The residue is the local coefficient that records the first-order obstruction to making a meromorphic differential holomorphic at an isolated singularity. It is small enough to be read from one Laurent coefficient, but strong enough to control contour integrals around the singularity. This chapter develops the residue from Laurent expansions to the residue theorem. The main setting is one complex variable. The notation $D(a,r)$ denotes the open disc with centre $a$ and radius $r$, and the punctured disc is $D(a,r)\setminus\{a\}$. A positively oriented circle around $a$ will be denoted by \begin{align*} \gamma_{a,\rho}(t)&=a+\rho e^{it}, & 0\leq t\leq 2\pi, \end{align*} where the parameter $\rho$ is chosen so that the circle lies in the domain under discussion. The differential notation $f(z)\,dz$ is used when the coordinate-invariant flavour matters. For functions on planar domains, the same information is contained in the integral of $f(z)\,dz$, and the residue of the function $f$ at $a$ is the residue of the differential $f(z)\,dz$ in the coordinate $z$. Residues sit between local singularity theory and global contour integration: they turn many integral computations into coefficient extraction, and they expose how poles, removable singularities, and essential singularities contribute to a contour integral. ## Definition ### Laurent Coefficient The first source of residues is the [Laurent expansion](/page/Laurent%20Series) on a punctured disc. If $f$ is holomorphic on $0<|z-a|<r$, then it has a Laurent series \begin{align*} \sum_{n\in\mathbb Z} c_n(z-a)^n \end{align*} converging normally on compact annuli. The coefficient $c_{-1}$ is singled out by integration. [example: Basic Residue] For \begin{align*} f(z)=\frac{1}{z-a}, \end{align*} we can write the Laurent expansion about $a$ as the single Laurent monomial \begin{align*} f(z)=1\cdot (z-a)^{-1}. \end{align*} Thus the coefficient multiplying $(z-a)^{-1}$ is $1$, so by the definition of residue, \begin{align*} \operatorname{Res}(f,a)=1. \end{align*} For \begin{align*} g(z)=\frac{1}{(z-a)^2}, \end{align*} the Laurent expansion about $a$ is \begin{align*} g(z)=1\cdot (z-a)^{-2}. \end{align*} This expansion has coefficient $1$ on $(z-a)^{-2}$ and coefficient $0$ on $(z-a)^{-1}$. Therefore \begin{align*} \operatorname{Res}(g,a)=0. \end{align*} The residue records only the coefficient of $(z-a)^{-1}$, not the coefficient of the most singular term. [/example] The example shows that the relevant coefficient is not the most singular term. The obstruction is that the principal part of a Laurent expansion may contain many negative powers, but only one of them contributes to a small contour integral around the singularity. The residue isolates exactly that coefficient, separating the integral contribution from the rest of the local singular behaviour. ### Formal Local Definition To turn the preceding contour intuition into a reusable local invariant, we need a definition that depends only on the Laurent expansion near the point. The formal point is to ignore all coefficients except the one attached to $(z-a)^{-1}$, because that is the unique term whose integral around a sufficiently small circle can survive. [definition: Residue] Fix $a\in\mathbb C$. Let $\mathcal{M}_a$ be the set of germs at $a$ of functions holomorphic on some punctured disc $D(a,r)\setminus\{a\}$. The residue at $a$ is the map \begin{align*} \operatorname{Res}_a: \mathcal{M}_a \to \mathbb C, \end{align*} defined by $\operatorname{Res}_a(f)=c_{-1}$, where $c_{-1}$ is the coefficient of $(z-a)^{-1}$ in the Laurent expansion \begin{align*} f(z)=\sum_{n\in\mathbb Z}c_n(z-a)^n \end{align*} about $a$. For a meromorphic differential $f(z)\,dz$ in the coordinate $z$, its residue at $a$ is denoted $\operatorname{Res}_a(f(z)\,dz)$ and is defined to be $\operatorname{Res}_a(f)$. [/definition] This definition is local: it depends only on the germ of $f$ near $a$, not on the behaviour of $f$ elsewhere in the domain. We will also use the two-variable notation $\operatorname{Res}(f,a)$ for the same number as $\operatorname{Res}_a(f)$. The subscript form emphasizes the local functional at the point $a$; the two-variable form is convenient in formulas where several singularities are being summed. ## Laurent Expansion and the Residue Changing $r$ does not change the coefficient, because Laurent expansions agree on overlaps of annuli. The coefficient can also be described by a small contour integral, which is often the most useful formulation. It explains why a local Laurent coefficient can be recovered without knowing the whole Laurent series, and it prepares the passage from small circles to larger contours. Concretely, if $\gamma_r(t)=a+re^{it}$ is a sufficiently small positively oriented circle around $a$ containing no other singularities, then the Laurent coefficient of $(z-a)^{-1}$ is recovered by \begin{align*} \operatorname{Res}_a(f)=\frac{1}{2\pi i}\int_{\gamma_r} f(z)\,dz. \end{align*} This is still a local statement: the contour is chosen inside the punctured neighbourhood where the Laurent expansion is valid. The reason only the residue survives is that the integrals of $(z-a)^n$ around a small circle vanish for all integers $n\ne -1$, while the integral of $(z-a)^{-1}$ is $2\pi i$. This small-circle formula is the local model for the later residue theorem. The global theorem will require additional language about winding numbers and homology; for now the point is simply that the residue is the unique Laurent coefficient detected by a tiny loop around the singularity. ## Simple Poles ### Local Limit Formula Many residues arise from simple [poles](/page/Pole). At a simple pole, the residue can be found by one limit. This is the most common computational situation, because many [meromorphic functions](/page/Meromorphic%20Function) are quotients with simple zeros in the denominator. [definition: Simple Pole] Let $U\subset\mathbb C$ be open with $a\in U$, and let $f:U\setminus\{a\}\to\mathbb C$ be holomorphic. The point $a$ is a simple pole of $f$ if $(z-a)f(z)$ extends to a [holomorphic function](/page/Holomorphic%20Function) $U\to\mathbb C$ whose value at $a$ is nonzero. [/definition] This definition says that the principal part has exactly one term. The residue is then that extended value. To compute residues in examples, it is useful to avoid building the whole Laurent series. The local question is whether the singularity has exactly one negative Laurent term, and if so how to read off its coefficient without expanding everything. The theorem is needed because simple poles occur whenever a denominator has a simple zero. It also explains why the residue is stable under multiplication by holomorphic nonzero factors. [quotetheorem:353] The three parts of this computation rule serve different purposes. The limit formula is the local test for a simple pole; the quotient formula is the version used most often in examples; and the nonvanishing hypotheses prevent accidental cancellation, which would make the singularity removable or change its order. Thus the theorem is not just a shortcut for one calculation: it separates the local pole condition from the algebraic data that produce it. ### Quotient Computations A frequent special case is a quotient $g/h$. The denominator has to vanish to first order, and the numerator should not cancel that zero if the quotient is to have a simple pole. The quotient part of the preceding theorem packages that common computation. The formula is often faster than expanding $g/h$ directly, because it uses only the numerator value and the first nonzero term of the denominator. [example: Reciprocal Quadratic] Consider \begin{align*} f(z)=\frac{1}{z^2+1}. \end{align*} Since \begin{align*} z^2+1=(z-i)(z+i), \end{align*} the denominator vanishes exactly at $z=i$ and $z=-i$. Write $f=g/h$ with \begin{align*} g(z)=1,\qquad h(z)=z^2+1. \end{align*} Then \begin{align*} h'(z)=2z. \end{align*} At $i$ we have \begin{align*} h(i)=i^2+1=-1+1=0,\qquad h'(i)=2i,\qquad g(i)=1\neq 0. \end{align*} Thus the hypotheses of *Quotient Formula for a Simple Pole* hold at $i$, and \begin{align*} \operatorname{Res}(f,i)=\frac{g(i)}{h'(i)}=\frac{1}{2i}. \end{align*} At $-i$ we similarly have \begin{align*} h(-i)=(-i)^2+1=-1+1=0,\qquad h'(-i)=2(-i)=-2i,\qquad g(-i)=1\neq 0. \end{align*} Applying the same formula at $-i$ gives \begin{align*} \operatorname{Res}(f,-i)=\frac{g(-i)}{h'(-i)}=\frac{1}{-2i}=-\frac{1}{2i}. \end{align*} The two simple poles have opposite residues, reflecting the opposite first-order factors $z-i$ and $z+i$ in the denominator. [/example] ## Higher-Order Poles A pole of order $m$ has a finite principal part with powers $(z-a)^{-m},\ldots,(z-a)^{-1}$. The residue is still only the coefficient of $(z-a)^{-1}$. [definition: Pole of Order $m$] Let $m\geq 1$. Let $U\subset\mathbb C$ be open with $a\in U$, and let $f:U\setminus\{a\}\to\mathbb C$ be holomorphic. The function $f$ has a pole of order $m$ at $a$ if $(z-a)^m f(z)$ extends to a holomorphic function $U\to\mathbb C$ whose value at $a$ is nonzero. [/definition] For higher-order poles, the coefficient can be extracted by differentiation. After clearing the pole, the desired coefficient becomes a Taylor coefficient of a holomorphic function. The result below records exactly which Taylor coefficient is needed. It converts a Laurent coefficient question into an ordinary derivative calculation at the singular point. [quotetheorem:3365] The formula is useful when the order is small. For large orders, direct Laurent multiplication can be more transparent. The choice depends on the expression and the desired simplification. [example: Second-Order Pole] Let \begin{align*} f(z)=\frac{e^z}{(z-a)^2}. \end{align*} Multiplying by $(z-a)^2$ clears the pole: \begin{align*} (z-a)^2f(z)=(z-a)^2\frac{e^z}{(z-a)^2}=e^z. \end{align*} Set $\phi(z)=(z-a)^2f(z)=e^z$. Since $\phi$ is holomorphic near $a$ and $\phi(a)=e^a\neq 0$, the function $f$ has a pole of order $2$ at $a$. By *Residue at a Pole of Order $m$* with $m=2$, \begin{align*} \operatorname{Res}(f,a)=\frac{\phi^{(2-1)}(a)}{(2-1)!}. \end{align*} Now $(2-1)!=1! =1$, and \begin{align*} \phi'(z)=\frac{d}{dz}e^z=e^z. \end{align*} Therefore \begin{align*} \operatorname{Res}(f,a)=\frac{\phi'(a)}{1}=\phi'(a)=e^a. \end{align*} The coefficient of $(z-a)^{-1}$ is therefore $e^a$, even though the pole itself has order $2$. [/example] ## Removable Singularities and Essential Singularities Residues do not detect every singular feature. A removable singularity has no negative Laurent terms, so its residue is zero. For instance, \begin{align*} \frac{\sin z}{z} \end{align*} has residue zero at $0$, because it extends holomorphically there. An essential singularity may have infinitely many negative Laurent terms, but only one of those terms contributes to the residue. [example: Essential Singularity] For $z\neq 0$, substitute $w=1/z$ into the [power series](/page/Power%20Series) for the exponential function: \begin{align*} e^w=\sum_{n=0}^{\infty}\frac{w^n}{n!}. \end{align*} This gives \begin{align*} e^{1/z}=\sum_{n=0}^{\infty}\frac{(1/z)^n}{n!}. \end{align*} Since $(1/z)^n=z^{-n}$ for every $n\geq 0$, the Laurent expansion at $0$ is \begin{align*} e^{1/z}=\sum_{n=0}^{\infty}\frac{z^{-n}}{n!}. \end{align*} The coefficient of $z^{-1}$ occurs when $-n=-1$, hence when $n=1$, and that coefficient is \begin{align*} \frac{1}{1!}=1. \end{align*} Therefore \begin{align*} \operatorname{Res}(e^{1/z},0)=1. \end{align*} Although $e^{1/z}$ has infinitely many negative powers in its Laurent expansion, the residue records only the coefficient of $z^{-1}$. [/example] The residue can vanish at an essential singularity. For example, $e^{1/z^2}$ contains only even negative powers of $z$, so the coefficient of $z^{-1}$ is absent. Residues therefore measure a precise integral contribution, not the severity of the singularity. ## The Residue Theorem The local contour formula becomes global when a contour surrounds finitely many isolated singularities. The complement of the singularities is a region where the function is holomorphic. [Cauchy's theorem](/theorems/797) then reduces the integral over the outer contour to the sum of small inner circles. For a closed contour $\gamma$ and a point $w$ not on its image, write $n(\gamma,w)$ for the [winding number](/page/Winding%20Number), or index, of $\gamma$ about $w$; it records how many times the contour winds around that point with orientation. [quotetheorem:5024] The theorem also holds with winding numbers for contours that are not simple. In that form, the coefficient of $\operatorname{Res}(f,a)$ is the winding number of the contour around $a$. The simple-contour version is enough for many applications and captures the main mechanism. ## Real Integrals from Residues Residues can evaluate real integrals by embedding them into contour integrals. The method has three steps: choose a contour whose boundary integral contains the desired real integral, estimate the added contour pieces, and compute the residues inside the contour. For rational functions on the real line, semicircles in the upper half-plane often work. [example: Gaussian-Style Rational Integral] Fix $R>1$, and let $\Gamma_R$ be the positively oriented boundary of the upper half-disc: first the interval $[-R,R]$, then the arc $C_R(t)=Re^{it}$ for $0\leq t\leq \pi$. We integrate $F(z)=1/(z^2+1)$ over $\Gamma_R$ in order to compute \begin{align*} \int_{-\infty}^{\infty}\frac{dx}{x^2+1}. \end{align*} The denominator factors as \begin{align*} z^2+1=(z-i)(z+i). \end{align*} Thus the singularities are $i$ and $-i$. Since $R>1$, the point $i$ lies inside the upper half-disc and $-i$ does not. At $i$, the pole is simple, and by *Residue at a Simple Pole*, \begin{align*} \operatorname{Res}\left(\frac{1}{z^2+1},i\right)=\lim_{z\to i}(z-i)\frac{1}{(z-i)(z+i)}. \end{align*} For $z\neq i$ and $z\neq -i$ the factor $z-i$ cancels, so \begin{align*} \lim_{z\to i}(z-i)\frac{1}{(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* applied to $\Gamma_R$, \begin{align*} \int_{\Gamma_R}\frac{dz}{z^2+1}=2\pi i\operatorname{Res}\left(\frac{1}{z^2+1},i\right). \end{align*} Substituting the residue gives \begin{align*} \int_{\Gamma_R}\frac{dz}{z^2+1}=2\pi i\cdot \frac{1}{2i}=\pi. \end{align*} Splitting the contour into the real segment and the arc gives \begin{align*} \int_{\Gamma_R}\frac{dz}{z^2+1}=\int_{-R}^{R}\frac{dx}{x^2+1}+\int_{C_R}\frac{dz}{z^2+1}. \end{align*} Therefore \begin{align*} \int_{-R}^{R}\frac{dx}{x^2+1}+\int_{C_R}\frac{dz}{z^2+1}=\pi. \end{align*} It remains to show that the arc integral vanishes as $R\to\infty$. If $z\in C_R$, then $|z|=R$, so by the [reverse triangle inequality](/theorems/2300), \begin{align*} |z^2+1|\geq ||z^2|-|1||=R^2-1. \end{align*} Hence \begin{align*} \left|\frac{1}{z^2+1}\right|\leq \frac{1}{R^2-1}. \end{align*} The arc $C_R$ has length $\pi R$, so the standard contour length estimate gives \begin{align*} \left|\int_{C_R}\frac{dz}{z^2+1}\right|\leq \frac{\pi R}{R^2-1}. \end{align*} Since \begin{align*} \lim_{R\to\infty}\frac{\pi R}{R^2-1}=0, \end{align*} we have \begin{align*} \lim_{R\to\infty}\int_{C_R}\frac{dz}{z^2+1}=0. \end{align*} Taking the limit in \begin{align*} \int_{-R}^{R}\frac{dx}{x^2+1}=\pi-\int_{C_R}\frac{dz}{z^2+1} \end{align*} therefore gives \begin{align*} \int_{-\infty}^{\infty}\frac{dx}{x^2+1}=\pi. \end{align*} The residue at the single upper-half-plane pole accounts for the whole real integral because the added semicircular arc contributes nothing in the limit. [/example] The contour estimate is as important as the residue computation. Without the estimate, the added arc could carry part of the limiting value. This is the point at which analysis enters the calculation. ## Logarithmic Derivatives Residues also count zeros and poles. The key expression is the logarithmic derivative $f'/f$. Here $\mathcal{M}(\Omega)$ denotes the field of meromorphic functions on the domain $\Omega$, so a function $f\in\mathcal{M}(\Omega)$ may have isolated poles but is holomorphic away from them. Near a zero or pole, the principal part of $f'/f$ has residue equal to the multiplicity with sign. [quotetheorem:3368] This theorem is the local input for the [argument principle](/page/Argument%20Principle). Integrating $f'/f$ around a contour counts zeros minus poles inside the contour. The count is algebraic and includes multiplicity. ## Coordinate Viewpoint In one complex variable, the expression $f(z)\,dz$ is often more natural than $f(z)$ alone. Under a holomorphic change of coordinate $z=\varphi(w)$ with $\varphi(b)=a$ and $\varphi'(b)\neq 0$, the differential becomes \begin{align*} f(\varphi(w))\varphi'(w)\,dw . \end{align*} The central point is that the residue belongs to the differential, not to the chosen coordinate name. If a local parameter changes, the coefficient changes in exactly the way needed for the product with the differential to keep the same contour integral. [quotetheorem:7948] This result is the bridge to residues on Riemann surfaces. A meromorphic differential can be described in many charts, but its residue at a point is a chart-independent complex number. On a compact surface there is no boundary contour at infinity. The residue theorem is replaced by a global balance law: every local singular contribution is cancelled somewhere else on the surface. [quotetheorem:7949] In the plane, the same balance reappears after adding one missing point. That missing point is infinity on the Riemann sphere. ## Residue at Infinity The point at infinity packages the behaviour of a meromorphic function for large $z$. For a meromorphic differential $f(z)\,dz$ on the Riemann sphere, define the residue at infinity by changing coordinate $w=1/z$. Then $dz=-w^{-2}\,dw$, so \begin{align*} f(z)\,dz=f(1/w)(-w^{-2})\,dw . \end{align*} [definition: Residue at Infinity] The residue at infinity is the map \begin{align*} \operatorname{Res}_{\infty}: \{f(z)\,dz \mid f \text{ is meromorphic near infinity}\} \to \mathbb C, \end{align*} defined by \begin{align*} \operatorname{Res}_{\infty}(f(z)\,dz)=\operatorname{Res}_{w=0}\left(-f(1/w)w^{-2}\,dw\right), \end{align*} where $w=1/z$. [/definition] For rational functions, the only singularities on the Riemann sphere are finitely many finite poles together with the possible pole at infinity. This creates a useful computational problem: an expansion at infinity can be awkward, while the finite poles may be easy to list, or conversely the finite residues may be numerous while the behaviour at infinity is short. The next theorem turns the global balance law into a bookkeeping identity that lets either side determine the other. [quotetheorem:3366] This identity can compute the residue at infinity from the finite residues, or the finite sum from the expansion at infinity. It is a compact expression of global balance. ## Common Mistakes A residue is not the value of a function at a pole; the function may not have a finite value there. A residue is not the coefficient of the highest negative power: for a pole of order $m$, the leading principal coefficient multiplies $(z-a)^{-m}$, while the residue multiplies $(z-a)^{-1}$. A residue is not determined by the absolute size of the function near the singularity, but by the angular coefficient that survives integration. When applying the residue theorem, the singularities on the contour must be excluded or handled by a separate limiting prescription. When using large contours, the limiting arc estimate must be proved; the residue computation alone does not justify discarding the arc. ## Beyond and Connected Topics Residues extend naturally from plane domains to meromorphic differentials on Riemann surfaces. On compact Riemann surfaces, the [global residue theorem](/theorems/7949) states that the sum of all residues is zero. In the plane, this same balance is already visible through the residue at infinity. The local ingredients of the theory point back to [Laurent series](/page/Laurent%20Series), [poles](/page/Pole), and [meromorphic functions](/page/Meromorphic%20Function): without Laurent expansions there is no coefficient to extract, and without meromorphic singularities there are no finite principal parts to measure. The global side points forward to contour integrals and the [argument principle](/page/Argument%20Principle), where residues convert winding information into counts of zeros and poles. In several complex variables, residue theory becomes more delicate because singular sets can have positive dimension. The one-variable residue remains the model case: a local coefficient controls a contour integral and participates in a global balance law. Residues also connect with distribution theory through principal value integrals and with algebraic geometry through duality and trace maps. For surrounding background in one variable, see [Cambridge IB Complex Analysis](/page/Cambridge%20IB%20Complex%20Analysis). For a later context where holomorphy and domains become higher-dimensional, see [Androma Several Complex Variables I: Domains and Holomorphy](/page/Androma%20Several%20Complex%20Variables%20I%3A%20Domains%20and%20Holomorphy). ## Summary Residues extract the coefficient of $(z-a)^{-1}$ in a Laurent expansion. That coefficient is equal to a small contour integral divided by $2\pi i$. Simple poles give residues by a limit, while higher-order poles give residues by differentiating the holomorphic numerator after clearing the pole. The residue theorem turns local coefficients into global contour integrals, and logarithmic derivatives show that residues can count zeros and poles. The coordinate viewpoint explains why residues belong naturally to meromorphic differentials. These ideas make residue calculus a central tool in complex analysis and a prototype for local-to-global methods in analysis. ## References Androma, [Cambridge III Functional Analysis](/page/Cambridge%20III%20Functional%20Analysis). Androma, [Several Complex Variables II: Sheaves and Stein Theory](/page/Several%20Complex%20Variables%20II%3A%20Sheaves%20and%20Stein%20Theory). 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, [Laurent Series](/page/Laurent%20Series). Androma, [Pole](/page/Pole). Androma, [Meromorphic Function](/page/Meromorphic%20Function). Androma, [Argument Principle](/page/Argument%20Principle). Ahlfors, *Complex Analysis* (1979). Conway, *Functions of One Complex Variable I* (1978).