Jacobi elliptic sine and $\wp$.

Suppose a nonconstant holomorphic map $f:X\to {\mathbb{P}}^1$ from a connected compact Riemann surface has degree of a holomorphic map two and exactly four distinct branch values of a holomorphic map. The valency theorem forces each of their fibres to contain one ramification point with $e_p=2$, a simple ramification point of a holomorphic map. HenceThis determines the genus without recognizing a torus in a drawing.

A concrete realization is the compact Riemann surface obtained by adding the two points at infinity to the even-degree hyperelliptic model given by the algebraic curveProjection to $z$ has degree of a holomorphic map two. At a simple zero $a$ of $P$, $w$ is a local coordinate and $z-a=w^2/P^{\prime }(a)+O(w^4)$, so $e_p=2$. At infinity set $u=1/z$ and $s=w/z^2$: then $s^2=(1-4u^2)(1-u^2)$. There are two points with $u=0$, $s=\pm 1$, each unramified since $u$ is a local coordinate. The four finite roots are therefore all the branch values of a holomorphic map.^[ref]

Choose a triangulation of a surface on the Riemann sphere with these four branch values of a holomorphic map as vertices, combinatorially the boundary of a tetrahedron. Its counts are $V=4,E=6,F=4$. Upstairs, each vertex has one preimage and each open edge and face has two lifts: $V_X=4,E_X=12,F_X=8$, giving Euler characteristic $4-12+8=0$. This is a cell schematic, not a geometric immersion of $X$. The Triangulation proof of the Riemann-Hurwitz formula applies the same counting idea after refining a general triangulation of a surface around the branch values of a holomorphic map.^[ref]

A planar domain whose complement in the Riemann sphere has exactly two connected components. An annulus is the standard example. If neither complementary component is a single point, the domain is conformally equivalent to a circular annulus with finite positive conformal modulus of an annulus; punctured-disc and punctured-plane cases account for degenerate complementary components.

Every elliptic function has a unique expression $A(\wp )+B(\wp ){\wp }^{\prime }$ with rational functions $A,B$. Its even part is rational in the Weierstrass elliptic function; its odd part divided by ${\wp }^{\prime }$ is even and meromorphic, so is also rational in $\wp$. The local even Laurent series at a half-period and at the pole justify meromorphic descent through the double cover of the Riemann sphere.

The Weierstrass elliptic function is the quotient map from the complex torus by $z\mapsto -z$. An even elliptic function descends through this map; its even Laurent series at each branch point makes the descended function meromorphic there. A meromorphic function on the Riemann sphere is rational, so $f=Q(\wp )$.

A meromorphic function on a punctured disk has finite pulled-back spherical area precisely when it extends to the Riemann sphere at the puncture. A removable singularity or pole gives a smooth spherical extension and finite area. At an essential singularity, the meromorphic version of the Great Picard theorem gives infinitely many preimages of all but at most two spherical targets. The area formula then makes the area infinite. The logarithmically weighted annular area divided by $\log (1/r)$ tends to the total unweighted area divided by $4\pi$, by averaging a nondecreasing area function.

For $n\in \mathbb{Z}[i]$ and a nonzero integer $r$, the displayed determinant-one matrix has trace two and represents a nonidentity parabolic Möbius transformation fixing $n/r$. Its fixed-point equation is $(rz-n{)}^2=0$. These parabolic fixed points are dense in the Riemann sphere, proving that the Gaussian-integer determinant-one Kleinian group has full limit set.