A first course in complex differentiation can give the misleading impression that complex analysis is a local theory: a derivative is tested by zooming in near a point, and the [Cauchy-Riemann equations](/page/Cauchy-Riemann%20Equations) are checked by comparing infinitesimal directions. Complex integration reveals the opposite force. Once a function is holomorphic, the value of an integral around a closed curve remembers holes, singularities, winding, and global geometry. The basic surprise is that integrating around a loop often gives zero, but not always. The function \begin{align*} f:\mathbb C\setminus\{0\} &\to \mathbb C \\ z &\mapsto \frac{1}{z} \end{align*} is holomorphic, and yet the unit circle detects the missing origin: [example: The Unit Circle Detects a Hole] Let $\gamma:[0,2\pi]\to \mathbb C\setminus\{0\}$ be the unit-circle path \begin{align*} \gamma(t)=e^{it}, \end{align*} and let \begin{align*} f:\mathbb C\setminus\{0\}&\to\mathbb C,\\ z&\mapsto \frac{1}{z}. \end{align*} Since $\gamma'(t)=ie^{it}$, the definition of the complex line integral gives \begin{align*} \int_\gamma f(z)\,dz &=\int_0^{2\pi} f(\gamma(t))\gamma'(t)\,dt\\ &=\int_0^{2\pi} \frac{1}{e^{it}}\,ie^{it}\,dt\\ &=\int_0^{2\pi} i\,\frac{e^{it}}{e^{it}}\,dt\\ &=\int_0^{2\pi} i\,dt\\ &=i(2\pi-0)\\ &=2\pi i. \end{align*} The nonzero value shows that this closed integral is not determined only by the local holomorphicity of $f$ along the curve: the path winds once around the missing point $0$, where $f$ is not defined. [/example] This computation is the gateway to the subject. Complex integrals are defined by parametrising curves, but the main theorems explain why many such integrals depend only on endpoints, singularities, or winding numbers. The theory turns path integrals into a method for proving [power series](/page/Power%20Series) expansions, evaluating real integrals, counting zeros, and measuring [analytic continuation](/page/Analytic%20Continuation) around holes. ## Curves and Orientation ### Parametrized Curves Before we can integrate, we need a controlled class of curves. A completely arbitrary continuous curve may move too wildly for a derivative vector to exist. The line integral needs a velocity along the path, but we still want to allow curves made from finitely many smooth arcs, such as rectangles and polygonal contours. [definition: Piecewise $C^1$ Path] Let $a,b\in\mathbb R$ with $a<b$. A piecewise $C^1$ path in $\mathbb C$ is a continuous map $\gamma:[a,b]\to\mathbb C$ for which there exist points \begin{align*} a=t_0<t_1<\cdots<t_n=b \end{align*} such that $\gamma|_{[t_{j-1},t_j]}$ is $C^1$ for each $j\in\{1,\ldots,n\}$. [/definition] The word path keeps track of parametrisation, not merely the geometric trace. Two different parametrisations can travel over the same set at different speeds, and a closed path is the first case where the integral has no endpoint displacement to measure. We therefore name the closed case separately before studying what a loop can detect. [definition: Closed Path] A piecewise $C^1$ path $\gamma:[a,b]\to\mathbb C$ is closed if \begin{align*} \gamma(a)=\gamma(b). \end{align*} [/definition] ### Closed Loops and Reversal Closed paths are often called contours when they are used as integration paths. Since a contour may be travelled clockwise or counterclockwise, we need a formal operation that changes the orientation while keeping the same trace. This operation is essential because contour integrals must remember direction. [definition: Reverse Path] Let $\mathcal P_{a,b}$ be the set of piecewise $C^1$ paths $\gamma:[a,b]\to\mathbb C$. The path reversal map is \begin{align*} R:\mathcal P_{a,b} &\to \mathcal P_{a,b} \\ \gamma &\mapsto \left(t\mapsto \gamma(a+b-t)\right) \end{align*} defined as follows. If $\gamma\in\mathcal P_{a,b}$, then the reverse path $-\gamma=R(\gamma):[a,b]\to\mathbb C$ is given by \begin{align*} (-\gamma)(t)=\gamma(a+b-t). \end{align*} [/definition] ## Definition The actual integral is now just the real-variable integral of the pullback of $f(z)\,dz$ along the path. This definition is computational, but it is carefully chosen so that the chain rule drives all later structure. [definition: Complex Line Integral] Let $\Omega\subset\mathbb C$ be open, let $f:\Omega\to\mathbb C$ be continuous, and let $\gamma:[a,b]\to\Omega$ be a piecewise $C^1$ path. The complex line integral of $f$ over $\gamma$ is \begin{align*} I_\Omega:\bigcup_{a<b}\{(f,\gamma): f\in C(\Omega;\mathbb C),\ \gamma:[a,b]\to\Omega\text{ is piecewise }C^1\} &\to \mathbb C \\ (f,\gamma) &\mapsto \int_\gamma f(z)\,dz, \end{align*} with value \begin{align*} \int_\gamma f(z)\,dz = \sum_{j=1}^n \int_{t_{j-1}}^{t_j} f(\gamma(t))\gamma'(t)\,dt, \end{align*} where $a=t_0<t_1<\cdots<t_n=b$ is any subdivision on which $\gamma$ is $C^1$ on each subinterval. [/definition] Here $dt$ denotes integration with respect to the ordinary real parameter. Some theorem cards write the same one-dimensional integral as $d\mathcal L^1(t)$, where $\mathcal L^1$ is one-dimensional [Lebesgue measure](/page/Lebesgue%20Measure) on the parameter interval; for the piecewise continuous integrands in this page, this is the same parameter integral written in measure notation. The notation hides the parametrisation, but the definition uses it. A useful contour integral must be independent of the speed at which the contour is traversed, otherwise geometric contour methods would depend on arbitrary bookkeeping. The next foundational fact supplies that invariance and records the effect of reversing orientation. [quotetheorem:5021] This result says that the integral belongs to the oriented curve, not to a particular speed chosen for travel along it. It justifies drawing contours geometrically while still computing them by a convenient parametrisation. A first computation beyond the unit circle shows how powers behave. It also foreshadows why the function \begin{align*} z\mapsto \frac{1}{z} \end{align*} is exceptional. [example: Powers on a Circle] Let $n\in\mathbb Z$, let $r>0$, and let $\gamma:[0,2\pi]\to\mathbb C$ be the circle $\gamma(t)=r e^{it}$. For $f(z)=z^n$, defined on a domain containing the circle and avoiding $0$ when $n<0$, we compute the contour integral from the parametrisation. Since \begin{align*} \gamma'(t)=i r e^{it}, \end{align*} the definition of the complex line integral gives \begin{align*} \int_\gamma z^n\,dz &=\int_0^{2\pi}(\gamma(t))^n\gamma'(t)\,dt\\ &=\int_0^{2\pi}(r e^{it})^n i r e^{it}\,dt\\ &=\int_0^{2\pi}r^n e^{int} i r e^{it}\,dt\\ &=i r^{n+1}\int_0^{2\pi}e^{i(n+1)t}\,dt. \end{align*} If $n\neq -1$, then $n+1\neq 0$, and \begin{align*} \int_0^{2\pi}e^{i(n+1)t}\,dt &=\left[\frac{e^{i(n+1)t}}{i(n+1)}\right]_{0}^{2\pi}\\ &=\frac{e^{i(n+1)2\pi}-e^0}{i(n+1)}\\ &=\frac{1-1}{i(n+1)}\\ &=0, \end{align*} because $n+1\in\mathbb Z$. Hence \begin{align*} \int_\gamma z^n\,dz=0 \end{align*} for $n\neq -1$. If $n=-1$, then $e^{i(n+1)t}=e^0=1$ and $r^{n+1}=r^0=1$, so \begin{align*} \int_\gamma z^{-1}\,dz &=i\int_0^{2\pi}1\,dt\\ &=i(2\pi-0)\\ &=2\pi i. \end{align*} Thus among the Laurent monomials about $0$, exactly $(z-0)^{-1}$ contributes a nonzero value to this positively oriented circular contour integral around $0$. [/example] This example is the algebraic shadow of residues. Before reaching residues, we need the estimates and endpoint principles that make integrals controllable. ## Length and Estimates ### Path Length Complex integrals are complex numbers, so cancellation can make their exact value subtle. The first robust estimate ignores cancellation and bounds the integral by the size of the function times the length of the path. This is the complex analogue of estimating work by force times distance. [definition: Length of a Piecewise $C^1$ Path] Let \begin{align*} \mathcal P=\bigcup_{a<b}\{\gamma:[a,b]\to\mathbb C:\gamma\text{ is piecewise }C^1\}. \end{align*} The length functional is the map \begin{align*} L:\mathcal P &\to [0,\infty) \end{align*} defined as follows. If $\gamma:[a,b]\to\mathbb C$ is a piecewise $C^1$ path, then \begin{align*} L(\gamma)=\sum_{j=1}^n\int_{t_{j-1}}^{t_j}|\gamma'(t)|\,dt, \end{align*} where $a=t_0<t_1<\cdots<t_n=b$ is any subdivision on which $\gamma$ is $C^1$ on each subinterval. [/definition] ### The $ML$ Estimate Length is the quantity that lets us turn pointwise control of $f$ along a curve into control of the integral. This estimate is used constantly: to prove convergence of integrals, shrink contours, discard large arcs, and justify limiting arguments. [quotetheorem:5022] The estimate is crude but powerful. Its strength comes from being geometric: the bound depends only on the maximum size of the integrand along the path and the length of that path. [example: Vanishing on Large Semicircles] Let $a>0$, and for $R>1$ let $\gamma_R:[0,\pi]\to\mathbb C$ be the upper semicircle \begin{align*} \gamma_R(t)=R e^{it}. \end{align*} We show that \begin{align*} \lim_{R\to\infty}\int_{\gamma_R}\frac{e^{iaz}}{z^2+1}\,dz=0. \end{align*} If $z=\gamma_R(t)=R e^{it}=R\cos t+iR\sin t$, then \begin{align*} iaz &=ia(R\cos t+iR\sin t)\\ &=iaR\cos t-aR\sin t, \end{align*} so \begin{align*} |e^{iaz}| &=|e^{iaR\cos t-aR\sin t}|\\ &=e^{-aR\sin t}. \end{align*} Since $0\le t\le \pi$ gives $\sin t\ge 0$, we have \begin{align*} e^{-aR\sin t}\le 1. \end{align*} On the same semicircle, $|z|=R$, so the [reverse triangle inequality](/theorems/2300) gives \begin{align*} |z^2+1| &\ge \bigl||z^2|-|1|\bigr|\\ &=\bigl|R^2-1\bigr|\\ &=R^2-1, \end{align*} because $R>1$. Hence \begin{align*} \left|\frac{e^{iaz}}{z^2+1}\right| &=\frac{|e^{iaz}|}{|z^2+1|}\\ &\le \frac{1}{R^2-1} \end{align*} for every $z\in\gamma_R([0,\pi])$. Also \begin{align*} \gamma_R'(t)=iR e^{it}, \end{align*} and therefore the length of $\gamma_R$ is \begin{align*} L(\gamma_R) &=\int_0^\pi |\gamma_R'(t)|\,dt\\ &=\int_0^\pi |iR e^{it}|\,dt\\ &=\int_0^\pi R\,dt\\ &=\pi R. \end{align*} By the *$ML$ Estimate*, \begin{align*} \left|\int_{\gamma_R}\frac{e^{iaz}}{z^2+1}\,dz\right| &\le \frac{1}{R^2-1}L(\gamma_R)\\ &=\frac{\pi R}{R^2-1}. \end{align*} Finally, \begin{align*} \frac{\pi R}{R^2-1} &=\frac{\pi}{R-\frac{1}{R}}, \end{align*} and $R-\frac{1}{R}\to\infty$ as $R\to\infty$, so \begin{align*} \lim_{R\to\infty}\frac{\pi R}{R^2-1}=0. \end{align*} Thus \begin{align*} \lim_{R\to\infty}\int_{\gamma_R}\frac{e^{iaz}}{z^2+1}\,dz=0. \end{align*} Here the denominator growth alone forces the arc integral to vanish; for slower-decaying denominators, one needs sharper estimates that use the factor $e^{-aR\sin t}$ near the upper semicircle. [/example] The preceding example is intentionally asymptotic: estimates often matter because we will later let a radius tend to infinity or a small circle shrink to a point. The next section explains why many integrals do not need estimation at all. ## Primitives and Path Independence In real calculus, an integral of $F'$ along an interval depends only on endpoint values of $F$. Complex integration has the same principle along paths. When a function has a holomorphic primitive, closed contour integrals vanish because the path returns to its starting point. [definition: Primitive] Let $\Omega\subset\mathbb C$ be open and let $f:\Omega\to\mathbb C$ be holomorphic. A primitive of $f$ on $\Omega$ is a [holomorphic function](/page/Holomorphic%20Function) $F:\Omega\to\mathbb C$ such that \begin{align*} F'(z)=f(z) \end{align*} for every $z\in\Omega$. [/definition] The existence of a primitive is stronger than holomorphicity of $f$. The unit-circle example shows the obstruction: the function \begin{align*} z\mapsto \frac{1}{z} \end{align*} is holomorphic on $\mathbb C\setminus\{0\}$ but cannot have a global primitive there, since a closed integral is nonzero. [quotetheorem:339] This theorem turns the search for vanishing closed integrals into the search for primitives. But in practice we often know integrals around loops before we know a formula for a primitive. To make that route available, we isolate the property that every route between the same two endpoints gives the same value. [definition: Path Independence] Let $\Omega\subset\mathbb C$ be open and let $f:\Omega\to\mathbb C$ be continuous. The complex line integral of $f$ is path independent on $\Omega$ if for every pair of piecewise $C^1$ paths $\gamma_0:[a_0,b_0]\to\Omega$ and $\gamma_1:[a_1,b_1]\to\Omega$ with \begin{align*} \gamma_0(a_0)=\gamma_1(a_1), \qquad \gamma_0(b_0)=\gamma_1(b_1), \end{align*} one has \begin{align*} \int_{\gamma_0} f(z)\,dz = \int_{\gamma_1} f(z)\,dz. \end{align*} [/definition] Path independence is the integral form of having no memory of the route. The next theorem is needed because it tells us that the three practical tests for this phenomenon are the same: construct a primitive, compare paths, or check all closed contours. This equivalence is the bridge between calculus and topology. [quotetheorem:340] The theorem explains why topology can obstruct primitives. If a loop cannot be filled inside the domain without crossing a singularity, the closed integral may detect that obstruction. [example: A Local Primitive Without a Global Primitive] On the slit plane $\Omega=\mathbb C\setminus(-\infty,0]$, write each point uniquely as \begin{align*} z=re^{i\theta},\qquad r>0,\qquad -\pi<\theta<\pi, \end{align*} and define the principal logarithm by \begin{align*} \operatorname{Log} z=\log r+i\theta. \end{align*} If $z=x+iy$, then \begin{align*} \operatorname{Log} z =\frac{1}{2}\log(x^2+y^2)+i\operatorname{Arg}(x+iy). \end{align*} On the slit plane the principal argument is smooth, and its partial derivatives are \begin{align*} \frac{\partial}{\partial x}\operatorname{Arg}(x+iy) =-\frac{y}{x^2+y^2}, \qquad \frac{\partial}{\partial y}\operatorname{Arg}(x+iy) =\frac{x}{x^2+y^2}. \end{align*} Thus, for $u(x,y)=\frac{1}{2}\log(x^2+y^2)$ and $v(x,y)=\operatorname{Arg}(x+iy)$, \begin{align*} u_x&=\frac{x}{x^2+y^2},& u_y&=\frac{y}{x^2+y^2},\\ v_x&=-\frac{y}{x^2+y^2},& v_y&=\frac{x}{x^2+y^2}. \end{align*} Hence $u_x=v_y$ and $u_y=-v_x$, so $\operatorname{Log}$ is holomorphic on $\Omega$, and its complex derivative is \begin{align*} (\operatorname{Log})'(z) &=u_x+iv_x\\ &=\frac{x}{x^2+y^2}-i\frac{y}{x^2+y^2}\\ &=\frac{x-iy}{x^2+y^2}\\ &=\frac{1}{x+iy}\\ &=\frac{1}{z}. \end{align*} Therefore $f(z)=1/z$ has a primitive on the slit plane. On $\mathbb C\setminus\{0\}$, suppose instead that a global primitive $F$ of $1/z$ existed. For the unit circle $\gamma(t)=e^{it}$, $0\le t\le 2\pi$, the *Fundamental Theorem for Complex Line Integrals* would give \begin{align*} \oint_\gamma \frac{1}{z}\,dz &=F(\gamma(2\pi))-F(\gamma(0))\\ &=F(1)-F(1)\\ &=0. \end{align*} But the same integral computed from the parametrisation is \begin{align*} \oint_\gamma \frac{1}{z}\,dz &=\int_0^{2\pi}\frac{1}{e^{it}}\,ie^{it}\,dt\\ &=\int_0^{2\pi}i\,dt\\ &=2\pi i, \end{align*} a contradiction. Thus $1/z$ is locally the derivative of a logarithm, but on the punctured plane the logarithm cannot be chosen as a single-valued holomorphic function around the missing point $0$. [/example] This failure motivates the next major theorem: under suitable geometric hypotheses on the domain, holomorphicity alone forces closed integrals to vanish. ## Cauchy Theory and Holomorphicity ### Cauchy's Theorem The central miracle of complex integration is that holomorphic functions behave like derivatives even when no primitive has been exhibited. [Cauchy's theorem](/page/Cauchy's%20Theorem) gives the integral consequence that closed contour integrals vanish under suitable topological hypotheses; combined with the path-independence criterion above, this is the route from holomorphicity to primitives on domains without holes. [definition: Simply Connected Domain] A domain $\Omega\subset\mathbb C$ is simply connected if for every closed continuous path $\gamma:[0,1]\to\Omega$, there exist a point $p\in\Omega$ and a continuous map \begin{align*} H:[0,1]\times[0,1]&\to\Omega \end{align*} such that \begin{align*} H(0,t)&=\gamma(t),\\ H(1,t)&=p,\\ H(s,0)&=H(s,1) \end{align*} for every $s,t\in[0,1]$. [/definition] Simple connectedness is the condition that removes homotopy obstructions for loops. It is the topological hypothesis that rules out the punctured plane example, where the unit circle cannot be contracted within the domain without crossing $0$. The next theorem answers the central question raised by the primitive obstruction: when every loop can be contracted in the domain, does every holomorphic function have vanishing integrals around closed contours? [quotetheorem:344] This theorem says that on a domain without holes, holomorphicity has a global integral consequence. The next form is more local and is often the one used in computations: integrals around the boundary of a triangle or disk vanish when the function is holomorphic inside. [quotetheorem:341] The local theorem is stronger than it may look. By subdividing regions into small triangles, it becomes the engine behind deformation of contours. It also raises a more precise question: if an integral of a holomorphic function around a closed boundary vanishes, what happens when the integrand has a controlled kernel singularity at an interior point? ### Integral Formulae A contour integral can recover function values. This is much stronger than vanishing: it says the values of a holomorphic function in the interior are encoded by boundary values. The singular kernel \begin{align*} z\mapsto \frac{1}{z-z_0} \end{align*} is allowed because the singularity is placed at the value being recovered, and the integral measures exactly that local contribution. We use a small amount of contour notation in the theorem cards below. For a path $\gamma:[a,b]\to\mathbb C$, write \begin{align*} \operatorname{im}(\gamma)=\gamma([a,b]) \end{align*} for its image, and write $\gamma^*$ for the same traced curve when a theorem uses that convention. If $\gamma$ is closed and $z_0\notin\operatorname{im}(\gamma)$, then $n(\gamma,z_0)$ denotes the [winding number](/page/Winding%20Number) of $\gamma$ about $z_0$, the integer \begin{align*} n(\gamma,z_0)=\frac{1}{2\pi i}\oint_\gamma \frac{dz}{z-z_0}. \end{align*} Some existing theorem cards write $I(\gamma,z_0)$ for the same index. A closed curve is null-homotopic in a domain when it can be continuously contracted to a point while staying inside that domain. [quotetheorem:3362] The formula should change how the reader thinks about holomorphic functions. A holomorphic function is not freely chosen point by point: once its boundary values on a circle are known, every interior value is determined. [example: Recovering a Value from a Boundary Integral] Let $f:\mathbb C\to\mathbb C$ be $f(z)=e^z$, let $z_0=0$, and let $\gamma:[0,2\pi]\to\mathbb C$ be the positively oriented circle \begin{align*} \gamma(t)=2e^{it}. \end{align*} The function $e^z$ is holomorphic on $\mathbb C$, and the closed disk $\overline{B}(0,2)$ is contained in $\mathbb C$, so *[Cauchy Integral Formula](/theorems/345)* applies with radius $2$. Therefore \begin{align*} f(0) &=\frac{1}{2\pi i}\oint_\gamma \frac{f(z)}{z-0}\,dz\\ &=\frac{1}{2\pi i}\oint_\gamma \frac{e^z}{z}\,dz. \end{align*} Since $f(0)=e^0=1$, we get \begin{align*} \frac{1}{2\pi i}\oint_\gamma \frac{e^z}{z}\,dz &=1, \end{align*} equivalently \begin{align*} \oint_\gamma \frac{e^z}{z}\,dz &=2\pi i. \end{align*} The integral samples $e^z$ only along the boundary circle $|z|=2$, but Cauchy's formula recovers the value of the holomorphic function at the centre. [/example] The value formula raises a new problem: boundary values should control not only $f(z_0)$ but also the rate at which $f$ changes near $z_0$. To get derivative estimates, power series expansions, and rigidity results such as [Liouville's theorem](/page/Liouville's%20Theorem), we need an integral formula for $f^{(n)}(z_0)$. The next theorem gives that formula by differentiating the kernel instead of asking for new boundary data. The formula below uses the compact notation $\mathcal{O}(\Omega)$ for the set of holomorphic functions on the [open set](/page/Open%20Set) $\Omega$. Thus $f\in\mathcal{O}(\Omega)$ means that $f$ is holomorphic on $\Omega$, and this notation prepares the derivative version of Cauchy's formula without changing the hypothesis. [quotetheorem:2570] The derivative formula is one of the reasons [complex differentiability](/page/Complex%20Differentiability) is so rigid. A single derivative implies infinitely many derivatives, and integral estimates become derivative estimates. [example: Cauchy Estimate on a Disk] Let $\Omega\subset\mathbb C$ be open, let $f:\Omega\to\mathbb C$ be holomorphic, let $z_0\in\Omega$, and choose $r>0$ such that $\overline{B}(z_0,r)\subsetneq\Omega$. Suppose that $|f(z)|\le M$ for every $z$ on the circle $|z-z_0|=r$, and let $n\ge 1$. Let $\gamma:[0,2\pi]\to\mathbb C$ be the positively oriented circle \begin{align*} \gamma(t)=z_0+r e^{it}. \end{align*} Since $\overline{B}(z_0,r)\subsetneq\Omega$ and $f$ is holomorphic on $\Omega$, *[Cauchy Integral Formula for Derivatives](/theorems/2570)* gives \begin{align*} f^{(n)}(z_0) =\frac{n!}{2\pi i}\oint_\gamma \frac{f(z)}{(z-z_0)^{n+1}}\,dz. \end{align*} Taking absolute values, \begin{align*} |f^{(n)}(z_0)| &=\left|\frac{n!}{2\pi i}\oint_\gamma \frac{f(z)}{(z-z_0)^{n+1}}\,dz\right|\\ &=\frac{n!}{2\pi}\left|\oint_\gamma \frac{f(z)}{(z-z_0)^{n+1}}\,dz\right|, \end{align*} because $|i|=1$. For every $z$ on the circle $\gamma([0,2\pi])$, we have $|z-z_0|=r$ and $|f(z)|\le M$, so \begin{align*} \left|\frac{f(z)}{(z-z_0)^{n+1}}\right| &=\frac{|f(z)|}{|z-z_0|^{n+1}}\\ &\le \frac{M}{r^{n+1}}. \end{align*} Also \begin{align*} \gamma'(t)=ir e^{it}, \end{align*} and hence \begin{align*} L(\gamma) &=\int_0^{2\pi}|\gamma'(t)|\,dt\\ &=\int_0^{2\pi}|ir e^{it}|\,dt\\ &=\int_0^{2\pi}r\,dt\\ &=2\pi r. \end{align*} By the *$ML$ Estimate* applied to $z\mapsto f(z)/(z-z_0)^{n+1}$ on $\gamma$, \begin{align*} \left|\oint_\gamma \frac{f(z)}{(z-z_0)^{n+1}}\,dz\right| &\le \frac{M}{r^{n+1}}L(\gamma)\\ &=\frac{M}{r^{n+1}}(2\pi r). \end{align*} Substituting this into the derivative formula estimate gives \begin{align*} |f^{(n)}(z_0)| &\le \frac{n!}{2\pi}\frac{M}{r^{n+1}}(2\pi r)\\ &=\frac{n!M(2\pi r)}{2\pi r^{n+1}}\\ &=\frac{n!M r}{r^{n+1}}\\ &=\frac{n!M}{r^n}. \end{align*} Thus boundary control of $f$ on the circle of radius $r$ gives explicit control of every derivative at the centre. [/example] Cauchy theory is strongest when no singularity lies inside the contour. The next section studies what remains when singularities are present but controlled. ## Singularities and Residues The unit-circle integral of \begin{align*} z\mapsto \frac{1}{z} \end{align*} suggests that a contour integral around an isolated singularity should depend on one special coefficient. [Laurent series](/page/Laurent%20Series) make this precise: all powers except $(z-z_0)^{-1}$ integrate to zero around small circles. [definition: Isolated Singularity] Let $\Omega\subset\mathbb C$ be open, let $z_0\in\Omega$, and let $f:\Omega\setminus\{z_0\}\to\mathbb C$ be holomorphic. The point $z_0$ is an isolated singularity of $f$ if there exists $r>0$ such that \begin{align*} B(z_0,r)\subset\Omega \end{align*} and $f$ is holomorphic on $B(z_0,r)\setminus\{z_0\}$. [/definition] To turn an isolated singularity into a number, we need a local invariant that a small circle can measure. The Laurent expansion supplies many coefficients, but the circle computation for powers tells us that all coefficients except the $(z-z_0)^{-1}$ coefficient vanish under integration. The next definition names precisely the coefficient that will become the singularity's contribution to any surrounding contour. [definition: Residue] Let $f:B(z_0,r)\setminus\{z_0\}\to\mathbb C$ be holomorphic and suppose $f$ has Laurent expansion \begin{align*} f(z)=\sum_{n=-\infty}^{\infty} a_n(z-z_0)^n \end{align*} on $B(z_0,r)\setminus\{z_0\}$. Let $\mathcal A_{z_0,r}$ be the set of holomorphic functions $g:B(z_0,r)\setminus\{z_0\}\to\mathbb C$ admitting a Laurent expansion \begin{align*} g(z)=\sum_{n=-\infty}^{\infty} b_n(g)(z-z_0)^n \end{align*} on $B(z_0,r)\setminus\{z_0\}$. The residue operation at $z_0$ is the map \begin{align*} \operatorname{Res}_{z_0}:\mathcal A_{z_0,r} &\to \mathbb C \\ g &\mapsto b_{-1}(g). \end{align*} The residue of $f$ at $z_0$ is \begin{align*} \operatorname{Res}(f,z_0)=a_{-1}. \end{align*} [/definition] The Laurent expansion on a punctured disk is unique, so the coefficient $a_{-1}$ is well-defined. The residue is not an arbitrary coefficient; it is chosen because contour integration filters out every other Laurent term on a small circle. More generally, residues localise the obstruction to [Cauchy's theorem](/theorems/797): if a contour encloses finitely many isolated singularities, the integral is controlled by the sum of their local contributions. [quotetheorem:352] The residue theorem is the mature form of the unit-circle computation. Instead of one missing point and one loop, we can handle finitely many singularities and any simple closed contour enclosing them. [example: A Rational Integral by Residues] Consider \begin{align*} I=\int_{-\infty}^{\infty}\frac{1}{x^2+1}\,d\mathcal L^1(x). \end{align*} For $R>1$, let $\sigma_R:[-R,R]\to\mathbb C$ be $\sigma_R(t)=t$, let $\eta_R:[0,\pi]\to\mathbb C$ be $\eta_R(t)=R e^{it}$, and let $\Gamma_R$ be the positively oriented contour obtained by following $\sigma_R$ and then $\eta_R$. Put \begin{align*} f(z)=\frac{1}{z^2+1}=\frac{1}{(z-i)(z+i)}. \end{align*} The poles are $i$ and $-i$, and for $R>1$ only $i$ lies inside $\Gamma_R$. The residue at $i$ is \begin{align*} \operatorname{Res}(f,i) &=\lim_{z\to i}(z-i)f(z)\\ &=\lim_{z\to i}\frac{z-i}{(z-i)(z+i)}\\ &=\lim_{z\to i}\frac{1}{z+i}\\ &=\frac{1}{i+i}\\ &=\frac{1}{2i}. \end{align*} By the *Residue Theorem*, \begin{align*} \oint_{\Gamma_R} f(z)\,dz &=2\pi i\,\operatorname{Res}(f,i)\\ &=2\pi i\cdot \frac{1}{2i}\\ &=\pi. \end{align*} Splitting the contour into the real segment and the upper semicircle gives \begin{align*} \oint_{\Gamma_R} f(z)\,dz &=\int_{\sigma_R} f(z)\,dz+\int_{\eta_R} f(z)\,dz\\ &=\int_{-R}^{R} f(\sigma_R(t))\sigma_R'(t)\,dt+\int_{\eta_R}\frac{1}{z^2+1}\,dz\\ &=\int_{-R}^{R}\frac{1}{t^2+1}\,dt+\int_{\eta_R}\frac{1}{z^2+1}\,dz. \end{align*} Hence \begin{align*} \int_{-R}^{R}\frac{1}{t^2+1}\,dt =\pi-\int_{\eta_R}\frac{1}{z^2+1}\,dz. \end{align*} It remains to show that the semicircle term vanishes as $R\to\infty$. If $z\in\eta_R([0,\pi])$, then $|z|=R$, so the reverse triangle inequality gives \begin{align*} |z^2+1| &\ge \bigl||z^2|-|1|\bigr|\\ &=|R^2-1|\\ &=R^2-1, \end{align*} because $R>1$. Therefore \begin{align*} \left|\frac{1}{z^2+1}\right| \le \frac{1}{R^2-1}. \end{align*} Also \begin{align*} \eta_R'(t)=iR e^{it}, \end{align*} so \begin{align*} L(\eta_R) &=\int_0^\pi |\eta_R'(t)|\,dt\\ &=\int_0^\pi |iR e^{it}|\,dt\\ &=\int_0^\pi R\,dt\\ &=\pi R. \end{align*} By the *$ML$ Estimate*, \begin{align*} \left|\int_{\eta_R}\frac{1}{z^2+1}\,dz\right| &\le \frac{1}{R^2-1}L(\eta_R)\\ &=\frac{\pi R}{R^2-1}\\ &=\frac{\pi}{R-\frac{1}{R}}. \end{align*} Since $R-\frac{1}{R}\to\infty$, this bound tends to $0$, and hence \begin{align*} \lim_{R\to\infty}\int_{\eta_R}\frac{1}{z^2+1}\,dz=0. \end{align*} Taking limits in the identity for the real segment gives \begin{align*} \lim_{R\to\infty}\int_{-R}^{R}\frac{1}{t^2+1}\,dt &=\pi. \end{align*} By monotone convergence for the nonnegative functions $\frac{1}{x^2+1}\mathbf 1_{[-R,R]}(x)$, \begin{align*} I=\int_{-\infty}^{\infty}\frac{1}{x^2+1}\,d\mathcal L^1(x)=\pi. \end{align*} The real integral is recovered from the single residue of the upper-half-plane pole, while the large semicircle contributes nothing in the limit. [/example] Residues also explain why the exact shape of a contour often does not matter. If two contours enclose the same singularities with the same orientation and can be deformed into each other without crossing singularities, they give the same integral. ## Deformation and Winding The phrase inside a contour is intuitive for simple closed curves, but many useful contours cross themselves or wind around a point several times. To state the correct general theorem, we need a number that records how often a closed curve winds around a point. [illustration:winding-number] [definition: Winding Integral] Let $\mathcal W$ be the set of pairs $(\gamma,z_0)$ where $\gamma:[a,b]\to\mathbb C$ is a closed piecewise $C^1$ path and $z_0\notin\gamma([a,b])$. The winding integral is the map \begin{align*} N:\mathcal W &\to \mathbb C \end{align*} defined by \begin{align*} N(\gamma,z_0)=\frac{1}{2\pi i}\oint_\gamma \frac{dz}{z-z_0}. \end{align*} [/definition] The formula is analytic, but its decisive feature is arithmetic: the value is always an integer. This integrality is what converts the analytic integral of \begin{align*} z\mapsto \frac{1}{z-z_0} \end{align*} into a topological invariant. Positively oriented simple closed contours produce the values $0$ and $1$ for points outside and inside, while negatively oriented ones produce $0$ and $-1$. General contours need arbitrary integer weights. [quotetheorem:5023] For residue calculus, the remaining obstruction is bookkeeping. The integral $N(\gamma,z_0)$ already gives the correct weight, but the residue theorem needs those weights as integer data attached to points, not as unnamed contour integrals. The following definition packages that integer-valued invariant so later statements can say precisely how much each singularity contributes. [definition: Winding Number] Let $\mathcal W$ be the set of pairs $(\gamma,z_0)$ where $\gamma:[a,b]\to\mathbb C$ is a closed piecewise $C^1$ path and $z_0\notin\gamma([a,b])$. The winding number is the map \begin{align*} n:\mathcal W &\to \mathbb Z \\ (\gamma,z_0) &\mapsto N(\gamma,z_0). \end{align*} The integer $n(\gamma,z_0)$ is called the winding number of $\gamma$ about $z_0$. [/definition] Now the residue theorem can be stated with winding weights, so it can handle contours that loop repeatedly or with mixed orientation. This matters because the phrase inside the contour no longer makes sense once a path crosses itself. A singularity should contribute according to how the whole path winds around it, while the contour must also be null-homologous in the analytic domain: it must have zero winding around every point outside that domain. In the weighted statement below, $n(\gamma,z_j)$ denotes the winding number of the contour $\gamma$ about the singularity $z_j$. [quotetheorem:5024] This form removes the need for the curve to be simple. The contour carries weights, and each singularity contributes according to its winding number. [example: Traversing a Circle Twice] Let $\gamma:[0,4\pi]\to\mathbb C$ be \begin{align*} \gamma(t)=e^{it}. \end{align*} Then $\gamma(0)=1=\gamma(4\pi)$, so $\gamma$ is closed, and $0\notin\gamma([0,4\pi])$ because $|e^{it}|=1$ for every $t$. We compute the winding integral of $\gamma$ about $0$. Since \begin{align*} \gamma'(t)=ie^{it}, \end{align*} the definition of the complex line integral gives \begin{align*} \oint_\gamma \frac{dz}{z} &=\int_0^{4\pi}\frac{1}{\gamma(t)}\gamma'(t)\,dt\\ &=\int_0^{4\pi}\frac{1}{e^{it}}\,ie^{it}\,dt\\ &=\int_0^{4\pi} i\,\frac{e^{it}}{e^{it}}\,dt\\ &=\int_0^{4\pi} i\,dt\\ &=i(4\pi-0)\\ &=4\pi i. \end{align*} Therefore, by the definition of winding number, \begin{align*} n(\gamma,0) &=\frac{1}{2\pi i}\oint_\gamma\frac{dz}{z}\\ &=\frac{1}{2\pi i}(4\pi i)\\ &=2. \end{align*} The same geometric circle, traversed twice counterclockwise, contributes winding number $2$ around $0$, so a residue at $0$ would be counted twice. [/example] Contour deformation is the operational version of winding. If a family of curves moves through a region where the integrand has a primitive, the integral remains fixed throughout the deformation because every closed member of the family has zero endpoint contribution. For holomorphic functions this primitive hypothesis is supplied on simply connected regions by Cauchy's theorem; for punctured regions, residues and winding numbers describe exactly what can change. To discuss ordinary contour deformations without introducing singular homology, we name the regular homotopies that appear in standard contour arguments. [definition: Homotopy of Closed Piecewise $C^1$ Paths] Let $\Omega\subset\mathbb C$ be open. A homotopy of closed piecewise $C^1$ paths in $\Omega$ is a continuous map \begin{align*} H:[0,1]\times[a,b]&\to\Omega \end{align*} such that $H(s,a)=H(s,b)$ for every $s\in[0,1]$ and, for every $s\in[0,1]$, there is a partition \begin{align*} a=t_0<t_1<\cdots<t_n=b \end{align*} for which $t\mapsto H(s,t)$ is $C^1$ on each interval $[t_{j-1},t_j]$. For each $s\in[0,1]$, write $\gamma_s:[a,b]\to\Omega$ for the closed piecewise $C^1$ path $\gamma_s(t)=H(s,t)$. [/definition] Homotopy is the language behind the instruction deform the contour. It records that a contour may slide through a region where the integral has no local obstruction, such as a region on which the integrand has a primitive. In the primitive case, the equality of integrals does not depend on the extra differentiability of the two-parameter deformation; that regularity only guarantees that every intermediate curve is an allowed path. [quotetheorem:343] This theorem explains why rectangular, circular, and keyhole contours can be interchanged after the relevant region has been reduced to one where the integrand has a primitive. In punctured domains, the residue theorem supplies the complementary rule: deformation preserves the integral exactly when the winding numbers around the singularities do not change. ## Applications and Consequences Complex integration becomes useful because it converts analytic information into algebraic or asymptotic information. The same contour integral can prove rigidity theorems, evaluate real integrals, and count zeros. The first consequence of Cauchy's estimates is a rigidity theorem: a bounded entire function cannot vary. The proof belongs elsewhere, but the statement belongs here because it is one of the cleanest examples of integral estimates controlling a function. [quotetheorem:346] [Liouville's theorem](/theorems/38) is a global conclusion from local holomorphicity plus an integral estimate on arbitrarily large circles. To count zeros and poles rather than force constancy, we need functions whose singularities are controlled enough that residues still make sense. This leads from holomorphic functions to meromorphic functions. [definition: Meromorphic Function] Let $\Omega\subset\mathbb C$ be open. A function $f:\Omega\to\mathbb C\cup\{\infty\}$ is meromorphic on $\Omega$ if the set of points where $f$ takes the value $\infty$ is discrete in $\Omega$, $f:\Omega\setminus f^{-1}(\{\infty\})\to\mathbb C$ is holomorphic, and every point $z_0\in f^{-1}(\{\infty\})$ is a pole of the ordinary holomorphic function obtained by restricting $f$ to a punctured neighbourhood of $z_0$. [/definition] Meromorphic functions are the natural class for residue calculus because their singularities are controlled by finitely many negative Laurent terms near each pole. If $f$ is meromorphic, then the logarithmic derivative \begin{align*} \frac{f'}{f} \end{align*} has simple poles whose residues encode zeros and poles of $f$. The next theorem packages that observation into a counting formula. The statement below uses standard order notation to turn zeros and poles into signed multiplicities. If $f$ is meromorphic near $z_0$ and not identically zero, then $\operatorname{ord}(f,z_0)=m$ means that \begin{align*} f(z)=(z-z_0)^m g(z) \end{align*} near $z_0$, where $g$ is holomorphic and $g(z_0)\ne 0$. Positive $m$ records a zero of order $m$, negative $m$ records a pole of order $-m$, and \begin{align*} \operatorname{ord}(f,z_0)^{-}=\max\{-\operatorname{ord}(f,z_0),0\} \end{align*} records only the pole contribution; this prepares [the argument principle](/theorems/356) as a weighted counting theorem. [quotetheorem:3359] This formula turns a contour integral into a counting device. It is the conceptual source of Rouche's theorem, zero-counting arguments, and many stability results in complex analysis. [example: Counting Zeros by a Logarithmic Derivative] Let \begin{align*} f(z)=z^3-2,\qquad \gamma(t)=2e^{it} \end{align*} for $0\le t\le 2\pi$. The path $\gamma$ is the positively oriented circle $|z|=2$. Since $f$ is a polynomial, it has no poles. If $a$ is a zero of $f$, then \begin{align*} a^3-2&=0,\\ a^3&=2,\\ |a|^3&=|2|=2,\\ |a|&=2^{1/3}. \end{align*} Because $2<8=2^3$, we have $2^{1/3}<2$, so every zero of $f$ lies inside $\gamma$ and no zero lies on the image of $\gamma$. By the *[Fundamental Theorem of Algebra](/theorems/347)*, $z^3-2$ has three zeros counted with multiplicity. Also \begin{align*} f'(z)=3z^2, \end{align*} so the logarithmic derivative is \begin{align*} \frac{f'(z)}{f(z)}=\frac{3z^2}{z^3-2} \end{align*} wherever $f(z)\ne 0$. Applying the *[Argument Principle](/page/Argument%20Principle)* to $f$ on the circle $\gamma$, with $N=3$ and $P=0$, gives \begin{align*} \frac{1}{2\pi i}\oint_\gamma \frac{3z^2}{z^3-2}\,dz &=N-P\\ &=3-0\\ &=3. \end{align*} Thus the contour integral counts the three zeros inside the circle without requiring explicit formulas for the roots. [/example] A final application is real integral evaluation. The method is not a single theorem but a strategy: choose a contour, identify singularities, show auxiliary arcs vanish, and read off the real integral from residues. [example: A Fourier Integral] Let $a>0$, and put \begin{align*} F(z)=\frac{e^{iaz}}{z^2+1} =\frac{e^{iaz}}{(z-i)(z+i)}. \end{align*} For $R>1$, let $\sigma_R:[-R,R]\to\mathbb C$ be $\sigma_R(t)=t$, let $\eta_R:[0,\pi]\to\mathbb C$ be $\eta_R(t)=Re^{it}$, and let $\Gamma_R$ be the positively oriented contour obtained by first following $\sigma_R$ and then following $\eta_R$. The only pole of $F$ inside $\Gamma_R$ is $i$, since $-i$ lies in the lower half-plane. Its residue is \begin{align*} \operatorname{Res}(F,i) &=\lim_{z\to i}(z-i)\frac{e^{iaz}}{(z-i)(z+i)}\\ &=\lim_{z\to i}\frac{e^{iaz}}{z+i}\\ &=\frac{e^{ia i}}{i+i}\\ &=\frac{e^{-a}}{2i}. \end{align*} By the *Residue Theorem*, \begin{align*} \oint_{\Gamma_R}F(z)\,dz &=2\pi i\,\operatorname{Res}(F,i)\\ &=2\pi i\cdot \frac{e^{-a}}{2i}\\ &=\pi e^{-a}. \end{align*} Splitting the contour gives \begin{align*} \oint_{\Gamma_R}F(z)\,dz &=\int_{\sigma_R}F(z)\,dz+\int_{\eta_R}F(z)\,dz\\ &=\int_{-R}^{R}F(\sigma_R(t))\sigma_R'(t)\,dt+\int_{\eta_R}\frac{e^{iaz}}{z^2+1}\,dz\\ &=\int_{-R}^{R}\frac{e^{iat}}{t^2+1}\,dt+\int_{\eta_R}\frac{e^{iaz}}{z^2+1}\,dz, \end{align*} because $\sigma_R(t)=t$ and $\sigma_R'(t)=1$. Hence \begin{align*} \int_{-R}^{R}\frac{e^{iat}}{t^2+1}\,dt =\pi e^{-a}-\int_{\eta_R}\frac{e^{iaz}}{z^2+1}\,dz. \end{align*} It remains to show that the semicircle integral tends to $0$. If $z=\eta_R(t)=Re^{it}=R\cos t+iR\sin t$ with $0\le t\le\pi$, then \begin{align*} iaz &=ia(R\cos t+iR\sin t)\\ &=iaR\cos t-aR\sin t, \end{align*} so \begin{align*} |e^{iaz}| &=|e^{iaR\cos t-aR\sin t}|\\ &=e^{-aR\sin t}\\ &\le 1, \end{align*} because $a>0$ and $\sin t\ge 0$ on $[0,\pi]$. Also $|z|=R$, so the reverse triangle inequality gives \begin{align*} |z^2+1| &\ge \bigl||z^2|-|1|\bigr|\\ &=|R^2-1|\\ &=R^2-1. \end{align*} Therefore, for every $z$ on $\eta_R$, \begin{align*} \left|\frac{e^{iaz}}{z^2+1}\right| =\frac{|e^{iaz}|}{|z^2+1|} \le \frac{1}{R^2-1}. \end{align*} Since $\eta_R'(t)=iRe^{it}$, \begin{align*} L(\eta_R) &=\int_0^\pi |\eta_R'(t)|\,dt\\ &=\int_0^\pi |iRe^{it}|\,dt\\ &=\int_0^\pi R\,dt\\ &=\pi R. \end{align*} By the *$ML$ Estimate*, \begin{align*} \left|\int_{\eta_R}\frac{e^{iaz}}{z^2+1}\,dz\right| &\le \frac{1}{R^2-1}L(\eta_R)\\ &=\frac{\pi R}{R^2-1}\\ &=\frac{\pi}{R-\frac{1}{R}}. \end{align*} Since $R-\frac{1}{R}\to\infty$ as $R\to\infty$, this bound tends to $0$, and hence \begin{align*} \lim_{R\to\infty}\int_{\eta_R}\frac{e^{iaz}}{z^2+1}\,dz=0. \end{align*} Taking the limit in the real-segment identity gives \begin{align*} \lim_{R\to\infty}\int_{-R}^{R}\frac{e^{iat}}{t^2+1}\,dt &=\pi e^{-a}. \end{align*} Since \begin{align*} \left|\frac{e^{iax}}{x^2+1}\mathbf 1_{[-R,R]}(x)\right| &=\frac{1}{x^2+1}\mathbf 1_{[-R,R]}(x)\\ &\le \frac{1}{x^2+1}, \end{align*} and $\frac{1}{x^2+1}$ is integrable over $\mathbb R$, the *[Dominated Convergence Theorem](/theorems/4)* identifies this limit with the [Lebesgue integral](/page/Lebesgue%20Integral): \begin{align*} \int_{-\infty}^{\infty}\frac{e^{iax}}{x^2+1}\,d\mathcal L^1(x) =\pi e^{-a}. \end{align*} Writing $e^{iax}=\cos(ax)+i\sin(ax)$ therefore gives \begin{align*} \int_{-\infty}^{\infty}\frac{\cos(ax)}{x^2+1}\,d\mathcal L^1(x) +i\int_{-\infty}^{\infty}\frac{\sin(ax)}{x^2+1}\,d\mathcal L^1(x) =\pi e^{-a}. \end{align*} Equating real and imaginary parts yields \begin{align*} \int_{-\infty}^{\infty}\frac{\cos(ax)}{x^2+1}\,d\mathcal L^1(x) &=\pi e^{-a},\\ \int_{-\infty}^{\infty}\frac{\sin(ax)}{x^2+1}\,d\mathcal L^1(x) &=0. \end{align*} The pole at $i$ supplies the whole value of the Fourier integral, while the upper semicircle contributes nothing in the limit. [/example] These examples show the same pattern in different forms. Complex integrals convert geometry into algebra: contour shape becomes winding number, singularity type becomes residue, and boundary data becomes interior analytic information. ## Beyond and Connected Topics The natural next step is [Cambridge IB Complex Analysis](/page/Cambridge%20IB%20Complex%20Analysis), where contour integration sits beside power series, [conformal maps](/page/Conformal%20Maps), isolated singularities, and the residue theorem in a first systematic course. Several complex variables changes the story sharply. In one complex variable, isolated singularities and winding numbers dominate contour integration. In [Androma Several Complex Variables I: Domains and Holomorphy](/page/Androma%20Several%20Complex%20Variables%20I%3A%20Domains%20and%20Holomorphy), domains, holomorphic functions, and integral phenomena interact with higher-dimensional geometry, where Hartogs-type extension phenomena replace many one-variable intuitions. Sheaf methods give a structural language for analytic continuation and local-to-global problems. The continuation from contour arguments to cohomological tools is developed in [Several Complex Variables II: Sheaves and Stein Theory](/page/Several%20Complex%20Variables%20II%3A%20Sheaves%20and%20Stein%20Theory). Integral estimates also lead toward analytic PDE methods in complex geometry. The course [Several Complex Variables III: L² Methods and Applications](/page/Several%20Complex%20Variables%20III%3A%20L%C2%B2%20Methods%20and%20Applications) develops $L^2$ techniques that replace elementary contour estimates in higher-dimensional problems. Residue theory continues into algebraic geometry and differential topology. The residue theorem becomes a model for duality, index formulas, and localisation principles: a global invariant is computed as a sum of local contributions. ## References Androma, [Cambridge IB Complex Analysis](/page/Cambridge%20IB%20Complex%20Analysis). Androma, [Androma Several Complex Variables I: Domains and Holomorphy](/page/Androma%20Several%20Complex%20Variables%20I%3A%20Domains%20and%20Holomorphy). Androma, [Several Complex Variables II: Sheaves and Stein Theory](/page/Several%20Complex%20Variables%20II%3A%20Sheaves%20and%20Stein%20Theory). Androma, [Several Complex Variables III: L² Methods and Applications](/page/Several%20Complex%20Variables%20III%3A%20L%C2%B2%20Methods%20and%20Applications). Ahlfors, *Complex Analysis* (1979). Conway, *Functions of One Complex Variable I* (1978). Stein and Shakarchi, *Complex Analysis* (2003).