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A branch of a multivalued analytic expression is a single-valued holomorphic function on a specified domain that consistently selects one of the expression's possible values. A branch cut is a convenient way to choose such a domain, rather than an intrinsic locus of singularities. On $D=\mathbb{C}\setminus (-\infty ,0]$, the principal logarithm supplies two square-root branchesFor $z=i$, these give $B_{+}(i)=e^{i\pi /4}=(1+i)/\sqrt{2}$ and $B_{-}(i)=-(1+i)/\sqrt{2}$. Their derivatives follow by differentiating $B_{\pm }^2=z$: $B_{\pm }^{\prime }=1/(2B_{\pm })$, so $B_{+}^{\prime }(i)=(1-i)/(2\sqrt{2})$. There are exactly two branches on $D$: for any other holomorphic square root $B$, the ratio $B/B_{+}$ is continuous, takes values in $\{1,-1\}$, and hence is constant on the connected domain.

The cut can move. Choosing $0<\theta <2\pi$ defines a holomorphic logarithm $L_{+}(z)=\log |z|+i\theta$ on $D_{+}=\mathbb{C}\setminus [0,\infty )$. At $-1$ this gives $L_{+}(-1)=i\pi$ and $\exp (L_{+}(-1)/2)=i$, both regular values despite $-1$ lying on the previous cut. In the complex upper half-plane, $L_{+}$ agrees with $\mathrm{Log}$; in the lower half-plane it equals $\mathrm{Log}+2\pi i$, so the corresponding square roots have opposite signs there. The overlap $D\cap D_{+}$ has two components, exactly why the identity theorem cannot force agreement between them. ^[ref]

For example, the limits of $B_{+}$ at $-r$, $r>0$, are $i\sqrt{r}$ from above and $-i\sqrt{r}$ from below. This jump records the chosen branch cut; each point $-r\ne 0$ still has a neighbourhood supporting a holomorphic square root. Moving the cut changes the bookkeeping, while the branch point at $0$ and its monodromy persist. Specifying a branch of a multivalued function is therefore essential before differentiating or performing contour integration with holomorphic square roots, complex logarithms or complex exponentiation.

For a nonzero complex number $z=r{e}^{i\theta }$ with $r>0$, its two square roots are $\pm \sqrt{r}{e}^{i\theta /2}$. They are distinct and differ by a sign. At $z=0$ the only square root is zero. A principal value convention chooses $\theta =\mathrm{Arg}z\in (-\pi ,\pi ]$ and the plus sign; continuity across the negative real axis then fails. A holomorphic square root is a different, domain-dependent notion: it requires a holomorphic function whose square is $z$ on the specified domain. It cannot exist on every punctured neighborhood surrounding zero.

The monodromy theorem gives equality of continuations that exist under its hypothesis. It does not assert that every holomorphic function admits analytic continuation everywhere, even within a simply connected domain.

For an explicit obstruction, take $f(z)=1/(1-z)$ near $a=0$ and attempt to apply the theorem on $D=\mathbb{C}$. The path $\eta (t)=t$ ends at $1$, where no terminal germ of a holomorphic function exists. If a holomorphic representative $G$ near $1$ continued $f$, then the identity theorem would extend $(1-z)G(z)=1$ to that neighbourhood. Substituting $z=1$ would give $0=1$. Thus the pole prevents the existence hypothesis, although $\mathbb{C}$ is a simply connected domain. The analytic continuation article explains why the same rational function does continue throughout $\mathbb{C}\setminus \{1\}$.

Existence at an endpoint by some route is also weaker than existence along every path from the base point. Along the real segment from $1$ towards $0$, a holomorphic square root of $z$ tends to $0$, yet cannot continue holomorphically through $0$: a local representative $G$ would satisfy $G^2=z$ by the identity theorem, and differentiating at $0$ would give $2G(0)G^{\prime }(0)=1$ with $G(0)=0$. A finite limiting value therefore does not remove a branch point. More severe obstructions appear at a natural boundary of a holomorphic function.

Exhaust a simply connected hyperbolic Riemann surface in potential theory by compact bordered Jordan domains $R_n$. Their Green functions on a Riemann surface $G_n(\cdot ,p)$ have logarithmic poles and zero boundary values. Exponentiating their harmonic conjugates produces proper degree-one maps $f_n:R_n\to \mathbb{D}$ with $f_n(p)=0$. If $G_n=-\log |\zeta |+c_n+o(1)$, normalize $f_n^{\prime }(p)=e^{-c_n}>0$. Domain monotonicity and a global positive Green function bound $c_n$ increasingly from above. Thus Montel theorem gives a nonconstant limit $F$, which is injective by the locally uniform limit of univalent functions theorem. The Schwarz lemma makes $F^{\prime }(p)$ extremal among normalized disk-valued maps. If $F$ omitted $a\ne 0$, taking a global holomorphic square root of a disk automorphism applied to $F$, then renormalizing at $p$, would multiply its derivative by $(1+|a|)/(2\sqrt{|a|})>1$. Thus $F$ is onto and gives the desired biholomorphism.

A holomorphic logarithm of a nowhere-zero holomorphic function $f$ on a domain $U$ is a holomorphic function $g$ such that $e^{g(z)}=f(z)$ throughout $U$. Differentiating with the chain rule gives $g^{\prime }=f^{\prime }/f$. For $f(z)=z$, a holomorphic logarithm is a branch of the complex logarithm, and its derivative is $1/z$.

The principal logarithm is holomorphic on the slit complex plane $D=\mathbb{C}\setminus (-\infty ,0]$, withHere $\mathrm{Arg}$ is the chosen complex argument, and $\mathrm{Im}\mathrm{Log}z\in (-\pi ,\pi )$. One may conventionally assign a principal value on the negative real axis, but that axis is excluded from this holomorphic domain. ^[ref] For example, $1+i=\sqrt{2}{e}^{i\pi /4}$ gives $\mathrm{Log}(1+i)=\frac{1}{2}\log 2+i\pi /4$ and $(\mathrm{Log}{)}^{\prime }(1+i)=(1-i)/2$; all its other complex logarithms differ by $2\pi ik$, $k\in \mathbb{Z}$.

There is no holomorphic logarithm of $z$ on the punctured complex plane $\mathbb{C}\setminus \{0\}$. Such a function would be a holomorphic primitive of $1/z$, so its contour integral around every closed path would vanish. On the counterclockwise unit circle, however,This is the period obstruction to a holomorphic antiderivative, and also exactly the logarithm change in the monodromy diagram.

Every nowhere-zero holomorphic function $f$ on a simply connected domain $U$ does admit a holomorphic logarithm. Choose $a\in U$ and $b\in \mathbb{C}$ with $e^b=f(a)$, and defineThe Cauchy integral theorem makes this contour integral independent of the path in $U$. Then $g^{\prime }=f^{\prime }/f$ and $(f{e}^{-g}{)}^{\prime }=e^{-g}(f^{\prime }-f{g}^{\prime })=0$; its value at $a$ is $1$, proving $e^g=f$. Any two choices on a connected domain differ by a constant $2\pi ik$, since their difference is continuous and takes values in the discrete set $2\pi i\mathbb{Z}$. ^[ref]

For a computed example, take $f(z)=1+z$ on the unit disc, with $a=b=0$. Expanding $f^{\prime }/f=1/(1+z)$ as a geometric series and integrating givesThe construction also produces holomorphic square roots: $h=e^{g/2}$ satisfies $h^2=f$, so here $h(1/2)=\sqrt{3/2}$. More generally, $e^{\alpha g}$ gives a consistent branch of a multivalued function for complex exponentiation by $\alpha$. A holomorphic logarithm converts multiplicative expressions into additive ones while making the domain restrictions explicit.

A holomorphic square root of a nonvanishing holomorphic function $f$ is a holomorphic function $g$ satisfying $g^2=f$. It exists locally. On a connected open domain it exists globally exactly when the winding number of $f\circ \gamma$ around zero is even for every closed path $\gamma$ in the domain. Continuing a local square root once along such a path multiplies it by $(-1{)}^m$, where $m$ is that winding number. Thus the obstruction concerns parity, whereas a holomorphic logarithm needs every such winding number to be zero.

On $\mathbb{C}\setminus \{0\}$, the function $f(z)=z$ has no global holomorphic square root: its value winds once on the unit circle, and the lift shown under monodromy changes sign. In contrast, $f(z)=z^2$ winds twice there and has global square roots $g(z)=z$ and $g(z)=-z$. It still has no holomorphic logarithm on that domain. This distinguishes the two global continuation conditions.

An even-degree polynomial with all roots in $|z|<R$ admits a holomorphic square root throughout $|z|>R$. Factor out its even power of $z$ and use the convergent logarithm of each $1-a_j/z$ to construct it. Even degree removes the annular winding obstruction. The two choices differ by a global sign.

The Joukowski map restricts to a conformal map from $\{\mathrm{Im}z>0,|z|>1\}$ onto the upper half-plane. Indeed, $\mathrm{Im}g(z)=\mathrm{Im}z(1-|z{|}^{-2})>0$ there. For any $\mathrm{Im}\zeta >0$, the roots of $z^2-\zeta z+1=0$ have product one and cannot lie on the unit circle; exactly one has modulus greater than one, and that root has positive imaginary part. This proves that the map is a bijection.

Its inverse function iswith each factor taken as the principal square root. This product is the holomorphic square root of ${\zeta }^2-4$ that behaves as $\zeta$ at infinity in the upper half-plane. The upper unit semicircle maps to $[-2,2]$; the exterior real rays map to the remaining real rays. The derivative $g^{\prime }=1-z^{-2}$ vanishes only at the boundary endpoints $z=\pm 1$.

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.

Start at $a=1$ with complex logarithm germ of a holomorphic function $L(1)=0$ and holomorphic square root germ of a holomorphic function $S(1)=1$. On the punctured plane $D=\mathbb{C}\setminus \{0\}$, both admit analytic continuation along every path. To construct it locally, for any $c\ne 0$ choose a small open disc avoiding $0$, select a holomorphic logarithm there, and select its additive multiple of $2\pi i$ to agree with the arriving germ of a holomorphic function; exponentiating half this logarithm supplies the arriving holomorphic square root. A finite chain of these open discs follows any given path.

Compare two paths from $1$ to $i$, parametrized by $0\le t\le 1$:The first follows a counterclockwise quarter-circle; the second follows a clockwise three-quarter-circle. Following ${\gamma }_A$ and then path reversal of ${\gamma }_B$ makes one positive turn about $0$. Its winding number is $1$, so the two paths cannot be homotopic with endpoints fixed in the punctured plane.

The continued values along the chosen paths areAt their common endpoint,Both complex logarithm values exponentiate to $i$, and both holomorphic square root values square to $i$. These give different terminal germ of a holomorphic function representatives, consistently obtained from the same starting germ of a holomorphic function. The monodromy theorem applies, but does not equate these two non-homotopic paths.

The closed path $\lambda (t)=e^{2\pi it}$ instead returns to $1$, with continued values $L_{\lambda }(t)=2\pi it$ and $S_{\lambda }(t)=e^{\pi it}$. One counterclockwise turn changes the complex logarithm by $2\pi i$ and flips the holomorphic square root from $1$ to $-1$. Hence neither initial germ of a holomorphic function can come from a global holomorphic logarithm or a global holomorphic square root of $z$ on the punctured plane. Repeating a loop of winding number $n$ gives the changes $2\pi in$ and $(-1{)}^n$. ^[ref]

Now restrict to the slit domain $D_{\mathrm{slit}}=\mathbb{C}\setminus (-\infty ,0]$. The principal logarithm and $\exp (\mathrm{Log}z/2)$ realize the same initial germ of a holomorphic function for each expression globally. To see that this is a simply connected domain, polar form identifies this domain with $(0,\infty )\times (-\pi ,\pi )$, a convex set; every loop contracts there. The simply connected corollary of the monodromy theorem now forces equality along every pair of paths with matching endpoints. In particular, ${\gamma }_A$ and the segment $\sigma (t)=1-t+it$ both give $L(i)=\pi i/2$ and $S(i)=(1+i)/\sqrt{2}$. The explicit homotopy $H(s,t)=(1-s){\gamma }_A(t)+s\sigma (t)$ stays in the closed first quadrant away from $0$, hence in $D_{\mathrm{slit}}$.

The other route ${\gamma }_B$ crosses the removed ray at $t=2/3$ and is no longer admissible. In the punctured plane, that crossing is locally regular: analytic continuation passes through $-1$ smoothly, although evaluating the principal logarithm would reset its argument. The branch cut changes the allowed domain; it does not create additional branch points or poles. That distinction is the beauty of the obstruction: perfectly compatible local holomorphic functions can fail to make a global choice solely because of the allowed based loops. The branches of a multivalued function article develops how moving the branch cut changes the selected values.

Monodromy records how a germ of a holomorphic function changes after analytic continuation around a closed path. The source point returns to itself, but the continued germ of a holomorphic function need not. For $z(t)=e^{it}$, $0\le t\le 2\pi$, starting with holomorphic square root value $1$ gives $w(t)=e^{it/2}$ and hence $w(2\pi )=-1$; starting with complex logarithm value $0$ gives $L(t)=it$ and hence $L(2\pi )=2\pi i$. After $m$ counterclockwise turns the values are $(-1{)}^m$ and $2\pi im$. Thus monodromy exchanges the two square-root branches of a multivalued function, whereas it shifts through infinitely many logarithm branches of a multivalued function. ^[ref]

The three panels use the same parameter $t$. They show analytic continuation through local holomorphic patches, including across a chosen branch cut; repeatedly evaluating the principal logarithm would instead reset its imaginary part at the cut. Monodromy makes the obstruction to a global branch of a multivalued function visible and motivates the use of Riemann surfaces as domains on which continued values stay distinct.

The quadratic formula gives two algebraic possibilities. To select one consistent holomorphic square root, setwhere each factor is the principal square root. Both arguments lie in the upper half-plane, so neither crosses its branch cut. This choice satisfies $S(\zeta {)}^2={\zeta }^2-4$ and $S(\zeta )\sim \zeta$ at large $\zeta$ in the upper half-plane. The required inverse function isFor $\zeta =iR$, $R>0$, this gives $z=i(R+\sqrt{R^2+4})/2$, whose modulus exceeds one. As $\zeta$ varies continuously in the upper half-plane, this root cannot cross $|z|=1$, since that would make $\zeta$ real. Hence the selected root always lies in the fluid region. The other root is its reciprocal and lies inside the unit disk. Simply writing the principal square root of ${\zeta }^2-4$ would not specify this branch correctly throughout the upper half-plane.

Use the Joukowski map with $|c|<a$, in particular its intended range $0\le c<a/2$. On the original circle,Define the exterior inversewhere the holomorphic square root is chosen so that $\sqrt{Z^2-4c^2}\sim Z$ at infinity. Then $f(Z)\sim Z$ is the branch mapping the exterior of the ellipse to $|z|>a$. Composition of the zero-circulation complex potential from part 4(c) with this inverse givesThe ellipse is a streamline because its inverse image is the circle, and $W(Z)\sim \beta e^{i\alpha }{Z}^2$ at infinity. An additive constant in the complex potential has no effect on the velocity.

The transformed complex velocity obeys the chain rule:Since $|c|<a$, the denominator is nonzero on and outside $|z|=a$. The four stagnation points therefore come fromTheir physical locations areThey all lie on the ellipse, and their positions are independent of $\beta >0$, which only sets the overall speed scale.

The printed restriction that $c$ is real and $c<a/2$ is insufficient literally: $c=-2a$ satisfies it but makes $dZ/dz=0$ at $z=2a$ in the exterior, so there is no exterior conformal map of the claimed type; the stated minor semiaxis would also be negative. The intended positive-parameter range, or more generally $|c|<a$, is needed for the above solution. The reference to a sphere in the source is a wording slip: the geometry is two-dimensional flow around the mapped cylinder's elliptical cross-section.

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.

Choose the holomorphic square root in the complex upper half-plane whose boundary value is positive on $0<x<1$. Its primitive has derivative $F_1^{\prime }(z)=1/\sqrt{z(1-z^2)}$. The integral from the boundary basepoint zero is well-defined by a limiting path: the local singularity is only $z^{-1/2}$.

The three finite prevertices $-1,0,1$ have derivative exponent $-1/2$, hence image angle $\pi /2$. At infinity $F_1^{\prime }(z)=O(z^{-3/2})$, so the local coordinate $1/z$ also gives image angle $\pi /2$. These are the four corners of a Schwarz-Christoffel mapping to a rectangle. To establish that it is a square rather than a general rectangle, putThe length along $(-1,0)$ is also $A$ by $x\mapsto -x$. The length along $(1,\infty )$ isby $x=1/t$, and the fourth side has the same length by symmetry. With the chosen branch, the boundary derivatives on the intervals $(-\infty ,-1),(-1,0),(0,1),(1,\infty )$ have phases $-1,-i,1,i$, respectively. ThereforeThe extended real boundary maps continuously and once around this square. Each side is traversed monotonically, and the four corner limits are finite. The argument principle, applied after small indentations around the boundary singularities, counts one inverse image of each interior value and zero of each exterior value. Since $F_1^{\prime }$ never vanishes in the complex upper half-plane, those interior inverse images are simple. Thus

The disc lies in the upper-right half-plane and avoids zero and the negative real axis. On the slit plane take the holomorphic logarithm $\mathrm{Log}z=\log |z|+i\arg z$ with $-\pi <\arg z<\pi$. Its derivative is $1/z$, and $e^{\mathrm{Log}z}=z$. Therefore the holomorphic square rootis analytic and satisfies $f^2=z$. One can also construct it without a global branch theorem: fix either scalar square root $a$ of $3+3i$ and expand $a(1+(z-3-3i)/(3+3i){)}^{1/2}$ by its convergent binomial series. The series converges on the entire disc because $|z-3-3i|/|3+3i|<1/\sqrt{18}$. Either sign gives an admissible choice.

For a plane domain, the following are equivalent descriptions of a simply connected domain: every based loop contracts in the domain with its basepoint fixed; any two paths with the same endpoints are homotopic while those endpoints stay fixed; its complement in the Riemann sphere is connected; every holomorphic function has a single-valued primitive; and every nowhere-zero holomorphic function has a holomorphic logarithm. The complement must be taken in the extended plane, including infinity. An equivalent winding number formulation is that every closed curve has index zero about every point outside the domain.

Here is a direct equivalence proof for the first two topological formulations. If all paths with given endpoints are homotopic relative to those endpoints, apply this to a loop based at $a$ and the constant path at $a$, obtaining a contraction. Conversely, let ${\gamma }_0,{\gamma }_1$ join $a$ to $b$. The loop ${\gamma }_0\ast {\gamma }_1^{-1}$ contracts with basepoint fixed. Concatenating this contraction with ${\gamma }_1$ produces a homotopy from ${\gamma }_0\ast {\gamma }_1^{-1}\ast {\gamma }_1$ to ${\gamma }_1$, relative to $a,b$. The middle retracing excursion contracts by shortening its traversal in both directions, giving a homotopy from that first path to ${\gamma }_0$. Hence ${\gamma }_0$ and ${\gamma }_1$ are homotopic relative to their endpoints. Planar domains are path-connected, so no extra path-component condition is needed.

The analytic descriptions follow from path independence of the holomorphic integral. Local primitives exist on discs by integrating power series; a homotopy can be subdivided into pieces lying in such discs, and local path integrals cancel along its common edges. Thus integrals over homotopic paths agree, giving a global primitive. Applied to $h^{\prime }/h$ for a nowhere-zero function, a primitive $H$ makes $h{e}^{-H}$ constant and hence supplies a holomorphic logarithm. These observations are also the tools used in the mapping proof below.

The Riemann mapping theorem states that every nonempty simply connected proper domain $D⊊\mathbb{C}$ admits a conformal bijection onto the unit disc. Given $z_0\in D$, there is exactly one such map with $f(z_0)=0$ and $f^{\prime }(z_0)>0$. The exclusion of the whole plane is necessary: an entire map into the disc would be constant by Liouville's theorem.

An extremal proof has the following concrete steps. First construct an injective bounded holomorphic map. Choose $a\notin D$ and a holomorphic square root $h$ of $z-a$. It is injective, since equality of its values implies equality of their squares. Its image is disjoint from its negative: $h(z)=-h(w)$ would imply $z=w$ and then $h(z)=0$, impossible. If $b=h(z_0)$, choose $r>0$ with $B(b,r)\subset h(D)$; the disc $B(-b,r)$ is omitted. Therefore $1/(h+b)$ is injective and bounded. Scaling it and applying an automorphism of the unit disk gives a map into the disc taking $z_0$ to zero; a rotation makes its derivative positive.

Let $\mathcal{U}$ consist of all injective maps $D\to \mathbb{D}$ with this normalization. It is nonempty. The derivative at $z_0$ is bounded above by the Cauchy integral formula on a small disc compactly contained in $D$. Choose $f_n\in \mathcal{U}$ with derivatives tending to their positive supremum $L$. Boundedness, the Cauchy estimates and a diagonal Arzelà-Ascoli theorem argument give a subsequence converging locally uniformly to $f$; this is the bounded-family form of the Montel theorem. Then $f(z_0)=0$, $f^{\prime }(z_0)=L>0$, and the maximum modulus principle puts its image in the open disc. The limit is injective: if distinct $z,w$ had $f(z)=f(w)$, apply Hurwitz's theorem near $w$ to $f_n(\cdot )-f_n(z)$, which has no zero there; the limit would vanish identically, contradicting $f^{\prime }(z_0)>0$.

Finally suppose $a\in \mathbb{D}$ is omitted by $f$. It is nonzero since $f(z_0)=0$. Let $T_a(w)=(w-a)/(1-\overline{a}w)$ and choose a holomorphic square root $h$ of $T_a\circ f$. It is injective and takes values in the disc. Put $b=h(z_0)$, so $|b|=\sqrt{|a|}$, and normalize $T_b\circ h$ by a rotation. The new map's derivative magnitude at $z_0$ isThis square-root improvement of a normalized conformal map contradicts extremality. Thus $f$ is onto. Two normalized bijections differ by a disc automorphism fixing zero; Schwarz lemma applied to it and its inverse makes it a rotation, and positivity of the normalized derivatives forces that rotation to be the identity. This completes the requested proof outline.

We first derive the exterior area theorem for univalent functions directly from Green's theorem, then obtain the coefficient bound needed for the Koebe quarter theorem.

Suppose $G$ is univalent on $|\zeta |>1$ and has a Laurent seriesFor $R>1$, the image of $R{e}^{it}$, $0\le t\le 2\pi$, is a regular simple closed curve. By the Jordan curve theorem the curve has one bounded and one unbounded complementary component. The image of $|\zeta |>R$ lies in the unbounded component, since it is connected, avoids the curve by injectivity and contains points near infinity. It fills that component: any finite boundary limit of the image has preimages bounded by $G(\zeta )/\zeta \to 1$, and a convergent preimage subsequence must land on $|\zeta |=R$ or else give an interior image point. Thus no boundary of the image lies inside that complementary component. The curve is positively oriented about its bounded interior, since the exterior is on the right of $R{e}^{it}$ and a holomorphic map with nonzero derivative preserves orientation. Green's formula gives its enclosed area asFor completeness, the positive Fourier mode is $R{e}^{it}$ and the negative modes are $b_n{R}^{-n}{e}^{-int}$. On multiplying by the conjugate series and integrating, all unequal-frequency products integrate to zero by Fourier orthogonality. The equal-frequency products give $i{R}^2$ and $-in|b_n{|}^2{R}^{-2n}$, respectively; the constant term gives zero. The Laurent series and its differentiated series converge uniformly on the circle, since $R>1$, so these operations are justified. Nonnegativity of $A_R$ yields $\sum n|b_n{|}^2{R}^{-2n}\le R^2$. Letting $R\downarrow 1$, or first doing so for every finite partial sum, givesNo regularity of the map on the unit circle has been assumed.

Now let $f(z)=z+a_2{z}^2+\cdots$ be a normalized univalent function. Its only zero is at zero, so $f(z)/z$, extended there by the value one, is holomorphic and nowhere zero. On the simply connected disc it has a holomorphic square root with value one at zero. Its odd square-root transform isThis is an odd function which is holomorphic and normalized. It is also univalent: $h(z)=h(w)$ gives $f(z^2)=f(w^2)$ and hence $z=\pm w$. If $z=-w$, oddness gives $h(z)=-h(z)$, so $h(z)=0$, forcing $z=w=0$. Thus $G(\zeta )=1/h(1/\zeta )$ is univalent on the exterior disc, with expansionApplying the area bound to its first negative coefficient proves the second coefficient bound for normalized univalent functions, $|a_2|\le 2$.

Let $w_0$ be any point omitted by $f$. It is nonzero. Postcomposition with a Möbius map gives another normalized univalent functionwhose denominator does not vanish. Applying the second coefficient bound both to $f$ and to $f_{w_0}$ givesEvery omitted point therefore has modulus at least $1/4$, provingThis establishes the Koebe quarter theorem from the allowed area formula.

Sharpness is witnessed by the Koebe function $k(z)=z/(1-z{)}^2$. It has the required value and derivative at zero. The equality $k(z)=k(w)$ simplifies to $(z-w)(1-zw)=0$, and $|zw|<1$ makes it injective. Moreover,Thus $-1/4$ is omitted, so no larger universal radius is possible. In fact $4k(z)+1=((1+z)/(1-z){)}^2$ shows its image is the plane slit along $(-\infty ,-1/4]$.

Without univalence take a real $\lambda >4$ andIt is entire, with $f_{\lambda }(0)=0$ and $f_{\lambda }^{\prime }(0)=1$, but it omits $-1/\lambda$, a point of modulus less than $1/4$, since the exponential never vanishes. It is genuinely not univalent on the disc: the distinct points $\pm \pi i/\lambda$ lie there and both give exponential value $-1$. Even its derivative never vanishes, illustrating why local conformality alone does not give the quarter-disc inclusion.

A hyperbolic Riemann surface in potential theory is a noncompact Riemann surface admitting a positive Green function on a Riemann surface. With pole $p$, this is a positive function harmonic off $p$, having local form $-\log |\zeta |$ plus a harmonic function in a coordinate $\zeta (p)=0$, and obtained as the minimal such positive function by exhaustion. This is a noncircular definition for the present proof. For a simply connected surface it is equivalent to disk conformal type; for arbitrary surfaces potential-theoretic hyperbolicity and disk universal-cover type should be distinguished.

The unit disk is an example. At pole zero its function is $G(z,0)=-\log |z|$: it is positive in the punctured disk, harmonic there, has the correct logarithmic singularity, and tends to zero at the boundary. If another positive function has the same singularity and is harmonic elsewhere, subtract $G$ and apply the maximum principle for harmonic functions on disks of radius $r<1$; the boundary lower bound is $\log r$. Letting $r\uparrow 1$ shows that the other function dominates $G$. Thus $G$ is indeed the positive Green function on a Riemann surface. At a general pole it is

Here is a Green-function exhaustion proof of disk uniformization. The main analytic idea is to build conformal maps on bordered disks from their Green functions and pass to an extremal limit. The positive global Green function prevents that limit from becoming constant.

First exhaust the noncompact simply connected surface $R$ by relatively compact domains $R_n$ containing $p$, each with a smooth Jordan boundary. This topological step does not use conformal uniformization: take regular neighborhoods of finitely many coordinate disks covering successive compact sets and fill bounded complementary components. Simple connectedness and the Jordan curve theorem make the filled neighborhoods topological disks; enlarge them to be nested.

Each $R_n$ has a zero-boundary Green function on a Riemann surface $G_n(\cdot ,p)$. One analytic construction, independent of conformal uniformization, is as follows. Equip the compact bordered domain with a smooth conformal metric. Multiply $-\log |\zeta |$ by a smooth cutoff function equal to one near $p$ and zero near the boundary; call the result $s$. Distributionallywhere $k$ is smooth and supported away from $p$. Solve $\Delta v=k$ with zero boundary data. This auxiliary Dirichlet problem can be obtained by minimizing $\frac{1}{2}\int |\nabla v{|}^2+\int kv$ over functions with zero boundary trace: the Poincaré inequality gives coercivity, minimization in a Hilbert space gives a weak solution, and interior and smooth-boundary elliptic regularity give a smooth solution. Then $G_n=s+v$ is harmonic away from $p$, has the logarithmic pole, and vanishes at the boundary. It is positive by the maximum principle for harmonic functions, applied after removing a sufficiently small circle around its positive pole. This construction uses ordinary linear elliptic existence, not the desired uniformization theorem. Equivalently, the same bordered-domain Green functions can be constructed by the Perron method for the Dirichlet problem using local logarithmic barriers.

On $R_n\setminus \{p\}$ choose local harmonic conjugates $G_n^{\ast }$. Their period around the pole is $-2\pi$, and every closed loop is homologous to an integer multiple of that loop. Thereforeis a single-valued holomorphic function. It extends across $p$ with one simple zero and has no others. Since $G_n>0$, $|f_n|<1$; since $G_n=0$ on the boundary, $|f_n|\to 1$ there. Thus $f_n:R_n\to \mathbb{D}$ is proper. For any $a\in \mathbb{D}$, choose a smooth interior contour sufficiently close to the boundary that $|f_n|>|a|$ everywhere on it. The contour surrounds $p$ and all possible preimages of $a$, because $|f_n|>|a|$ throughout a thin boundary collar. By Rouché's theorem, $f_n-a$ has the same number of zeros inside as $f_n$, namely one. Hence $f_n$ is onto and one-to-one, with nonzero derivative: it is a biholomorphism from $R_n$ to the disk.

Write near $p$Multiply $f_n$ by a unit constant so that, in the fixed coordinate, $f_n^{\prime }(p)=e^{-c_n}>0$. Domain monotonicity gives $G_n\le G_{n+1}$: their difference is harmonic even at $p$ and nonnegative on $\partial R_n$. The global positive Green function on a Riemann surface $G_R$ similarly majorizes every $G_n$. Comparing the regular parts shows that $c_n$ increases and is bounded above by the finite regular part of $G_R$ at $p$. Consequently

On each fixed compact subdomain, the $f_n$ are eventually bounded holomorphic functions. Montel theorem and a diagonal subsequence therefore give a locally uniform convergence limit $F:R\to \mathbb{D}$ with $F(p)=0$ and $F^{\prime }(p)=d>0$. The maximum modulus principle keeps its values strictly inside the disk. It is injective: for fixed $q$, the functions $f_n-f_n(q)$ have no zeros away from $q$; Hurwitz's theorem implies that their nonconstant limit has no other zeros. This is the locally uniform limit of univalent functions argument.

There is also an extremality property. For any holomorphic function $g:R\to \mathbb{D}$ with $g(p)=0$, the function $g\circ f_n^{-1}$ is a disk self-map fixing zero. The Schwarz lemma gives $|g^{\prime }(p)|\le f_n^{\prime }(p)$. Passing to the limit yields $|g^{\prime }(p)|\le d=F^{\prime }(p)$.

Finally suppose $F$ omits $a\in \mathbb{D}$. Necessarily $a\ne 0$. Apply the disk Möbius transformation ${\varphi }_a(w)=(w-a)/(1-\overline{a}w)$. The nowhere-zero function ${\varphi }_a\circ F$ has a global holomorphic logarithm because $R$ is simply connected, hence a holomorphic square root $h$. It maps into the disk and $|h(p)|=\sqrt{|a|}$. Normalize it by setting $g={\varphi }_{h(p)}\circ h$, so $g(p)=0$. Direct differentiation givesThe strict inequality follows from $(1-\sqrt{|a|}{)}^2>0$. This contradicts extremality. Therefore $F$ omits no point, and we conclude

The Riemann mapping theorem asserts that a nonempty simply connected domain $\Omega ⊊\mathbb{C}$ is biholomorphic to the unit disk $\mathbb{D}$. More precisely, for each $z_0\in \Omega$ there is a unique biholomorphism $f:\Omega \to \mathbb{D}$ satisfying $f(z_0)=0$ and $f^{\prime }(z_0)>0$. We prove existence through an extremal derivative, deriving the compactness and injectivity steps rather than assuming special lemmas for the theorem.

First construct a bounded univalent function on $\Omega$. Choose $a\notin \Omega$ and a holomorphic logarithm of $z-a$, which exists by the permitted logarithm result on the translated simply connected domain. PutThen $u(z{)}^2=z-a$, so $u$ is univalent and never zero. Its image $G=u(\Omega )$ is open, and $G\cap (-G)=\varnothing$: equality $u(x)=-u(y)$ would imply $x=y$ after squaring and then $u(x)=0$.

Fix $w_0\in G$ and choose $r>0$ with $B(w_0,r)\subset G$. It follows that $B(-w_0,r)\subset -G$ is disjoint from $G$, so $|u(z)+w_0|\ge r$ throughout $\Omega$. Henceis a univalent function with $|h|\le 1/2$. For $b\in \mathbb{D}$, defineThe identityand the inverse $T_b^{-1}(w)=(w+b)/(1+\overline{b}w)$ show directly that $T_b$ is an automorphism of the unit disk. Apply $T_{h(z_0)}$ to $h$ and then a rotation to make its nonzero derivative at $z_0$ positive. The family $\mathcal{A}$ of univalent maps $\Omega \to \mathbb{D}$ with value zero and positive derivative at $z_0$ is therefore nonempty.

LetIt is positive. Choose $\rho >0$ with $\overline{B(z_0,\rho )}\subset \Omega$. The Cauchy estimate gives $v^{\prime }(z_0)\le 1/\rho$ for every $v\in \mathcal{A}$, so $M$ is finite. Take a sequence $v_j\in \mathcal{A}$ with $v_j^{\prime }(z_0)\to M$.

We now justify a locally uniform convergence subsequence. On any compact subset of $\Omega$, the bound $|v_j|\le 1$ and the Cauchy estimate on slightly larger interior disks give a uniform derivative bound. They therefore give equicontinuity. Apply the Arzelà-Ascoli theorem on a compact exhaustion of $\Omega$, then take a diagonal subsequence for locally uniform convergence. Its limit $v$ is holomorphic: pass to the limit in the Cauchy integral formula on every small interior circle. The same formula for derivatives givesThus $v$ is nonconstant. Initially $|v|\le 1$, and the maximum modulus principle improves this to $|v|<1$ everywhere.

For completeness, the needed locally uniform limit of univalent functions argument is short. If $x\ne y$ and $v(x)=v(y)$, the nonconstant holomorphic function $v-v(y)$ has an isolated zero at $x$. Choose a small closed disk around $x$ excluding $y$, with no zero on its boundary. Uniform convergence there gives $v_j-v_j(y)\to v-v(y)$. The Rouche theorem then makes $v_j-v_j(y)$ have a zero inside that disk for all sufficiently large $j$. Its only possible zero, by injectivity of $v_j$, is $y$, outside the disk. This contradiction proves injectivity of $v$. Therefore $v\in \mathcal{A}$ and attains the supremum $M$.

It remains to prove surjectivity. Suppose $c\in \mathbb{D}$ is omitted by $v$; since $v(z_0)=0$, necessarily $c\ne 0$. The domain $T_c(v(\Omega ))$ is simply connected, because both constituent maps are homeomorphisms onto their images, and it excludes zero. The permitted holomorphic logarithm gives a holomorphic square rootThis maps $\Omega$ into $\mathbb{D}$ and is univalent, since equality of its values implies equality after squaring and hence equality of the original arguments. Write $\gamma =g(z_0)$. Then ${\gamma }^2=T_c(0)=-c$ and $|\gamma |=\sqrt{|c|}$. Normalize again with $V=T_{\gamma }\circ g$ and a rotation. Its derivative magnitude isThe strict inequality follows from $(1-\sqrt{|c|}{)}^2>0$. This square-root improvement of a normalized conformal map contradicts extremality. Thus $v(\Omega )=\mathbb{D}$. An injective holomorphic map has nonzero derivative and a holomorphic local inverse; these inverses agree, so the resulting bijection is biholomorphic. We have proved existence.

For uniqueness, two normalized maps give an automorphism of the unit disk $v_2\circ v_1^{-1}$ fixing zero. Apply the Schwarz lemma, proved in the next part, to this automorphism and its inverse. Both inequalities force equality, so the equality case makes it a rotation. Its derivative at zero is the positive real number $v_2^{\prime }(z_0)/v_1^{\prime }(z_0)$, forcing that rotation to be the identity. ThereforeThe proper-domain hypothesis is necessary: a bounded entire holomorphic function has derivative at most $1/R$ on disks of arbitrarily large radius by the Cauchy estimate, and hence is constant. Conversely, a domain biholomorphic to the disk is simply connected because the map is a homeomorphism.