Residue Theorem

From a Local Coefficient to a Contour Integral

The previous tutorial showed that a small counterclockwise circle around an isolated singularity contributes $2\pi i$ times the residue there. The Residue Theorem extends this local calculation: when a contour encloses several isolated singularities, its integral is determined by the sum of their residues. Singularities outside the contour do not contribute.

We will work with a positively oriented, simple, closed, piecewise $C^1$ contour $\Gamma$. “Positively oriented” means that the bounded interior lies to the left as the contour is traversed. The contour must not pass through a singularity. These hypotheses give an unambiguous inside and outside, and the orientation fixes the sign of the integral.

The Boundary-Integral Fact We Need

The proof uses a version of Cauchy’s theorem for a region with more than one boundary component. Its content is that if a function is holomorphic throughout such a region, the integral over the whole oriented boundary is zero. The outer boundary is traversed counterclockwise and boundaries of holes are traversed clockwise.

Lemma (Cauchy’s Theorem for a Finitely Connected Region): Let $\Omega$ be a bounded region whose boundary consists of finitely many disjoint, piecewise $C^1$ simple closed curves. Orient the outer boundary counterclockwise and any inner boundary curves clockwise. If $h$ is holomorphic on an open set containing $\overline{\Omega}$, then the sum of its integrals over all boundary components is zero.

Proof. First suppose the region has polygonal boundary. Subdivide it into finitely many triangles, choosing the subdivision fine enough that every triangle lies in the open set where $h$ is holomorphic. Cauchy’s Theorem for Triangles gives zero for the integral around each triangle. Add these equalities. Every edge inside the region occurs twice, once in each direction, so the two integrals along that edge cancel. The uncancelled edges are exactly the oriented boundary edges. Their integrals therefore sum to zero.

For piecewise $C^1$ boundary curves, approximate each boundary component by inscribed polygonal curves. Since there are finitely many components and they are disjoint, the approximations can be taken sufficiently fine that they remain disjoint and bound polygonal regions with the same outer and inner boundary orientations. Apply the polygonal result to each approximating region. The function $h$ is uniformly continuous on a compact neighborhood of the boundary. As the polygonal curves converge to the piecewise $C^1$ curves, their line integrals converge to the corresponding boundary integrals: on each smooth piece, the polygonal parametrizations and their tangents approximate the original curve and tangent, and the finitely many corner points do not affect the integrals. Taking the limit in the polygonal boundary identity proves that the sum of the integrals over the original boundary components is zero. $\square$

The Residue Theorem

Theorem (Residue Theorem): Let $\Gamma$ be a positively oriented, simple, closed, piecewise $C^1$ contour. Suppose $f$ is holomorphic on an open set containing the closure of the interior of $\Gamma$, except at finitely many isolated singularities $a_1,\ldots,a_n$ in the interior. Then $$ \int_{\Gamma} f(z)\,dz =2\pi i\sum_{k=1}^{n}\operatorname{Res}(f,a_k). $$

Proof. Choose small closed disks centered at $a_1,\ldots,a_n$, each lying inside $\Gamma$, with pairwise disjoint closures and containing no other singularities. Let their radii be $r_1,\ldots,r_n$. Remove the interiors of these disks from the region enclosed by $\Gamma$, leaving a region on which $f$ is holomorphic on a neighborhood of its closure.

The boundary of this region consists of $\Gamma$, oriented counterclockwise, and the small circles around the $a_k$, oriented clockwise as boundaries of holes. The lemma therefore gives

\int_{\Gamma} f(z)\,dz -\sum_{k=1}^{n}\int_{|z-a_k|=r_k}f(z)\,dz=0,

where each circle in the displayed sum is now written counterclockwise. By the Small-Circle Residue Formula from the previous tutorial,

\int_{|z-a_k|=r_k}f(z)\,dz =2\pi i\,\operatorname{Res}(f,a_k).

Substituting these equalities into the boundary identity proves the formula. The radii may be chosen independently, so the argument applies to every singularity in the finite list. $\square$

The minus sign in the boundary identity records the orientation of the inner circles: as boundaries of the region with holes, they run clockwise. The small-circle formula uses counterclockwise orientation, so converting to that convention introduces the minus sign. Rearranging then gives a plus sign for each residue in the final theorem.

Using the Theorem in Practice

To apply the theorem, first check that the contour avoids every singularity and that the function is holomorphic on the enclosed region apart from finitely many isolated singularities. Then identify which singularities are inside. Compute only their residues and add them. A pole outside the contour is not included, even if it is close to the contour.

Locate the singularities.
Find the points where the function fails to be holomorphic, and verify that none lies on the contour.

Select the enclosed points.
Use the geometry of the contour to determine which singularities lie in its interior.

Compute and add residues.
Use the residue formulas from the previous tutorial, then multiply their sum by $2\pi i$.

Worked Example: One Pole Inside the Unit Circle

Let $f(z)=\dfrac{z+1}{z(z-3)}$, and let $\Gamma$ be the unit circle oriented counterclockwise. The singularities are $0$ and $3$; only $0$ lies inside $\Gamma$. It is a simple pole, and

\operatorname{Res}(f,0) =\lim_{z\to0}z\frac{z+1}{z(z-3)} =\frac{1}{-3} =-\frac{1}{3}.

The Residue Theorem therefore gives

\int_{|z|=1}\frac{z+1}{z(z-3)}\,dz =2\pi i\left(-\frac{1}{3}\right) =-\frac{2\pi i}{3}.

The pole at $3$ contributes nothing to this integral because it is outside the unit circle.

Worked Example: Two Enclosed Simple Poles

Consider $f(z)=\dfrac{e^z}{(z+1)(z-2)}$ on the circle $|z|=3$, oriented counterclockwise. Both poles, $-1$ and $2$, lie inside. At $-1$, cancel the factor $z+1$ and evaluate the remaining holomorphic factor:

\operatorname{Res}(f,-1) =\left.\frac{e^z}{z-2}\right|_{z=-1} =-\frac{e^{-1}}{3}.

At $2$, cancel $z-2$ instead:

\operatorname{Res}(f,2) =\left.\frac{e^z}{z+1}\right|_{z=2} =\frac{e^2}{3}.

Their sum is $\dfrac{e^2-e^{-1}}{3}$, so

\int_{|z|=3}\frac{e^z}{(z+1)(z-2)}\,dz =\frac{2\pi i}{3}\left(e^2-e^{-1}\right).

Both contributions matter, including the negative residue at $-1$; the theorem uses their algebraic sum, not the sum of their absolute values.

Worked Example: Opposite Residues Can Cancel

Let $f(z)=\dfrac{1}{z^2-1}$ and take the counterclockwise circle $|z|=2$. The poles at $1$ and $-1$ are both inside. Since $z^2-1=(z-1)(z+1)$, the simple-pole residues are

\operatorname{Res}(f,1) =\left.\frac{1}{z+1}\right|_{z=1} =\frac{1}{2}, \qquad \operatorname{Res}(f,-1) =\left.\frac{1}{z-1}\right|_{z=-1} =-\frac{1}{2}.

Their sum is zero. Consequently,

\int_{|z|=2}\frac{1}{z^2-1}\,dz =2\pi i\left(\frac{1}{2}-\frac{1}{2}\right) =0.

The function has poles inside the contour, but the integral is still zero because the residues cancel. The theorem does not say that every enclosed singularity makes a nonzero contribution.

Orientation, Scope, and a Common Pitfall

Reversing the orientation of a contour changes the sign of its line integral. For the same geometric curve traversed clockwise, the theorem therefore gives $-2\pi i$ times the sum of the enclosed residues. The positive orientation in the theorem is essential to its stated sign.

Another common mistake is to sum residues at every singularity in the formula, including those outside the contour. The theorem concerns only the singularities in the interior. For instance, a rational function can have poles both inside and outside a circle; only the inside poles enter its contour integral. A further caution is that the theorem as stated assumes a simple closed contour. For self-intersecting contours, the relevant generalization uses winding numbers; the simple inside/outside rule is not sufficient.

The main advantage is that a global integral can be evaluated using local information. The contour may be large or geometrically complicated, while the residue calculations can be short. Once the enclosed singularities and their residues are known, the integral follows from their sum without parametrizing the contour.

Check Your Understanding

Use the hypotheses and formula of the Residue Theorem to answer the following questions.

Why must the contour avoid every singularity of the function?
In the stated theorem, which singularities are included in the residue sum?
Explain why the boundaries of the removed disks are clockwise when applying the boundary-integral lemma.
For a clockwise traversal of the same simple closed curve, how does the contour integral compare with the counterclockwise integral?
Can an integral around a contour enclosing poles be zero? Explain how the residues can produce that result.

What You'll Learn