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An analytic continuation extends a holomorphic function beyond its original domain without changing its existing values. Formally, if $U\subset V$ are domains and $f$ is holomorphic on $U$, a holomorphic function $F$ on $V$ with $F{|}_U=f$ is an analytic continuation. More locally, holomorphic functions on overlapping open sets continue one another when they agree on the overlap, allowing a chain of patches along a path. The identity theorem for holomorphic functions forces agreement throughout a connected overlap once they agree on a nonempty open set in it. It also makes an extension to a fixed connected domain unique. ^[ref]

The geometric series illustrates why a formula can reach farther than the power series that first represents it. For $z\ne 1$, its partial sum isWhen $|z|<1$, $z^{N+1}\to 0$, so the series converges to $f(z)=1/(1-z)$. The rational function $F(z)=1/(1-z)$ is holomorphic on $\mathbb{C}\setminus \{1\}$ and agrees there with the sum on the unit disc. It therefore continues $f$ to every point except its simple pole at $1$. In particular,Thus analytic continuation gives a value at $2$ without making the original geometric series converge there. Its radius of convergence remains $1$: on $|z|\ge 1$ the terms $z^n$ do not even tend to zero.

The pole at $1$ limits the original Taylor series, although most points of its convergence circle are regular for $F$. Recentring at $-1$ gives another genuinely convergent power series,At $z=-3/2$ this yields $\frac{1}{2}{\sum }_{n\ge 0}(-1/4{)}^n=2/5$, even though the original powers of $-3/2$ grow. This is the useful idea: overlapping local descriptions can represent one holomorphic function on a larger region.

Uniqueness on a connected overlap does not imply that every path leads to the same terminal germ of a holomorphic function. Continuing a complex logarithm around $0$ changes its value by $2\pi i$, giving monodromy; the return patch can carry a different branch of a multivalued function. If two patches have a disconnected overlap, agreement on one component imposes no agreement on the others. ^[ref] Continuation using the rational function $1/(1-z)$ is single-valued: any path avoiding $1$ gives the same terminal germ of a holomorphic function, even if it winds around the pole.

For an injective holomorphic function $f(z)={\sum }_{n\ge 0}{c}_n{z}^n$ on the unit disc, its real Jacobian determinant is $|f^{\prime }(z){|}^2$. Change of variables gives image area ${\int }_{|z|<r}|f^{\prime }(z){|}^{2}dA$. Angular orthogonality removes cross terms of the differentiated power series; integrating $2\pi n^2|c_n{|}^2{\rho }^{2n-1}$ radially gives the title formula. For $r<1$, uniform convergence on a slightly larger closed disk justifies termwise operations. Injectivity is essential for this to be image area rather than multiplicity of a parametrized surface integral.

For $|x|<1$, the binomial series is

For a complex number $\alpha$, set $\left(\genfrac{}{}{0ex}{}{\alpha }{0}\right)=1$ and $\left(\genfrac{}{}{0ex}{}{\alpha }{n}\right)=\alpha (\alpha -1)\cdots (\alpha -n+1)/n!$ for $n\ge 1$. The displayed power series represents $\exp (\alpha \mathrm{Log}(1+z))$ on the unit disc, using the branch of the complex logarithm with value zero at $1$. It terminates for a nonnegative integer $\alpha$. Otherwise the ratio test gives absolute convergence for $|z|<1$. Termwise differentiation gives $(1+z)F^{\prime }(z)=\alpha F(z)$ and $F(0)=1$, which determine that holomorphic function. With $\alpha =-1/2$ and argument $-z$, its coefficients are the normalized central binomial coefficients.

The Blaschke condition for a sequence in the unit disc is ${\sum }_n(1-|a_n|)<\infty$. It characterizes the zero sequences, counted with multiplicity, of nonzero bounded holomorphic functions. Sufficiency follows by constructing the Blaschke product. For necessity, Jensen's formula gives ${\sum }_{|a_n|<r}\log (r/|a_n|)\le \log \Vert f{\Vert }_{\infty }-\log |f(0)|$ when $f(0)\ne 0$. Letting $r\uparrow 1$ and using $-\log |a|\ge 1-|a|$ proves the condition. A finite zero at zero is factored out first. Equivalently, for the curvature-minus-one hyperbolic metric, ${\sum }_n{e}^{-\rho (0,a_n)}<\infty$, since $e^{-\rho (0,a)}=(1-|a|)/(1+|a|)$.

Every nonzero bounded holomorphic function on the unit disc has this factorization, with $B$ the Blaschke product of its zeros and $g$ zero-free and holomorphic. The Blaschke condition follows from Jensen's formula. Dividing by any finite partial zero product gives a holomorphic function; the maximum modulus principle and the finite product's boundary modulus one bound that quotient by $\Vert f{\Vert }_{\infty }$. Locally uniform convergence of the partial products gives the same bound for $g$, with removable extensions at their zeros. This factorization alone does not make $g$ constant or a full inner-outer factorization.

For zeros $a_n$ in the unit disc satisfying ${\sum }_n(1-|a_n|)<\infty$, with $m$ occurrences at zero, a Blaschke product isFor $|z|\le R<1$ and $a\ne 0$, the normalized factor $b_a$ satisfies $|1-b_a(z)|\le (1-|a|)(1+R)/(1-R)$. The product therefore converges locally uniformly; outside its prescribed zeros the logarithms of its tail factors converge absolutely, so it has no additional zeros. Every finite product has modulus at most one in the disc, and the limit does too. Since $|b_a(w)|$ is pseudohyperbolic distance from $w$ to $a$, its modulus can be estimated geometrically.

A sequence in the unit disc, counted with multiplicity, is a Blaschke sequence when it satisfies the Blaschke condition. The zeros of every nonzero bounded holomorphic function form such a sequence: Jensen's formula bounds ${\sum }_{|a_n|<r}\log (r/|a_n|)$ uniformly as $r\uparrow 1$, and $1-|a|\le -\log |a|$ gives the assertion. There are only finitely many zeros at the origin, which can be factored out first.

For a conformal bijection $f:D\to \tilde{D}$ between proper planar domains, put $d=\mathrm{dist}(z,\partial D)$ and $\tilde{d}=\mathrm{dist}(f(z),\partial \tilde{D})$. Apply the Koebe quarter theorem to $(f(z+dw)-f(z))/(d{f}^{\prime }(z))$ on the unit disc to obtain $d|f^{\prime }(z)|/4\le \tilde{d}$. Apply it again to $f^{-1}$ to get the opposite bound. Thus $\boxed{\tilde{d}/(4d)\le |f^{\prime }(z)|\le 4\tilde{d}/d}$. The domains need not be simply connected domains. For the whole plane, the boundary distances are infinite and these ratios are undefined.

A holomorphic function on the unit disc has bounded characteristic when ${\sup }_{r<1}\int {\log }^{+}|f(r\xi )|dm(\xi )<\infty$, where ${\log }^{+}t=\max (0,\log t)$. Thus it belongs to the Nevanlinna class. A general meromorphic function requires the additional pole-counting term in its characteristic; the displayed analytic definition has no such term.

For nonzero prescribed zeros $a_n$ with $|a_n|\to 1$, the Weierstrass elementary factors of increasing order give the displayed holomorphic function on the unit disc; $m$ handles finitely many zeros at zero. On any compact disc, eventually $|z/a_n|\le q<1$, and $|\log E_n(z/a_n)|\le q^{n+1}/[(n+1)(1-q)]$. The infinite product convergence from logarithmic tails gives a nonvanishing tail. The finite initial factors supply exactly the prescribed zeros and multiplicities. No Blaschke condition is required unless boundedness is imposed.

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 $F(z)=i\mathrm{Im}F(0)+\mathrm{Re}F(0)(1+\omega z)/(1-\omega z)$.

A boundary neighborhood over an arc $I$ of the unit circle, with radial depth equal to its normalized angular length $m(I)$. Changing that depth by any fixed positive factor gives an equivalent Carleson measure condition. Large arcs and central mass can be handled separately by finitely many fixed neighborhoods covering the unit disc.

A positive Borel measure on the unit disc satisfies $\int |f|d\mu \le N\Vert f{\Vert }_1$ for every H1 space on the unit disk function exactly when it is a Carleson measure, with comparable optimal constants. The normalized squared Cauchy kernel tests necessity. For sufficiency, the boundary shadow of every point with $|f(z)|>t$ lies inside a component arc of $\{Nf>t\}$. The geodesic-cap and Carleson-box comparison bounds the measure of each shadow region by its component length, giving $\mu \{|f|>t\}\le C\Vert \mu {\Vert }_{\mathrm{Carl}}m\{Nf>t\}$. The layer cake representation and H1 non-tangential maximal inequality on the disk give the embedding. The broader kernel and covering mechanism is discussed in Terence Tao's harmonic-analysis notes, Exercise 2 after Lemma 4.1; the analytic endpoint here uses the fractional-power argument just given.

A positive Borel measure near a domain boundary is a Carleson measure when its mass in boundary neighborhoods is bounded by a constant times their boundary measure. On the unit disc, this becomes the displayed bound for every Carleson box. The equivalent hyperbolic geodesic cap version follows from the geodesic-cap and Carleson-box comparison. The Carleson embedding theorem on the disk explains why this geometric growth condition controls integration of analytic Hardy spaces.

Suppose $f$ is a holomorphic function on an open set containing the closed disc $|z-a|\le R$, with $R>0$. For any nonnegative integer $n$ and $|w-a|<R$, the counterclockwise boundary contour satisfiesThe case $n=0$ is the Cauchy integral formula. On a smaller closed disc, the denominator stays uniformly away from zero, allowing repeated differentiation under the contour integral. This proves that a holomorphic function has complex derivatives of every order. ^[ref]

At the centre, if $M={\max }_{|z-a|=R}|f(z)|$, taking absolute values gives the Cauchy estimate $|f^{(n)}(a)|\le n!M/R^n$. Boundary size therefore limits interior variation. For $f(z)=z^2$ on the unit disc, $M=1$, and the $n=2$ estimate is exact: $|f^{\prime \prime }(0)|=2$. This is one quantitative reason that holomorphic functions are so rigid.

For a start $Z_0=r_0{e}^{i{\alpha }_0}$ in the punctured unit disc, the logarithmic lift $-i\log Z$ starts at ${\alpha }_0-i\log r_0$ in the upper half-plane. Local conformal invariance of planar Brownian motion changes its clock by $\int |Z_s{|}^{-2}ds$. That clock is finite at the disc exit because the stopped path is compact and avoids zero. Its terminal real coordinate is therefore the usual upper-half-plane Cauchy exit law from a Brownian half-plane. Subtracting ${\alpha }_0$ gives the displayed winding law. Its first absolute moment diverges logarithmically, even though the final point modulo angle lies on a compact circle.

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

Every conformal automorphism of the upper half-plane is a Möbius transformation with real coefficients and positive determinant, and conversely every such map is a conformal bijection of the upper half-plane to itself:The real-coefficient and sign conditions matter. The complex Möbius transformation $z\mapsto iz$ sends $i$ to $-1$ on the boundary, while the real Möbius transformation $z\mapsto 1/z$ has negative determinant and sends $i$ to $-i$ in the lower half-plane. ^[ref]

For two concrete upper half-plane automorphisms, takeBoth matrices have determinant one. Here $T(i)=1+i$, $S(i)=i$, and $S(1+i)=(-1+i)/2$. On the extended real line, $T$ fixes infinity and $S$ exchanges zero and infinity. On the unit disc, the same maps become $C\circ T\circ C^{-1}$ and $C\circ S\circ C^{-1}$; the second is simply $w\mapsto -w$.

Multiplying all four coefficients by a nonzero real scalar leaves $f$ unchanged. Dividing them by $\sqrt{\Delta }$ normalizes the determinant to one; the two remaining representatives $A$ and $-A$ give the same map. Thus the group under function composition is the projective special linear groupwhere $S{L}_2(\mathbb{R})$ is the real special linear group of degree two. Composition of the Möbius transformations corresponds to multiplication of their matrices. ^[ref]

On the boundary,So $f$ preserves the cyclic boundary orientation of the extended real line; when $c\ne 0$, its pole at $-d/c$ prevents treating it as a single increasing function across that point on the ordinary real line. Three distinct boundary images determine a Möbius transformation uniquely, but an upper half-plane automorphism can realize only triples with matching cyclic boundary orientation.

A brief proof outline starts with the identitiesThey show that $f$ and its inverse function preserve the upper half-plane. For completeness of the classification, conjugate an arbitrary conformal bijection by the Cayley transform. Normalize its image of zero with an automorphism of the unit disk, then apply the Schwarz lemma to the resulting map and its inverse function; both inequalities force a rotation. Conjugating back gives the real Möbius transformation above.

Finally, $|f^{\prime }(z)|=\Delta /|cz+d{|}^2$ and the displayed imaginary part identity show that $|df(z)|/\mathrm{Im}f(z)=|dz|/\mathrm{Im}z$. Every upper half-plane automorphism is therefore also an isometry of the Poincaré half-plane model. The same formulas describe both holomorphic coordinate symmetries and motions of the hyperbolic plane.

The map $C(z)={\tan }^2(\pi \sqrt{z}/4)$ is a conformal map from $\{x+iy:y^2<4(1-x)\}$ onto the unit disc. A principal square root first maps the slit parabolic region to a vertical strip, and a scaled tangent followed by squaring maps the strip to a slit disc. Evenness in the square-root variable gives a holomorphic extension across the removed axis, filling the target slit.

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.