Dictionary Thesaurus Word Dynamo Quotes Reference Translator Spanish

Residue (complex analysis)

In complex analysis, the residue is a complex number which describes the behavior of line 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.

Definition

The residue of a meromorphic function

f

at an isolated singularity

a

, often denoted

Res(f,a)

is the unique value

R

such that

f(z)-{R over (z-a)}

has an analytic antiderivative in a punctured disk

0<|z-a| . Alternatively, residues can be calculated by finding Laurent series expansions, and are sometimes defined in terms of them.

Motivation

As an example, consider the contour integral

oint_C {e^z over z^5},dz

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

oint_C {1 over z^5}left(1+z+{z^2 over 2!} + {z^3over 3!} + {z^4 over 4!} + {z^5 over 5!} + {z^6 over 6!} + cdotsright),dz.

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

oint_C left({1 over z^5}+{z over z^5}+{z^2 over 2!;z^5} + {z^3over 3!;z^5} + {z^4 over 4!;z^5} + {z^5 over 5!;z^5} + {z^6 over 6!;z^5} + cdotsright),dz =

oint_C left({1 over;z^5}+{1 over;z^4}+{1 over 2!;z^3} + {1over 3!;z^2} + {1 over 4!;z} + {1over;5!} + {z over 6!} + cdotsright),dz.

The integral now collapses to a much simpler form. Recall that

oint_C {1 over z^a} ,dz=0,quad a in mathbb{Z},mbox{ for }a ne 1.

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

oint_C {1 over 4!;z} ,dz={1 over 4!}oint_C{1 over z},dz={1 over 4!}(2pi i) = {pi i over 12}.

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

mathrm{Res}_0 {e^z over z^5}, mathrm{or} mathrm{Res}_{z=0} {e^z over z^5}, mathrm{or} mathrm{Res}(f,0).

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:

operatorname{Res}(f,c)=lim_{zto c}(z-c)f(z).

According to Cauchy's integral formula, we have:

operatorname{Res}(f,c) =

{1 over 2pi i} int_gamma f(z),dz

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 where ε is as small as we desire.

The residue of a function f(z)=g(z)/h(z) at a simple pole c, where g and h are holomorphic functions in a neighborhood of c with h(c) = 0 and g(c) ≠ 0 is given by

operatorname{Res}(f,c) = frac{g(c)}{h'(c)}.

More generally, the residue of f around z = c, a pole of order n, can be found by the formula:

mathrm{Res}(f,c) = frac{1}{(n-1)!} cdot lim_{z to c} left(frac{d}{dz}right)^{n-1}left(f(z)cdot (z-c)^{n} right).

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.

The latter formula can be very useful in determining the residues for low-order poles. For higher order poles, series expansion is usually easier.

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, then calculating the residue is significantly simpler than by other methods.

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

f(z)={sin{z} over z^2-z}

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