In complex analysis, not every singularity is equally bad. A function such as $1/z$ fails to be [holomorphic](/page/Holomorphic%20Function) at $0$, but its failure is completely organized: multiplying by $z$ repairs it. By contrast, $e^{1/z}$ has infinitely many negative powers near $0$, and no finite correction captures its behaviour. A pole isolates the first kind of singularity: a point where a holomorphic function blows up like a finite negative power. Poles are the singularities that make [Meromorphic Function](/page/Meromorphic%20Function) theory possible. They are also the local source of residues, the obstruction measured by contour integrals, and the singularities controlled by a [Laurent Series](/page/Laurent%20Series). The concept is useful because it turns local blow-up into finite algebraic data: an integer order and finitely many principal coefficients. ## Definition Throughout this page, $B(a,r)=\{z\in\mathbb{C}: |z-a|<r\}$ denotes the open disk of radius $r$ centered at $a$. A punctured neighbourhood of $a$ means a set of the form $0<|z-a|<r$, obtained from such a disk by deleting its center. This is the natural local domain for studying an isolated singularity: the function is defined near $a$, but not necessarily at $a$ itself. The defining feature of a pole is not merely that a function is undefined at one point, but that the failure is repaired by multiplying by a finite power of the distance from that point. This distinguishes a pole from both a removable singularity, where no negative power is needed, and an essential singularity, where no finite power is enough. [definition: Pole] Let $U \subset \mathbb{C}$ be open, let $a \in U$, and let $f: U \setminus \{a\} \to \mathbb{C}$ be holomorphic. The point $a$ is a pole of $f$ if there exists a positive integer $m$, a radius $r > 0$ with $B(a,r) \subset U$, and a holomorphic function $g: B(a,r) \to \mathbb{C}$ such that $g(a) \ne 0$ and \begin{align*} f(z) &= \frac{g(z)}{(z-a)^m} \end{align*} for all $z \in B(a,r) \setminus \{a\}$. [/definition] This definition already contains the main local data: the singularity is finite, and the exponent $m$ is a positive integer measuring the amount of finite blow-up. The supporting notions used to compute that exponent are collected next, so the primary definition remains separate from the surrounding bookkeeping. [example: Simple Pole at the Origin] The function $f(z)=1/z$ on $\mathbb{C}\setminus\{0\}$ has a pole at $0$. Indeed, take $m=1$ and $g(z)=1$ in the definition, so \begin{align*} f(z) &= \frac{g(z)}{z} \end{align*} for every $z\ne 0$. The principal part is the single term $z^{-1}$, so this is a simple pole with residue $1$. [/example] Here the residue means the coefficient of $(z-a)^{-1}$ in the Laurent expansion about the pole $a$. This page uses that coefficient-level convention when discussing contour integrals; no separate global residue theory is needed for the local computations below. ## Supporting Local Notions Laurent expansions and local contour integrals require a punctured disc around the point with no other singularities interfering. When singularities accumulate, there is no annulus on which the local Laurent machinery can be applied to one point at a time, so isolation must be stated relative to the whole singular set. [definition: Isolated Singularity] Let $U \subset \mathbb{C}$ be open, let $S \subset U$, let $a \in S$, and let $f: U \setminus S \to \mathbb{C}$ be holomorphic. The point $a$ is an isolated singularity of $f$ if there exists $r > 0$ such that $B(a,r) \subset U$ and \begin{align*} B(a,r) \cap S &= \{a\}. \end{align*} [/definition] The word isolated matters because if singular points accumulate at $a$, the local annulus on which Laurent theory works may not exist. For a pole, the punctured disc is the stage on which the finite blow-up can be factored cleanly. To compare two poles, or to compute residues and contour integrals, it is not enough to know that some exponent works. The exact exponent is the local invariant that measures the strength of the singularity. The function $g$ is the holomorphic part left after the finite blow-up has been factored out, and the condition $g(a) \ne 0$ prevents overestimating the exponent. [definition: Order of a Pole] Let $U \subset \mathbb{C}$ be open, let $a \in U$, and let $f: U \setminus \{a\} \to \mathbb{C}$ be holomorphic. Suppose $a$ is a pole of $f$. The order of the pole at $a$ is the unique positive integer $m$ for which there exists a radius $r > 0$ with $B(a,r) \subset U$ and a holomorphic function $g: B(a,r) \to \mathbb{C}$ with $g(a) \ne 0$ such that \begin{align*} f(z) &= \frac{g(z)}{(z-a)^m} \end{align*} for all $z \in B(a,r) \setminus \{a\}$. [/definition] To use a pole in calculations, the order alone is not always enough; contour integrals and local subtraction formulas require the actual negative-power terms. Laurent expansion gives a compact way to collect exactly the terms responsible for the singularity and separate them from the holomorphic remainder. A pole of order $1$ is called a simple pole, while higher order poles have a longer finite singular part. [definition: Principal Part at a Pole] Let $U \subset \mathbb{C}$ be open, let $a \in U$, and let $f: U \setminus \{a\} \to \mathbb{C}$ be holomorphic. Suppose $a$ is a pole of order $m$ and suppose the Laurent expansion of $f$ about $a$ is \begin{align*} f(z) &= \sum_{k=-m}^{\infty} c_k (z-a)^k \end{align*} for $0 < |z-a| < r$, where $r > 0$ and $B(a,r) \subset U$. The principal part of $f$ at $a$ is \begin{align*} \sum_{k=-m}^{-1} c_k (z-a)^k. \end{align*} [/definition] Quotients force a second local measurement: not only how a function blows up, but how another holomorphic function vanishes. The cancellation between numerator and denominator is controlled by the order of a zero, so pole computations need a precise definition of that order. [definition: Zero of Order] Let $U \subset \mathbb{C}$ be open, let $a \in U$, and let $h: U \to \mathbb{C}$ be holomorphic. The function $h$ has a zero of order $m$ at $a$ if $m$ is a positive integer and there exists a radius $r > 0$ with $B(a,r) \subset U$ and a holomorphic function $q: B(a,r) \to \mathbb{C}$ such that $q(a) \ne 0$ and \begin{align*} h(z) &= (z-a)^m q(z) \end{align*} for all $z \in B(a,r)$. [/definition] For global complex analysis, singularities usually occur at many points rather than just one, so the next useful class consists of functions whose singularities are all controlled in the finite Laurent sense. This class should allow algebraic operations while excluding essential singularities. The definition of a meromorphic function supplies that global framework. [definition: Meromorphic Function] Let $U \subset \mathbb{C}$ be open. A function $f$ is meromorphic on $U$ if there exists a subset $P \subset U$ with no accumulation point in $U$ such that $f: U \setminus P \to \mathbb{C}$ is holomorphic and every point of $P$ is a pole of $f$. [/definition] Thus a meromorphic function is not arbitrary singular data added to a holomorphic function. Its singularities are exactly of the finite Laurent type, which makes algebra and integration behave predictably. ## Equivalent Characterisations The definition through division by $(z-a)^m$ is often the most economical. Laurent series give the same idea in coefficient form and are usually the best way to compare poles with removable and essential singularities. [quotetheorem:7870] The Laurent criterion turns the definition into something directly visible in the local series. A removable singularity has no negative-power terms at all, while an essential singularity has infinitely many negative-power terms. A pole is exactly the middle case: the negative part is present but finite, and the order is the largest negative exponent that occurs. This is why the coefficient of the lowest nonzero negative power is decisive; it records both that the singularity is not removable and that the blow-up stops after finitely many powers. In practice, this criterion is most useful when a Laurent expansion is already available, but it may be inefficient when the function is presented as a quotient whose cancellation has not yet been simplified. Many candidate poles arise from quotients, where the visible singularity may disappear after cancellation. Directly estimating $f$ can obscure whether the singularity is really a pole, but the reciprocal turns blow-up into ordinary vanishing. The obstruction is to distinguish genuine pole behaviour from removable zeros in the denominator and from more irregular singularities; measuring the zero order of $1/f$ resolves exactly that local question. [quotetheorem:7871] The reciprocal criterion is useful because it converts blow-up into vanishing, a phenomenon that holomorphic functions handle very rigidly. The requirement that $f$ be nonzero on a punctured disk is not cosmetic: without it, $1/f$ would not even be holomorphic on the punctured neighborhood, so its zero order at $a$ would not be a well-defined local invariant. When the reciprocal extends holomorphically and has a zero of order $m$, the original function has exactly a pole of order $m$. This is often the fastest test for quotients, since after cancellation one can read the pole order from the remaining zero order of the denominator relative to the numerator. Sometimes the first question is even more basic than the exact order: does the function approach infinity near the missing point? The growth formulation gives a geometric criterion in the extended complex plane, while Laurent or reciprocal methods provide the sharper local invariant. [quotetheorem:7872] The growth criterion is conceptually simple, but it does not reveal the order without more work. Laurent or reciprocal methods provide the sharper local invariant. ## Model Computations and Boundary Cases ### Pure Powers and Quotient Cancellation The model example is a pure negative power. Every pole is this example multiplied by a nonvanishing holomorphic function. [example: Pure Power Pole] Let $a \in \mathbb{C}$ and let $m$ be a positive integer. Define $f: \mathbb{C} \setminus \{a\} \to \mathbb{C}$ by \begin{align*} f(z) = \frac{1}{(z-a)^m}. \end{align*} To match the definition of a pole, take $g(z)=1$ on any disc $B(a,r)$. This function is holomorphic and satisfies $g(a)=1 \ne 0$, and for every $z \ne a$, \begin{align*} f(z) = \frac{g(z)}{(z-a)^m}. \end{align*} Thus $a$ is a pole of order $m$. The Laurent expansion about $a$ is the single term \begin{align*} f(z) = (z-a)^{-m}. \end{align*} Equivalently, the Laurent coefficients satisfy $c_{-m}=1$ and $c_k=0$ for every $k \ne -m$. Therefore the principal part is \begin{align*} (z-a)^{-m}. \end{align*} The residue is the coefficient of $(z-a)^{-1}$. If $m=1$, then $(z-a)^{-m}=(z-a)^{-1}$, so the residue is $1$. If $m \ne 1$, then the coefficient of $(z-a)^{-1}$ is $0$, so the residue is $0$. [/example] This example shows that the order is not just the fact of blow-up. The contour integral around a small circle detects only the coefficient of $(z-a)^{-1}$, so higher-order poles can have zero residue. Quotients of holomorphic functions are the most common source of poles. The relevant question is whether the denominator vanishes to higher order than the numerator. [example: Cancellation in a Quotient] Consider \begin{align*} f(z) &= \frac{z^2-1}{(z-1)^3} \end{align*} as a function on $\mathbb{C} \setminus \{1\}$. Factoring the numerator gives \begin{align*} z^2-1 &= (z-1)(z+1). \end{align*} Therefore, for $z \ne 1$, \begin{align*} f(z) &= \frac{(z-1)(z+1)}{(z-1)^3}. \end{align*} Since $(z-1)^3=(z-1)(z-1)^2$ and $z-1 \ne 0$, cancellation gives \begin{align*} f(z) &= \frac{z+1}{(z-1)^2}. \end{align*} Now write the numerator in powers of $z-1$: \begin{align*} z+1 &= (z-1)+2. \end{align*} Substituting this into the quotient gives \begin{align*} f(z) &= \frac{(z-1)+2}{(z-1)^2}. \end{align*} Splitting the fraction term by term, \begin{align*} f(z) &= \frac{z-1}{(z-1)^2}+\frac{2}{(z-1)^2}. \end{align*} Since $z \ne 1$, the first term reduces to \begin{align*} \frac{z-1}{(z-1)^2} &= \frac{1}{z-1}. \end{align*} Hence \begin{align*} f(z) &= \frac{1}{z-1}+\frac{2}{(z-1)^2}. \end{align*} This Laurent expansion has lowest power $(z-1)^{-2}$ with nonzero coefficient $2$, so $1$ is a pole of order $2$, not order $3$. The principal part is \begin{align*} \frac{2}{(z-1)^2}+\frac{1}{z-1}, \end{align*} and the residue at $1$ is the coefficient of $(z-1)^{-1}$, namely $1$. [/example] The example also shows why a formula alone can mislead: the displayed denominator may overstate the pole order when the numerator vanishes at the same point. ### Removable Boundary A removable singularity can look like a quotient with a zero denominator, but after cancellation no negative Laurent powers remain. [example: Removable Singularity Is Not a Pole] Define $f: \mathbb{C} \setminus \{0\} \to \mathbb{C}$ by \begin{align*} f(z) &= \frac{\sin z}{z}. \end{align*} The Taylor series for sine is \begin{align*} \sin z = \sum_{n=0}^{\infty} \frac{(-1)^n z^{2n+1}}{(2n+1)!}. \end{align*} For $z \ne 0$, division by $z$ gives \begin{align*} \frac{\sin z}{z} = \sum_{n=0}^{\infty} \frac{(-1)^n z^{2n}}{(2n+1)!}. \end{align*} Writing the first terms of this series, \begin{align*} \frac{\sin z}{z} = 1-\frac{z^2}{3!}+\frac{z^4}{5!}-\frac{z^6}{7!}+\cdots. \end{align*} Every exponent appearing in this Laurent expansion at $0$ is one of $0,2,4,6,\ldots$, so no term of the form $c_k z^k$ with $k<0$ occurs. Define \begin{align*} F(z) = \sum_{n=0}^{\infty} \frac{(-1)^n z^{2n}}{(2n+1)!}. \end{align*} This [power series](/page/Power%20Series) is holomorphic on $\mathbb{C}$ and satisfies $F(z)=\sin z/z$ for every $z \ne 0$. Its value at $0$ is the constant term of the power series, namely \begin{align*} F(0) = 1. \end{align*} Thus assigning $f(0)=1$ gives a holomorphic extension across $0$. Since the Laurent expansion has no negative powers, *Laurent Characterisation of Poles* implies that $0$ is not a pole; the singularity is removable. [/example] This boundary case is important because both removable singularities and poles arise from quotients. The difference is the remaining order after cancellation. ### Essential Boundary At the other boundary, an essential singularity has too many negative Laurent terms to be controlled by a finite order. [example: Essential Singularity Is Not a Pole] Define $f: \mathbb{C} \setminus \{0\} \to \mathbb{C}$ by \begin{align*} f(z) = e^{1/z}. \end{align*} The exponential power series is \begin{align*} e^w = \sum_{k=0}^{\infty} \frac{w^k}{k!}. \end{align*} Substituting $w=1/z$ for $z \ne 0$ gives \begin{align*} e^{1/z} = \sum_{k=0}^{\infty} \frac{(1/z)^k}{k!}. \end{align*} Since $(1/z)^k=z^{-k}$ for every integer $k \ge 0$, this becomes \begin{align*} e^{1/z} = \sum_{k=0}^{\infty} \frac{1}{k!}z^{-k}. \end{align*} For every positive integer $k$, the coefficient of $z^{-k}$ is $1/k!$, and $1/k! \ne 0$. Thus the Laurent expansion at $0$ contains the nonzero negative-power terms $z^{-1}, z^{-2}, z^{-3}, \ldots$. If $0$ were a pole of finite order $m$, then by *Laurent Characterisation of Poles* its Laurent expansion would have no terms with powers below $-m$. But the displayed expansion contains the term $\frac{1}{(m+1)!}z^{-(m+1)}$, whose exponent is below $-m$ and whose coefficient is nonzero. Therefore no finite pole order exists, so $0$ is not a pole of $e^{1/z}$; it is an essential singularity. [/example] This example marks the limit of the pole concept. A pole has finite principal data; an essential singularity does not. ## Algebra of Pole Orders The order of a pole behaves like an integer valuation. Multiplication adds orders, division subtracts zero orders, and cancellation is controlled by the first nonzero coefficient. [quotetheorem:7873] The theorem explains most pole computations for rational and meromorphic functions because it reduces a quotient to two ordinary zero orders. Its necessity is visible when both numerator and denominator vanish at the same point: the displayed denominator alone can exaggerate the pole order, or even suggest a pole when cancellation leaves a removable singularity. Its limitation is also local. One must first know the numerator and denominator are holomorphic near the point and have finite zero orders there; it is not a test for arbitrary isolated singularities or for functions with essential behaviour already present. In practice, this result is the fast route from a formula such as $p(z)/q(z)$ to the order of the pole before any Laurent expansion is written. The complementary case is when the numerator vanishes at least as strongly as the denominator. Then the quotient no longer has finite blow-up at the point, but it is still important to know whether the singularity disappears cleanly or whether another obstruction remains. The cancellation theorem isolates precisely this situation: equal orders leave a nonzero finite limit, while a numerator of higher order leaves a zero after extension. [quotetheorem:7874] Together, the pole and cancellation rules say that the local behaviour of a quotient is governed by the difference between the two vanishing orders. After the order is known, the next local invariant usually needed for integration is the residue. At a simple pole, the residue can be extracted by removing the one negative power. [quotetheorem:353] This is the formula used constantly in contour integration. It turns a local Laurent coefficient into a limit of holomorphic data. In applications, simple poles often arise from quotients $p/q$ where $q$ has a simple zero, and this same limit often evaluates the residue without writing a full Laurent expansion. To connect poles with integration, the key point is that a small loop around an isolated pole records exactly the residue. This observation is the local engine behind the global residue theorem: the entire principal part affects the function near $a$, but contour integrals detect only the residue coefficient. ## Beyond and Connected Topics Poles sit inside the classification of isolated singularities. A removable singularity has no negative Laurent coefficients, a pole has finitely many, and an essential singularity has infinitely many. This trichotomy is why [Laurent Series](/page/Laurent%20Series) are the natural language for singularities. The reciprocal relation connects poles to zeros: the reciprocal characterisation above is the precise theorem-level form of this correspondence. This is why pole order and zero order are best learned together rather than as unrelated local notions. In the theory of [Meromorphic Function](/page/Meromorphic%20Function), poles are allowed singularities rather than defects to be removed. Rational functions on the Riemann sphere are the basic examples: their finite singularities are poles, and their behaviour at infinity is also described in terms of zeros and poles after the change of variable $w=1/z$. For residue theory, poles are the singularities where computations are most explicit. Simple poles produce residues through a single limit; higher-order poles require more coefficient extraction but remain finite calculations. This is why the residue theorem converts many real integrals and contour integrals into sums of local data. The same bookkeeping reappears geometrically in the language of divisors and valuations. On a Riemann surface or algebraic curve, a zero contributes positive order and a pole contributes negative order, so a meromorphic function can be studied through the formal sum of its local orders. ## References Ahlfors, *Complex Analysis* (1979). Conway, *Functions of One Complex Variable I* (1978). Stein and Shakarchi, *Complex Analysis* (2003). [Laurent Series](/page/Laurent%20Series). [Meromorphic Function](/page/Meromorphic%20Function).