Complex analysis

A holomorphic square root of a nonvanishing holomorphic function $f$ is a holomorphic function $g$ satisfying $g^2=f$. It exists locally, and globally exactly when the winding obstruction around every closed path vanishes.

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.

Monodromy is the change in a locally defined analytic object after analytic continuation around a closed path.

The fractional linear mapsmap the right half-plane and unit disk conformally onto one another.

A holomorphic logarithm of a nonvanishing holomorphic function $f$ is a holomorphic function $g$ with $e^g=f$. On a simply connected domain, every nonvanishing holomorphic function has one.

The complex upper half-plane is $\mathbb{H}=\{z\in \mathbb{C}:\mathrm{Im}z>0\}$.

A conformal bijection of the complex upper half-plane to itself is a real Möbius transformation with $ad-bc>0$. Three distinct boundary values determine it uniquely. Its real derivative is positive away from its pole, and $\mathrm{Im}\varphi (z)=(ad-bc)\mathrm{Im}z/|cz+d{|}^2$.

For $|z|<1$, the polylogarithm is ${\mathrm{Li}}_s(z)={\sum }_{n\ge 1}{z}^n/n^s$; contour formulas analytically continue it to a slit plane.

The dilogarithm is the order-two polylogarithm, defined for $|z|<1$ by ${\mathrm{Li}}_2(z)={\sum }_{n\ge 1}{z}^n/n^2$. Its derivative is $-\mathrm{Log}(1-z)/z$, with a removable value at zero; a chosen complex logarithm branch gives its analytic continuation.

A complex-valued function is holomorphic on an open set when it has a complex derivative at every point of that set.

For a holomorphic function nonzero at zero and without boundary zeros, the formula relates its interior zeros, counted by multiplicity, to the circle mean of its logarithmic modulus. Divide out the finite interior zero factors; the remaining logarithmic modulus is harmonic, and each zero contributes $\log (R/|a_j|)$ by the harmonic mean-value formula. Boundary radii follow by limits where appropriate.

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 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.

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

An analytic continuation extends a holomorphic function through overlapping connected open sets while preserving its values on their overlap. Continuation along different paths can produce different germs when the domain contains branch points.

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.

If a function element can be analytically continued along every path in a domain, continuation along two endpoint-fixed homotopic paths gives the same terminal germ.

For a covering $p:X\to Y$ and base point $y$, lifting loops based at $y$ permutes the fibre $p^{-1}(y)$. The image ofis the monodromy group.

A connected three-sheeted cover has transitive monodromy. If it also has a simple branch point, its monodromy contains a transposition and must be the full symmetric group on three letters. A local triple branch supplies a three-cycle explicitly.

The degree-four rational map $F(z)=z^2+z^{-2}$ has branch values $-2,2,\infty$. Away from their inverse images, the transformationsare commuting deck involutions. They generate a Klein four-group acting transitively on each fibreConsequently the monodromy group is $V_4$, acting by the identity and the three double transpositions.

If a parameter-dependent contour integral has a moving pole, deforming the integration contour continuously so that the pole never crosses it preserves a holomorphic branch. Returning the contour to its original path after crossing a pole adds or subtracts the corresponding residue contribution.

The solutions of $\sinh w=z$ arefor any chosen local branch $\mathrm{Arcsinh}$.

An entire function is a holomorphic function whose domain is the whole complex plane.

A nonzero entire function can be described by its discrete zero locations and multiplicities, a product of Weierstrass elementary factors and a zero-free exponential factor. Conversely, consistent prescribed multiplicities at distinct points escaping every compact subset can be realized by choosing factor orders large enough for infinite product convergence from logarithmic tails. Repeated locations with incompatible exact multiplicities are not admissible data, and a finite accumulation of distinct zeros is excluded by the identity theorem.

An entire function of finite order factors as its canonical zero product times the exponential of a polynomial. For order at most one and nonzero value at zero, it has the form $f(z)=f(0)e^{Bz}{\prod }_j(1-z/a_j)e^{z/a_j}$. Genus-one factors converge when ${\sum }_j|a_j{|}^{-2}<\infty$. A zero-free function of this order is therefore the exponential of an affine polynomial.

If two nonzero entire functions have identical zeros and orders, their quotient extends to a nowhere-zero entire function. Its logarithmic derivative has a primitive on the simply connected plane, producing a holomorphic logarithm $h$ and the displayed relation. Any two choices differ by one constant integer multiple of $2\pi i$. On a multiply connected domain the logarithm can fail to exist globally.

Suppose that for every $z\in \mathbb{C}$, at least one derivative $f^{(n)}(z)$ of an entire function vanishes. The closed sets $E_n=\{z:f^{(n)}(z)=0\}$ cover $\mathbb{C}$. By the Baire category theorem, some $E_N$ has nonempty interior. The identity theorem gives $f^{(N)}=0$ everywhere, so $f$ is a polynomial of degree below $N$.

Every nonconstant entire function takes every complex value with at most one exception.

Near an essential singularity, a holomorphic function takes every complex value, with at most one exception, infinitely often.

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.

A domain is simply connected when every closed path in it can be continuously contracted to a point while remaining in the domain.

A crosscut of a planar domain is a simple open arc inside the domain whose two endpoints are distinct boundary points, interpreted as prime ends when necessary. In a simply connected domain it divides the domain into two components. In the half-plane, a bounded crosscut joining two real points separates the intervening real interval from infinity; a Loewner chain completing that crosscut swallows this interval.

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.

The principal complex inverse sine onis the branchwhose derivative equals one at zero.

If $G$ is the principal complex inverse sine, all values reached by analytic continuation areThis follows from the monodromy reflections about the branch values $-\pi /2$ and $\pi /2$.

Let $w$ be a local value of the complex inverse sine. Since the limiting values at the two branch points are $\pi /2$ at $z=1$ and $-\pi /2$ at $z=-1$, monodromy reflection at a square-root branch point givesFollowing a counterclockwise loop that encloses both branch points in the convention where $\sqrt{1-z^2}$ is positive just above $(-1,1)$ applies the two reflections in the order that givesThe reverse orientation gives $w+2\pi$. Thus the translational part of the monodromy is generated by $2\pi$.

If a continuous complex-valued function on a domain has zero integral around the boundary of every triangle contained in that domain, then it is holomorphic.

If a function is holomorphic on a simply connected domain, its integral around every closed piecewise smooth contour in that domain vanishes. In particular, contours with common endpoints may be deformed through the domain without changing the integral.

A holomorphic function on an open neighbourhood of a closed triangle has zero contour integral around its boundary. Goursat's subdivision proof uses complex differentiability alone: repeatedly choose a quarter-size triangle carrying at least one quarter of the integral, and subtract the affine expansion at the limiting point. No continuity of the derivative is assumed.

A contour deformation changes the path of a contour integral through a region where its integrand is a holomorphic function. The Cauchy integral theorem preserves the integral when the endpoints are fixed and the connecting boundary terms vanish. Crossing a pole instead contributes the appropriately oriented residue; a branch cut limits which deformations are permitted on one analytic sheet.

For $a>0$ and real $c$, integrate $e^{-a{z}^2}$ around a wide rectangle between the real axis and $\mathrm{Im}z=c$. The vertical contributions vanish as the width tends to infinity, so

The principal value uses symmetric truncation at infinity and symmetric deletion around real singularities.

For $0<s<a$,

For real $a$ and nonzero real $\omega$,

For real $x>0$, the decaying exponential integral isThe other standard real convention satisfies $\mathrm{Ei}(-x)=-E_1(x)$. The NIST definitions specify the analytic continuation and branch conventions.

For $J(\epsilon )={\int }_0^{\infty }{e}^{-x-\epsilon /x}dx/x$, reciprocity $x\mapsto \epsilon /x$ and the small-argument expansion of the exponential integral give $J=-\log \epsilon -2\gamma +o(1)$. Differentiation and integration by parts give $\epsilon J^{\prime \prime }+J^{\prime }=J$. Its Frobenius method yieldsThe logarithm is a singular contribution that a naive termwise expansion at $x=0$ misses.

For $I={\int }_{\epsilon }^{\infty }{e}^{-x-\epsilon /x}dx/x$, the omitted interval becomes $J-I={\int }_1^{\infty }{e}^{-t-\epsilon /t}dt/t$. A uniform Taylor expansion gives $J-I=E_1(1)-\epsilon [e^{-1}-E_1(1)]+O({\epsilon }^2)$. Subtracting this correction from the logarithmic expansion of a symmetric exponential integral preserves both the constant and first-order coefficient.

As $x\downarrow 0$, the exponential integral haswhere $\gamma$ is the Euler--Mascheroni constant. Integrating the Taylor series of $E_1^{\prime }(x)=-e^{-x}/x$ gives every nonconstant coefficient; the constant follows from the limiting definition of $\gamma$. A logarithm of a stretched variable can create a switchback term in a matched asymptotic expansion.

$\Gamma (z)={\int }_0^{\infty }{t}^{z-1}{e}^{-t}dt$ for $\Re z>0$, and $\Gamma (z+1)=z\Gamma (z)$ gives meromorphic continuation.

The Gamma function recurrence and $\Gamma (z)\sim 1/z$ at zero imply a simple pole at every $z=-n$, $n\ge 0$, with residue $(-1{)}^n/n!$.

For $z$ away from the poles, the gamma function satisfiesIts finite product is $m!(m+1{)}^z/[z(z+1)\cdots (z+m)]$. Splitting even and odd factors gives the gamma duplication formula.

The Gamma function recurrence gives $\Gamma (s)=\Gamma (s+j+1)/(s(s+1)\cdots (s+j))$. Its numerator equals one at $s=-j$, so the residue there is $(-1{)}^j/j!$ for every nonnegative integer $j$.

For $a,s>0$, the change of variable $u=at$ in the defining integral for the Gamma function givesIn particular, $\Gamma (1/2)=\sqrt{\pi }$ implies

For $t>0$, contour rotation through angle $-\pi /6$ givesTaking real parts yields

Integration by parts gives

For $\mathrm{Re}s>1$, expansion of $(e^x-1{)}^{-1}$ as a geometric series and termwise integration give

The Euler--Mascheroni constant is

The reciprocal gamma function has the entire-product representation

The digamma function is the logarithmic derivativeLogarithmically differentiating the Weierstrass product gives

The trigamma function is ${\psi }^{\prime }(z)$. For real $z>0$,

The digamma function is strictly increasing on the positive real axis, whileIt therefore has exactly one positive zero, and that zero lies in $(1,2)$.

It follows by making the logarithm of the quotient entire and periodic; growth forces its periodic remainder to be constant.

For $\mathrm{Re}p,\mathrm{Re}q>0$, Euler's beta function is

For $c=1-a-b$ with positive real parts $\mathrm{Re}a,\mathrm{Re}b,\mathrm{Re}c$, the integral converges at its two finite singularities and at infinity, and equalsTo prove it, use Schwinger parameterization with exponents $1-a$ and $1-b$. Integrating the resulting planar Gaussian integral gives $\pi /(\lambda +\rho )$ times $\exp [-\lambda \rho /(\lambda +\rho )]$. Set $q=\lambda +\rho$ and $x=\lambda /q$; the $q$ integral is a gamma function and the $x$ integral is $B(a,b)$, giving the stated ratio. This domain justifies the interchanges; elsewhere the answer is interpreted by analytic continuation. The string measure $d^{2}z=2dxdy$ multiplies the answer by two.

The sum-and-ratio change of variables in a product of gamma integrals gives

For $s,t>0$, put $r=s+t$ and $u=t/(s+t)$. Then $(s,t)=(r(1-u),ru)$ and $dsdt=rdrdu$, separating radial gamma and ratio beta integrals.

The Beta function obeysThese identities follow immediately from the beta--gamma identity and the Gamma function recurrence.

For $0<a<2$,Differentiating at $a=1$ and expanding the secant gives