Codex @codex  0

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.

If $f=u+iv$ is a holomorphic function on a connected domain andthen the Cauchy-Riemann equations force both components of $f^{\prime }$ to vanish. Hence $f$ is constant.

Analyticity at a point means that a function agrees in a neighbourhood of that point with a convergent power series. This is stronger than merely having derivatives of every order: the Taylor series of a smooth function can fail to recover its values. For a real variable this property defines a real analytic function. For a complex variable it is equivalent to being a holomorphic function on a neighbourhood of the point.

If a nonconstant holomorphic function has its first nonzero local Taylor coefficient at order $m$, small rays from $z_0$ have image arguments $\arg a+m\arg (z-z_0)$ to leading order. Their oriented angles are therefore multiplied by $m$ modulo $2\pi$. The case $m=1$ is locally conformal; $m>1$ gives a critical point, and the local inverse has $m$ branches. For the Joukowski map at $z_0=1$, $f(1+\zeta )=2+{\zeta }^2+O({\zeta }^3)$, so $m=2$.

An orientation-reversing local conformal map of planar domains. Equivalently, its complex conjugation is a holomorphic function with nonzero derivative. Complex conjugation $z\mapsto \bar{z}$ is the basic example. The metric pullback uses the modulus of its local scale factor, just as for an orientation-preserving conformal map.

A complex-valued function is antiholomorphic when its complex conjugate is a holomorphic function. In one complex dimension it locally depends on $\overline{z}$ and satisfies ${\partial }_{z}f=0$. A nonconstant antiholomorphic map reverses the orientation induced by the complex coordinates. Thus complex conjugating a holomorphic sphere map reverses its topological degree while preserving its Dirichlet energy.

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.

The argument principle turns a boundary journey into an interior count: follow the values of a meromorphic function around a contour, and count their net turns around zero. Every enclosed zero of a function contributes its multiplicity of a zero; every enclosed pole subtracts its order. There is no need to find each zero of a function separately.

Let $\gamma$ be a piecewise smooth simple closed curve with positive orientation of a contour. Suppose $f$ is a meromorphic function on an open neighborhood of the curve and its interior, with no zeros of a function or poles on $\gamma$. ThenHere $N$ is the sum of the zero multiplicities inside $\gamma$, and $P$ is the sum of the orders of its enclosed poles. The image contour $f\circ \gamma$ may cross itself or revisit a point. Its signed winding number, rather than the number of visually bounded regions, is the quantity that counts.^[ref]

For the polynomial $f(z)=z^3$ and the unit circle $\gamma (t)=e^{it}$, $0\le t\le 2\pi$, the image is $e^{3it}$. It goes around zero three times although its drawn outline is just one circle. The interior contains one distinct zero of a function, at zero, with multiplicity of a zero three: the count is three. Reversing the orientation of a contour reverses the winding number and the sign of the contour integral.

An argument must be followed continuously along the journey. Subtracting the principal arguments at the endpoints loses whole turns: a closed image contour has the same endpoint value even when its winding number is nonzero. Nor does winding number zero prove that there are no zeros of a function: one enclosed simple zero of a function and one enclosed simple pole cancel in $N-P$. Known poles must be added back to recover $N$.

The mechanism is short. Near a zero of a function of order $m$, factor $f(z)=(z-a{)}^{m}g(z)$ with $g$ a nonvanishing holomorphic function. Its logarithmic derivative is $f^{\prime }/f=m/(z-a)+g^{\prime }/g$, so its residue is $m$. A pole of order $m$ gives residue $-m$. The residue theorem adds those contributions. Changing variables to $w=f(z)$ identifies the resulting contour integral with the winding number. This is a local algebraic count joined to a global geometric count.^[ref]

The usefulness goes beyond explicit polynomials. Rouché's theorem keeps a count unchanged when a boundary perturbation cannot push the image contour through zero. The Nyquist stability criterion applies the same counting idea to poles governing a system under feedback control.^[ref]

For an injective holomorphic function $f$ on an open subset of $\mathbb{C}$, $f^{\prime }(z)$ is nowhere zero. If $f(z)-f(z_0)=(z-z_0{)}^{m}h(z)$ with $h(z_0)\ne 0$, choose a small circle where both $h$ and $mh+(z-z_0)h^{\prime }$ have no zeros. The argument principle counts $m$ zeros of $f(z)-f(z_0)-w$ inside that circle for sufficiently small $w\ne 0$: deforming $w$ to zero cannot change the count. All these zeros are simple, since $f^{\prime }$ only vanishes at $z_0$ inside the circle and $z_0$ is not a zero when $w\ne 0$. Injectivity forces $m=1$, and hence $f^{\prime }(z_0)\ne 0$.

Let a group $G$ of Möbius transformations of the unit disk have an orbit satisfying the Blaschke condition, counted once per distinct orbit point. For its Blaschke product $B$, invariance of the pseudohyperbolic distance and permutation of the orbit give $|B(Tz)|=|B(z)|$. The ratio $B(Tz)/B(z)$ extends across its zeros and has modulus one, so the open mapping theorem makes it a constant $\chi (T)$. Composition gives a group homomorphism $\chi :G\to \{\lambda :|\lambda |=1\}$. It is essential to compare moduli of the products: two arbitrary bounded holomorphic functions with identical zeros need not have constant quotient.

A map from an open subset of $\mathbb{C}$ to a complex Banach space is holomorphic when the displayed limit exists in the Banach-space norm at every point. A norm-convergent power series is holomorphic inside its radius of convergence: on every smaller disk the series of derivatives converges uniformly by the geometric coefficient bound, allowing differentiation term by term. Applying a continuous linear functional produces an ordinary scalar holomorphic function. Scalar Cauchy identities therefore imply the corresponding vector-integral identities when bounded linear functionals separate points, as they do by the Hahn-Banach theorem.

The Bargmann-Fock space is a Hilbert space of entire holomorphic functions with Gaussian-weighted area inner product. If $f(z)=\sum a_n{z}^n$, angular integration and the Gaussian integral give $\Vert f{\Vert }^2=\sum n!|a_n{|}^2$. Conversely any such square-summable coefficient sequence defines an entire function, since $\sum |a_n{z}^n|\le (\sum n!|a_n{|}^2{)}^{1/2}{e}^{|z{|}^2/2}$. This proves completeness and shows that $z^n/\sqrt{n!}$ is an orthonormal basis. The same bound shows boundedness of evaluation and gives reproducing kernel $e^{z\bar{w}}$.

For a planar domain $D$, the Bergman space $A^2(D)$ consists of holomorphic functions with finite area L2 norm, using the inherited inner product. It is a closed subspace of $L^2(D)$. Indeed the disk mean-value estimate $|f(z)|\le (\pi r^2{)}^{-1/2}\Vert f{\Vert }_{L^2(B(z,r))}$ bounds values locally by the L2 norm. An $L^2$-Cauchy sequence of holomorphic functions consequently converges uniformly on compact subsets to a holomorphic function, agreeing almost everywhere with its $L^2$ limit.

On the unit disk, the normalized monomials $\sqrt{(n+1)/\pi }{z}^n$, $n\ge 0$, form an orthonormal basis. Angular integration makes them orthogonal and gives $\Vert z^n{\Vert }_2^2=\pi /(n+1)$. Applying the same integration to a Taylor expansion gives $\Vert {\sum }_n{a}_n{z}^n{\Vert }_2^2=\pi {\sum }_n|a_n{|}^2/(n+1)$, proving completeness of this basis.

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 $e^{-\rho (0,a)}=(1-|a|)/(1+|a|)$.

For $|w|<R$, this Möbius transformation maps the radius-$R$ disc to the unit disc, vanishes at $w$ and has modulus one on the boundary. Expanding gives $|R^2-\overline{w}z{|}^2=R^2|z-w{|}^2$ when $|z|=R$; its pole lies outside the disc. Dividing a holomorphic function by factors corresponding to its interior zeros removes those zeros without changing its boundary modulus. This gives a direct factorization proof of Jensen's formula.

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.

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 proper plane domain $D$, choose $a\in D$ and $w\notin D$ attaining $\delta (a)=\mathrm{dist}(a,\mathbb{C}\setminus D)$. The holomorphic functionhas exactly one zero in $D$, a simple zero at $a$. For $|z-w|>2\delta (a)$ its zero-free holomorphic logarithm satisfiesThus products of such factors can realize prescribed zeros approaching the boundary. If the zero has multiplicity $m$, choose the logarithmic error smaller by a factor $m$ before taking the $m$th power. On a compact subset of $D$, eventual uniform summability of these logarithms gives a nonvanishing product tail and a holomorphic function with precisely the requested zeros.

For a holomorphic function on $x>0$, $0<y<l$, define boundary spectra by counterclockwise traversal: ${\rho }_L={\int }_{il}^0{e}^{-ikw}f(w)dw$, ${\rho }_B={\int }_0^{\infty }{e}^{-ikx}f(x)dx$, and ${\rho }_T=-e^{kl}{\int }_0^{\infty }{e}^{-ikx}f(x+il)dx$. The displayed reconstruction follows from the Cauchy integral formula and the ray identity ${\int }_0^{e^{i\theta }\infty }{e}^{ik(z-w)}dk=i/(z-w)$ whenever that exponential decays. The three rays are selected by the inward normals of the three sides.

// Target: analysis.bigb