Residue (complex analysis)

Template:Short description Template:Complex analysis sidebar In mathematics, more specifically complex analysis, the residue of a function at a point of its domain is a complex number proportional to the contour integral of a meromorphic function along a path enclosing one of its singularities. (More generally, residues can be calculated for any function Template:Tmath that is holomorphic except at the discrete points Template:Tmath, which may include essential singularities.) Residues are typically readily computed and, once known, allow the determination of general contour integrals via the residue theorem.

Definition

The residue of a meromorphic function $f$ at an isolated singularity $a$, often denoted Template:Tmath, Template:Tmath, Template:Tmath or Template:Tmath, is the unique value Template:Tmath such that Template:Tmath has an analytic antiderivative in a punctured disk Template:Tmath.

Alternatively, residues can be calculated by finding Laurent series expansions, and one can define the residue as the coefficient Template:Tmath of a Laurent series.

The concept can be used to provide contour integration values of certain contour integral problems considered in the residue theorem. According to the residue theorem, for a meromorphic function Template:Tmath, the residue at point Template:Tmath is given as:

$\mathrm{Res}(f,a_k)=\frac{1}{2\pi i}{\oint }_{\gamma }f(z)dz.$

where $\gamma$ is a positively oriented simple closed curve around Template:Tmath and not including any other singularities on or inside the curve.

The definition of a residue can be generalized to arbitrary Riemann surfaces. Suppose Template:Tmath is a 1-form on a Riemann surface. Let $\omega$ be meromorphic at some point Template:Tmath, so that we may write Template:Tmath in local coordinates as Template:Tmath. Then, the residue of Template:Tmath at Template:Tmath is defined to be the residue of Template:Tmath at the point corresponding to Template:Tmath.

Contour integration

Contour integral of a monomial

Computing the residue of a monomial

${\oint }_C{z}^{k}dz$

makes most residue computations easy to do. Since path integral computations are homotopy invariant, we will let $C$ be the circle with radius $1$ going counter clockwise. Then, using the change of coordinates $z\to e^{i\theta }$ we find that

$dz\to d(e^{i\theta })=i{e}^{i\theta }d\theta$

hence this integral now reads as

${\oint }_C{z}^{k}dz=\underset{0}{\overset{2\pi }{\int }}i{e}^{i(k+1)\theta }d\theta =\{\begin{array}{ll}2\pi i& \text{if }k=-1,\\ 0& \text{otherwise}.\end{array}$

Thus, the residue of Template:Tmath is Template:Tmath if integer Template:Tmath and Template:Tmath otherwise.

Generalization to Laurent series

If a function is expressed as a Laurent series expansion around Template:Tmath as follows: $$f(z)=\sum _{n=-\mathrm{\infty }}^{\mathrm{\infty }}{a}_n(z-c{)}^n.$$ Then, the residue at the point Template:Tmath is calculated as: $$\mathrm{Res}(f,c)=\frac{1}{2\pi i}{\oint }_{\gamma }f(z)dz=\frac{1}{2\pi i}\sum _{n=-\mathrm{\infty }}^{\mathrm{\infty }}{\oint }_{\gamma }{a}_n(z-c{)}^{n}dz=a_{-1}$$ using the results from contour integral of a monomial for counter clockwise contour integral Template:Tmath around a point Template:Tmath. Hence, if a Laurent series representation of a function exists around Template:Tmath, then its residue around Template:Tmath is known by the coefficient of the term Template:Tmath.

Application in residue theorem

For a meromorphic function Template:Tmath, with a finite set of singularities within a positively oriented simple closed curve $C$ which does not pass through any singularity, the value of the contour integral is given according to residue theorem, as: $${\oint }_{C}f(z)dz=2\pi i\sum _{k=1}^n\mathrm{I}(C,a_k)\mathrm{Res}(f,a_k).$$ where Template:Tmath, the winding number, is Template:Tmath if Template:Tmath is in the interior of Template:Tmath and Template:Tmath if not, simplifying to: $${\oint }_{\gamma }f(z)dz=2\pi i\sum \mathrm{Res}(f,a_k)$$ where Template:Tmath are all isolated singularities within the contour Template:Tmath.

Calculation of residues

Suppose a punctured disk Template:Tmath in the complex plane is given and Template:Tmath is a holomorphic function defined (at least) on Template:Tmath. The residue Template:Tmath of Template:Tmath at Template:Tmath is the coefficient Template:Tmath of Template:Tmath in the Laurent series expansion of Template:Tmath around Template:Tmath. 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 the residue theorem, we have:

$\mathrm{Res}(f,c)=\frac{1}{2\pi i}{\oint }_{\gamma }f(z)dz$

where Template:Tmath traces out a circle around Template:Tmath in a counterclockwise manner and does not pass through or contain other singularities within it. We may choose the path Template:Tmath to be a circle of radius Template:Tmath around Template:Tmath. Since Template:Tmath can be as small as we desire it can be made to contain only the singularity of Template:Tmath due to nature of isolated singularities. 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 Template:Tmath can be continued to a holomorphic function on the whole disk Template:Tmath, then Template:Tmath. The converse is not in general true.

Simple poles

If Template:Tmath is a simple pole of Template:Tmath, the residue of Template:Tmath is given by:

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

If that limit does not exist, then Template:Tmath instead has an essential singularity at Template:Tmath. If the limit is Template:Tmath, then Template:Tmath is either analytic at Template:Tmath or has a removable singularity there. If the limit is equal to infinity, then the order of the pole is higher than Template:Tmath.

It may be that the function Template:Tmath can be expressed as a quotient of two functions, Template:Tmath, where Template:Tmath and Template:Tmath are holomorphic functions in a neighbourhood of Template:Tmath, with Template:Tmath and Template:Tmath. In such a case, L'Hôpital's rule can be used to simplify the above formula to:

$\begin{array}{rl}\mathrm{Res}(f,c)& =\underset{z\to c}{\lim }(z-c)f(z)=\underset{z\to c}{\lim }\frac{zg(z)-cg(z)}{h(z)}\\ & =\underset{z\to c}{\lim }\frac{g(z)+z{g}^{\prime }(z)-c{g}^{\prime }(z)}{h^{\prime }(z)}=\frac{g(c)}{h^{\prime }(c)}.\end{array}$

Limit formula for higher-order poles

More generally, if Template:Tmath is a pole of order Template:Tmath, then the residue of Template:Tmath around Template:Tmath can be found by the formula:

$\mathrm{Res}(f,c)=\frac{1}{(p-1)!}\underset{z\to c}{\lim }\frac{d^{p-1}}{d{z}^{p-1}}((z-c{)}^{p}f(z)).$

This formula 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. For essential singularities, no such simple formula exists, and residues must usually be taken directly from series expansions.

Residue at infinity

In general, the residue at infinity is defined as:

$\mathrm{Res}(f(z),\mathrm{\infty })=-\mathrm{Res}(\frac{1}{z^2}f\left(\frac{1}{z}\right),0).$

If the following condition is met:

$\underset{|z|\to \mathrm{\infty }}{\lim }f(z)=0,$

then the residue at infinity can be computed using the following formula:

$\mathrm{Res}(f,\mathrm{\infty })=-\underset{|z|\to \mathrm{\infty }}{\lim }zf(z).$

If instead

$\underset{|z|\to \mathrm{\infty }}{\lim }f(z)=c\ne 0,$

then the residue at infinity is

$\mathrm{Res}(f,\mathrm{\infty })=\underset{|z|\to \mathrm{\infty }}{\lim }{z}^2{f}^{\prime }(z).$

For functions that are meromorphic on the entire complex plane with finitely many singularities, the sum of the residues at the (necessarily) isolated singularities plus the residue at infinity is zero, which gives:

$\mathrm{Res}(f(z),\mathrm{\infty })=-\sum _k\mathrm{Res}(f(z),a_k).$

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. The residue of the function is simply given by the coefficient of Template:Tmath in the Laurent series expansion of the function.

Examples

Residue from series expansion

Example 1

As an example, consider the contour integral

${\oint }_C\frac{e^z}{z^5}dz$

where Template:Tmath is some simple closed curve about Template:Tmath.

Let us evaluate this integral using a standard convergence result about integration by series. Substituting the Taylor series for Template:Tmath into the integrand, the integral becomes

${\oint }_C\frac{1}{z^5}(1+z+\frac{z^2}{2!}+\frac{z^3}{3!}+\frac{z^4}{4!}+\frac{z^5}{5!}+\frac{z^6}{6!}+\cdots )dz.$

Let us bring the term in Template:Tmath into the series. The contour integral of the series then writes

$\begin{array}{rl}& {\oint }_C(\frac{1}{z^5}+\frac{z}{z^5}+\frac{z^2}{2!z^5}+\frac{z^3}{3!z^5}+\frac{z^4}{4!z^5}+\frac{z^5}{5!z^5}+\frac{z^6}{6!z^5}+\cdots )dz\\ =& {\oint }_C(\frac{1}{z^5}+\frac{1}{z^4}+\frac{1}{2!z^3}+\frac{1}{3!z^2}+\frac{1}{4!z}+\frac{1}{5!}+\frac{z}{6!}+\cdots )dz.\end{array}$

Since the series converges uniformly on the support of the integration path, we are allowed to exchange integration and summation. The series of the path integrals then collapses to a much simpler form because of the previous computation. So now the integral around Template:Tmath of every other term not in the form Template:Tmath is zero, and the integral is reduced to

${\oint }_C\frac{1}{4!z}dz=\frac{1}{4!}{\oint }_C\frac{1}{z}dz=\frac{1}{4!}(2\pi i)=\frac{\pi i}{12}.$

The value 1/4! is the residue of Template:Tmath at Template:Tmath, and is denoted

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

Example 2

As a second example, consider calculating the residues at the singularities of the function$$f(z)=\frac{\sin z}{z^2-z}$$which may be used to calculate certain contour integrals. This function appears to have a singularity at Template:Tmath, but if one factorizes the denominator and thus writes the function as $$f(z)=\frac{\sin z}{z(z-1)}$$ it is apparent that the singularity at Template:Tmath is a removable singularity and then the residue at Template:Tmath is therefore Template:Tmath. The only other singularity is at Template:Tmath. Recall the expression for the Taylor series for a function Template:Tmath about Template:Tmath: $$g(z)=g(a)+g^{\prime }(a)(z-a)+\frac{g^{″}(a)(z-a{)}^2}{2!}+\frac{g^{‴}(a)(z-a{)}^3}{3!}+\cdots .$$ So, for Template:Tmath and Template:Tmath we have $$\sin z=\sin 1+(\cos 1)(z-1)+\frac{-(\sin 1)(z-1{)}^2}{2!}+\frac{-(\cos 1)(z-1{)}^3}{3!}+\cdots ,$$ and for Template:Tmath and Template:Tmath 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 Template:Tmath gives us$$\frac{\sin z}{z(z-1)}=\frac{\sin 1}{z-1}+(\cos 1-\sin 1)+(z-1)(-\frac{\sin 1}{2!}-\cos 1+\sin 1)+\cdots .$$ So the residue of Template:Tmath at Template:Tmath is Template:Tmath.

Example 3

The next example shows that, computing a residue by series expansion, a major role is played by the Lagrange inversion theorem. Let $$u(z):=\sum _{k\ge 1}{u}_k{z}^k$$ be an entire function, and let $$v(z):=\sum _{k\ge 1}{v}_k{z}^k$$ with positive radius of convergence, and with Template:Tmath. So Template:Tmath has a local inverse Template:Tmath at Template:Tmath, and $u(1/V(z))$ is meromorphic at 0. Then we have: $${\mathrm{Res}}_0(u(1/V(z)))=\sum _{k=0}^{\mathrm{\infty }}k{u}_k{v}_k.$$ Indeed, $${\mathrm{Res}}_0(u(1/V(z)))={\mathrm{Res}}_0(\sum _{k\ge 1}{u}_{k}V(z{)}^{-k})=\sum _{k\ge 1}{u}_k{\mathrm{Res}}_0(V(z{)}^{-k})$$because the first series converges uniformly on any small circle around 0. Using the Lagrange inversion theorem $${\mathrm{Res}}_0(V(z{)}^{-k})=k{v}_k,$$ and we get the above expression. For example, if $u(z)=z+z^2$ and also $v(z)=z+z^2$, then $$V(z)=\frac{2z}{1+\sqrt{1+4z}}$$ and $$u(1/V(z))=\frac{1+\sqrt{1+4z}}{2z}+\frac{1+2z+\sqrt{1+4z}}{2z^2}.$$ The first term contributes Template:Tmath to the residue, and the second term contributes Template:Tmath since it is asymptotic to Template:Tmath.

With the corresponding stronger symmetric assumptions on Template:Tmath and Template:Tmath, it also follows that $${\mathrm{Res}}_0(u(1/V))={\mathrm{Res}}_0(v(1/U)),$$ where Template:Tmath is a local inverse of Template:Tmath at Template:Tmath.

See also

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

External links

• Template:Springer