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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

For $0<r<R$, suppose $f_n,f$ are holomorphic on a neighborhood of the closed radius-$R$ disc. Subtract their Cauchy integral formula expressions on that circle. Its length is $2\pi R$, and the distance from an interior point with modulus at most $r$ to the circle is at least $R-r$. The estimation lemma gives the displayed bound. Uniform convergence on that circle therefore implies uniform convergence of each fixed-order derivative on the smaller disc. The positive separation from the contour is essential for a uniform bound.

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.

If $f$ is holomorphic on a neighbourhood of the closed disc $|z-a|\le R$ and $|f(z)|\le M$ on its boundary, differentiating the Cauchy integral formula gives

For a continuously differentiable function on a bounded planar domain, the Cauchy-Pompeiu formula supplements the Cauchy integral formula with an area integral of its Wirtinger derivative. Apply the Generalized Stokes theorem to $f(\zeta )d\zeta /(\zeta -z)$ on a punctured domain; the small-circle integral supplies $2\pi if(z)$. If the boundary term vanishes at infinity, it gives the Cauchy-Green operator representation $f(z)={\pi }^{-1}\int f_{\bar{\zeta }}(\zeta )/(z-\zeta )dA$.

Suppose $w$ is holomorphic off $[0,1]$, tends to zero at infinity, has sufficiently regular boundary values in the interval interior, and has endpoint growth $O(|z-a{|}^{-\alpha })$ for some $\alpha <1$ at each endpoint $a$. If its jump is integrable, the displayed reconstruction follows by deforming a Cauchy integral formula contour onto the two sides of the interval. An endpoint circle of radius $\epsilon$ contributes $O({\epsilon }^{1-\alpha })$, which vanishes. The two oppositely oriented banks leave precisely the jump. Without decay at infinity, an entire remainder can survive; a nonzero constant is the simplest example.

Complex analysis studies functions of a complex number, especially holomorphic functions. Its distinctive feature is that a single complex derivative controls variation in every direction of the complex plane. The map $f(z)=z^2$ gives a first picture: straight coordinate lines become curved, yet small perpendicular arrows remain perpendicular wherever $f^{\prime }(z)\ne 0$.

The bottom panels show displacements rather than absolute positions. Multiplication by $f^{\prime }(1+i)=2+2i$ predicts the image of each small arrow; the exact image differs by $h^2$. The origin is exceptional because $f^{\prime }(0)=0$. This distinction between local conformal maps and a critical point is central: complex differentiability imposes strong geometric structure, without making every holomorphic function globally one-to-one.

The remarkable next step is that holomorphic functions are determined inside a circle by their values on its boundary. Precisely, let $R>0$, let $D=\{z:|z-a|<R\}$, and suppose $f$ is a holomorphic function on an open set containing the closed disc $\overline{D}$. Orient its boundary counterclockwise. The Cauchy integral formula states that, for every $w\in D$,The contour integral combines boundary values with a kernel depending on the requested interior point. It is a reconstruction formula, not merely a trick for evaluating an integral. The hypothesis about the closed disc ensures that all boundary values and the contour integral are well defined. ^[ref]

At the centre, substituting $z=a+R{e}^{i\theta }$ cancels the denominator against $dz=iR{e}^{i\theta }d\theta$:Thus the centre value is the angular average of the boundary values. For the polynomial $f(z)=z^2$, $a=1+i$ and $R=1/2$, expand the boundary data asBoth oscillating terms have zero average, leaving $f(a)=2i$. Away from the centre the weights change: on the unit circle,With $f(z)=z^2$ and $w=1/2$, expanding the denominator as a geometric series leaves only its second-power term after integration, giving $f(1/2)=1/4$. This also explains how the Cauchy integral formula recovers an entire Taylor series from boundary data.

When a holomorphic function has isolated singularities, a different boundary calculation detects their residues. Suppose $\Gamma$ is a counterclockwise simple closed contour and $f$ is holomorphic on a neighbourhood of it and its interior except at finitely many interior isolated singularities $a_j$. The residue theorem givesA residue is the coefficient of $(z-a_j{)}^{-1}$ in the Laurent series. A short proof idea is to deform the contour to small circles around the isolated singularities using the Cauchy integral theorem. Integrating the Laurent series on each circle kills every power except $-1$. ^[ref]

Here is how a complex pole determines a real improper integral. Set $g(z)=1/(1+z^2)$ and use the contour consisting of the segment from $-R$ to $R$, followed by the upper semicircle $A_R$, with $R>1$. Only the simple pole $i$ lies inside, andThe arc must actually disappear. On $A_R$, the reverse triangle inequality gives $|1+z^2|\ge R^2-1$, and its length is $\pi R$, soThe real integrand is bounded by $x^{-2}$ for $|x|\ge 1$, so both tails converge separately. Taking the limit therefore proves the ordinary improper integralFor example, at $R=10$ the real segment contributes $2\arctan (10)\approx 2.942255$, and the arc contributes $0.199337$, below the bound $10\pi /99\approx 0.317333$. The usefulness of the residue theorem is that the same strategy works for many rational functions whose real antiderivatives are cumbersome: find the enclosed poles, compute their residues, and control the connecting arc. ^[ref]

Historically, Cauchy's 1825 memoir on integrals with imaginary limits develops complex contour integration and its use in transforming and evaluating definite integrals. The opening sections define complex path integrals, discuss changing the path, and examine what happens near singular points. These are recognizable ancestors of the modern link between holomorphic functions, contours and residues. ^[ref]

Choose a cycle $\Gamma$ surrounding $\sigma (x)$ and a cycle $\Delta$ surrounding $f(\sigma (x))$, with $f(\Gamma )$ inside the region of winding number one for $\Delta$. The algebra-homomorphism property givesSubstitute this into the contour formula for $g(f(x))$, interchange integrals, and apply the Cauchy integral formula to the inner $\zeta$ integral. This proves composition as an equality of algebra elements, not merely an equality on characters, which need not separate a nonsemisimple algebra.

If $f$ is holomorphic on a neighborhood of the closed unit disc and $|a|<1$, this boundary integral is independent of $a$. Parametrizing the unit circle by $z=e^{it}$ and taking a complex conjugate giveswhere the last equality is the Cauchy integral formula applied to the holomorphic function $f(z)/(1-\overline{a}z)$ at zero. The conjugation of the whole integral includes the factor $1/i$ and the differential; parametrization keeps their signs consistent. Equivalently, the conjugated boundary values have only nonpositive Fourier frequencies, so the Cauchy kernel detects their constant coefficient.

The Abelian Ward splitting by Laurent series gives this gauge potential through the Cauchy integral formula, with $f_{,B}=\partial f/\partial {\omega }^B$ evaluated on the twistor line. The integrand has total projective degree zero. Holding the contour fixed locally and differentiating gives $F_{A{A}^{\prime }B{B}^{\prime }}={ϵ}_{A^{\prime }{B}^{\prime }}(2\pi i{)}^{-1}{\oint }_{\Gamma }{f}_{,AB}{\pi }_{D^{\prime }}d{\pi }^{D^{\prime }}$, using ${\pi }_{A^{\prime }}{\iota }_{B^{\prime }}-{\pi }_{B^{\prime }}{\iota }_{A^{\prime }}={ϵ}_{A^{\prime }{B}^{\prime }}({\iota }_{C^{\prime }}{\pi }^{C^{\prime }})$. Its unprimed two-component spinor is symmetric, so the primed gauge curvature vanishes and the field obeys the anti-self-dual Maxwell equations. These signs use ${ϵ}_{0^{\prime }{1}^{\prime }}=1$, ${ϵ}^{0^{\prime }{1}^{\prime }}=-1$ and raising by left multiplication; other consistent raising conventions can reverse both displayed overall signs.

Let $S$ be a closed square and let $f$ be holomorphic on a neighbourhood of a slightly larger square. The Cauchy integral formula on the larger boundary is uniformly approximated on $S$ by Riemann sums of resolvents whose poles lie outside $S$. Far-away resolvents have convergent geometric-series expansions in $z$, and resolvent propagation across a complementary component moves this polynomial approximability to every pole outside the square. Approximating the finitely many resolvents in each Riemann sum by polynomials proves polynomial approximation on $S$ without invoking the general Runge theorem.

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.

For $f(z)=z+{\sum }_{n\ge 2}{a}_n{z}^n$ in the normalized univalent function class, the integral mean bound for an odd square-root transform gives $(2\pi {)}^{-1}\int |f(r{e}^{i\theta })|d\theta \le r/(1-r)$. The Cauchy integral formula for Taylor coefficients therefore yields $|a_n|\le r^{1-n}/(1-r)$. Choosing $r=(n-1)/n$ and applying the logarithm tangent bound gives the displayed bound for every $n\ge 2$.

Every nonconstant entire function takes every complex value with at most one exception. Equivalently, its image is either $\mathbb{C}$ or $\mathbb{C}\setminus \{a\}$ for one complex number $a$. The hypothesis of being nonconstant matters: a constant function has only one value. The Little Picard theorem guarantees existence of preimages, without requiring infinitely many preimages for each attained value. ^[ref]

For a nonconstant polynomial $p$, the fundamental theorem of algebra applied to $p(z)-w$ supplies a preimage for every $w\in \mathbb{C}$. The complex exponential function has precisely one omitted value, zero, as the explicit formula for its preimages shows. In contrast, the complex sine function has no omitted value. To solve $\sin z=w$, set $u=e^{iz}$ and rearrange to the quadratic equationIts roots have product $-1$, so they are nonzero. Every such root has a preimage under the complex exponential function, giving a solution for $z$. Adding $2\pi k$ supplies further solutions by the periodicity of the complex sine.

The force of the Little Picard theorem is that it applies to every nonconstant entire function, without an explicit formula. The Liouville theorem already gives a dense image of a nonconstant entire function, but a dense subset could still leave out many individual points. The Little Picard theorem limits these missing points to one. In particular, establishing two omitted values proves that an entire function is constant.

For a nonconstant meromorphic function on the complex plane, the corresponding bound is two omitted values in the Riemann sphere $\mathbb{C}\cup \{\infty \}$, where poles take the value infinity. ^[ref] For example,omits zero and one, and has poles at $2\pi ik$ for $k\in \mathbb{Z}$. Every other finite value $w$ is attained by solving $e^z=1-1/w$. For an entire function, infinity is already an omitted value in the Riemann sphere, leaving at most one additional finite omitted value.

Proof outline: use the version of Montel's theorem for two fixed finite omitted values. ^[ref] It makes the rescaled holomorphic functions $g_n(\zeta )=f(a+n\zeta )$ a normal family on the unit disc. The fixed value $g_n(0)=f(a)$ gives pointwise anchoring of a holomorphic normal family, hence a subsequence with a finite holomorphic function as its locally uniform convergence limit. The Cauchy integral formula bounds its derivatives at zero, whereas $g_n^{\prime }(0)=n{f}^{\prime }(a)$. Consequently $f^{\prime }(a)=0$; arbitrary $a$ makes $f$ constant.

A locally uniform convergence limit of holomorphic functions is holomorphic, and its derivatives converge locally uniformly as well. The limit is continuous. On each small triangle contained in the domain, uniform convergence allows passage to the limit in its contour integral; the integrals vanish by the Cauchy integral theorem. The Morera theorem therefore makes the limit holomorphic. Applying the Cauchy integral formula on circles in a slightly larger compact subset proves locally uniform convergence of every fixed-order derivative.

Use the boundary form of Cauchy's integral theorem: the integral of a holomorphic function around the oriented boundary of a compact region on whose neighborhood it is holomorphic is zero, including annular regions with the inner boundary oppositely oriented. Apply it to $f(z)/(z-a)$ between the circles of radii $r$ and $\rho$, both around $a$. It givesas $\rho \downarrow 0$, by continuity. This proves the Cauchy integral formula.

For a bounded entire function with $|f|\le M$, apply that formula at variable points inside a fixed circle and differentiate its kernel under the integral. At its center $a$, this givesThe circle can have any radius because the function is entire. Letting $R\to \infty$ gives $f^{\prime }(a)=0$ at every $a$, so $f$ is constant. This proves Liouville's theorem.

Finally, if a nonconstant entire function omitted all points within distance $ϵ$ of some $w$, then $g=1/(f-w)$ would be entire and bounded by $1/ϵ$. Liouville's theorem would make $g$, and hence $f$, constant. Therefore every such neighborhood contains a value of $f$, proving the dense image of a nonconstant entire function property.

If $f$ is holomorphic on a neighborhood of the closed disc, the differentiated Cauchy integral formula iswith positive orientation. On the circle the integrand has modulus at most $M/r^{k+1}$ and the contour has length $2\pi r$. Thus

Choose a small positively oriented circle $C_{\epsilon }$ about $z_0$ inside the given region. Cauchy's theorem on the region between the two contours givesContinuity makes the last integrand converge uniformly to $f(z_0)$ as $\epsilon \downarrow 0$. Its limit is $2\pi if(z_0)$, proving the Cauchy integral formula.

The same formula holds with an interior point $w$ near $z_0$. Subtract its value at $z_0$ and divide by $w-z_0$:The contour is a positive distance from $z_0$, so the integrands converge uniformly as $w\to z_0$. Passing to the limit gives the differentiated formulaOn the unit circle, $\bar{z}=1/z$ and $\mathrm{Im}z=(z-\bar{z})/(2i)$, proving the requested identity. Hence the requested integral isThe first integral is $2\pi i/3$. For the second useThe differentiated formula applied to the constant function gives $\int z^{-2}dz=0$; the remaining two terms integrate to $-6\pi i+6\pi i=0$. Therefore

Use the following boundary version of Cauchy's integral theorem: if a function is holomorphic on a neighborhood of a compact region bounded by finitely many piecewise smooth simple closed curves, its integral around the positively oriented boundary is zero. On an annulus this means that its integrals around the two circles, both taken counterclockwise, are equal.

For $0<\rho <r$, the function $f(z)/(z-w)$ is holomorphic on a neighborhood of $\rho \le |z-w|\le r$. The theorem therefore givesAs $\rho \downarrow 0$, continuity of $f$ makes the integrand converge uniformly to $f(w)$. The right side tends to $f(w)$, proving Cauchy integral formula without assuming that formula in an intermediate step.

Let $I$ denote the requested boundary integral. Parametrize the unit circle by $z=e^{it}$. Since $dz=i{e}^{it}dt$,Taking complex conjugates and using $\overline{e^{it}}=e^{-it}$ givesThis is the conjugated Cauchy integral on a circle identity. The function $h(z)=f(z)/(1-\overline{a}z)$ is holomorphic on a neighborhood of the closed unit disc, because $|a|<1$. Applying Cauchy integral formula at zero gives $\overline{I}=h(0)=f(0)$. ThusThe complex conjugates in these formulas are essential; the boundary integrand contains $\overline{f(z)}$.

For a holomorphic function on $|z-a|<R$, the complex Taylor theorem stateswith uniform convergence on each smaller closed disc. Here is a proof. Choose $r<\rho <R$. For $|z-a|\le r$ and $|\zeta -a|=\rho$, the geometric seriesconverges uniformly. Substitute it into Cauchy integral formula and integrate term by term to getIf $M_{\rho }={\max }_{|\zeta -a|=\rho }|f(\zeta )|$, then $|c_n|\le M_{\rho }{\rho }^{-n}$. These bounds justify differentiating the power series any fixed number of times on a smaller closed disc, by comparison with convergent differentiated geometric series. Evaluation at $a$ gives $f^{(n)}(a)=n!c_n$. Since every $|z-a|<R$ belongs to some such smaller disc, the formula holds throughout the stated domain.

The identity theorem says that two holomorphic functions on a connected open subset of $\mathbb{C}$ are equal everywhere if their equality set has an accumulation point in that subset. Equivalently, a holomorphic function with an interior accumulation point of zeros is identically zero there.

The proposed Taylor coefficients at $i$, including the specified value for $n=0$, giveThe function $1/z$ is holomorphic on $\mathbb{C}\setminus \{0\}$ and has exactly the given derivatives, so it supplies existence. Any other admissible holomorphic function has the same Taylor series on $|z-i|<1$. The punctured plane is connected, so the identity theorem gives uniqueness: