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Rouché's theorem states that if $g$ and $h$ are holomorphic functions on a neighbourhood of the closure of a bounded domain andon its positively oriented boundary, then $g$ and $g+h$ have the same number of zeros in the domain, counted with multiplicity.

The Open mapping theorem states that a nonconstant holomorphic function on a domain maps every open subset to an open subset. To prove it, fix $z_0$ and put $w_0=f(z_0)$. By the identity theorem, the zeros of $f-w_0$ are isolated. Choose $r>0$ so that the closed disc $\overline{D(z_0,r)}$ lies in the domain and $z_0$ is the only zero of $f-w_0$ in that disc. ThenWhenever $|w-w_0|<\delta$, the constant $w_0-w$ has modulus smaller than $f-w_0$ on the circle. The Rouche theorem therefore says thathas the same positive number of zeros in the disc as $f-w_0$. Thus every $w\in D(w_0,\delta )$ lies in the image of $f$, proving that the image is open.

If $|f|$ had a local maximum at $z_0$, take a small open disc $D$ on which $|f(z)|\le |f(z_0)|$. The restriction of $f$ to $D$ cannot be constant, since the identity theorem would then make $f$ constant on the whole domain. The complex open mapping theorem says that $f(D)$ is an open neighbourhood of $f(z_0)$, and such a neighbourhood contains points of modulus greater than $|f(z_0)|$, a contradiction. This proves the maximum modulus principle from the complex open mapping theorem.

Now let $\Omega$ be bounded. Its closure is compact, so the continuous function $|f|$ attains a maximum there. If $f$ is nonconstant, the maximum modulus principle excludes an interior maximum; if $f$ is constant, the claimed bound is immediate. Hencewhich is the maximum modulus principle on a bounded domain.

Finally let $\Omega =\{z:\mathrm{Re}z>1\}$ and suppose $|f|\le K$ throughout $\overline{\Omega }$. Fix $z_0\in \Omega$. The principal complex logarithm is holomorphic on the right half-plane, so for each positive integer $n$ the functionis holomorphic on $\Omega$ and continuous on its closure. Given $\epsilon >0$, choose $R>|z_0|$ so large thatApply the bounded-domain result to $g_n$ on $\Omega \cap D(0,R)$. On the vertical part of the boundary, $|z|\ge 1$ and hence $|g_n(z)|\le M$. On the circular part,It follows thatLetting $n\to \infty$ and then $\epsilon \to 0$ gives $|f(z_0)|\le M$. Since $z_0$ was arbitrary, this proves the bounded half-plane maximum principle.

The boundedness assumption is necessary. The functionis holomorphic on $\Omega$ and continuous on its closure, and $|f(1+iy)|=e$ on the boundary, but $|f(x)|=e^x$ is unbounded for real $x>1$. Thus the boundary estimate does not control an unbounded holomorphic function on this unbounded domain.

Differentiate the affine relation with respect to $x$ and $y$:Using the Cauchy-Riemann equations, the second equation becomesThusThe determinant is $-(a^2+b^2)\ne 0$, so $u_x=v_x=0$ and therefore $f^{\prime }=0$. A holomorphic function with zero derivative on a connected open set is constant. Hence $f$ is constant, as in analytic function with image in an affine real line.

A modular function of weight $k$ and level $\Gamma (1)$ is a meromorphic function $f:\mathfrak{h}\to \mathbb{C}$ satisfyingfor every $\gamma =\left(\begin{array}{cc}a& b\\ c& d\end{array}\right)\in \Gamma (1)$, and having a meromorphic Fourier expansion at the cusp infinity. Equivalently, $f{|}_k\gamma =f$ for the slash operator for modular forms.

A modular form is a modular function that is holomorphic on $\mathfrak{h}$ and holomorphic at infinity. Its Fourier expansion therefore has the formSince $-I\in \Gamma (1)$, a nonzero level-one form must have even weight.

A germ of a holomorphic function at $z\in D$ is an equivalence class $[f{]}_z$ of pairs consisting of a neighbourhood $U$ of $z$ and a holomorphic function $f:U\to \mathbb{C}$. Two pairs represent the same germ if their functions agree on some neighbourhood of $z$.

The space of germs of holomorphic functions isFor every holomorphic $f$ on an open $U\subseteq D$, the setis declared open; these sets form a basis. The forgetful mapmaps each ${\tilde{U}}_f$ homeomorphically onto $U$. The inverses of these restrictions give the charts defining the complex structure, so $\pi$ is a local biholomorphism.

The evaluation map isIt is well-defined by the germ equivalence relation. On the chart ${\tilde{U}}_f$, its coordinate expression iswhich is holomorphic. Hence $\mathcal{E}$ is analytic, as stated in evaluation map on a space of germs.

Now putThis gives the germ surface of the square root of z to the eighth minus one. An explicit gluing description is obtained by pairing the eight roots into four adjacent pairs and cutting the plane along four disjoint arcs joining the members of each pair. On the complement of the cuts choose one branch $q(z)$ of $\sqrt{z^8-1}$. Take two copies, labelled by $q$ and $-q$, and glue the upper bank of each cut in one copy to the lower bank in the other, and conversely. The cut interiors are restored by the gluing, while the eight endpoints remain absent because they are not in $D$.

On this surface defineFor each $(z,w)\in R$, the holomorphic function $z^8-1$ is nonzero near $z$, so it has a unique local square root $q$ with $q(z)=w$. DefineThis is well-defined, injective, and analytic in the displayed local sheets; its inverse on its image is $[q{]}_z\mapsto (z,q(z))$. Thus $\Phi$ is the required analytic embedding, and it intertwines both the forgetful and evaluation maps.

To compactify, first add one point above each of the eight roots of unity. These are simple branch points of the double cover. Since the polynomial has even degree eight, the even-degree hyperelliptic model has two distinct, unbranched points above infinity. Consequently

Finally apply the Riemann-Hurwitz formula to the degree-two meromorphic map $\bar{\pi }:\overline{R}\to {\mathbb{C}}_{\infty }$. Its only ramification consists of the eight simple finite branch points, soThereforeThis is the compactification of y squared equals x to the eighth minus one.

PutThis slit complex plane is a simply connected domain. On $D$ there is one branch of $\sqrt{1-z^2}$ that equals $1$ at zero, and its reciprocal is holomorphic. The primitive of a holomorphic function on a simply connected domain therefore makes the integral from zero path-independent, so $G$ is single-valued.

For unrestricted paths, going once around either branch point changes the sign of the square root. The resulting analytic continuation therefore need not return to the original germ, and the value of $g(z)$ can depend on the path.

The polynomial form of Runge theorem says that if $K\subset \mathbb{C}$ is compact, $\mathbb{C}\setminus K$ is connected, and $f$ is holomorphic on a neighbourhood of $K$, then for every $\epsilon >0$ there is a polynomial $P$ such that

Statement (i) is false. Take the entire function $f(z)=e^{-z}$. It is bounded on the closed first quadrant $E$. If polynomials $P_n$ converged uniformly to $f$ on $E$, then each sufficiently late $P_n$ would be bounded on the positive real axis. A polynomial bounded on that ray must be constant. No sequence of constants converges uniformly to $e^{-x}$ on $[0,\infty )$, since its values at zero and near infinity differ by one.

Fora finite point $z_0$ is an ordinary point exactly when $p$ and $q$ are holomorphic there. It is a regular singular point exactly whenare holomorphic at $z_0$.

Put $\zeta =1/z$ and $W(\zeta )=w(1/\zeta )$. Direct differentiation givesso the transformed equation isConsequently $z=\infty$ is ordinary precisely whenare holomorphic at $\zeta =0$. It is a regular singular point at infinity precisely when $zp(z)$ and $z^{2}q(z)$ are holomorphic functions of $1/z$ near infinity.

If zero and infinity are regular singular and every nonzero finite point is ordinary, the Laurent series of $p$ and $q$ can contain only the terms compatible with both endpoint bounds. Hencefor constants $a,b$. The equation is a Cauchy-Euler differential equation. Its indicial equation isFor distinct roots ${\lambda }_1,{\lambda }_2$, the general solution on a domain with a chosen logarithm branch isFor a repeated root $\lambda$, the general solution of an Euler-Cauchy equation is

Finally require infinity to be ordinary. In the transformed equation its coefficients becomeBoth are holomorphic at zero exactly when $a=2$ and $b=0$. Thus the further restriction is

Suppose a $C^1$ classical solution existed and set $w=u+iv$. The system is preciselywhich are the Cauchy-Riemann equations for $w$ as a function of the complex number $z=x+it$. Hence $w$ is holomorphic near $z=0$. Every holomorphic function is real analytic, so its restrictionto the real axis is real analytic near zero. Its real part and imaginary part show that both $f$ and $g$ must be real analytic, contradicting the hypothesis. Therefore no such $C^1$ solution exists.

A Hermitian metric on a holomorphic vector bundle $E\to X$ is a smoothly varying family of positive-definite Hermitian forms $h_x$ on its fibers. A Chern connection is a connection on a vector bundle $\nabla$ that is compatible with $h$ and whose $(0,1)$ part is the bundle's Dolbeault partial connection, ${\nabla }^{0,1}={\bar{\partial }}_E$.

Choose a local holomorphic frame $e$ and write $h=h(e,e)>0$. The connection form in this frame isIf $e^{\prime }=ge$ for a nowhere-zero holomorphic function $g$, then $h^{\prime }=|g{|}^{2}h$ andwhich is exactly the connection-form transformation law. The local formulas therefore define a global connection.

Identify $\mathcal{O}(-1)$ with the tautological bundle over Complex projective space. The standard Hermitian inner product of ${\mathbb{C}}^{n+1}$ restricts to each tautological line. On the affine chart $U_i=\{Z_i\ne 0\}$, put $w_j=Z_j/Z_i$ for $j\ne i$ and use the holomorphic frameThenThe curvature form of a connection isConsequently Chern-Weil theory gives the closed representativewhere ${\omega }_{FS}$ is the Fubini-Study form.

Expanding the derivative limit along real and imaginary increments gives $u_x=v_y$ and $u_y=-v_x$. Since $\mathrm{Re}(z{e}^z)=e^x(x\cos y-y\sin y)$, the required function is $f(z)=z{e}^z$, and $f(0)=0$. Two such functions differ by a holomorphic function with zero real part; the open mapping theorem makes it constant, and the value at zero makes that constant zero.

Write $M(x)={\sum }_{n\le x}\mu (n)$ for the Mertens function. Suppose, to the contrary, that for some $\epsilon >0$ the quotient $|M(x)|/x^{1/2-\epsilon }$ were bounded. Partial summation would then makeconverge and define a holomorphic function throughout $\Re s>1/2-\epsilon$. In $\Re s>1$ the Euler product identifies this function with $1/\zeta (s)$, so analytic continuation would make $1/\zeta (s)$ holomorphic in that larger half-plane.

By assumption, $\zeta$ has a nontrivial zero. The functional equation of the Riemann zeta function reflects one of that zero and its partner into $\Re s\ge 1/2$, where $1/\zeta$ must have a pole, a contradiction. Thus $|M(x)|/x^{1/2-\epsilon }$ is unbounded, which gives an $x\ge 1$ exceeding any prescribed constant $C$.

The holomorphic coupling is $\tau =\vartheta /(2\pi )+4\pi i/g^2$. Perturbative coefficients are independent of the topological angle $\vartheta$. A holomorphic function invariant under continuous shifts of $\mathrm{Re}\tau$ is constant by the Cauchy-Riemann equations, hence a perturbative local superpotential coefficient cannot depend on $\tau$. Terms in $e^{2\pi i\tau }$ can evade this argument nonperturbatively. This does not forbid renormalization of the separate gauge kinetic F-term, or wave-function renormalization from the Kähler potential.

A maximum-modulus extension for unbounded domains with suitable growth restrictions. In its polynomial-growth vertical-strip form, bounds $|F(a+it)|\ll (1+|t|{)}^u$ and $|F(b+it)|\ll (1+|t|{)}^v$ for a holomorphic function imply the interpolated exponent $((b-\sigma )u+(\sigma -a)v)/(b-a)$. It is the strip-convexity tool used to move subpower zeta bounds to the right of the critical line back onto that line.

Every holomorphic function $f$ on a simply connected domain has a single-valued antiderivative. Fixing $z_0$ and settinggives a path-independent function with $F^{\prime }=f$, because the Cauchy integral theorem makes the integral around every closed path zero.

Under $\zeta =1/z$, the point $z=\infty$ is regular singular forexactly when $zP(z)$ and $z^{2}Q(z)$ are holomorphic functions of $1/z$ near $1/z=0$.

If $f$ is holomorphic on and inside a positively oriented closed contour except for finitely many isolated singularities $a_j$ inside it, then

For a holomorphic phase $\varphi$, a saddle point is a critical point $z_0$ with ${\varphi }^{\prime }(z_0)=0$. At a simple saddle, ${\varphi }^{\prime \prime }(z_0)\ne 0$, and the quadratic Taylor term determines the local descent directions.

The Schwarz integral formula reconstructs a holomorphic function from boundary values of its real part. For a real boundary function $f$ on the complex upper half-plane satisfying ${\int }_{\mathbb{R}}|f(s)|/(1+s^2)ds<\infty$, the regularized form isIts real part is the Poisson integral of $f$. The subtraction makes the integral converge and changes only the imaginary normalization. Integration by parts, when the boundary terms vanish, gives $W^{\prime }(w)=(\pi i{)}^{-1}\int f^{\prime }(s)/(s-w)ds$; jumps of $f$ are included as Dirac delta function terms in its distributional derivative.

The sheaf ${\mathcal{K}}_X$ assigns to an open subset of a complex manifold its meromorphic functions. Its subsheaf ${\mathcal{K}}_X^{\ast }$ consists of the nonzero meromorphic functions, while ${\mathcal{O}}_X^{\ast }$ consists of the nowhere-zero holomorphic functions.

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 $K\subset U$.