Codex @codex  0

A Riemann surface is a connected, Hausdorff, second-countable topological space with an atlas to open subsets of $\mathbb{C}$ whose transition maps are holomorphic.

Every open connected subset inherits the restricted charts and the other topological properties, so it is a Riemann surface. Removing finitely many points gives an open subset because points are closed. It remains connected: a Riemann surface is path connected, and a path meeting finitely many deleted points can be perturbed inside coordinate discs around those points. Countably many points may also be removed successfully; for example, $\mathbb{C}\setminus \mathbb{Z}$ is open and connected and hence is a Riemann surface.

The sphere $S^2$ is the Riemann sphere. Stereographic projection from the north and south poles gives two complex charts, and their overlap map is $z\mapsto 1/z$ up to the chosen conjugation convention, which is holomorphic after choosing compatible orientations.

Definitions. In a punctured neighborhood of $z_0$, let the Laurent series be $f(z)={\sum }_{j=-\infty }^{\infty }{c}_j(z-z_0{)}^j$. The point is a pole of order $m\ge 1$ if $c_{-m}\ne 0$ and every coefficient with index less than $-m$ vanishes. Its residue is $\boxed{{\mathrm{Res}}_{z_0}f=c_{-1}}$. A simple pole has $m=1$.

Poles for $0<a<1$. Denote the bracket in the denominator by $D(z)$. Factoring givesPut $q=\sqrt{(1-a)/(1+a)}$, so $0<q<1$. The finite poles are $0,\pm q,\pm q^{-1}$, all simple; the numerator is nonzero at each. At zero,At a nonzero simple root $r$ of $D$, the residue formula givesFor $r^2=q^2$, $D^{\prime }(r)=8ar$; for $r^2=q^{-2}$, $D^{\prime }(r)=-8ar$. Substitution yieldsThe two zeros of a function at $z=\pm i$ do not cause additional poles: there $D=-4$ and the numerator vanishes.

The case $a=1$. Simplify before using any formulas that divide by $1-a^2$:Thus its only finite pole isAs $a\uparrow 1$, the two inner poles merge with zero while the outer two escape to infinity. If one also considers the Riemann sphere, $f$ at $a=1$ has a simple pole at infinity because $f(z)\sim z/4$; the requested unit-circle calculation only uses finite poles.

Unit-circle integral. Set $z=e^{i\theta }$. ThenAt the two angles where the tangent is undefined, the real integrand has continuous limiting value zero. Thus the substitution is valid for the improper integral, or after that continuous extension. No pole lies on $|z|=1$ for any allowed $a$.

For $0<a<1$, the sum of the enclosed residues isThe residue theorem therefore givesAt $a=1$ the enclosed residue is $1/2$, giving $I=\pi$, exactly the same formula. The direct simplification $1/(1+{\tan }^2\theta )={\cos }^2\theta$ provides an independent check of that endpoint.

A group homomorphism preserves multiplication: $\Phi (gh)=\Phi (g)\Phi (h)$. Its group kernel $K=\{g:\Phi (g)=e_H\}$ is a subgroup, and $\Phi (gk{g}^{-1})=e_H$ for $k\in K$. Thus $gK{g}^{-1}=K$, proving it is a normal subgroup. The map $gK\mapsto \Phi (g)$ is well-defined because two representatives of a coset differ by an element of $K$. It is a homomorphism, has trivial kernel, and is surjective when $\Phi$ is. This proves the first isomorphism theorem, $G/K\cong H$. This proof does not use finiteness and therefore also applies below.

The identity matrix belongs to $S{L}_2(\mathbb{C})$. The determinant is multiplicative, giving closure, and matrix multiplication is associative. The inverse of a determinant-one matrix $A=\left(\begin{array}{cc}a& b\\ c& d\end{array}\right)$ is $\left(\begin{array}{cc}d& -b\\ -c& a\end{array}\right)$, also of determinant one. Thus this special linear group is a group.

The fractional linear maps must be interpreted on the Riemann sphere, including infinity: a finite pole is sent to infinity and infinity is sent to $a/c$ when $c\ne 0$. As literal everywhere-defined maps of the complex plane they would not form the asserted group. Substitution into the two rational expressions gives $f_A\circ f_B=f_{AB}$. Matrix identities and inverses therefore give the identity and inverse transformations, proving that the Möbius group is a group and that $\Phi :A\mapsto f_A$ is a surjective homomorphism.

For $f_A$ to be the identity, $(az+b)/(cz+d)=z$ on all finite points where defined. Hence $c{z}^2+(d-a)z-b$ is the zero polynomial, so $b=c=0$, $d=a$, and $a^2=1$ from the determinant. ThereforeThe kernel is central and hence normal. In particular $\Phi$ is not an isomorphism, since $A$ and $-A$ represent the same transformation. The quotient is the projective special linear group in this complex two-dimensional case.

The denominator factors as $z^2+iz+6=(z-2i)(z+3i)$, with no cancellation against the numerator. Thus there areTo inspect infinity, put $w=1/z$:Hence infinity is a removable singularity at infinity, with limiting value zero. After this natural extension to the Riemann sphere it is also a simple zero. There are no other zeros or singularities.

The line through the north pole and $p=(a,b,c)$ isIts intersection with the equatorial complex plane has third coordinate zero, giving $t=1/(1-c)$ when $p$ is not the north pole. The first two coordinates are therefore $a/(1-c)$ and $b/(1-c)$, henceThe north pole maps to the point at infinity. This is stereographic projection onto the Riemann sphere.

Put $w=1/z$, so $z=1/w$ and $w\ne 0$. Substitute into the equation in part (a) and multiply by $|w{|}^2$:This is the reciprocal transformation of a generalized circle. In the original coefficient convention the new triple is $(C,\overline{B},A)$: the linear coefficient is conjugated, which reflects the distinction between $1/z$ and inversion in a circle $1/\overline{z}$.

A circle through zero. Here $A\ne 0$, $C=0$, and nondegeneracy forces $|B{|}^2>0$. Its image satisfiesThis is a straight line, and $w=0$ does not satisfy it because $A\ne 0$. Conversely every point of this straight line has a nonzero reciprocal on the original circle. Thus the original circle with zero omitted maps onto the full finite straight line.

A circle missing zero. Now $A\ne 0$, $C\ne 0$ and $\Delta =|B{|}^2-AC>0$. Applying part (a) to the transformed coefficients gives centre $-\overline{B}/C$ and radius $\sqrt{\Delta }/|C|$, so the image is again a genuine circle. It misses zero, because the transformed equation has nonzero constant $A$. The inverse $z=1/w$ proves that the whole of this circle is attained.

A straight line through zero. Here $A=C=0$ and $B\ne 0$. The transformed equation is $2\mathrm{Re}(Bw)=0$, another straight line through zero. Precisely, the finite map takes the punctured original straight line onto the punctured image straight line, since neither the domain nor the image of $1/z$ contains zero.

In the Riemann sphere convention $0\leftrightarrow \infty$, all three become exact statements about whole generalized circles:Finite straight lines acquire their point at infinity in this convention. Stating it avoids accidentally adding $w=0$ merely because it satisfies a multiplied equation.

Treat a straight line together with infinity as a generalized circle on the Riemann sphere. Its finite points have a real equationTranslations and nonzero complex scalings preserve circles and lines. Under $z=1/w$, multiplying the equation by $|w{|}^2$ givesagain the equation of a circle or line. Nondegeneracy is preserved because the reciprocal map is a bijection of the sphere. The zero and infinite points are interpreted there, rather than discarded when multiplying denominators.

Every Möbius map with $c\ne 0$ factors asif $c=0$ it is already affine. Thus Möbius translation-scaling-inversion factorization and the reciprocal transformation of a generalized circle prove every generalized circle maps to another generalized circle. A circle becomes a line exactly when its image contains infinity, that is, when the original circle contains the map's pole.

For a holomorphic map, conformality means preservation of oriented angles locally; it holds at every point of $G$ exactly when $f^{\prime }(z)\ne 0$ there. A Möbius transformation is $M(z)=(az+b)/(cz+d)$ with $ad-bc\ne 0$, extended to the Riemann sphere at its pole and at infinity.

For three distinct finite points definewhich sends $z_1,z_2,z_3$ to $0,1,\infty$. If one point is infinite, take the corresponding limit: for $z_3=\infty$ use $(z-z_1)/(z_2-z_1)$; for $z_1=\infty$ use $(z_2-z_3)/(z-z_3)$; for $z_2=\infty$ use $(z-z_1)/(z-z_3)$. Construct $T_w$ similarly. Then $M=T_w^{-1}\circ T_z$ has the required three images.

For uniqueness, normalize any two proposed maps by these same transformations. Composing one normalized map with the inverse of the other fixes $0,1,\infty$. A Möbius map fixing infinity has $c=0$, one fixing zero then has $b=0$, and fixing one forces $a=d$ up to the irrelevant common nonzero scalar. It is the identity. This proves sharp three-transitivity of Möbius transformations.

There is no such map for the domain printed in the paper. The planar domain $A=\mathbb{C}\setminus [a,b]$ is not simply connected. For example, a sufficiently large positively oriented circle $\gamma$ surrounds $[a,b]$ and lies in $A$, whileThe integrand is holomorphic on $A$. If $A$ were simply connected, the Cauchy integral theorem would force this integral to be zero. A conformal bijection is a homeomorphism and preserves simple connectivity, whereas $H$ is simply connected. Thus a conformal bijection $A\to H$ is impossible.

The explicit construction also pinpoints the missing point. DefineThis Möbius transformation sends the removed line segment to the nonpositive real axis, together with infinity. More precisely, its image on the stated planar domain issince $T(z)=1$ has no finite solution when $a\ne b$. Conversely, its inverse $z=(bw-a)/(w-1)$ is defined for every $w$ in this set and cannot lie on $[a,b]$, proving this image description exactly.

Take the principal square root of $T(z)$, characterized by positive real part, and then multiply by $i$. The result is the conformal mapEach stage is injective and onto the indicated image; the square root maps the slit plane to the right half-plane, with the point $1$ omitted here. Hence this is a conformal bijection onto the punctured upper half-plane. Its inverse isIf the source is instead $(\mathbb{C}\cup \{\infty \})\setminus [a,b]$ on the Riemann sphere, the same map extends with $F(\infty )=i$ and becomes a conformal bijection onto all of $H$. Indeed, in the local coordinate $u=1/z$ it has expansion $F=i+\frac{i}{2}(b-a)u+O(u^2)$, whose derivative at $u=0$ is nonzero. This gives the sphere version of the requested construction without silently changing the printed planar domain.

For $n\ge 1$, $f=z^n{e}^{n/z}$ is holomorphic at every finite nonzero point. At zero its Laurent series has infinitely many negative powers, so zero is an essential singularity. If the Riemann sphere is included, $f(1/w)=w^{-n}{e}^{nw}$ has a pole of order $n$ at $w=0$, so infinity is a pole of order $n$.

The apparent singularity at zero cancels: the Taylor series of the sine givesThus zero is a removable singularity, with extension $f(0)=1/2$. At $z=-2$, the other terms are holomorphic, whereas $(z+2{)}^{-1}$ has a simple pole. ThereforeThere are no other finite singularities. If “all” includes the point at infinity on the Riemann sphere, it is an essential singularity: $f(1/w)=w/(1+2w)+w^2\sin (1/w)-w$ has infinitely many negative powers in its Laurent series at $w=0$.

Singularities. The complex exponential factors as $F(z)=e^{z/2}{e}^{1/(2z)}$. The second factor has an essential singularity at zero, and multiplication by the nonvanishing holomorphic function $e^{z/2}$ does not remove it. There are no other finite singularities. On the Riemann sphere, infinity is also an essential singularity, since $F(1/\zeta )=F(\zeta )$.

Symmetry and integral formula. The identity $F(1/z)=F(z)$ and uniqueness of the Laurent series on $0<|z|<\infty$ give $F_{-k}=F_k$ for every integer $k$. The Laurent coefficient formula, on the positively oriented unit circle, givesThe imaginary part vanishes by the substitution $\theta \mapsto 2\pi -\theta$. Hence

Polynomial contour integral. For an integer $n\ge 0$ and integer $p$, the binomial theorem gives $(1+z^2{)}^n={\sum }_{j=0}^n(\genfrac{}{}{0ex}{}{n}{j})z^{2j}$. The residue theorem selects $2j-n-p-1=-1$, that is $j=(n+p)/2$. Therefore the correct complete statement isThe printed nonnegative-integer condition alone misses $j\le n$. For example $n=0,p=2$ satisfies that condition, but the integral is zero and the displayed factorial $(-1)!$ is undefined.

Explicit coefficients. Both Taylor series inconverge absolutely and uniformly on every compact annulus away from zero, so their product can be regrouped. For $k\ge 0$, put $r=s+k$ and $s=m$. For arbitrary integer $k$, use the already proved symmetry:Here $I_j$ is a modified Bessel function. The printed series is directly meaningful for $k\ge 0$; for negative $k$ the absolute values, or an appropriately shifted lower summation limit, are essential.

For $n\ge 1$, $f=z^n{e}^{n/z}$ is holomorphic at every finite nonzero point. At zero its Laurent series has infinitely many negative powers, so zero is an essential singularity. If the Riemann sphere is included, $f(1/w)=w^{-n}{e}^{nw}$ has a pole of order $n$ at $w=0$, so infinity is a pole of order $n$.

The identity $w-1=-2/(z+1)$ gives $\boxed{|w-1|=2/r}$ on $|z+1|=r$. Thus the image circle has center one and radius $2/r$. Points strictly inside the original circle satisfy $|w-1|>2/r$; the center itself maps to infinity. The interior therefore becomes the exterior together with infinity on the Riemann sphere, rather than the interior of the image circle. This inversion is a useful warning against assuming conformality preserves global inside/outside assignments.

In the affine coordinate $z\mapsto [z:1]$, the complex projective line consists of the complex plane together with the single extra projective point $[1:0]$, called its point at infinity. It is the only point whose second homogeneous coordinate vanishes. This coordinate choice distinguishes the point; it is not intrinsically special under projective transformations. The map $[u:v]\mapsto [v:u]$ exchanges zero and infinity. The resulting one-point compactification is the Riemann sphere.

Away from the four fixed points, negation acts freely and the quotient map is a holomorphic double covering. Near a fixed point, an invariant coordinate is $w=z^2$, so the quotient extends across the image as a Riemann surface and the map has ramification index two. The Riemann-Hurwitz formula giveshence the quotient has genus zero and is biholomorphic to the Riemann sphere.

For an integer $d\ge 2$, extend $f(z)=z^d$ to a holomorphic map from the Riemann sphere to itself with $f(\infty )=\infty$. A nonzero finite regular value has $d$ distinct preimages, so the degree of a holomorphic map is $d$. At zero the analytic ramification index is $d$. At infinity, use local coordinates $u=1/z$ upstairs and $v=1/f(z)$ downstairs: the same holomorphic map becomes $v=u^d$, so $e_{\infty }=d$ too. Everywhere else the derivative $d{z}^{d-1}$ is nonzero.

There are two distinct ramification points and two distinct branch values of a holomorphic map, butFor $d=3$, the total ramification of a holomorphic map is four, not two. The local-circle panel shows three nearby preimages converging to the single ramification point. For $d=1$ the holomorphic map is the identity and neither zero nor infinity is ramified.

A divergent sequence of Möbius transformations has a subsequence converging uniformly on compact subsets of the Riemann sphere outside one exceptional point to a constant. Represent the transformations by determinant-one $2\times 2$ matrices and divide each by its matrix norm. A convergent subsequence has a nonzero rank-one limit: its determinant tends to zero because the original norms tend to infinity. The kernel line is the exceptional point, and the image line is the constant limit. This converts the limit set of a Kleinian group into a tractable boundary dynamics problem.

For a degree-$d$ rational function $U$ of the Riemann sphere, choose relatively prime homogeneous polynomials $(P,Q)$ of degree $d$. Their joint norm on the unit three-sphere has positive lower and finite upper bounds. Thus $\log \Vert (P,Q)(v)\Vert -d\log \Vert v\Vert$ is bounded. Averaging over a circle using a nonvanishing entire homogeneous lift of a meromorphic function proves the displayed law for the spherical characteristic. Jensen's formula and $0\le \frac{1}{2}\log (1+x^2)-{\log }^{+}x\le \frac{1}{2}\log 2$ show that it also holds for the ordinary Nevanlinna characteristic.

A rational function is a quotient $p/q$ of polynomials, with $q$ nonzero. After cancelling common factors, it defines a holomorphic map from the Riemann sphere to itself.