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 $\mathrm {Res} (f,a)$ is the unique value $R$ such that $f(z)-{R \over (z-a)}$ has an analytic antiderivative in a punctured disk $0<\left|z-a\right|<\delta$. 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 simple closed 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 factor into the series, 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

$\oint _{C}{1 \over z^{n}}\,dz=0,\quad n\in \mathbb {Z} ,{\mbox{ for }}n\neq 1.$

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

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

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. Various methods exist for calculating this value, and the choice of which method to use depends on the function in question, and on the nature of the singularity.

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. This may be used for calculation in cases where the integral can be calculated directly, but it is usually the case that residues are used to simplify calculation of integrals, and not the other way around.

Removable singularities

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.

Residues at low-order poles

At a simple pole c, the residue of f is given by:

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

It may be that the function f can be expressed as a quotient of two functions, f(z)=g(z)/h(z), where g and h are holomorphic functions in a neighborhood of c, with h(c) = 0 and g(c) ≠ 0. In such a case, the limit formula for a simple pole simplifies to:

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

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

$\mathrm {Res} (f,c)={\frac {1}{(n-1)!}}\lim _{z\to c}{\frac {d^{n-1}}{dz^{n-1}}}\left((z-c)^{n}f(z)\right).$

These formulas can be very useful in determining the residues for low-order poles. For higher order poles, the calculations can become unmanageable, and series expansion is usually easier. Also for essential singularities, residues often must be taken directly from series expansions.

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

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

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

The only other singularity is at z = 1. Recall the expression for the Taylor series for a function g(z) about z = a:

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

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

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

and for g(z) = 1/z and a = 1 we have

${\frac {1}{z}}={\frac {1}{(z-1)+1}}=1-(z-1)+(z-1)^{2}-(z-1)^{3}+\cdots .$

Multiplying those two series and introducing 1/(z − 1) gives us

${\frac {\sin {z}}{z(z-1)}}={\sin {1} \over z-1}+(\cos {1}-\sin 1)+(z-1)\left(-{\frac {\sin {1}}{2!}}-\cos 1+\sin 1\right)+\cdots .$

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

See also

• Cauchy's integral formula
• Cauchy integral theorem
• Methods of contour integration
• Morera's theorem
• Partial fractions in complex analysis

External links

• Weisstein, Eric W. "Complex Residue". MathWorld.
• Module for Residues by John H. Mathews

References

• Ahlfors, Lars (1979). Complex Analysis. McGraw Hill.
• Marsden & Hoffman, Basic complex analysis (Freeman, 1999).