Residues

The Coefficient That Controls a Small-Circle Integral

The Laurent expansion in the previous tutorial records the behavior of a function near an isolated singularity. Among its coefficients, the coefficient of $(z-a)^{-1}$ has a special role: it is exactly the coefficient that remains when the function is integrated once around a circle centered at $a$. This local quantity is called the residue.

Residues do not classify singularities by themselves. A pole can have a nonzero residue or a zero residue, and an essential singularity can also have a residue. The residue is one particular Laurent coefficient, not a measure of the total size or type of the singularity. In this tutorial, we define it and establish formulas that make it useful to compute. The general theorem for integrating around contours enclosing singularities comes next.

Definition and First Consequences

Definition: Suppose $f$ is holomorphic on a punctured disk $0<|z-a|<R$, and its Laurent expansion there is $f(z)=\sum_{n=-\infty}^{\infty}c_n(z-a)^n$. The residue of $f$ at $a$, denoted $\operatorname{Res}(f,a)$, is the Laurent coefficient $c_{-1}$.

The center $a$ is part of the definition: the relevant term is $(z-a)^{-1}$, not $z^{-1}$ unless $a=0$. By the uniqueness of Laurent coefficients, the residue does not depend on which sufficiently small annulus is used to represent the function.

The definition applies to every isolated singularity, whether removable, a pole, or essential. If the singularity is removable, every negative-power coefficient vanishes, so its residue is zero. For poles, the residue may or may not vanish. To see why the $-1$ coefficient is distinguished, integrate the Laurent terms around a circle. For an integer $n$, the parametrization $z=a+re^{it}$, $0\leq t\leq 2\pi$, gives

\int_{|z-a|=r}(z-a)^n\,dz =ir^{n+1}\int_0^{2\pi}e^{i(n+1)t}\,dt = \begin{cases} 2\pi i,&n=-1,\\ 0,&n\neq -1. \end{cases}

Thus every integral power except $(z-a)^{-1}$ contributes zero. The next theorem makes this observation precise for an entire Laurent expansion.

The Small-Circle Integral Formula

Theorem (Small-Circle Residue Formula): Suppose $f$ is holomorphic on $0<|z-a|<R$, and let $0<r<R$. Orient the circle $|z-a|=r$ counterclockwise. Then $$ \int_{|z-a|=r} f(z)\,dz=2\pi i\,\operatorname{Res}(f,a). $$

Proof. Write the Laurent expansion of $f$ on the annulus as $f(z)=\sum_{n=-\infty}^{\infty}c_n(z-a)^n$. The circle $|z-a|=r$ lies within this annulus, and the Laurent series converges uniformly on that circle. We can therefore integrate the series term by term along the circle. Using the integral of each power computed above, all terms integrate to zero except the one with $n=-1$. Consequently,

\int_{|z-a|=r} f(z)\,dz =\sum_{n=-\infty}^{\infty}c_n \int_{|z-a|=r}(z-a)^n\,dz =2\pi i c_{-1} =2\pi i\,\operatorname{Res}(f,a).

The argument holds for every $r$ with $0<r<R$. In particular, the integral is independent of the radius as long as the circle stays in the punctured disk. $\square$

The formula is local: it concerns a circle centered at one isolated singularity, with no other excluded points in the disk. It is not yet a formula for an arbitrary contour. Its practical message is that a contour integral around this small circle can be found by computing a single Laurent coefficient.

Worked Example: Reading the Residue from a Laurent Expansion

Let $f(z)=(e^z-1)/z^2$ for $z\neq 0$. The Taylor expansion of the exponential gives

e^z-1=z+\frac{z^2}{2!}+\frac{z^3}{3!}+\frac{z^4}{4!}+\cdots, \qquad f(z)=\frac{1}{z}+\frac{1}{2!}+\frac{z}{3!}+\frac{z^2}{4!}+\cdots.

The coefficient of $z^{-1}$ is $1$, so $\operatorname{Res}(f,0)=1$. The Small-Circle Residue Formula now gives, for every sufficiently small $r>0$,

\int_{|z|=r}\frac{e^z-1}{z^2}\,dz=2\pi i.

The coefficient of $z^0$ is $1/2$, but it is not the residue. Only the coefficient of $z^{-1}$ determines this integral.

Computing Residues at Poles

The Laurent expansion is not always the quickest way to find a residue. At a simple pole, multiplication by $z-a$ cancels the singularity and leaves a function whose value at $a$ is the residue. More generally, a pole of order at most $m$ can be handled using a derivative.

Theorem (Residue Formula for a Pole): Suppose $g$ is holomorphic near $a$, and $$ f(z)=\frac{g(z)}{(z-a)^m} $$ for $z\neq a$ near $a$, where $m$ is a positive integer. Then $$ \operatorname{Res}(f,a)=\frac{g^{(m-1)}(a)}{(m-1)!}. $$ If $g(a)\neq 0$, the singularity is a pole of order $m$. If $g(a)=0$, the expression still gives the residue, although cancellation may make the pole order smaller or remove the singularity.

Proof. Since $g$ is holomorphic near $a$, it has a Taylor expansion there: $g(z)=\sum_{n=0}^{\infty}g^{(n)}(a)(z-a)^n/n!$. On the punctured neighborhood, division by $(z-a)^m$ gives

f(z)=\sum_{n=0}^{\infty}\frac{g^{(n)}(a)}{n!}(z-a)^{n-m}.

A term has exponent $-1$ exactly when $n-m=-1$, or $n=m-1$. Its coefficient is $g^{(m-1)}(a)/(m-1)!$, proving the formula. If $g(a)\neq0$, then $(z-a)^m f(z)=g(z)$ extends holomorphically with nonzero value at $a$, so the singularity is a pole of order $m$ by the definition in the previous tutorial. If $g(a)=0$, the coefficient computation remains valid, but the factor $g$ may cancel some or all of the denominator’s powers. $\square$

For $m=1$, the formula reduces to a particularly useful limit rule. If $a$ is a simple pole and $f(z)=g(z)/(z-a)$ with $g$ holomorphic near $a$, then

\operatorname{Res}(f,a)=g(a)=\lim_{z\to a}(z-a)f(z).

Worked Example: A Residue at a Simple Pole

Consider $f(z)=(z^2+1)/((z-1)(z+2))$ near $a=1$. Write $f(z)=g(z)/(z-1)$, where $g(z)=(z^2+1)/(z+2)$. This function is holomorphic near $1$, since its denominator is nonzero there, and

g(1)=\frac{1^2+1}{1+2}=\frac{2}{3}\neq 0.

Therefore $1$ is a simple pole and its residue is $2/3$. Equivalently, the limit rule gives

\operatorname{Res}(f,1) =\lim_{z\to1}(z-1)\frac{z^2+1}{(z-1)(z+2)} =\lim_{z\to1}\frac{z^2+1}{z+2} =\frac{2}{3}.

Worked Example: A Pole with Zero Residue

Let $f(z)=(z+1)/(z-2)^3$. Here $a=2$, $m=3$, and $g(z)=z+1$. The pole formula gives

\operatorname{Res}(f,2)=\frac{g''(2)}{2!}=\frac{0}{2}=0.

This is still a pole of order three: $g(2)=3\neq0$. Directly setting $w=z-2$ gives $z+1=w+3$, and hence

f(z)=\frac{w+3}{w^3}=\frac{1}{w^2}+\frac{3}{w^3}.

There is no $w^{-1}$ term, confirming that the residue is zero. This example shows why a pole’s order alone does not determine its residue.

Residues at Essential Singularities

The pole formula requires a representation with a finite power $(z-a)^{-m}$. At an essential singularity there are infinitely many negative powers, so the direct method is to identify the $(z-a)^{-1}$ term in the Laurent expansion. The definition of residue applies without change.

Worked Example: An Essential Singularity with Zero Residue

Consider $h(z)=\cos(1/z)$ for $z\neq0$. Substitution into the Taylor series for cosine gives

\cos(1/z) =1-\frac{1}{2!z^2}+\frac{1}{4!z^4}-\frac{1}{6!z^6}+\cdots.

There are infinitely many nonzero negative-power coefficients, so $0$ is an essential singularity by the classification theorem from the previous tutorial. But every negative power in this expansion has an even exponent. In particular, the coefficient of $z^{-1}$ is zero, and $\operatorname{Res}(h,0)=0$. The Small-Circle Residue Formula therefore gives $\int_{|z|=r}\cos(1/z)\,dz=0$ for every $r>0$.

How to Use the Residue

The residue is useful because it compresses the part of a local Laurent expansion that contributes to an integral around a centered circle. To calculate it, first identify the center and the local form of the function. Then choose a method that matches the singularity:

For a Laurent series already available, read off the coefficient of $(z-a)^{-1}$.
At a simple pole, use $\lim_{z\to a}(z-a)f(z)$, after verifying that the remaining factor is holomorphic at $a$.
For $f(z)=g(z)/(z-a)^m$, use $g^{(m-1)}(a)/(m-1)!$.
At an essential singularity, inspect the Laurent expansion; the residue is still just its $(z-a)^{-1}$ coefficient.

A common error is to confuse the residue with the coefficient of the constant term, or to assume that every pole has a nonzero residue. Another is to apply the simple-pole limit rule before checking that the singularity is simple. The order of a pole determines which derivative appears in the higher-order formula, while the residue itself can vanish. These local calculations are the ingredients for the contour-integral result developed in the next tutorial.

Check Your Understanding

Use the Laurent coefficient definition, the small-circle formula, and the pole formulas to answer the following questions.

Which Laurent coefficient defines $\operatorname{Res}(f,a)$?
If the residue at $a$ is $c$, what is the integral of $f$ around a sufficiently small counterclockwise circle centered at $a$?
Suppose $f(z)=g(z)/(z-a)^4$, with $g$ holomorphic near $a$. Which derivative of $g$ gives the residue, and what factorial appears in the denominator?
Can a pole of order three have residue zero? Give the reason based on the Laurent coefficient.
Why does $\cos(1/z)$ have residue zero at $0$, even though its singularity there is essential?

What You'll Learn