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.

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:

$\oint_C {1 \over z^5}\left(1+z+{z^2 \over 2!} + {z^3\over 3!} + {z^4 \over 4!} + {z^5 \over 5!} + {z^6 \over 6!} + \ldots\right)\,dz$

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

$\oint_C {1 \over z^5}+{z \over z^5}+{z^2 \over 2!\;z^5} + {z^3\over 3!\;z^5} + {z^4 \over 4!\;z^5} + {z^5 \over 5!\;z^5} + {z^6 \over 6!\;z^5} + \ldots\,dz$

$\oint_C {1 \over\;z^5}+{1 \over\;z^4}+{1 \over 2!\;z^3} + {1\over 3!\;z^2} + {1 \over 4!\;z} + {1\over\;5!} + {z \over 6!} + \ldots\,dz$

The integral now collapses to a much simpler form. Recall

$\oint_C {1 \over z^a} \,dz=0,\quad a \in \mathbb{R}, \ a \ne 1$

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

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

The value 1/4! is known as 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:

$\operatorname{Res}(f,c)=\lim_{z\to c}(z-c)f(z).$

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

$\operatorname{Res}(f,c) = {1 \over 2\pi i} \int_\gamma f(z)\,dz$

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 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:

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

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.

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

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

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

$f(z)={\sin{z} \over z(z-1)}$

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

$g(z) = g(a) + g'(a)(z-a) + {g''(a)(z-a)^2 \over 2!} + {g'''(a)(z-a)^3 \over 3!}+ \cdots$

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

$\sin{z} = \sin{1} + \cos{1}(z-1)+{-\sin{1}(z-1)^2 \over 2!} + {-\cos{1}(z-1)^3 \over 3!}+\cdots$

Introducing 1/(z-1) gives us

${\sin{z} \over z-1} = {\sin{1} \over z-1} + {\cos{1}(z-1) \over z-1}+{-\sin{1}(z-1)^2 \over 2!(z-1)} + {-\cos{1}(z-1)^3 \over 3!(z-1)}+\cdots$

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

Categories: Complex analysis