Residue (complex analysis)

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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.

1 Definition
2 Motivation
3 Calculating residues
- 3.1 Series methods
4 See also
5 External links
6 References

[edit] Definition

The residue of a meromorphic function $f$ at an isolated singularity $a$ , often denoted $R e s (f, a)$ is the unique value $R$ such that $f(z)-{R \over (z-a)}$ has an analytic antiderivative in a punctured disk $0 < | z - a | < δ$ . Alternatively, residues can be calculated by finding Laurent series expansions, and are sometimes defined in terms of them.

[edit] 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

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

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

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

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

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

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

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

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

[edit] 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 Cauchy's integral formula, 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 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.

Due to its simplicity, the latter formula is more than useful in the computation of residues at first order poles.

[edit] 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

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