Holomorphic function ($f\in \mathcal{O}(U)$)

A complex-valued function $f:U\to \mathbb{C}$ is holomorphic on an open set $U$ when it has a complex derivative at every point of $U$. Requiring a derivative throughout a neighbourhood is much stronger than complex differentiability at a point. It forces locally convergent Taylor series and derivatives of every order, as follows from the Cauchy integral formula and Cauchy derivative formula.

To see the restriction in real coordinates, write $z=x+iy$ and $f(z)=u(x,y)+iv(x,y)$. The real and imaginary increment quotients must agree, giving the Cauchy-Riemann equationsConversely, if $u,v$ have continuous partial derivatives on $U$ and satisfy the Cauchy-Riemann equations there, then $f$ is a holomorphic function. The equations at a lone point, without a differentiability hypothesis for the real coordinate map, are insufficient. Their role is to force the Jacobian matrix to have the special formSuch a matrix represents multiplication by a complex number; it has only two free real entries. ^[ref]

For $f(z)=z^2$, $u=x^2-y^2$ and $v=2xy$, so $u_x=v_y=2x$ and $u_y=-v_x=-2y$. Directly,At $z_0=1+i$, multiplication by $2+2i$ scales lengths by its modulus $2\sqrt{2}$ and rotates by its argument $\pi /4$. A horizontal increment $h=0.01$ gives the exact displacement $0.0201+0.02i$; a vertical increment $h=0.01i$ gives $-0.0201+0.02i$. The first-order predictions $0.02+0.02i$ and $-0.02+0.02i$ are perpendicular, and the errors have modulus $10^{-4}$.

At every nonzero $z_0$, this local rotation and scaling makes $z^2$ a local conformal map. At zero, the first-order map vanishes: $r{e}^{i\theta }$ becomes $r^2{e}^{2i\theta }$, so the argument doubles and the origin is a critical point. Globally $z$ and $-z$ have the same square. Local preservation of angles therefore coexists with global folding, a useful warning when using holomorphic functions as coordinate changes.

A map from an open subset of $\mathbb{C}$ into a complex Banach space is weakly holomorphic when every composition with a bounded complex-linear functional is a scalar holomorphic function. Norm holomorphy requires its difference quotients to converge in the Banach norm. Weak holomorphy is initially a condition on scalar observations; the weak holomorphy implies norm holomorphy argument recovers an actual vector derivative. A closed target subspace containing values on a sequence with an interior accumulation point contains every value on the connected domain: annihilate the subspace, apply the scalar identity theorem, and use Hahn-Banach separation.

For a weakly holomorphic Banach-valued function, choose a closed disk contained in its domain. Each continuous functional is bounded on the image of that disk by scalar continuity. The Uniform boundedness principle, applied to its evaluation functionals in the bidual, gives a common bound $M$ on the norms of the image vectors. Cauchy's formula on the disk's boundary then bounds the scalar difference between difference quotients at increments $h,k$ by a constant times $M\Vert \phi \Vert |h-k|$, uniformly for sufficiently small increments. Explicitly the difference kernel isThe Hahn-Banach theorem dual norm formula converts this into the same norm bound for the vector difference quotients. They are Cauchy, so completeness gives a norm derivative. This also explains why weak holomorphy suffices without a separately assumed norm-continuity hypothesis.

A holomorphic function $F$ on a larger open set $V$ whose restriction to $U$ is the original holomorphic function $f$. A removable singularity is a point across which such an extension exists. On a connected larger domain, a holomorphic extension is unique by the identity theorem, provided $U$ is nonempty.

The rational function $F(z)=z^2/(1-z{)}^2$ is holomorphic on the unit disc and maps it onto the entire complex plane. For any $w\in \mathbb{C}$, choose a square root $r$ with $\mathrm{Re}r\ge 0$ and put $z=r/(1+r)$. The identity $|1+r{|}^2-|r{|}^2=1+2\mathrm{Re}r>0$ gives $|z|<1$, while $F(z)=r^2=w$. It is not injective: the two choices $r=\pm i$ give distinct points in the disc with image $-1$.

A holomorphic bijection in this direction is impossible: its inverse would be a bounded entire function, contradicting the Liouville theorem. Thus a surjection onto the complex plane can exist even when a conformal equivalence cannot.

If $f$ is holomorphic on an open set $U$, its conjugate reflection $f^{\ast }$ is holomorphic on $U^{\ast }=\{z:\overline{z}\in U\}$, with $(f^{\ast }{)}^{\prime }(z)=\overline{f^{\prime }(\overline{z})}$. The identity follows by conjugating the difference quotient with increment $\overline{h}$; the two conjugations restore complex differentiability.

If $U$ is invariant under complex conjugation and is connected, and $f$ is real at a set of real points accumulating inside $U$, then $f=f^{\ast }$ by the identity theorem. In particular, an entire function real at every $1/n$ has this symmetry. This is a symmetry statement on an already defined domain; Schwarz reflection also constructs extensions across a boundary.

A complex-valued function is complex differentiable at $z_0$ ifexists. The same limit must hold for every approach, including curved paths. Equivalently,The error divided by $|h|$ tends to zero. Thus the derivative is a single complex multiplier, not an arbitrary real linear map of the plane. If the real coordinate map is differentiable, complex differentiability is equivalent to its satisfying the Cauchy-Riemann equations at that point.

The simplest failure is complex conjugation, $f(z)=\overline{z}$. At every $z_0$ its increment quotient is $\overline{h}/h$. For real $h=t\ne 0$ it is $1$, while for $h=it$ it is $-1$. More generally, $h=t{e}^{i\theta }$ gives $e^{-2i\theta }$. Although complex conjugation is a smooth real reflection and has every real directional derivative, these incompatible quotients rule out a complex derivative everywhere.

Agreement on two axes is necessary, but does not by itself establish a limit over the whole complex plane. The Cauchy-Riemann equations need the real differentiability condition above, or a sufficient regularity condition such as continuous partial derivatives near the point. There is a second distinction: $f(z)=|z{|}^2$ has complex derivative zero at zero because its quotient there is $\overline{h}\to 0$, but its Cauchy-Riemann equations fail at every nonzero point. It is complex differentiable at zero without being a holomorphic function on any neighbourhood of zero.

At zero, $[f(h)-f(0)]/h\to 1$, so the complex derivative exists. At a nonzero point, write $r=|z|$. The Cauchy-Riemann equations for $x/(1+r)$ and $y/(1+r)$ force $x^2=y^2$ and $xy=0$, impossible when $r>0$. Hence the function is complex differentiable only at zero and is holomorphic on no nonempty open subset. A complex derivative at one point gives no neighborhood of complex differentiability.

The analytic Nevanlinna class consists of holomorphic functions on the unit disc with uniformly bounded means of their positive logarithmic modulus. Here $m$ is normalized angular measure. A bounded holomorphic function belongs to this class, but the class also permits unbounded functions. It is the analytic version of bounded characteristic.

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.

A meromorphic function on the unit disc has bounded characteristic if and only if it has this representation. Its pole Blaschke product $B$ makes $g=Bf$ analytic with bounded positive logarithmic means. The least harmonic majorant by expanding disk lifts gives $H\ge {\log }^{+}|g|$. A harmonic conjugate gives an analytic $A$ with $\mathrm{Re}A=H$, so $u=g{e}^{-A}$ and $v=B{e}^{-A}$ are bounded. Conversely the Nevanlinna first main theorem bounds the characteristic of $u/v$ by that of $u$ and $1/v$, hence by a constant.

A holomorphic function that is not identically zero has a zero of multiplicity $m$ at $z_0$ if $f(z)=(z-z_0{)}^{m}h(z)$ for a holomorphic function $h$ with $h(z_0)\ne 0$. Its Taylor series identifies $m$ as the first nonzero coefficient index. Logarithmic differentiation gives $f^{\prime }/f=m/(z-z_0)+h^{\prime }/h$, so the residue of $f^{\prime }/f$ is $m$. Counting zeros with zero multiplicity repeats that zero $m$ times.

If a holomorphic function of $w$ is periodic with period one, then $F((\log z)/(2\pi i))$ is independent of the choice of a local logarithm: two choices differ by $2\pi ik$, hence their arguments differ by an integer. Consequently it defines a holomorphic function on the image of the original domain under $w\mapsto e^{2\pi iw}$, provided that domain is invariant under integer shifts. Local logarithms prove holomorphy across an arbitrary global branch cut. Singularities on an integer lattice all map to $z=1$.

A holomorphic function with constant modulus on a connected domain is constant. If $h=u+iv$ and $u^2+v^2=c^2>0$, differentiation and the Cauchy-Riemann equations give $u{u}_x+v{v}_x=0$ and $-u{v}_x+v{u}_x=0$. Their coefficient determinant is $-c^2$, so $u_x=v_x=0$, and all first derivatives vanish. When $c=0$, the conclusion is immediate. Consequently two nonvanishing holomorphic functions with equal moduli differ by a single constant phase on a connected domain.

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.

For a Banach algebra-valued holomorphic function, set $p_n(w)=\Vert f(w{)}^{2^n}{\Vert }^{1/2^n}$. These continuous functions decrease to the spectral radius by the spectral radius formula. If $M={\sup }_{\partial K}r(f(w))$, the continuous functions $\max \{p_n,M\}$ on $\partial K$ decrease to the continuous constant $M$. The Dini theorem gives uniform convergence. Apply the norm maximum principle for a Banach-space-valued holomorphic function to each power, and then let $n\to \infty$. This proof does not assume continuity of spectral radius.

For a compact set $K$ in the domain, a norming bounded linear functional at $f(z)$ reduces the assertion to the scalar maximum modulus principle. If $z$ lies inside $K$, use its connected component of the interior; that component's boundary lies in $\partial K$. If $z\in \partial K$, the inequality is immediate. No smoothness or connectedness assumption on $K$ is required.

Every closed discrete subset of a plane domain is the zero set of a holomorphic function, with assigned finite positive multiplicities if desired. Exhaust the domain by relatively filled compact sets. For a prescribed point outside an earlier compact, choose a rational factor with that zero and its possible pole outside the domain in the same complementary component. A slit joining zero and pole allows a holomorphic logarithm near the earlier compact. Approximate that logarithm by Runge theorem and multiply the rational factor by the negative exponential of the approximant. The resulting factors tend to one with summable errors on each compact. Their infinite product has the prescribed zeros and no additional zeros.

// Destination: analysis.bigb

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 compact $K$ in a plane domain $\Omega$, its holomorphic convex hull isIt is $K$ together with the components of $\hat{\mathbb{C}}\setminus K$ lying entirely in $\Omega$. Inclusion of those components follows from the maximum modulus principle. For exclusion, suppose $w$ lies in a complementary component meeting $\hat{\mathbb{C}}\setminus \Omega$. Choose an allowed pole there. The open-and-closed resolvent argument in the proof of Runge theorem approximates $(z-w{)}^{-1}$ uniformly on $K$ by functions $h$ holomorphic on $\Omega$. Then $1-(z-w)h(z)$ has value one at $w$ and arbitrarily small modulus on $K$, separating $w$ from the hull. For $\Omega =\mathbb{C}$, this is the polynomial hull in one complex variable.

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 finite Blaschke product is holomorphic across the unit circle and has modulus one there. For a constant $a$ with $|a|<1$, Rouché's theorem makes $B-a$ and $B$ have the same number of interior zeros, counting multiplicity. Each factor contributes its prescribed zero; the poles are outside the closed disk.

Use curvature-minus-one Hyperbolic distance in the Poincare disc. Each Blaschke factor has modulus $\tanh (\rho (w,a)/2)=(1-t)/(1+t)$, with $t=e^{-\rho (w,a)}$. For $0\le t<1$, differentiating shows $\log [(1+t)/(1-t)]\ge 2t$. Sum this logarithmic inequality over the zero factors and pass to the Blaschke product by locally uniform convergence. At a prescribed zero, its modulus vanishes and the bound is immediate. All sums count multiplicities.

For a holomorphic function on the unit disc, this boundary condition characterizes Blaschke products, including unimodular constants. The nondecreasing radial means of the nonnegative subharmonic function ${\log }^{+}|f|$ must vanish, so $|f|\le 1$. In the Blaschke factorization of a bounded holomorphic function $f=Bg$, both logarithmic means tend to zero. The mean value property for harmonic functions applied to $\log |g|$ gives $|g(0)|=1$, and the maximum modulus principle makes $g$ unimodular constant. The converse follows from the radial logarithmic mean of a Blaschke product. Omitting the outer absolute value changes the condition: $e^z$ has zero mean logarithmic modulus, but is not a Blaschke product.

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 $B(z)=\lambda z^m{\prod }_n{b}_{a_n}(z)$ with nonzero zeros $a_n$ counted with multiplicity, Jensen's formula gives the displayed identity. The negative logarithms of partial products increase, so the monotone convergence theorem permits termwise integration. The Blaschke condition implies ${\sum }_n-\log |a_n|<\infty$, which dominates the absolute summands. The dominated convergence theorem therefore gives a limit of zero as $r\uparrow 1$. Since $|B|\le 1$, the mean of $|\log |B||$ tends to zero as well.

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.

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

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.

Let $R>0$ and let $f$ be a holomorphic function on an open neighbourhood of the closed disk $\{z:|z|\le R\}$. Assume $f(0)\ne 0$ and that $f$ has no zeros of a function on the boundary circle $|z|=R$. List its interior zeros of a function as $a_1,\dots ,a_N$, repeating each according to its multiplicity of a zero. With the natural logarithm, Jensen's formula statesThe empty sum is zero. This boundary-zero-free version is the holomorphic special case of the general identity in the source.^[ref]

Every interior zero of a function makes a positive contribution; a zero nearer the centre contributes more. Consequently boundary growth relative to $|f(0)|$ limits how many zeros of a function can lie well inside the disk. The Jensen zero-count bound makes this quantitative, while the integrated zero count retains the individual radial weights. The boundary statistic here is the circle log-mean of a holomorphic function; exponentiating gives a circle geometric mean of a holomorphic function.

Under the hypotheses of Jensen's formula, the circle log-mean is the angular average of the natural logarithm of the complex modulus. The factor $1/(2\pi )$ normalizes the integral to a mean. It averages values at equally weighted angles on the circle, and Jensen's formula computes it from $|f(0)|$ and the interior zeros of a function.

The continuous angular geometric mean of the positive boundary complex modulus isHere the empty product is one. This differs from the angular arithmetic mean $A_f(R)=(2\pi {)}^{-1}{\int }_0^{2\pi }|f(R{e}^{i\theta })|d\theta$. In general $G_f(R)\le A_f(R)$ by the concavity of the logarithm; equality requires constant boundary complex modulus. For a holomorphic function with no zeros of a function in the closed disk, Jensen's formula gives $G_f(R)=|f(0)|$, even when the boundary complex modulus varies with angle. Thus replacing the mean of a logarithm by the logarithm of an arithmetic mean loses the exact zero identity.

For a polynomial $p(z)=c{\prod }_j(z-a_j)$ with $c\ne 0$, no polynomial root at zero, and $R>0$ different from every $|a_j|$, Jensen's formula gives the displayed expression. The polynomial roots are repeated according to their multiplicity of a zero: an exterior root contributes $\log |a_j|$, whereas an interior root contributes $\log R$.

For an explicit example, take the polynomialIts two simple zeros are $1/4$ and $-1/2$. The circle log-mean of a holomorphic function is thereforeAt $R=0.2$ the open disc contains no zeros of a function, at $R=0.4$ it contains $1/4$ only, and at $R=0.75$ it contains both. These three circles avoid the polynomial roots; the radii $1/4$ and $1/2$ are excluded from the displayed version.

The corresponding values of the circle log-mean of a holomorphic function and the circle geometric mean of a holomorphic function areFor example, at $R=0.4$ the contribution of the sole interior zero of a function is $\log (0.4/0.25)=\log (8/5)$; adding it to $\log (3/8)$ gives $\log (3/5)$. The second polynomial root still affects $f(0)$, but contributes no additional interior-zero term at this radius.

Fix a holomorphic function on a neighbourhood of a closed disk, with $f(0)\ne 0$. On an interval of positive radii whose circles contain no zeros of a function, differentiating Jensen's formula with respect to $x=\log R$ givesEach interior zero of a function contributes slope one for each unit of its multiplicity of a zero. Between consecutive zero moduli, $L_f(e^x)$ is an affine function; passing the modulus of a zero of multiplicity of a zero $m$ increases the adjacent slopes by $m$. The finite sum $\log |f(0)|+{\sum }_j\max (0,x-\log |a_j|)$ defines a continuous convex function of $x$ whose values agree with the circle log-mean of a holomorphic function at every boundary-zero-free radius. This describes the adjacent line segments without asserting a boundary-singularity extension of the integral.

For a nonzero holomorphic function, $n_f(r)$ counts its zeros of a function in the centred open disc of radius $r$, with multiplicity of a zero. Thus a double zero counts twice, even though it is a single point. On every smaller closed disk in the domain the count is finite. The strict inequality $|a|<r$ specifies the convention at a zero modulus; elsewhere it agrees with counting in the closed disk. In the polynomial circle log-mean example the count is zero, one and two on the three displayed intervals.

Under the hypotheses of Jensen's formula, $f(0)\ne 0$ ensures that the radial zero-counting function vanishes for sufficiently small positive $t$. Integrating each zero's contribution separately givesThe values of $n_f$ at its finitely many jump points do not change the integral. For $0<r<R$, every zero in the smaller open disc supplies at least $\log (R/r)$ per unit of multiplicity of a zero, soThe second inequality bounds the circle log-mean of a holomorphic function by the logarithm of the largest boundary complex modulus, recovering the Jensen zero-count bound. An upper bound on boundary growth and a fixed nonzero centre value therefore constrain interior zeros. The factor $\log (R/r)$ also explains why the same estimate becomes weaker as the smaller radius approaches the boundary.

Divide $f$ by its finitely many interior zero factors, with each factor repeated according to its multiplicity of a zero. The resulting holomorphic function $g$ has no zeros of a function in the closed disk, so $\log |g|$ is a harmonic function and its circle average is $\log |g(0)|$ by the mean value property for harmonic functions. For each removed factor $z-a$, with $0<|a|<R$, the circle average of $\log |R{e}^{i\theta }-a|$ is $\log R$: factor out $R{e}^{i\theta }$ and apply the mean value property for harmonic functions to $\log |1-(a/R)w|$ on a neighbourhood of the closed unit disk. Reassembling the factors gives $\log (R/|a|)$ for each zero, proving Jensen's formula.

For a meromorphic function with no zero or pole on $|w|=R$, set $G_R(z,a)=\log |(R^2-\overline{a}z)/(R(z-a))|$. At source points that are neither zeros nor poles, the displayed identity expresses its logarithmic modulus as the Poisson integral of its boundary values, minus its zero Green potentials plus its pole Green potentials. The Poisson kernel is $(R^2-|z{|}^2)/|R{e}^{i\theta }-z{|}^2$ with normalized angular measure. Subtracting the corresponding logarithmic singularities leaves a harmonic function, so the harmonic Poisson formula proves the identity. At zero it reduces to Jensen's formula.

For a meromorphic function near $0<|z|\le 1$, and a target $a$ off its image of the unit circle, let $u_a=-\log k(f,a)$ for the chordal metric, $M_a(s)$ its normalized angular mean, and $I(a)=M_a^{\prime }(1)$. The weighted spherical area $T(r)=(4\pi {)}^{-1}{\int }_{r<|z|<1}\log (|z|/r)(f_{\mathrm{round}}^{\#}{)}^{2}dA$ satisfies the displayed formula, where $N$ counts $a$-points with multiplicity and weight $\log (|z|/r)$ and $\overline{m}=M_a(r)-M_a(1)$. Indeed $\Delta u_a=(f_{\mathrm{round}}^{\#}{)}^2/2-2\pi \sum {\mathrm{deg}}_z(f){\delta }_z$, and Green's second identity with the radial logarithmic weight proves the formula. The angular mean must include $1/(2\pi )$.

Every zero in the smaller disk contributes at least $\log (R/r)$ in Jensen's formula. Bounding the outer circle average by its maximum proves the estimate. With $R=2r$, an exponential $O(r\log r)$ maximum bound gives an $O(r\log r)$ zero count.

A nonzero entire function vanishing at every nonzero point of the integer lattice has at least a constant times $R^2$ zeros in the disc of radius $R/2$. Each contributes at least $\log 2$ to Jensen's formula at radius $R$. Therefore the circle mean of $\log |F|$, and hence the logarithm of its maximum modulus, is bounded below by $c{R}^2$ for all sufficiently large boundary-zero-free radii. Choosing maximizing points yields $|F(z_j)|>e^{c^{\prime }|z_j{|}^2}$ for some $c^{\prime }>0$ and $|z_j|\to \infty$. If zero is also a zero, first factor out its finite multiplicity; the additional $m\log R$ term does not weaken the quadratic lower bound.

A holomorphic primitive of $f$ is a holomorphic function $F$ with $F^{\prime }=f$. A holomorphic $f$ on a domain has such a primitive exactly when its integrals around all closed piecewise smooth curves vanish. Integrating from a fixed starting point then constructs a path-independent primitive.

A function is complex differentiable at $p$ whenIf $f^{\prime }(p)\ne 0$, the real derivative is multiplication by a nonzero complex number, hence a rotation and scaling; this is the local source of angle preservation by holomorphic maps.

The complex derivative exists when this limit has the same finite value for increments $h$ approaching zero from every complex direction. For $f=u+iv$, real and imaginary increments give the necessary Cauchy-Riemann equations. Existence of those partial derivatives at one point alone is not sufficient for complex differentiability at a point; continuous partial derivatives in a neighbourhood, together with the Cauchy-Riemann equations there, give a standard sufficient condition for $f$ to be a holomorphic function.

For an open set $U\subseteq \mathbb{C}$, the space $\mathcal{O}(U)$ consists of all holomorphic functions on $U$. Its standard topology is locally uniform convergence, equivalently the compact-open topology defined by the seminorms $f\mapsto {\sup }_{z\in K}|f(z)|$ for compact $K\subset U$.

Let $R,\alpha >0$ and ${\mathcal{C}}_{\alpha }=\{(x,t)\in {\mathbb{C}}^2:|x|<R,|t|<\alpha (1-|x|/R)\}$. The displayed norm makes the holomorphic functions with finite norm a Banach space. Finiteness forces $u(x,0)=0$: for fixed $x$, the weight grows like $1/|t|$ near zero. A Cauchy sequence in this norm converges uniformly on every compact subset of ${\mathcal{C}}_{\alpha }$, including at $t=0$. Its limit is holomorphic by locally uniform convergence of holomorphic functions, and the pointwise norm bounds show convergence in the original norm. The shrinking domain compensates for loss of an $x$-derivative in a Cauchy-Kovalevskaya theorem argument.

Write $N=\Vert u{\Vert }_{\alpha }$, $r=|t|$ and $A=\alpha (1-s)>r$. Along the straight integration segment, choose $\sigma (\tau )=s+(A-\tau )/(2\alpha )$. A Cauchy estimate on a disc of radius $R(\sigma -s)$ bounds the integrand by $4\alpha N\tau /[R(A-\tau {)}^2]$. ThereforeThe integral is taken at fixed $x$, and the disc and segment lie inside the shrinking domain. Consequently $u\mapsto {\int }_0^t(i{u}_x+f)dz$ is a contraction mapping when $0<\alpha <R/4$ and the holomorphic source $f$ is bounded on the relevant closed polydisc. The source term has weighted norm at most $\alpha \sup |f|$, so the Banach fixed-point theorem gives a holomorphic solution with zero initial data.

If a nonzero holomorphic function has expansionthen $m$ is the order or multiplicity of its zero at $z_0$.

A nonzero holomorphic function on the unit disc can have exactly a prescribed sequence of zeros, counted with multiplicity, if and only if that sequence is locally finite in the disc. Necessity is the identity theorem for holomorphic functions. For sufficiency, remove a finite number $m$ of zeros at zero and enumerate the others as $a_n$, so $|a_n|\to 1$. Use the Weierstrass elementary factors $E_n(t)=(1-t)\exp ({\sum }_{j=1}^n{t}^j/j)$. On every compact subdisc, eventually $|z/a_n|\le q<1$ and $|\log E_n(z/a_n)|\le q^{n+1}/((n+1)(1-q))$. Hence $z^m{\prod }_n{E}_n(z/a_n)$ converges locally uniformly and has exactly the prescribed zeros. Finite sequences simply give finite products.

A simple zero has order one, equivalently $f(z_0)=0$ and $f^{\prime }(z_0)\ne 0$.

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.

A boundary of a domain of a holomorphic function is natural if the function has no analytic continuation through any of its points. For $f(z)={\sum }_{n\ge 1}{z}^{2^n}$, each root of unity of order a power of two is a singular boundary point: along $z=r\zeta$ all sufficiently late terms equal $r^{2^n}$ and their sum tends to infinity as $r\uparrow 1$. These roots are dense on the unit circle, so no boundary arc admits analytic continuation.

Zeta function regularization assigns a finite value to a divergent spectral sum by first forming a convergent complex-power series and then using its analytic continuation. For positive frequencies ${\omega }_n$, continue ${\sum }_n{\omega }_n^{-s}$ to $s=-1$ to define a regulated frequency sum, when that continuation is regular there. This does not turn the divergent positive-term sum into an ordinary convergent sum. It is a useful subtraction convention for the normal-ordering constant of a string and for vacuum energies.

For $r=n+1/2$, $n\ge 0$, the convergent frequency zeta function, expressed through the Riemann zeta function, is ${\sum }_r{r}^{-s}=(2^s-1){\zeta }_R(s)$ when $\mathrm{Re}s>1$. Its analytic continuation at $s=-1$ gives $(1/2-1)(-1/12)=1/24$. A common exponential frequency cutoff gives ${\sum }_{r}r{e}^{-\epsilon r}={\epsilon }^{-2}+1/24+O({\epsilon }^2)$, so the same finite part follows by subtracting the leading divergence. In contrast, integer frequencies have regulated sum $-1/12$.

A meromorphic continuation extends a holomorphic function from a connected open set to a larger connected domain as a meromorphic function, allowing isolated poles. Equality on the initial open set determines the extension uniquely by the identity theorem. Integral formulas whose integrands depend holomorphically on a parameter often supply such an extension after contour deformation or subtraction of singular terms.

As distributions on the real line,The Sokhotski--Plemelj formula relates the two boundary values of a Cauchy transform across its contour and fixes the spectral delta functions generated by retarded and advanced prescriptions.