Home > Residue (complex analysis)

In complex analysis, the residue is a complex number which describes the behavior of path integrals of a meromorphic function around a singularity. Residues can be computed quite easily and, once known, allow the determination of more complicated path integrals via the residue theorem.

1 Motivation

As an example, consider the contour integral

where C is some Jordan curve about 0.

Let us evaluate this integral without using standard integral theorems that may be available to us. Now, the Taylor series for e^z is well-known, and we substitute this series into the integrand. The integral then becomes:

Let us bring the 1/z⁵ term into the series, and so, we obtain

The integral now collapses to a much simpler form. Recall

So now the integral around C of every other term not in the form cz⁻¹ becomes zero, and the integral is reduced to

The value 1/4! is known as the residue of e^z/z⁵ at z=0, and is notated as

2 Calculating residues

Suppose a punctured disk D = {z : 0 < |z − c| < R} in the complex plane is given and f is a holomorphic function defined (at least) on D. The residue Res(f, c) of f at c is the coefficient a₋₁ of (z − c)⁻¹ in the Laurent series expansion of f around c. At a simple pole, the residue is given by:

According to the integral formula given in the Laurent series article we have:

where γ traces out a circle around c in a counterclockwise manner. We may choose the path γ to be a circle in radius ε around c were ε is as small as we desire.

To calculate the residue of a function around z = c, a pole of order n, one may use the following formula:

If the function f can be continued to a holomorphic function on the whole disk { z : |z − c| < R }, then Res(f, c) = 0. The converse is not generally true.

2.1 Series methods

If parts or all of a function can be expanded into a Taylor series or Laurent series, which may be possible if the parts or the whole of the function has a standard series expansion, calculating the residue is significantly simpler than by other methods.

As an example, consider calculating the residues at the singularities of the function

which may be used to calculate certain contour integrals. This function appears to have a singularity at z=0, but if one factorizes the denominator and thus writes the function as

it is apparent that the singularity at z=0 is a removable singularity and thus the residue at z=0 is therefore 0.

The only other singularity is at z=1. Recall

about z=a, so, for g(z)=sin z and a=1 we have

Introducing 1/(z-1) gives us

So the residue of f(z) at z=1 is sin 1.